Calculus.manifolds Spivak(Solutions)

Exercises 1.1 file:///home/aluno/Desktop/sol.spivak/ch1a.htm Exercises 1.1 1. Prove that One has Taking the square root of both sides gives the result. 2. When does equality hold in Theorem 1-1 (3)? Equality holds precisely when one is a nonnegative multiple of the other. This is a consequence of the analogous assertion of the next problem. 3. Prove that . When does equality hold? . . The first assertion is the triangle inequality. I claim that equality holds precisely when one vector is a non-positive multiple of the other. If for some real , then substituting shows that the inequality is equivalent to and clearly equality holds if a is non-positive. Similarly, one has equality if for some real . Conversely, if equality holds, then , and so . By Theorem 1-1 (2), it follows that and are linearly dependent. If for some real , then substituting back into the equality shows that must be non-positive or must be 0. The case where is treated similarly. 4. Prove that . If , then the inequality to be proved is just which is just the triangle inequality. On the other hand, if the result follows from the first case by swapping the roles of 5. The quantity is called the distance between interpret geometrically the ``triangle inequality": . The inequality follows from Theorem 1-1(3): and , then and . . Prove and 1 de 6 16-03-2010 16:53 Exercises 1.1 file:///home/aluno/Desktop/sol.spivak/ch1a.htm Geometrically, if , , and are the vertices of a triangle, then the inequality says that the length of a side is no larger than the sum of the lengths of the other two sides. 6. Let and be functions integrable on . . 1. Prove that Theorem 1-1(2) implies the inequality of Riemann sums: Taking the limit as the mesh approaches 0, one gets the desired inequality. 2. If equality holds, must continuous? for some ? What if and are No, you could, for example, vary at discrete points without changing the values of the integrals. If and are continuous, then the assertion is true. In fact, suppose that for each and are continuous. So , there is an with since . . Then the inequality holds true in an open neighborhood of always non-negative and is positive on some subinterval of Expanding out gives for all quadratic has no solutions, it must be that its discriminant is negative. 3. Show that Theorem 1-1 (2) is a special case of (a). Let , , and for all in since the integrand is . Since the for . Then part (a) gives the inequality of Theorem 1-1 (2). Note, however, that the equality condition does not follow from (a). 7. A linear transformation is called norm preserving if , and inner product preserving if 1. Show that is norm preserving if and only if is inner product preserving. . 2 de 6 16-03-2010 16:53 Exercises 1.1 file:///home/aluno/Desktop/sol.spivak/ch1a.htm If is inner product preserving, then one has by Theorem 1-1 (4): Similarly, if is norm preserving, then the polarization identity together with the linearity of T give: . 2. Show that such a linear transformation same sort. Let be norm preserving. Then implies , i.e. the kernel of is trivial. So T is 1-1. Since is a 1-1 linear map of a finite dimensional vector space into itself, it follows that is also onto. In particular, has an inverse. Further, given , there is a with , and so preserving. Thus product preserving. 8. If and in are both non-zero, then the angle between and , , is defined to be which makes denoted sense by Theorem 1-1 (2). The linear transformation is angle , one has . preserving if is 1-1 and for 1. Prove that if is norm preserving, then is angle preserving. Assume is norm preserving. By Problem 1-7, preserving. So . 2. If there is a basis , prove that equal. The assertion is false. For example, if , and , then , but , , , . Now, of and numbers such that are is inner product , since is norm is norm preserving, and hence also inner is 1-1, and that is of the is angle preserving if and only if all 3 de 6 16-03-2010 16:53 any of this form is angle preserving as the composition of two angle preserving linear transformations is angle preserving. suppose that is angle preserving. Since the are pairwise orthogonal and is angle preserving. Let be .e. and the operation is functional composition. To correct the situation.Exercises 1. Now define to be the linear transformation is norm .1 file:///home/aluno/Desktop/sol. V is norm preserving. So. . For the converse. Clearly. Then one has because all the cross because all the suffices to make Now suppose that are equal and cancel out. for some and and that . So. i.htm showing that T is not angle preserving. this condition suffices to make not be angle 3. Then since preserving. 4 de 6 16-03-2010 16:53 . What are all angle preserving ? The angle preserving are precisely those which can be expressed in the form where U is angle preserving of the kind in part (b). the are also pairwise orthogonal. Suppose all the are equal in absolute value. because the cross terms all cancel out.spivak/ch1a. This proves that preserving. Define to be the linear transformation an orthogonal basis of such that for each . add the condition that the be pairwise orthogonal. for all . this means that: terms are zero. Using bilinearity. In particular. this condition be angle preserving. This completes the characterization. 12. since . Further. If such that is a linear transformation. One has . By Problem 8(a). . Further. it follows that 10. is angle . show that there is a number for . Define unique . is norm preserving since by the Pythagorean Theorem.1 file:///home/aluno/Desktop/sol. it follows that preserving. then .spivak/ch1a. Let be the maximum of the absolute values of the entries in the matrix of and . If that . Further. Let denote the dual space of the vector space .Exercises 1. This is a perfectly straightforward computation in terms of the coordinates of using only the definitions of inner product and norm. 11. Show is angle preserving and that if The transformation is 1-1 by Cramer's Rule because the determinant of its matrix is 1. If . so is a map of the type in part (b). then one has is norm preserving. By the definition of angle. maps each to a scalar multiple of itself. Show that is . If . .htm Then clearly and is angle preserving because it is the composition of two angle preserving maps. Note that and denote points in and . 9. by . For and . let have the matrix . show that . define is a for a 1-1 linear transformation and conclude that every 5 de 6 16-03-2010 16:53 . Since the dual space has dimension n. If and are perpendicular. then and are called perpendicular (or orthogonal ) if . the proofs are is a linear transformation.spivak/ch1a. This proves the last assertion. 13. it follows that onto. By bilinearity of the inner product. prove that . If .1 file:///home/aluno/Desktop/sol. one has for perpendicular and : 6 de 6 16-03-2010 16:53 . is a linear tranformation .Exercises 1. Together these imply that Since for . These follow from bilinearity. has no non-zero vectors in its kernel is also and so is 1-1.htm One needs to verify the trivial results that (a) and (b) omitted. . ) containing and contained in (resp. and boundary of the sets: 1 de 4 16-03-2010 16:55 . Find the interior. Since the intersection of two open rectangles is an open rectangle (Why?). Prove that If . If . Prove that the union of any (even infinite) number of open sets is open. The assertion about finitely many sets follows by induction. Let and be open. Since is open. If . be the open rectangle centered at . Section 2 http://www. 16.uky. we have . exterior. and so .. Give a counterexample for infinitely many open sets. so is open. and so the intersection of even countably many open sets is not necessarily open. then with sides of . then let length is open.edu/~ken/ma570/homework/hw. Exercises: Chapter 1. and .Exercises: Chapter 1. Section 2 14. Let be a collection of open sets. 15. and be their union. If then there is an with . Prove that the intersection of two (and hence finitely many) open sets is open. This proves that is open. there is an open rectangle containing . then there are open rectangles (resp. ). The intersection of the open intervals is the set containing only . So is open.ms. neither has yet been used. etc. 18. The interior of is the empty set . If it is . show that the boundary of 2 de 4 16-03-2010 16:55 . . To do the construction. then the sixteenths. the proofs are straightforward and omitted.uky. and the boundary is the set In each case. such that contains at most one point 17. Then make a list of all the quarters. etc. etc. To show that this works. 1]. To show this. could be made by listing all pairs (a. . Hint: It suffices to ensure that contains points in each quarter of the square and also in each sixteenth. Similarly. both and are in the list . For example. it follows that is in the boundary of . and such that the latter occurring (in ) of them is earliest possible.edu/~ken/ma570/homework/hw. one could make by listing first the quarters.Exercises: Chapter 1. and . and amongst those with same value of in increasing lexicographical order. . choose any open rectangle containing . sixteenths. choose the point such that is in the portion. Section 2 http://www. the exterior is the empty set . Then there is some portion of the square in which is entirely contained within the rectangle and containing .. first make a list of all the rational numbers in the interval [0. sixteenths.ms. traverse the list : for each portion of the square. non-negative. Construct a set on each horizontal and each vertical line but the boundary of is . the exterior is . and the boundary is the set The interior of is the empty set . of the unit sqare. it suffices to show that every point in the square is in the boundary of . let . b) of integers with positive. Now. the exterior is .. The interior of is the set the boundary is the set . If number in is the union of open intervals is contained in some such that each rational . in increasing order of . Since this part of the square contains an element of the set A and elements not in A (anything in the portion with the same x-coordinate works). and amongst such the other one is the earliest possible. with an obvious lexicographical order amongst the quarters. Let be chosen so that . then simply eliminate those pairs for which there is an earlier pair with the same value of . etc. is closed.e. Since is compact. Let . Since is open. is be a is an of such that Such an is in the exterior of . then and so is bounded. it suffices to show that nothing in is in the exterior of . 20. which is a contradiction. Suppose is in the exterior of . If is a closed set that contains every rational number that . Suppose not in . Now contains a non-empty open subinterval of and this is necessarily disjoint from . and contains all rational numbers in .Exercises: Chapter 1. Let be an open interval containing and disjoint from . a. prove that there is a number . there is an open interval containing and disjoint from . Suppose is compact. 19. Since the boundary is the complement of the union of the interior and the exterior. Section 2 http://www.edu/~ken/ma570/homework/hw.uky. and so there is an open rectangle containing and disjoint from . Then open neighborhood of which is disjoint from . Let finite subcover. which is a contradiction. . the interior of is itself since it is a union of open sets. Prove the converse of Corollary 1-7: A compact subset of bounded. also the exterior of clearly contains as . there is a finite subcover . 21. This was chosen 3 de 4 16-03-2010 16:55 . Then there is a which contains . Let is closed and be the open cover consisting of rectangles for all positive integers . i.ms. Clearly. If is closed and for all .. To show that is closed. is . If . So the complement is open. it suffices its complement is open.. Then the collection where is an open cover of . show Suppose . Let be a rational number in contained in . Let . But every non-empty open subinterval of contains rational numbers. Section 2 http://www. by the triangle inequality. Clearly.uky. If is open and is compact. 4 de 4 16-03-2010 16:55 . is also true. and let Then. so is compact.Exercises: Chapter 1. It is straightforward to verify that is bounded and closed.ms. prove that there is a and . which shows the assertion. Let . we know that satisfies the assertion. this means that no can be . Finally. show that there is a compact set such that is contained in the interior of and . c. is the graph of the if and are required both to be . If is closed.edu/~ken/ma570/homework/hw. Give a counterexample in closed with neither compact. 22.. b. to be as in part (a). and for all . .. such that For each . Let be as in Problem 1-21 (b) applied with and . choose is compact. A counterexample: is the x-axis and exponential function. Then is an open cover of Let be a finite subcover. so that is entirely contained within the open rectangle. So T is continuous at . Choose for each . Let satisfies . there is an and .. then .. then . then the whole line is disjoint from . Let around which is in . it follows by the intermediate value theorem that 1 de 3 16-03-2010 16:56 . Let such that . if at and . . Section 3 23. then the line intersects the graph of and nowhere else. if is in and . for each i. and is chosen as in . for each i. suppose that the definition of satisfies . 25. Prove that and for each Suppose that positive such that for every . Let . Then are is continuous and .a if and only if This is an immediate consequence of Problem 1-23 and the definition of continuity.edu/~ken/ma570/homework/hw. Let with . So .Exercises: Chapter 1. if . Then. . Then. Exercises: Chapter 1. Since the only roots of at 0 and . one has . . So. Prove that is continuous at if and only if each is. .ms. Show that every straight line through Let the line be . 26. Section 3 http://www. If . 24. Prove that a linear transformation By Problem 1-10. show that . Let . for all . Conversely. On the other hand. then . contains an interval a.uky. If satisfies is continuous. Exercises: Chapter 1. every is clearly continuous at 0. Section 3 http://www. Clearly. If . one has . Then for any with where we have used Problem 1-4 in the simplification. is identically zero in a neighborhood of zero by part (a). 27.edu/~ken/ma570/homework/hw. it follows that is open by Theorem 1-8. one has . So. This proves that is continuous. one has: . In particular.ms. let and choose ... but by by is not continuous at if and . let and . for all with . is not closed. Define define 0. To show it is continuous at . As suggested. . 29.uky. if cannot intersect . This shows that continuous at . In fact. For is continuous at For each . the line anywhere to the left of . choose to be a boundary point of which is not in . Since . prove that every continuous function is takes on a 2 de 3 16-03-2010 16:56 . If is compact. and let . So. is open by considering the function is continuous. Let . If which is unbounded. cannot be continuous at because every open rectangle containing contains points of and for all those points . show that there is a continuous function 28. On the other hand. Show that each . then by Problem 1-4. Prove that with The function . this is unbounded. b. it is an upper bound for the sum of the and the right hand side ``telescopes" and is bounded above by the difference of the two end terms which in turn is bounded above by . Now assume that the have been re-ordered so that they are in increasing order.uky.ms. maximum and a minimum value. 30. The function on the right is an increasing function of . Section 3 http://www. let . Then and are boundary points of . ) be the greatest lower bound (respectively least upper bound) of . Clearly these are the minimum and maximum values of .. is compact. Let be an increasing function. If . By Theorem 1-9. and hence is closed and bounded. Now add up all the inequalities with this value of .Exercises: Chapter 1. and hence are in since it is closed. 3 de 3 16-03-2010 16:56 . are distinct. and they are taken on since they are in .edu/~ken/ma570/homework/hw. in particular. show that One has .. is bounded above by the quantity on the right for any . Let (resp. Exercises Chapter 2. then So. a 1 x 2 matrix. Show that independent of the second variable if and only if there is a function such that .. 3.ms. Just as before. we need only show that immediately from Problem 1-10. Define when a function find for such is the Jacobian. A function . What is in terms of ? The first assertion is trivial: If can let be defined by . Section 1 http://www. then it is continuous If is differentiable at . but this follows is said to be independent of the second variable is if for each we have for all . is differentiable at . 2. . it would be . Which functions are independent of the first variable and also of the second variable? The function is independent of the first variable if and only if for all . this is equivalent to 1 de 5 16-03-2010 16:58 .uky. Prove that if at . more proper to say that with matrix. i.. you . if . even though one is a linear transformation and the other is a is independent of the first variable and . Conversely. Section 1 1. but I will often confound .edu/~ken/ma570/homework/hw. So. If is independent of the second variable. Exercises Chapter 2. then because: Note: Actually.e. then is independent of the second variable. An a. More generally. and so we see that this just . . Then one must have: and so .uky. b. Show that is a function of the kind considered in Problem 2-4. Thus is identically zero. Similarly.ms. One has both cases. . Show that Suppose when and is linear and hence differentiable. 0) unless is differentiable at with. is not differentiable at (0.. show that is differentiable.. so that 2 de 5 16-03-2010 16:58 . their being a function such that for all argument similar to that of the previous problem shows that . But says that 5. . Let be a continuous real-valued function on the unit circle such that define by and . using the definition of derivative. But otherwise. If and is defined by .edu/~ken/ma570/homework/hw. In . say. 4. Let be defined by for all .Exercises Chapter 2. we get for fixed : . Section 1 http://www. one gets . Define by for all . Then it is trivial to show that satisfies all the properties of Problem 2-4 and that the function obtained from this is as in the statement of this problem. if were differentiable at 0. 3 de 5 is differentiable at with the 16-03-2010 16:58 . On the other hand. is not differentiable at . one can show that. if the are differentiable at inequality derived from Problem 1-1: the squeeze principle to conclude that desired derivative. by the squeeze principle. 8. So. by approaching zero along the 45 degree line in the first quadrant. Let be a function such that .edu/~ken/ma570/homework/hw. 7. Show that is not differentiable at 0. one would then have: clearly 1.ms. Prove that is differentiable at if and only if . Let by the squeeze principle using . must be differentiable at with . In fact. . Section 1 http://www.Exercises Chapter 2. and are..uky. 6. Then one has the inequality: .. Let be defined by . Just as in the proof of Problem 2-4. then would be the zero map. and in this case Suppose that . Show that is in spite of the fact that the limit is differentiable at 0. then use the and On the other hand. The converse is not true. Section 1 http://www. If is differentiable at .1 times to the limit to see that the value of the limit is .edu/~ken/ma570/homework/hw. then the function works by the definition of derivative. Indeed. add the assumption that at : If there is a of the specified form with and first order. Since is continuous. Show that is differentiable at if and only if there is a function of the form such that and are equal up to first order at . On the other hand. one has: 4 de 5 16-03-2010 16:58 . . you can change the value of at without changing whether or not and are equal up to first order. To make the converse true. this . But clearly changing the value of at changes whether or not is differentiable at . defined by are equal up to order at a. show that with and the function .uky.. If be continuous equal up to . Two functions are equal up to order at if a. But then the condition is equivalent to the is differentiable at exist.. Apply L'Hôpital's Rule n .ms. then we see that means that assertion that b. Multiplying this by . 9.Exercises Chapter 2. ..edu/~ken/ma570/homework/hw.uky. Subtracting these two results gives shows that to order at . and are equal up 5 de 5 16-03-2010 16:58 . Section 1 http://www.ms.Exercises Chapter 2. b.. . Use the theorems of this section to find a. then . one has: .e. Using Theorem 2-3 (3) and part (a).Exercises: Chapter 2. One has .uky. Section 2 http://www. i. and so by the chain rule: for the following: d. Exercises: Chapter 2.ms. The chain rule gives: f. We have and so by the chain rule. .edu/~ken/ma570/homework/hw.. 1 de 6 16-03-2010 16:59 . Using the chain rule. If is the function of part (c). we get: e. Section 2 10. If derivative of . c. then and we know the from part (a). one has: . If . Using the last part: j. .Exercises: Chapter 2. then . h. 2 de 6 16-03-2010 16:59 . then . So one gets: .ms.uky. If for the following (where . . (c). and so: . then . The chain rule gives: i. If . is as in part (a). b. One has . If . and so: c. Using parts (h). g. Find a.. 11. . one gets .edu/~ken/ma570/homework/hw. Section 2 http://www. then is continuous): . So one gets: . and (a).. Exercises: Chapter 2. Prove that if is bilinear. So . a.. By Problem 2-12 and the fact that is bilinear. Define a. is defined by b. b. But. one has . then Let have a 1 in the place only. . where given in parts (d) and (a) above.. Then we have depending only on . Find and by . This follows by applying (b) to the bilinear function 13.edu/~ken/ma570/homework/hw. we have and have derivatives as is bilinear if for . Section 2 http://www. Since such by an obvious induction using bilinearity. Prove that One has by bilinearity and part (a). c.uky. it was verified in the proof of Theorem 2-3 (5). Show that the formula for (b). A function and . 12. It follows that there is an that: . . If are differentiable.ms. . we see that it suffices to show the result in the case where and the bilinear function is the product function. in Theorem 2-3 is a special case of . and 3 de 6 16-03-2010 16:59 . in this case. Exercises: Chapter 2, Section 2 http://www.ms.uky.edu/~ken/ma570/homework/hw... , show that (Note that is an matrix; its transpose .) is an matrix, which we consider as a member of Since c. If , one can apply the chain rule to get the assertion. is differentiable and . for all , show that Use part (b) applied to to get . This shows the result. such that the function |f| d. Exhibit a differentiable function defined by |f|(t) = |f(t)| is not differentiable. Trivially, one could let 14. Let the function . Then is not differentiable at 0. be Euclidean spaces of various dimensions. A function is called multilinear if for each choice of defined by , show that for , with is a , linear transformation. a. If is multilinear and we have This is an immediate consequence of Problem 2-12 (b). b. Prove that . This can be argued similarly to Problem 2-12. Just apply the definition expanding the numerator out using multilinearity; the remainder looks like a sum of terms as in part (a) except that there may be more than two type arguments. These can be expanded out as in the proof of the bilinear case to get a sum of terms that look like constant multiples of 4 de 6 16-03-2010 16:59 Exercises: Chapter 2, Section 2 http://www.ms.uky.edu/~ken/ma570/homework/hw... where is at least two and the are distinct. Just as in the bilinear case, this limit is zero. This shows the result. matrix as a point in the -fold product 15. Regard an . considering each row as a member of is differentiable and a. Prove that by This is an immediate consequence of Problem 2-14 (b) and the multilinearity of the determinant function. b. If are differentiable and , show that This follows by the chain rule and part (a). c. If for all and are differentiable, let 5 de 6 16-03-2010 16:59 Exercises: Chapter 2, Section 2 http://www.ms.uky.edu/~ken/ma570/homework/hw... be the functions such that of the equations: differentiable and find . for are the solutions . Show that is Without writing all the details, recall that Cramer's Rule allows you to write down explicit formulas for the where is the matrix of the coefficients and the are obtained from by replacing the column with the column of the . We can take transposes since the determinant of the transpose is the same as the determinant of the original matrix; this makes the formulas simpler because the formula for derivative of a determinant involved rows and so now you are replacing the row rather than the column. Anyway, one can use the quotient formula and part (b) to give a formula for the derivative of the . 16. Suppose is differentiable and has a differentiable inverse . Show that This follows immediately by the chain rule applied to . . 6 de 6 16-03-2010 16:59 Find the partial derivatives of the following functions: a.Exercises: Chapter 2. d. Section 3 http://www... e. . 1 de 7 16-03-2010 17:00 . c. g. f. b. . .edu/~ken/ma570/homework/hw.uky. . h. . Exercises: Chapter 2. Section 3 17.ms. i. ..edu/~ken/ma570/homework/hw. Find the partial derivatives of a. 2 de 7 16-03-2010 17:00 .ms. in terms of the derivatives of and if 20.. Section 3 http://www. . . d. b. 19. . .Exercises: Chapter 2. Find the partial derivatives of the following functions (where continuous): a. 18. one has .uky. . . c. If is find Since . such that . How should One could let c. . show that is independent of the second .. ? By the mean value theorem. e. Define by a. c. d.ms. Section 3 http://www.edu/~ken/ma570/homework/hw. Let be continuous. Find a function One could let Find one such that One could let 22. show that . . Show that True since the first term depends only on b. and . If and . and .Exercises: Chapter 2. 21.uky.. . b. If variable. . be defined so that . . one knows that if a function of one variable 3 de 7 16-03-2010 17:00 . 4 de 7 16-03-2010 17:00 . . and from to are all contained in . Show that Using part (a). Let a.. Then the line segment from to .. has zero derivative on a closed interval . Show that One has . . show that and is a constant. from to . and for all and for all when . one has and . a. Both assertions are immediate consequences of this result.Exercises: Chapter 2.uky. Further. Section 3 http://www. b.ms. If . One could let 24. and . ). Find a function of the second variable. it follows that . Define by such that but is not independent . Suppose and are arbitrary points of . So . By the proof of Problem 2-22. b. one has for all and (because follow immediately by substituting in the other. then it is constant on that interval. The assertions into one formula and . since and so . 23.edu/~ken/ma570/homework/hw. edu/~ken/ma570/homework/hw. Then it follows for all . In particular. the result is obvious. and Now. Section 3 http://www. At each of the exceptional points. that . Let c. By induction on . Show that is a function. For points other than 1 and -1.uky.ms. define that for all . To show that this limit is zero. b. it suffices to show that integer for each . Define by a. But this is an easy induction using L'Hôpital's rule: . Consider the function where is a polynomial. and defined by for all . We have . Show that 0 elsewhere.. suppose by induction on . function is . one has for . consider the derivative from the left and from the is a function which is positive on (-1.Exercises: Chapter 2. Then. 1) and 5 de 7 16-03-2010 17:00 .. 25. let be the open rectangle centered at with sides of length . and choose For each . Since the set of these rectangles is an open cover of the compact set .edu/~ken/ma570/homework/hw. If is open and function is compact. This follows from part (a).. . be as in Problem 2-25 and such that for Following the hint. Show that there is a and for function .Exercises: Chapter 2. . Let be the function defined for this rectangle as in part (c). we have positive on . let Now use Problem 2-25 to prove that e. a finite number of them also cover . define by Show that is a function which is positive on and zero elsewhere. Section 3 http://www.ms. let Since we have a subcover of . The choice 6 de 7 16-03-2010 17:00 . say the rectangles corresponding to form a subcover. If works. right.. d. using Problem 2-25 on the side closest to the origin. f. show that there is a non-negative such that for and outside . of some closed set contained in Let be the distance between and the complement of .uky. Finally. Since is compact.Exercises: Chapter 2. This proves the assertion. is either the . replace with where is the function of part (b). g. As suggested in the hint.edu/~ken/ma570/homework/hw.. Define by Show that the maximum of maximum of on on . or the maximum of This is obvious because 7 de 7 16-03-2010 17:00 .uky. 26.. so that and Let be as in part (d). Show that we can choose such an for .ms. We know that for all . one knows that is attains its minimum (Problem 1-29). Section 3 http://www. of guarantees that the union of the closures of the rectangles in the subcover is contained in and is clearly zero outside of this union. It is easy to verify that this new satisfies the required conditions. Exercises: Chapter 2. .. Section 4 http://www.uky. . is denoted . 1 de 4 and called the directional derivative of at 16-03-2010 17:01 . Find expressions for the partial derivatives of the following functions: a. in the direction .. 29. the limit if it exists.edu/~ken/ma570/homework/hw. Let . Section 4 28.Exercises: Chapter 2. b. d. . c. For .ms. a.edu/~ken/ma570/homework/hw. c. . This is obvious from the definitions. Show that . exists By Problem 1-26 (a). Then .. show that . With the notation of Problem 2-4. 32. exists for all . part (a) of that problem says that exists for all . Let be defined by 2 de 4 16-03-2010 17:01 . Show that but if . Section 4 http://www. Let be defined as in Problem 2-4. for all . Now suppose . Show that . If is differentiable at . Show that . b.Exercises: Chapter 2. and therefore One has which shows the result whenever trivially true. although is not even continuous at (0. then and all ..ms. 31. Let for all be defined as in Problem 1-26. But . The case when is The last assertion follows from the additivity of the function 30.0).uky. a. is not true for all . 0) but that is not continuous at The derivative at (0. and so they cannot be 3 de 4 16-03-2010 17:01 . is differentiable at (0. So. It follows from the differentiability of . At is not continuous at 0. But the second term takes on all values between -1 and 1 in every open neighborhood of . Let be defined by Show that .edu/~ken/ma570/homework/hw. for where is as in part (a). Show that Clearly.ms. one has since . just as in part (a).uky. 0) is the zero linear transformation because . Section 4 http://www. that and are defined for . one has .Exercises: Chapter 2. does not even exist. b. The first term has limit 0 as approaches 0... (The argument given above also shows that they are defined and 0 at . up to a sign. However. is differentiable at 0 but is differentiable at .) Further the partials are equal to continuous at 0. For . . Then . 34. If is differentiable and such that ..edu/~ken/ma570/homework/hw. In the case. Substituting in these two formulas show the result. .ms. A function and . at may be eliminated from the Proceed as in the proof of Theorem 2-8 for all suffices to note that from the definition of proof. Section 4 http://www. prove that there exist Following the hint. we have the result with . So. 33. let Theorem 2-9 gives . show that Applying Theorem 2-9 to other hand. On the other hand.uky.. 35.Exercises: Chapter 2. On the . 4 de 4 16-03-2010 17:01 . If . it follows . gives and so . This is all that is needed in the rest of the if for all is homogeneous of degree is also differentiable. Show that the continuity of hypothesis of Theorem 2-8. Exercises: Chapter 2, Section 5 http://www.ms.uky.edu/~ken/ma570/homework/hw... Exercises: Chapter 2, Section 5 36. Let set and open set . . By Theorem 2-11, such that and is open. Furthermore is differentiable at is differentiable. in place of , we see that . be an open set and for all a continusously differentiable . Show that is an open is open for any 1-1 function such that is differentiable. Show also that For every , there is an with there is an open set and an open subset . Since clearly , this shows that is differentiable. It follows that Since was arbitrary, it follows that By applying the previous results to the set is open. 37. a. Let is not 1-1. be a continuously differentiable function. Show that We will show the result is true even if non-empty open subset of case where neighborhood function is only defined in a is (the . Following the hint, we know that is analogous). Then there is an open with for all . The not constant in any open set. So, suppose we have of for all defined by . Assuming that satisfies and hence are 1-1, we can apply Problem 2-36. The inverse function is clearly of the form and so for all . Now is open but each horizontal line intersects at most once since is 1-1. This is a contradiction since is non-empty and open. b. Generalize this result to tthe case of a continuously differentiable function with . By replacing with a vector of variables, the proof of part (a) generalizes to the case where is a function defined on an 1 de 3 16-03-2010 17:01 Exercises: Chapter 2, Section 5 http://www.ms.uky.edu/~ken/ma570/homework/hw... open subset of where . where is an open subset , For the general case of a map of with , if with then we replace and drop out is constant in a non-empty open set reducing the value of for some by one. On the other hand, if the function like have made defined by . Replace with , then consider . Just as in part (a), and its inverse will look . Note that we this will be invertible on an open subset of . Again, by restricting to an appropriate and get a 1-1 function rectangle, we can simply fix the value of defined on on a rectangle in one less dimension and mapping into a space of dimension one less. By repeating this process, one eventually gets to the case where is equal to 1, which we have already taken care of. 38. a. If of . for some . By the mean value between and such that . Since both factors on the right are non-zero, this is impossible. b. Define by for all Clearly, but for all for all 39. Use the function . is not 1-1. . The function is not 1-1 since . Show that satisfies for all , show that is 1-1 on all Suppose one has theorem, there is a defined by 2 de 3 16-03-2010 17:01 Exercises: Chapter 2, Section 5 http://www.ms.uky.edu/~ken/ma570/homework/hw... to show that continuity of the derivative cannot be eliminated from the hypothesis of Theorem 2-11. Clearly, is differentiable for . At , one has So for satisfies the conditions of Theorem 2-11 at . except that it is not continuously differentiable at 0 since Now and it is straightforward to verify that for all sufficiently large positive integers . By the intermediate value theorem, there is a between and where . By taking n larger and larger, we see that is not 1-1 on any neighborhood of 0. 3 de 3 16-03-2010 17:01 Section 6 http://www. An explicit formula can be obtained by using Cramer's rule.. For each . By the uniqueness of the solutions in the last paragraph.uky. Note that this is of the same form as the set of equations for except that the right hand side functions have changed. In particular. call that solution by for all for . By differentiating the relation . Now. the uniqueness condition guarantees that is the same as the function provided by the implicit function theorem applied to .Exercises: Chapter 2. let be this be differentiable. the Implicit function theorem says that there is a defined in an open neighborhood of and such that function and is differentiable. it must be that for all in the domain of . In particular. was meant to be continuously is differentiable and defined with by there is a unique In this problem. Section 6 40.edu/~ken/ma570/homework/hw. is differentiable and differentiating this last relation gives . Define One has condition guarantees that for each . Suppose that for each . If for all . one gets for . show that Just as in the last problem. the various functions all glue together into a single function defined on all of and differentiable everywhere. Exercises: Chapter 2. 41. Let . for each . there is exactly one solution of . Solving for gives the 1 de 3 16-03-2010 17:02 . it is assumed that differentiable. The determinant .. a. .ms. Use the implicit function theorem to re-do Problem 2-15(c). Also. if and there is a unique is achieved at . Then we need to maximize . and (since both and are positive). Then . but just for those in the interval . So we need to go back and show that the earlier parts of the problem generalize to this case. Now actually. the minimum of occurs at by the second derivative test. Suppose . b.. so the maximum occurs at .. This follows immediately from part (a).ms. and so for fixed . This function has no zeros in the interval because has derivative which is always negative in the interval and the value of the function 2 de 3 16-03-2010 17:02 . Suppose function is . at We will find where the maximum of the minimum's are located in each of the three cases. then for some we have and Note that is defined only when and are positive. The derivative of this . Let . Show that if .Exercises: Chapter 2. One has so the hypothesis of part (a) is true and precisely when . c. we are not looking for the minimum over all . Find . The derivative of this function is negative throughout the interval. The maximum value is . The derivative for when fixed .uky.edu/~ken/ma570/homework/hw. For . Section 6 http://www. Further where if . precisely . and at if . result. there are no difficulties in doing this. By the result of the previous paragraph the maximum over the entire interval must therefore occur at .Exercises: Chapter 2. In view of the last paragraph. . and the value of the maximum is . Section 6 http://www. The maximum value is . Now on the interval. Suppose . and so the maximum must occur at the left hand end point. This is a decreasing function and so the maximum occurs at the left hand endpoint. Then . that means that the maximum over the entire interval occurs at .edu/~ken/ma570/homework/hw...ms.uky. 3 de 3 16-03-2010 17:02 . is positive at the right end point. In fact. Let be defined by Show that is integrable and that . . Section 1 http://www. Exercises: Chapter 3.ms. lower) . . if you partition by dividing each side into equal sized subintervals and . Let Show that be integrable and let is integrable and For any . 2. and .. Let than and be a partition as in Theorem 3-3 applied to such that every subrectangle of where . and so In fact. then any common has the same property. if refinement of this partition and is any partition. The intuitive iddea of the proof is that the worst case is when the point is in a `corner'. then any point is an element of at If most of the subrectangles of . because the and is any upper bound for the volume of the subrectangles of terms of the sum can differ only on those subrectangles which contain at 1 de 6 16-03-2010 17:02 . Then the hypotheses of Theorem is integrable. Let be a refinement of has volume less is the number of points where and have values which differ. then the volume of each subrectangle is precisely which is less than as soon as . except at finitely many points. where .Exercises: Chapter 3.uky. Section 1 1.. Furthermore. there is a partition of in which every subrectangle has volume less than .edu/~ken/ma570/homework/hw. (resp. the real proof is of course an induction on m. and is a partition of . and bounds for the values for all 3-3 are satisfied by and . where . Apply Theorem 3-3 to the partition For this partition. ) are upper (resp. uky..edu/~ken/ma570/homework/hw. Let (resp. one has and since greatest lower bounds are lower bounds. Show that is integrable and . Taking differences gives 3. ) and . 2 de 6 16-03-2010 17:02 .Exercises: Chapter 3. For any partition of and any subrectangle and therefore and . A similar argument shows the result for . and are just positively weighted sums of the . b. and so it is at most equal to the greatest lower bound of these values. a.. A similar argument shows the result for . Let be a common refinement of and . Since .ms. . . . By Theorem 3-3. Section 1 http://www. show that and For each . Further is integrable. Then by part (a) and Lemma 3-1. Let be integrable. least one of the points where and differ. Adding these inequalities shows that is a lower bound for . of . and the result for can be obtained by summing (with weights) the inequalities for the . ) be a partition as in Theorem 3-3 applied to (resp. For any constant . the other case being proved in a similar manner. Section 1 http://www... Since and for each subrectangle of . By the squeeze principle. its integral is 4. We will show the result in the case where .ms. one concludes that c. Let and be a partition of .uky. Let be a partition as in Theorem 3-3 applied to and . Show that is integrable if and 3 de 6 16-03-2010 17:02 .edu/~ken/ma570/homework/hw. applied to and . is integrable. show that . by the squeeze principle. we have By Theorem 3-3.Exercises: Chapter 3. . the function . only if for each subrectangle restricted to the function . Then . Let be the partition of A obtained by of the whose . taking the union of all the subsequences defining the partitions of the (for each dimension). by the squeeze principle. the function is integrable.ms. . 4 de 6 16-03-2010 17:02 .Exercises: Chapter 3. Let be a common refinement of and . and that in this case Suppose that is integrable and . Then there is a partition of whose subrectangles are precisely the subrectangles of which are contained in . it follows that is integrable. By Theorem 3-3. which consists of .uky. and. Section 1 http://www. Let be a partition of as in Theorem 3-3 applied to and . Then there are refinements rectangles are the set of all subrectangles of One has which are contained in By Theorem 3-3. Let be a partition as in Theorem 3-3 applied to and where is the number of rectangles in . Suppose that all the are integrable where is any subrectangle of ...edu/~ken/ma570/homework/hw. is integrble. If is integrable and in which is the only rectangle. But then by is integrable. it Similarly.. But then. so if . . 5. if . Since follows that 7. Then this implies that So. Let be integrable and suppose . Further.ms. is integrable. be defined by 5 de 6 16-03-2010 17:02 . Let . it follows that integrable. For any rectangle contained in . By Problem 3-3. we have . Problem 3-5 implies that by Problem 3-3 (c). Section 1 http://www.uky. If .Exercises: Chapter 3. . Show that . is integrable by Theorem 3-3. one can show that Problem 3-3. the function Using the trivial partition since 6. is integrable and .edu/~ken/ma570/homework/hw. since . is integrable. On the other hand. it has the desired value. Let be a and .. then partition as in Theorem 3-3 applied to . This proves the result. we have and then . show that Consider the function . Since there are at most such pairs . Show that is integrable and . the value of and their total volume is. of course.uky.ms.edu/~ken/ma570/homework/hw. such a exists and the total volume of all the rectangles containing points of this type is at most . no larger than 1. Section 1 http://www. It follows that . 6 de 6 16-03-2010 17:02 . the contribution to from these rectangles is also at most . Let be any partition of such that every point with lies in a rectangle of of height (in the direction) at most .. so their contribution to is at most . For the remaining rectangles .Exercises: Chapter 3. Since . By Theorem 3-3. is integrable and the squeeze principle implies that its integral is 0.. Choose a positive integer so that . Let . Prove that . say . if contains an interior point not in for some . it is of measure zero. Exercises: Chapter 3. also has . Then contains all the is bounded. for . Section 2 8. If C is a set of content 0. contrary to hypothesis. contains all the natural numbers and the total volume of a. Let . The sum of the volumes of the rectangles of is the volume of .ms. 9. then is a subset of for some since the union of the is .. are rectangles. if has non-empty interior. On the other hand. The set of natural numbers is unbounded. if .uky. Indeed. then contains only boundary points of .edu/~ken/ma570/homework/hw. So is not of content 0 as it cannot be covered with rectangles of total area less than the volume of . a.Exercises: Chapter 3. Show that an unbounded set Suppose where and . show that the boundary of content 0. with . Section 2 http://www. Suppose for By replacing the is not of content 0 if for b. Choose a partition which refines all of the partitions where Note that is a rectangle of the cover . Give an example of a closed set of measure 0 which does not have content 0. But then cannot have content 0.. then the union of the open intervals all the intervals is 10. . So. are closed rectangles which form a cover for . Let and hence also where . Let be any rectangle in with non-empty interior. and hence not of content 0 by part (a). Since the intersection of any two rectangles of a partition is contained in their boundaries. one can assume that for all . Suppose a finite set of open rectangles 1 de 3 16-03-2010 17:04 . which is at most equal to the sum of the volumes of the . Then the union of the cover the boundary of and have total volume less than the boundary of is also of content 0. and hence has measure 0 by Theorem 3-4. But then the sum of the lengths of the intervals in this finite subcover would be less than 1. Section 2 http://www. 11. but its boundary (by Theorem 3-6 and Problem 3-8). 13. But then there is a finite collection of open intervals which cover the set and have total volume less than . we know by Problem 1-30 that the set of where if finite for every .ms. a. show that there is a 2 de 3 16-03-2010 17:04 .. it must be countable too. 1]. Show that is a set of measure 0. Using the hint. show that the boundary of The set closed and bounded. So . . Since the set of these intervals is just a subset of this set. is of measure 0 (cf is not of measure 0 . Let be the set of Problem 1-18. form a countable set). and cartesian products of countable sets are countable.edu/~ken/ma570/homework/hw. If it were also of measure 0.uky. there is a finite subcover of . 12. and hence compact. The set of rational numbers in the interval Proof of Problem 3-9 (b)). Let and have total volume less than where where . so is the set of all -tuples of rational numbers. If does not have measure 0. b.. then it would be of content 0 by Theorem 3-6. If is any set and is an open cover of .e. Show that the set of all rectangles where each and each are rational can be arranged into a sequence (i. Since the set these open intervals together with the set of form an open cover of [0. Let be an increasing function. Give an example of a bounded set of measure 0 such that the boundry of does not have measure 0.Exercises: Chapter 3. b. Since the set of rational numbers is countable. cover of . contrary to Theorem 3-5. Hence the set of discontinuities of is a countable union of finite sets. Section 2 http://www. By part (a).Exercises: Chapter 3.ms. 3 de 3 16-03-2010 17:04 . there is a rectangle B of the type in part (a) such that has non-zero volume. sequence of members of which also cover . In particular.. contains and is contained in some in . In fact. Following the hint. for each in .edu/~ken/ma570/homework/hw. the union of the interiors of the rectangles (where is allowed to range throughout ) is a cover of . and hence so are the set of corresponding 's. this set of corresponding 's cover .uky. the set of these are countable. we can even assume that is in the interior of the rectangle .. for some closed rectangle and is Jordan-measurable and If has content 0. so it is a subset has content 0. so . Let be the set of rational numbers in . by Problem 3-5. exists.Exercises: Chapter 3. so is . But then the boundary of is contained in the closure of . which is not of measure 0. it follows that 1 de 3 16-03-2010 17:05 . Section 3 http://www. Further. show that . The set of where is not continuous is contained in the union of the sets where and are not continuous. show that has . Show that if has content 0. let be a partition of where is a closed rectangle containing . then it is bounded by Problem 3-9 (a). Show that if are integrable. Section 3 14.ms.. Exercises: Chapter 3. so . so theee first set is also of measure 0. So . But then there is a point of outside of . Then is not of measure 0 by Problem 3-8. If is a bounded set of measure 0 and Using the hint. one has for of an closed rectangle . 18. But then the boundary of must be of content 0. Give an example of a bounded set exist. Since this is true of all of . one has . one has which can be made as small as desired.uky.. Since this holds for all partitions if the integral exists. which is contained in the union of the closures of the (since this union is closed). Then the boundary of is does not exist by Theorem 3-9. then .edu/~ken/ma570/homework/hw. of measure 0 such that does not 16. 17. 15. and so . But then is integrable by Theorem 3-8. These last two sets are of measure 0 by Theorem 3-8. and so is Jordan measurable by Theorem 3-9. Then let be a rectangle of of positive volume. Since some open rectangles the sum of whose volumes can be made as small as desired. If is non-negative and . where intersecting and is Jordan measurable if and of such that consists of all subrectangles . 19. This satisfies the condition in the statement of the problem. If the set where is not continuous is not of measure 0. 20. This is an immediate consequence of Problem 3-12 and Theorem 3-8. we have . show that only if for every there is a partition . then taking the union of this set with the set of measure 0 consisting of differ gives a set of measure 0 which contains the points where and the set of points where is not continuous. So . the sum of whose volumes is at most ..Exercises: Chapter 3. Show that an increeasing function is integrable on . Let be a partition of such that is a rectangle of which intersects . Show that if . Let and choose a finite set for of open rectangles the sum of whose volumes is less than and such that the form a cover of the boundary of . then is not integrable by Theorem 3-8. Now apply Theorem 3-4 to conclude that has measure 0. So this set is also of measure 0.uky. If is a closed rectangle. Following the hint. Then if . Let be a partition of such that every subrectangle of is either contained within each or does not intersect it. . So has content 0.edu/~ken/ma570/homework/hw. there is a partition as in the statement. consists of allsubrectangles contained in Suppose is Jordan measurable. 2 de 3 16-03-2010 17:05 . Suppose for every . which is a contradiction.ms. By replacing the closed rectangles with slightly larger open rectangles. let be a positive integer and Let . On the other hand. if it is of measure 0. one can obtain the same result except now one will have in place of and the will be open rectangles.. one gets an open rectangular cover of with sets. This shows that the boundary of is of content 0. Then by replacing the rectangles with slightly larger ones. 21. Then its boundary is of content 0 by Theorem 3-9. Section 3 http://www. Let be the open set of Problem 3-11. which is not of except on a set of measure 0. measure 0. then f is not integrable on The set of where is not continuous is measure 0. is a Jordan measurable set and such that Jordan measurable set Let be a closed rectangle containing . Define . Apply Problem 3-21 with as the Jordan measurable set. .Exercises: Chapter 3.. hence 22.edu/~ken/ma570/homework/hw. If is Jordan measurable by Theorem 3-9. 3 de 3 16-03-2010 17:05 . Further . Then and clearly is Jordan measurable by Theorem 3-9.uky. show that there is a compact .ms. Section 3 http://www. Let be the partition as in Problem 3-21.. because the integral of its characteristic function does not exist. 26. and hence is not of measure 0. Use to show that the word ``measure" in Problem 3-23 cannot be replaced with ``content". Then the set can be covered by the rectangles and for each in . is not a set of measure 0 . Show that Following the hint. The sum of the areas of these rectangles is less than 25. Let has area One has . Section 4 23.. This follows from Problem 3-8 and Theorem 3-6. but not of content 0. Section 4 http://www. Both of these having integral 0 implies by Problem 3-18 that the sets where their integrand is non-zero are of is also of measure 0.Exercises: Chapter 3. and let . but that is not an induction. measure 0. Let be the union of all where is a rational number in written in lowest terms. We have condition that either or . let . Let be the set of all such that is not of content 0. be integrable and non-negative. Show by induction on (or content 0) if that for each .edu/~ken/ma570/homework/hw. Exercises: Chapter 3. To see that the set has content 0.ms. and so by Fubini. Show that is Jordan measurable and 1 de 6 16-03-2010 17:05 . Fubini's Theorem and induction on show that and so does not have content 0. is integrable with .. and so 24. by Problem is equivalent to the 3-15 and Fubini's Theorem.uky. Let be a set of content 0. Let be such that . Now is a set of measure 0. The set is the set of rational numbers in which is of measure 0. the rectangle where lowest terms with . uky. then we may assume (by replacing with if necessary that it is positive at . if is not zero for some point . where 27. show that where the upper bounds need to be determined... one has 2 de 6 16-03-2010 17:05 . If is an upper bound on the image of .edu/~ken/ma570/homework/hw. But then continuity implies that it is positive on a rectangle containing . On the other hand. is continuous. But then its integral over is also positive. shows that this integral is equal to 28. Section 4 http://www. if these are . By Fubini.ms.Exercises: Chapter 3. using Fubini on gives: Similarly. Use Fubini's Theorem to give an easy proof that continuous. Following the hint. the left hand iterated integral is just where Applying Fubini again. uky. obtained 29.. let be the original Jordan measurable set in the -plane. for in the interior of .Exercises: Chapter 3. The problem has a typo in it. that. Subtracting gives: which is a contradiction. Assuming that we know how to do . the result becomes 30. define by What is Let . We have 3 de 6 16-03-2010 17:05 . fix ? be in the interior of .edu/~ken/ma570/homework/hw.. arguments of The iterated integrals are zero because the inside integral is the zero function.ms. 31. To avoid overlap. Use Fubini's Theorem to derive an expression for the volume of a set in by revolving a Jordan measurable set in the -plane about the -axis. Section 4 http://www. The last integral cannot exist by Theorem 3-9 and Problem 1-17. and replace it with . the author should not have switched the order of the as that trivializes the assertion. The problem appears to be premature since we really want to be able to do a change of variables to cylindrical coordinates. If and is continuous. Let be the set in Problem 1-17. To do this. it is convenient to keep the set in the positive half plane. Show that but that does not exist. Theorem 3-9 can be used to show that is Jordan measurable if is. Let be continuous and suppose . .ms.uky. and is continuous.. Define . Let Problem 2-21.. and so by the chain rule one has 34. One has 33. Prove Leibnitz' Rule: Using the hint. define where the second assertion used . Find One has Problem 3-32.edu/~ken/ma570/homework/hw. by Fubini's Theorem. If We have and is continuous and . b. a. If . As in 4 de 6 16-03-2010 17:05 . find .Exercises: Chapter 3. 32. let be continuously differentiable and suppose . Section 4 http://www. we have is continuous. . Let be a linear transformation of one of the following types: If is a rectangle. 5 de 6 16-03-2010 17:05 . show that the volume of is . Show that One has .uky. Section 4 http://www.Exercises: Chapter 3.ms. 35..edu/~ken/ma570/homework/hw. a. This is an immediate consequence of Fubini's Theorem since the inside integrals are equal. the result follows in this case. then set in this proper subspace. Show that and have the same volume. This shows that the compact portion of the hyperplane is of volume 0. the hyperplane is the image of a linear transformation from into made up of a composition of maps of the first two types.ms. the parallelogram base is in the and directions and has corners . Suppose that each and are Jordan measurable and have the same area.edu/~ken/ma570/homework/hw. In particular. then is in the first case. So the volumes do not change in the second and third case and get multiplied by in the first case. In the second case. is a proper subspace of and is a compact If is singular. Section 4 http://www.. Prove that . If is non-singular. Since is multiplicative. (Cavalieri's principle). 6 de 6 16-03-2010 17:05 . Let is the volume of for any linear transformation and define similarly. By choosing the coordinate properly.uky.Exercises: Chapter 3. rectangle is . Let and be Jordan measurable subsets of . In the three cases. 36. is contained in a hyperplane. is a cylinder with a parallelogram base in the second case. If the original . This shows the result. this shows the result in this case too. b.. 1. and 1 respectively. Since the determinant is also 0. then it is a composition of linear transformations of the types in part (a) of the problem. and places are and is the same rectangle except that the intervals in the swapped in the third case. Consider a partition of unity subordinate to the cover . Suppose that for all . But . we only know the values of the integral of on the sets . but Take a partition of unity exists. one can assume that there is only one . Exercises: Chapter 3. Suppose that Show that For is a non-negative continuous function.so it could be that may not even exist for all To correct the situation. a. we . If not necessrily true. Let and not exist. exists if and only if exists. It follows that the sum in the middle does not converge as The assertion that and so does not exist. By summing the with the same in condition (4) of Theorem 3-11. but don't know how behaves on other intervals -.. .. So the sum converges if and only if b.uky. Show that satisfies does . Consider the convergence of where . define and . let us assume that 1 de 2 is of constant sign and bounded on 16-03-2010 17:06 . As in part (a). let it be . a natural number. From the hypothesis. Now exists if and only function for each converges.ms. One has where as in condition (4) of Theorem .edu/~ken/ma570/homework/hw.Exercises: Chapter 3. Section 5 37. subordinate to the cover for can assume there is only one 3-11. Section 5 http://www. Finally.uky. Let be a closed set contained in and and such that outside and . 38. Find two partitions of converge .. By using arrangements with different limiting values. By forming open covers each set of which consists of intervals for the sum of terms added to each of these partial sums. the integral in the old sense is monotone on each interval of . and the limit is just satisfies unity absolutely to different values. The sums and have terms of the same sign and . one gets the result. one gets covers of . Clearly. by re-ordering the terms of make the sum approach any number we want..ms. 2 de 2 16-03-2010 17:06 . Because is zero outside . one can are each divergent.edu/~ken/ma570/homework/hw. the integral in the extended sense is same as the that in the old sense. Then is bounded on each interval and so by Theorem 3-12. each set .Exercises: Chapter 3. one can `fatten' up the covering sets so that they are a cover of the real numbers no smaller than 1 without adding any points where is non-zero. Suppose that . one can take a partition of unity subordinate to this cover. further this can be done so that there are sequences of partial sums which converge monotonically to the limit value. So. Section 5 http://www. and so terms are identical. we get We use the same idea as in the proof of Theorem 3-13. If we can write we can write and . Let unity subordinate to an admissible cover is a partion of unity subordinate to the cover . inverses of and is a linear transformation. Combining results gives 1 de 5 16-03-2010 17:06 . . we know that Theorem 3-13.Exercises: Chapter 3. and .edu/~ken/ma570/homework/hw.. prove that in some open set containing . Let applies with Then in place of the of is open and Theorem 3-13 be a partition of . Then . Combining results. Then successively smaller open neighborhoods of . where . Show that is absolutely also converges since the . Section 6 http://www.. be a point .uky. Now convergent. Define for . So we can define on . 40. Section 6 39. Use Theorem 3-14 to prove Theorem 3-13 without the assumption that .ms. Let . One then can verify that . Let where . and if and only if is of the form is a diagonal matrix. Exercises: Chapter 3. So. Then in its statement. . By Theorem 3-14. the converse is false. . is a diagonal matrix. consider the function . compute . Suppose implies for all . Show that is not a diagonal matrix. Since is linear. it follows that has the same value. to show that the function and is 1-1.Exercises: Chapter 3. Define a. One has and imply implies that .. and so Now. so 41. and so in the domain of and . For example. If . But then contrary to hypothesis. then replace . i. But then . So. of Problem . Since suffices to show that . . If .edu/~ken/ma570/homework/hw. So. Show that for all 2-23. . is 1-1. it . then 2 de 5 16-03-2010 17:06 .e. Section 6 http://www. Suppose (or is 1-1. and show that is the set .uky.ms. for have the same form as the On the other hand. by . if and .. Then ). Then the with and . Clearly. show that (Here denotes the inverse of the function .0) through b. .pital's Rule allows one to calculate when by checking separately for the limit from the left and the limit from the right. If is integrable and .ms. c. One has is called the follow from the last paragraph of . The formulas for and the solution of part (a). and the ray from (0. then let -axis be the angle between the positive . Section 6 http://www..) Find P'(x. Further.edu/~ken/ma570/homework/hw. where . . show that 3 de 5 16-03-2010 17:06 . if and let . If . Then . contrary to hypothesis.uky. Let be the region between the circles of radii and and the half-lines through 0 which make angles of and with the -axis.y). On the other hand. The function polar coordinate system on . This is trivial from the formulas except in case .. For example.Exercises: Chapter 3. L'H@ocirc. So the first identity holds. apply part (c) with .uky. e.Exercises: Chapter 3. Apply Theorem 3-13 to the map . d. Then and conclude that 4 de 5 16-03-2010 17:06 .. If . One has . show that and For the first assertion. Applying (c) gives . Section 6 http://www. If .edu/~ken/ma570/homework/hw. Prove that . show that Assume that and and by .. The second identity is a special case of the first.ms. The second assertion follows from Fubini's Theorem. Since part (d) implies that . Section 6 http://www. also.uky.edu/~ken/ma570/homework/hw.ms.. the squeeze principle implies that But using part (d) again. One has So and the integrands are everywhere positive.Exercises: Chapter 3. we get (since the square root function is continuous).. also exists and is 5 de 5 16-03-2010 17:06 . If the factor were not in the definition of . Show that side be if the factor ? . Assume therefore that the are distinct. Show that is the determinant of thee minor of obtained by selecting columns . in that case. One has using Theorem 4-1(3): because all the summands except that corresponding to the identity permutation are zero. Assume as in part (a) that the are all distinct.edu/~ken/ma570/homework/hw. a. Section 1 http://www. then the right hand side would have been .uky. Let be the usual basis of and let be the dual basis. Consider the linear map defined by . b.. it follows that we need only verify the result when the are in the subspace generated by the for .Exercises: Chapter 4. A computation similar to that of part (a) shows that if some for all . What would the right hand did not appear in the definition of The result is false if the are not distinct. Then 1 de 8 16-03-2010 17:07 . By multilinearity. Exercises: Chapter 4.ms. the value is zero.. Section 1 1. then . If . 2..edu/~ken/ma570/homework/hw. One has for all : This shows the result.Exercises: Chapter 4. 2 de 8 16-03-2010 17:07 is a linear transformation and . Show that must be multiplication by some constant .ms..uky. Section 1 http://www. . 6..uky. and is .ms. show that is continuous and each . show that where Let . 3. If is the volume element of determined by . a. the right hand sides are just the entries of and so substituting gives the result. If . and . it follows that the image of consists of numbers all of the same sign. and . So all the have the same orientation. So .edu/~ken/ma570/homework/hw. whose image does not contain 0 since is a basis for every t. 5. be an orthonormal basis for V with respect to where . By Theorem 4-6. . . Then we have by blinearity: . 4. Taking absolute values and an isomorphism such that . by the definition of is the volume element with respect to for some because and .Exercises: Chapter 4. If is the volume element determined by . Then by Theorem 4-6.. what is ? is the cross product of a single vector. If . Section 1 http://www. and so desired. and let . show that One has and the fact that Further. is of dimension 1. By the intermediate value theorem. Combining. we have and such that The function is a continuous function. 3 de 8 16-03-2010 17:07 . one has for . is a basis for as and . Let . i. Show that every non-zero is the volume element determined by some inner product and orientation for . Then to . for every . Let . such that for all The cross product is the . we have .uky. Since the are linearly independent. . and are an orthonormal basis of such that for with respect is a volume element. If are linearly independent. Let be the volume element determined by some inner product and orientation .. . the definition of cross product with completing the basis shows that the cross product is not zero. There is a scalar . By the definition.ms. and let be an orthornormal basis (with respect to ) such that . Section 1 http://www. define a ``cross product" .. 9. This shows that 8.Exercises: Chapter 4. and . 7. show that is the usual orientation of . If in terms of is the volume element of determined by and . Substitution shows that b. it is the vector such that works.edu/~ken/ma570/homework/hw. which shows the result. So in fact. the determinant is positive. . Deduce the following properties of the cross product in : 4 de 8 16-03-2010 17:07 .e. The second assertion follows from the definition since the determinant of a square matrix with two identical rows is zero.ms. c.. where . All of these follow immediately from the definition. Expanding out the determinant shows that: .uky. e. The result is true if either or both non-zero. the first identity is just . By Problem 1-8. This is easily verified by substitution using part (b).edu/~ken/ma570/homework/hw. and is zero. To show that . . 5 de 8 16-03-2010 17:07 . a. and since .. Suppose that and are . Section 1 http://www. b.g.Exercises: Chapter 4. note that for all . edu/~ken/ma570/homework/hw. But this can be easily verified by expanding everything out using the formula in part (b).ms.Exercises: Chapter 4. For the first assertion. one needs to show that for all .. one has and . Section 1 http://www. The third assertion follows from the second: 6 de 8 16-03-2010 17:07 ..uky. d. one has: So. For the second assertion. ms. This proves the result in the case where is not zero. 10. If .edu/~ken/ma570/homework/hw. and the bilinearity of inner product imply that too...uky. one has: since the matrix from Problem 4-3 has the form . Section 1 http://www. the are linearly dependent. See the proof of part (c). a linear transformation is called 7 de 8 16-03-2010 17:07 . show that . If is an inner product on . e. When it is zero. 11. Using the definition of cross product and Problem 4-3. where .Exercises: Chapter 4. which shows the result. But .uky.Exercises: Chapter 4. Use Problem 2-14 to derive a are differentiable. show that One has one has: for each for all is the matrix of . If formula for . self-adjoint (with respect to ) if is an orthogonal basis and respect to this basis. Section 1 http://www. Since the cross product is multilinear.edu/~ken/ma570/homework/hw. define when by .. 12.ms. . one can apply Theorem 2-14 (b) and the chain rule to get: 8 de 8 16-03-2010 17:07 . Using the orthonormality of the basis.. If with . tangent vector to This is an immediate consequence of Problem 4-13 (a). Exercises: Chapter 4. Here is the interpretation: . show that and The second assertion follows from: b. One has by the definition and the product rule: 14. a.. The notation does not fully elucidate the meaning of the assertion.ms.Exercises: Chapter 4. a differentiable function . show that . Section 2 13. show that the at is . If .edu/~ken/ma570/homework/hw. If . that is. Section 2 http://www. Define the tangent vector of at as . 1 de 5 16-03-2010 17:07 ..uky. Let be a differentiable curve in . If and . The tangent vector of vector of at is at is lies on the tangent line to the graph of . . be a curve such that for all . . define a vector field by 18. If . If is the tangent vector to . prove that is the direction in which is 2 de 5 16-03-2010 17:07 .uky. By Problem 2-29. Given such an . Show that every vector field on A vector field is just a function element by b. Show that the end point of the tangent vector of at .Exercises: Chapter 4.ms.edu/~ken/ma570/homework/hw. If For obvious reasons we also write and conclude that changing fastest at . Let at . Then . Differentiating where 17. . gives at . Show that One has .e.. i. . 15. . The end point of the tangent which is certainly on the tangent line to the graph of 16. Show that and the tangent vector to at are perpendicular. Section 2 http://www. Let and define at by . define a vector field by a. define . which assigns to each an . is of the form for some .. uky. define the forms a. If is a vector field on .. The direction in which is changing fastest is the direction given by a unit vector such thatt is largest possible.e. Since where .. in the direction of .Exercises: Chapter 4. 19. i. Prove that The first equation is just Theorem 4-7. one has: 3 de 5 16-03-2010 17:07 . Section 2 http://www. For the second equation.edu/~ken/ma570/homework/hw. this is clearly when .ms. e. Similarly. by part (a) and by part (a) and Theorem 4-10 (3). if show that for some vector field on .edu/~ken/ma570/homework/hw. c. By part (a). show . is exact. For the third assertion: b. So .. so Also.ms. Use (a) to prove that One has Theorem 4-10 (3).uky. i. if 4-11. . If is a vector field on a star-shaped open set and that for some function . .Exercises: Chapter 4. 4 de 5 16-03-2010 17:07 . then . . so the second assertion is also true. By the Theorem . Section 2 http://www.. e. Then such that and so be a differentiable function with a differentiable inverse . it must then be exact. 21.ms. If every closed form on is exact. i. Prove that on the set where is defined. Section 2 http://www. In case . Similarly.e.uky. For example.Exercises: Chapter 4. L'H^{o}pital's Rule allows one to calculate when by checking separately for the limit from the left and the limit from the right..edu/~ken/ma570/homework/hw. the assertion is immediate from the definition of in Problem 2-41. Further. show that the same is Suppose that the form . 20. But then and so there is a form is also exact. if . 5 de 5 16-03-2010 17:07 . then and so is closed.. one has trivially because is constant when and (or ). on is closed. on . By Theorem 4-11. i. . So as desired. for some . we have Except when . as desired. Let true of . Show that there is . . integer Given for some 2-chain is the function of Problem 3-41 -axis. as desired. The boundary of 24. Show that every -chain can be written and singular -cubes Since . Let . One has . show that there is an . Define and by and . Define . On the and .. and and are and for .ms. Exercises: Chapter 4. An -chain 22. Section 3 be the set of all singular -cubes. let where with where and is easily seen to be are . Section 3 http://www.edu/~ken/ma570/homework/hw. If is a singular 1-cube in such that . For and an integer. Show that and are -chains if and are. the functions if and are. . positive real numbers. let also denote the function such that . and the integers..uky. 1 de 1 16-03-2010 17:07 .Exercises: Chapter 4. for some integers -chains The second assertion is obvious since 23. define the singular 1-cube by a singular 2-cube Define by such that . . If . Let is a function such that for all but finitely many . So so that extended so that it is 0 on the positive is an integer because and other boundaries. and use Stokes Theorem to conclude that for any 2-chain in (recall the definition of in 1 de 11 16-03-2010 17:08 . Independence of parametrization). the chain rule. Section 4 25. Show that Problem 4-23)..uky.. Using the definition of the integral. Section 4 http://www. One has .ms. show that be a singular -cube and and Suppose . and Theorem 3-13 augmented by Problem 3-39: 26. Exercises: Chapter 4. Let a 1-1 function such that for . If is a -form.edu/~ken/ma570/homework/hw. Theorem 4-9.Exercises: Chapter 4. 28. the letting Theorem. one gets which is a contradiction. We assume these changes have been made. So . This integer is called the winding number of around 0. Then. and when . Define . if 2 de 11 16-03-2010 17:08 .ms. So .Exercises: Chapter 4. is the curve .edu/~ken/ma570/homework/hw. it is the curve . Recall that the set of complex numbers . .uky. When .as the sum of the coefficients of a boundary is always 0. Let . it is . Note however that no curve is the boundary of any two chain -. and that if is is simply with ... and the where . one has and and are . Show that large enough. and If 2-chains. Section 4 http://www. If . 27. Show that the integer of Problem 4-24 is unique. If let be by . Using Stokes . The problem statement is flawed: the author wants to be defined to . When the curve or . This would make the boundary be . thee singular 1-cube singular 2-cube by a. then Stokes Theorem gives because is closed. So there is a unique 3 de 11 . prove that be this value . we have assumed that is a 2-chain with values in . The differential is of the type for which there is 16-03-2010 17:08 considered in the last problem. and so This contradicts the result of the last paragraph.Exercises: Chapter 4. So the boundary of is . and when or . is a 1-form 29. when . then such that and . we get the curve . we have for all and all with . show that theere is a for some function with has no complex root. we see by part (a) and Stokes' Theorem that . when . Letting be sufficiently large. if we let .edu/~ken/ma570/homework/hw.ms. If unique number . and so Now consider the 2-chain defined by . we get the constant curve with value . Suppose that as above has no complex root. .uky. for some and . prove the Fundamental Theorem of Algebra: Every polynomial with has a root in . Using Problem 4-26. on with such that . we get Following the hint.. we see that cannot be zero since it is the sum of a number of length and one which is smaller in absolute value. applying Stokes' Theorem. . On the other hand. Section 4 http://www. Further. and so . b.. and so and 30. If is a 1-form on implies is unique. Again. Now. . Since where . we get the curve . Use this fact. By Problem 4-24. Suppose . Then since the integrand is continuous and of the same sign throughout the region of integration.uky.. By Stokes' 4 de 11 16-03-2010 17:08 . denote . .ms. show that there is a chain to prove .0) to and postpend a path from to (1. ..Exercises: Chapter 4. So . if you have two singular 1-cubes and with and . In fact. define the singular 2-cube . then prepend a curve from (1. By Stokes' . and so the integrals over and are equal. Let be a chain such that .edu/~ken/ma570/homework/hw. a such that and .0) to get two paths as in the last paragraph. by For positive of the last problem. The two integrals are both 0. If . Then in a closed rectangle of positive volume centered at . . Now the result follows from Problem 4-32 below. By Stokes' Theorem. let . Henceforth. Take for the k-cube defined in an obvious way so that its image is the part of the closed rectangle with for all different from the for . Note that Let be a singular 1-cube with and an such that . Section 4 http://www. and in particular. Suppose such that . By the proof . 31. integrating is independent of path. From the result of the last paragraph. So there is a 2-chain Theorem. it follows that this common value. Stokes' theorem and One has for some and some choice of . we have . then because constant. . we (since is closed). . Now. if interior of the subset. Then is the curve with constant if is exact. Conclude that if is merely closed. . Note that because and differentible.. and hence closed. and shows that there is no independence of path in b. show that for all is exact. and . that is. then are continuously is in the .Exercises: Chapter 4.. is calculated with a is well defined. because 32. a. we can assume that path that ends in a segment with Similarly. Further. Then by Stokes' Theorem. This is a contradiction. For every in the set. The example: . we would have: . where and are degenerate . it follows that is closed since . by treating each component separately. and are points.ms. Show that there is a singular 2-cube such that . we assume that the subset is pathwise connected. Because of independence of path. If with is a 1-form on a subset of and for closed forms.edu/~ken/ma570/homework/hw. Fix a point from to in the subset.uky. Theorem. let be any curve . Clearly. . One has 5 de 11 16-03-2010 17:08 . We want to check differentiability of . and set . we assume that the subset is open. and Although it is not stated. Let be singular 1-cubes in with and . and so . is exact. Section 4 http://www. Give a counter-example on Let value Suppose have be defined by where and similarly for . and so the last integral is also .Exercises: Chapter 4.edu/~ken/ma570/homework/hw. subtraction. (A first course in complex variables.uky. and so does not have a limit as . The first pair of terms is because . and quotient of analytic functions are analytic... because .ms. if you use the identities: 6 de 11 16-03-2010 17:08 . define to be exists. So is differentiable at . It is straightforward to check that the complex addition. Show that ).) If is differeentiable at every point analytic on in an open set . The assertion that being analytic is preserved under these operations as well as the formulas for the derivatives are then obvious. product. then is called a. (This quotient involves two complex numbers and this definition is completely different from the one in Chapter 2. continuity of implies that the integrand is . Show that the sum. 33. and division operations are continuous (except when the quotient is zero). This establishes the assertion. On the other hand. similarly the second pair of terms is . Finally. but . is analytic and is not (where and is continuous on . multiplication. Section 4 http://www.) If differentiable at if the limit . 16-03-2010 17:08 . . we must have: are continuously differentiable.ms. Comparing the real and imaginary parts gives the Cauchy-Riemann equations. Let vector space over ).) Following the hint. . has a derivative which is multiplication by a complex number. If Riemann} equations: . be a linear transformation (where is considered as a c. Section 4 http://www.Exercises: Chapter 4. Part (b) shows that an analytic function . What complex number is this? Comparing gives 7 de 11 and . show that and satisfy the Cauchy- (The converse is also true. considered as a function . show that is multiplication by a complex number if and only if and .. If the matrix of with respect to the basis is . is analytic on b..uky.edu/~ken/ma570/homework/hw. if and this is more difficult to prove. Section 4 http://www. One has for if and only if that satisfies the Cauchy-Riemann Clearly this is zero if and only if the Cauchy-Riemann equations hold true for .uky. the complex number is and d. and . then ) (singular 1-cube with By parts (b) and (d). where From the last paragraph. 8 de 11 16-03-2010 17:08 . Prove the Cauchy Integral Theorem: If for every closed curve such that for some 2-chain in is analytic in . By Stokes' Theorem. and Show that equations.edu/~ken/ma570/homework/hw. e.. it . .ms.. So. the 1-form follows that is closed. Define .Exercises: Chapter 4. and exist if and only if and . . The first assertion follows from part (e) applied to the singular 2-cube defined by .Exercises: Chapter 4. Use (f) to evaluate and conclude: Cauchy Integral Formula: If is analytic on closed curve in with winding number then and is a around 0.edu/~ken/ma570/homework/hw. If is analytic on analytic in if for . (or in classical notation) . . Show that if equals . use the fact that to show that is This then gives g.uky. f.. Section 4 http://www. Conclude that One has if is defined by . then for some function . 9 de 11 16-03-2010 17:08 ..ms. and Problem 5-33 prove this for (b) but the proof for (c) requires different techniques. define by If each is a closed curve. Suppose and are homotopies of closed curves..uky. The Cauchy integral formula follows from this and the result of the last paragraph. Section 4 http://www. we conclude that . and the pair of (b) or (c). By a trivial modification of Problem 4-24 (to use Theorem. Further.ms.Exercises: Chapter 4. If and .edu/~ken/ma570/homework/hw. It is intuitively obvious that there is no such homotopy with the pair of curves shown in Figure 4-6 (a). If are nonintersecting closed curves. ) and Stokes' if is chosen so that for all with . the pair is called a homotopy between the non-intersecting closed curves and . a. 34. define by 10 de 11 16-03-2010 17:08 . It follows that . The present problem.. Using part (f). if for each the closed curves and do not intersect. is called a homotopy between the closed curve and the closed curve . for with . When (respectively ). one gets the singular 2-cube (respectively ). show that By Stokes' Theorem and part (a).uky.Exercises: Chapter 4. b. one gets the same singular 2-cube . one has .. one gets the same singular 2-cube . similarly.. 11 de 11 16-03-2010 17:08 .ms. If is a closed 2-form on . Section 4 http://www. If is a homotopy of nonintersecting closed curves define by Show that When .edu/~ken/ma570/homework/hw. when . So which agrees with the assertion only up to a sign. must satisfy . If is a -dimensional manifold with boundary. Show that is an -dimensional manifold with boundary.uky. say with the function . Following the hint. then the same point in such that . Then condition holds except for part (3) since 3. then following example: if is a manifold with boundary. . Section 1 http://www.Exercises: Chapter 5.edu/~ken/ma570/homework/hw. Find a counter-example to Theorem 5-2 if condition (3) is omitted. Let be one of the half-planes or Suppose there is a sequence of points of such that the all lie in and converge to . If there is no . . it follows that 2. Exercises: Chapter 5.ms. 1 de 4 16-03-2010 17:10 . So is a manifold of dimension . The boundary of Let be as in condition . is the set of points which satisfy condition works for every .. Let be an open set such that boundary is an -dimensional manifold. Then satisfies with .) Since is open. each such is in . and because those points which don't satisfy is a manifold of dimension . (It is well to bear in mind the . Section 1 1. a. In particular. prove that is a -dimensional manifold and is a =dimensional manifold. Further.. also is map which shows that condition is satisfied for each such . . consider defined by Let . but . each of its points satisfies condition with Let . one gets back into the case where one is contained within . Prove a partial converse to Theorem 5-1: If manifold and . Section 1 http://www.edu/~ken/ma570/homework/hw. This proves the be as in condition be defined by . One can verify that so that by satisfies the condition is an is differentiable. 5.. the map defined by easily verified to be a coordinate system around all points of the graph of f. Then the function satisfies all the desired conditions. . which is absurd since the points of U in the boundary of all map to points with last coordinate 0. If of If . It follows that h restricted to an appropriately small open subset of either satisfies condition assertion. b.. then there is an open set differentiable function has rank Let when . 6. 2 de 4 16-03-2010 17:10 . applied to . and choose together form a basis for .Exercises: Chapter 5. Prove a similar assertion for an open subset of an n-dimensional manifold. 4. But then the line segments from to must contain a point o the boundary of . By working in the set of condition . then there is a sequence of points of such that the sequence converges to . is -dimensional (vector) subspace of is a -dimensional be a basis for the subspace.uky. Define a map .ms. the graph of is -dimensional manifold if and only if . The generalization to manifolds is proved in the same way. and such that is a -dimensional containing and a and or condition . except you need to restrict attention to a coordinate system around . so the graph is a manifold of dimension n. open neighborhood of such that . Let all the . Show that the graph is differentiable. and the same argument applies. Prove that a manifold. By choosing the thickness of the plate sufficiently small. This shows the result. Section 1 http://www.Exercises: Chapter 5. For each and . show that Example: the torus (Figure 5-4). or is non-zero. Take a thin plate including the image of . Let is as in condition be the projection on the last for some point coordinates. Consider the case where n = 3. it is not obvious what one means by ``rotate". 8.. Let -dimensional manifold and . . b. Let be the domain of one of these functions. If . we for the can guarantee that this value is no more than element of the cover. Give a counter-example if . where we can choose to be a ball with center at rational coordinates and rational radius. Conversely. a. Since either easy to see that the Jacobian has the proper rank. it is In the case where n > 3. one has condition holding for some function .. If is a -dimensional manifold in measure 0. If is obtained by revolving is a around the axis -dimensional manifold. show is not 3 de 4 16-03-2010 17:10 .edu/~ken/ma570/homework/hw. then is defined by is defined in some open set by .uky. If is a closed -dimensional manifold with boundary in that the boundary of is . 7. show that has . its inverse image has volume which can be bounded by where is the volume of the plate (by the change of variables formula). Now each maps points of in to points with the last coordinates 0.ms. suppose in the graph. The Jacobian is is a . The Then apply the Implicit function theorem to differentiable function obtained from this theorem must be none other than since the graph is the set of points which map to zero by . Then is a countable cover of . edu/~ken/ma570/homework/hw. If is in the boundary of . it follows that is of measure 0.. since is bounded. Section 1 http://www. every element of condition closed. The open unit interval in to be closed. the boundary of is . By part (b). By Problem 5-1. By part (a). Finally.Exercises: Chapter 5. 4 de 4 16-03-2010 17:10 . it must satisfy condition . is an -dimensional manifold contained in .ms. closed. then by the since is is in is in the boundary of . is a counter-example if we do not require c. So if the interior of . Clearly. But then because the dimension of is n. If is a compact -dimensional manifold with boundary in show that is Jordan-measurable. .uky.. the definition of Jordan measurable is satisfied. Section 2 9. Since tangent vector of 10. Let be such that .edu/~ken/ma570/homework/hw. we know that implies that where . Show that there is a unique orientation is orientation-preserving for all . that is a coordinate system about . suppose that is a curve in with . it follows that the is a collection of coordinate systems for such that (1) For there is which is a coordinate system around . 16-03-2010 17:10 .Exercises: Chapter 5. consists of the tangent vectors at of curves in with Let be a coordinate system around in . Show that . and with . by replace with a subset. Section 2 http://www. Conversely. For and . where . (2) if .e. We know by the proof of Theorem 5-2. we must show that we get the same orientation if we use use and . Then ranges through out as and vary. But analogous to the author's observation of p. then such that Define the orientation to be the . Exercises: Chapter 5. Then let be as in condition for the point . In order for this to be well defined.. . let be the curve . Suppose each of is in .uky. 119. 1 de 7 as desired.ms. one can assume that is a rectangle centered at . for every .. Then we have i. Define and . be the projection on the first coordinates. by definition.uky. Let perpendicular to the usual orientation of then 12. 11. Section 2 http://www. show that the two definitions of given be a coordinate system about with where . define . choose a . is the induced orientation on is is .Exercises: Chapter 5. show that there is an on with open set and a differentiable vector field is the dimension of . let to . the definition makes orientation preserving for all this is only orientation which could satisfy this condition.ms. Let and and . For . and so. define . Then is a to which extends the restriction of be a partition of unity subordinate to with non-zero only for elements of Finally.edu/~ken/ma570/homework/hw. If (the orientation so defined is the usual . and as the usual orientation of orientation of . Note that . If is an -dimensional manifold-with-boundary in . But is the unit normal in the second sense. For every satisfying condition where differentiable vector field on .. and . is a differentiable vector field on for Let . Then is a differentiable extension of 2 de 7 16-03-2010 17:10 . If above agree. . For .. there is a diffeomorphism . Clearly. where . Let . a. If is closed. to all of . we have now assured that in a neighborhood of any point. by . In the construction of part (a). We have changed the order of the arguments to correct an apparent typographical error in the problem statement. and clearly it cannot be extended to the point in a differentiable manner. let . Section 2 http://www. one might by . This assures that there are at most finitely which intersect any given bounded set. Let . Let and .edu/~ken/ma570/homework/hw. b. But now we see many that the resulting is a differentiable extension of . Let a. In is a fact.uky. This is a have a vector field defined on vector field of outward pointing unit vectors. If be as in Theorem 5-1. be defined by . Also. . diffeomorphism such that Define the vectors by is 1-1 so that are linearly independent.. Let be defined.ms. 13.. Since is closed. let be the essentially unique and . one can assume that the are open rectangles with sides at most 1. For example. as in the proof of the implicit function theorem. is compact. We can then replace with the union of all these finite subcovers for all . and The notation will be changed. Now 3 de 7 16-03-2010 17:10 . sum of finitely many differentiable vector fields Note that the condition that was needed as points on the boundary of the set of part (a) could have infinitely many intersecting every open neighborhood of .Exercises: Chapter 5. Show that . . show that we can choose . and so we can choose a finite subcover of . Then and so . ms. Since it is 1-1. b. which is 1-1 because is a diffeomorphism. must map to linearly independent vectors. .edu/~ken/ma570/homework/hw. it maps its domain onto a space of dimension and so the vectors. Let be the set of . this follows can be defined consistently. Then 4 de 7 16-03-2010 17:10 . show that the components of the outward normal at some multiple of . As the hint says. If is an orientable are and so by considering This shows that as desired. Each is non-zero only inside some with sums of the be defined by . has rank 1 for Choose an orientation for . -dimensional manifold. and we can assume are distinct for satisfies the by replacing the distinct . we can assume locally that the orientation imposed by is the given orientation .. we have a defined in an open neighborhood of . we get is perpendicular to 14. Let be an -dimensional manifold in .uky. So for each . Problem 5-4 does the problem locally. Let . Show that the orientations is orientable. 15. If . and be a partion of unity subordinate to . that the . ince is a coordinate system about every point of from Problem 5-10 with . .Exercises: Chapter 5. we can assume that takes on non-negative values. Further. By replacing with its square.. show that there so that is an open set and and a differentiable . We have the components. being a basis. Section 2 http://www. using Problem 5-13. so that c. Let desired conditions. the . and the useful but irrelevant is called a Lagrangian multiplier. Then we have a coordinate systems of the form and of the form . show that there are be as (respectively . which is often very simple if we leave the equation for last. Section 2 http://www. Let of be a coordinate system in a neighborhood of the extremum at . we must solve subject to the constraints the system of equations. Choose an orientation on each piece so that adding ) gives the usual orientation on . If on occurs at . In the case of the M"{o}bius strip. This is Lagrange's method.uky. Then and so . In particular. The following problem gives a nice theoretical use for Lagrangian multipliers. and so the row of 5 de 7 16-03-2010 17:10 .Exercises: Chapter 5. Let in Problem 5-13. This is an orientation for is equivalent to a single ring maximum (or minimum) of The maximum on on is sometimes called the maximum of . end-points of normal vectors (in both directions) of length and suppose is small enough so that is also an -dimensional manifold.ms. Now the image is just the tangent space . such that is differentiable and the . if equations in unknowns . One can attempt to find by solving . What is if is the M"{o}bius strip? Let .edu/~ken/ma570/homework/hw. and be as in Problem 5-4 in a neighborhood of . 16. Show that is orientable (even if is not)... Let be as in Theorem 5-1. But. and so the rows of are also perpendicular to . Suppose .. . But we also have for all near .ms. This shows the result. By show that there is Apply Problem 5-16 with . which is precisely the condition to be proved. is a Lagrangian multiplier precisely when and so the maximum has a for which the Lagrangian multiplier equations are true. . is of rank and is of dimension . In particular. Since is compact. is perpendicular to the tangent space . In particular. this is true at . be self-adjoint with matrix . If considering the maximum of and One has with . Let .edu/~ken/ma570/homework/hw. show that and is 6 de 7 16-03-2010 17:10 . show that on . a. b. Section 2 http://www. 17. and so the rows of generate the entire subspace of vectors perpendicular to . is in the subspace generated by the . Then . so that . and so . In this case..Exercises: Chapter 5. so that the manifold is takes on a maximum on .uky. If self-adjoint. 7 de 7 16-03-2010 17:10 . Suppose It is true for dimension . All the together is the basis of eigenvectors for . Section 2 http://www. So. is of dimension .. Since as a map of as a map of is also is Proceed by induction on . has a basis of eigenvectors with eigenvalues respectively. Then apply part (a) to find the eigenvector with eigenvalue .uky. . and so . This shows that self-adjoint and . Now. it is clear that self-adjoint (cf p.Exercises: Chapter 5.ms. the case has already been shown. c. Show that has a basis of eigenvectors.edu/~ken/ma570/homework/hw.. 89 for the definition). if we let but is not required to be compact. as defined in Chapter 3. One can also let b. one has and provided that . Show that can be defined. the set of positive . One has with . prove that is empty. If is an -dimensional manifold (or manifold-with-boundary) in . the two integrals give the same value. An absolute -tensor on as -dimensional manifold is not compact. it also works under this assumption because all but finitely many summands are zero if vanishes outside of a compact subset of .. as defined in this We can assume in the situation of Chapter 3 that has the usual orientation. An absolute absolute orientable. . with the usual orientation. a.edu/~ken/ma570/homework/hw.ms. Show that Theorem 5-5 holds for noncompact vanishes outside of a compact subset of . that . The singular -cubes with can be taken to be linear maps where and are scalar constants.. even if 16-03-2010 17:10 . .Exercises: Chapter 5. So.uky. Give a counter-example if One has real numbers. The compactness was used to guarantee that the sums in the proof were finite. 1 de 2 -tensor on . of the form such that for is an is not . one has with 21. is the same as . 20. Section 3 18. Section 3 http://www. With that is a function -form on is a function . Exercises: Chapter 5. Show that Theorem 5-5 is false if For example. 19. show that section. If is a -form on a compact . be the open interval . Because the outward directed normals at points of are in opposite directions for and .. 2 de 2 16-03-2010 17:10 . let . Section 3 http://www. and there the crucial step was to replace with its absolute value so that Theorem 3-13 could be applied.. Following the hint. prove that is an where is an -form on . except don't require the manifold be orientable. the orientation of are opposite in the two cases. we have . we need to have the argument of Theorem 5-4 work. Make the definition the same as done in the section.uky. and and orientations induced by the usual orieentations of have the and . In our case.Exercises: Chapter 5. but. for example.edu/~ken/ma570/homework/hw. as stated. So the result is equivalent to . this is automatic because Theorem 4-9 gives . is not correct. it would be true if were closed. and are compact.ms. 22. If -dimensional manifold-with-boundary and is an -dimensional manifold with boundary. nor that the singular -cubes be orientation preserving. In order for this to work. Then is an -dimensional manifold-with-boundary and its boundary is the union of and . So. By Stokes' Theorem. the result. Let be an orthonormal basis with the usual orientation. .edu/~ken/ma570/homework/hw. (Note that this depends on the numerical factor in the definition of Let be an orientation-preserving -cube. is the volume as defined . Generalize Theorem 5-6 to the case of an oriented -dimensional manifold in . is orientation preserving. i.ms. as desired. But then must have the usual orientation. show that . Exercises: Chapter 5. so that the volume of .) in Chapter 3. Section 4 23. it is the volume element. choose an orthonormal basis where .uky. as defined in this section. Since this is also equal to by orthonormality. 1 de 9 16-03-2010 17:10 . we have and so applying this to gives Since Now..Exercises: Chapter 5. Consider the 1-form defined by solution since .e. Section 4 http://www. with the usual orientation.e. 25. it must be that . 24. Then and where if and only if . To see this.. so . This is the form which matches the proposed Furthermore. it follows that the value is precisely 1. If is an oriented one-dimensional manifold in show that and is orientation-preserving. . Now. is orientation preserving. By Theorem 4-9. i. If is an -dimensional manifold in . and so it is the volume element in the sense of this section. . Section 4 http://www.ms. So is the volume element Expanding in terms of cofactors of the last row gives: As in the 2-dimensional case.Exercises: Chapter 5. As in the 2-dimensional case.. Letting . was the orientation used to . if the are an orthonormal basis with orientation (where determine the outward normal). we get 2 de 9 16-03-2010 17:10 .uky. for some scalar and so for all . The generalization is defined by where is the unit outward normal at .edu/~ken/ma570/homework/hw. Exercises: Chapter 5. and . 26.uky.edu/~ken/ma570/homework/hw. If . . So the surface area is and is a norm preserving linear transformation and . is orientable.. show that has the same volume as . and . Section 4 http://www. it is assumed that 3 de 9 16-03-2010 17:10 . is a .. Compute the area of Apply part (a) with . show that the area of One can use singular 2-cubes of the form . . If -axis in is non-negative and the graph of to yield a surface in the is -plane is revolved around the . -dimensional manifold in Although it is not stated. One has . a.ms. The quantities . and calculate out to . So 27. b. say values are of the same sign regardless of which non-zero k-form in is used. To find the area. If there is a nowhere-zero Suppose . just use the formulas for an orientable surface. we have fixed and the path is a line segment traversed from to . Further. the intermediate value theorem would where the function were zero. If is a singular .bius strip and find its area. By Problem 1-7. even if . then because the space is of dimension 1. If is a -dimensional manifold. a. For every into .. So the area is .edu/~ken/ma570/homework/hw. Section 4 http://www. Choose a that the value is positive for some p. for fixed . and . But then the This is well defined. both maps 4 de 9 16-03-2010 17:10 . which is just slightly larger than circular ring of radius 2 and height 2. In particular.e. 29. one gets that is a line segment of length 2. then is a singular -cube which is a coordinate system for in a neighborhood of .uky. 28. for fixed . Define and . This corresponds to twisting the paper 180 degrees as it goes around the ring. So. .Exercises: Chapter 5. if is a singular -cube which is a coordinate system for in a neighborhood of . can just remove the line at that .bius strip. the new -cube is either orientation preserving or reversing. so that the volume of This was already done in Problem 5-21. note that in cylindrical coordinates. since one . b.ms. which is absurd. the area of a is orientable is the nowhere-zero -form on -cube. then for every -cube so . we have -form on a -dimensional manifold . Depending on the sign of . show that . one can actually. if we had two such -cubes. it follows that the volumes of and are equal. then the ) are positive for . the line segment in the -plane has slope . Numerical evaluation of the integral yields the approximation . the equations are: . In thatt case one can verify.. . Then if it were negative at another point because this is a continuous function from guarantee a point . is not orientable. is the volume element of if is the volume element of (which is also orientable). To see that it is a M@ouml. Again. i. Indeed. the line segment starts vertically at and reduces in slope until it becomes vertical again at . choose a -cube of this type with (where in its image. Since the volume is just the integral of the volume element and the integral is calculated via the -cubes. preferably with machine help. So the value is positive for all . In particular. If is defined by can be defined as show that is a M@ouml. is inner product preserving and so it maps orthonormal bases to orthonormal bases.bius strip. Calculating the length of the line. which is the M@ouml. show that an absolute -tensor |dV| can be defined. Note that this varies from down to as ranges from 0 to . http://www. b. Consider the 2-form defined on by a.ms. We will need to be continuously differentiable. we get an increasing sequence of values with limit the value of the integral. so the integral is the least upper bound of all these lengths. The other two terms give similar results. shows that these approach the integral. Following the hint. Summing over gives a Riemann sum for . if for some . then by the mean value theorem. Starting from any partition. Show that this length is the least upper bound of lengths of inscribed broken lines. and taking successively finer partitions of the interval with the mesh approaching zero.Exercises: Chapter 5. Show that For is on let . we denote since. If . times the volume element and that by . is defined by .edu/~ken/ma570/homework/hw. the value of by expanding can be evaluated using cofactors of the third row. is differentiable and has length . show that This is an immediate consequence of Problem 5-23. as we shall see. is the analogue of the 1-form As in the proof of Theorem 5-6 (or Problem 5-25). Nevertheless. 5 de 9 16-03-2010 17:10 . and the sum is zero. . Section 4 define the same orientation on 30. For example. .. Taking the limit as the mesh approaches 0.uky. This is a straightforward calculation using the definition and Theorem 4-10. b. not just differentiable. Show that restricted to the tangent space of . 31. Conclude that is not exact. Show that is closed.. a. union of segments of rays through the origin. we have by the interpretation of 32. one has for any point on the generalized cone. (Actually. then using part (b).) will be a manifold-with-corners. Since the value is c. Prove that the We take the orientations induced from the usual orientation of Following the hint.g) of by (cf. But then . show that If . The orientation of the is opposite to that of the orientation of the same set as a part of part of the boundary on . choose small enough so that there is a three dimensional manifoldis the union of and . 6 de 9 16-03-2010 17:10 .. Let and the last integral is the solid angle subtended by of part (b). The solid angle subtended by is defined as the area of . The union of those rays through 0 which intersect . and a with-boundary N (as in Figure 5-10) such that part of a generalized cone. the end of the next section. we conclude that is not exact. Show that if (F. . Applying Stokes' Theorem because is closed by part (a). . Problem 4-34 } a. or equivalently as solid angle subtended by is times the area of . By part (c).ms. show that .uky. So.Exercises: Chapter 5. that is. then Stokes' Theorem would imply that . for some in . Let . If a two-dimensional manifold is part of a generalized cone. be a compact two-dimensionaal manifold-with-boundary such that every ray through 0 intersects at most once (Figure 5-10). . b. By Problem 5-9. for . If is a tangent vector such that .. the integral over gives the part of the boundary making up part of a generalized cone is zero.G) is a homotopy of nonintersecting closed curves. One by Problem 5-26. and be nonintersecting closed curves. for all is the . see the remarks at Note that this is essentially the same situation as in Problem 5-22. If show that . then This follows immediately from Problem 4-34 (b) and Problem 5-31 (a). the line through and the origin lies on the . is a solid cone . Section 4 The second assertion follows from directed normal can be taken to be has If http://www. Define the linking number l(f.edu/~ken/ma570/homework/hw. But then is identically 0 for all surface and so its tangent line (the same line) is in points d. and the fact that the outward be an appropriate choice of orientation. where is a compact manifold-with-boundary of .. then by Stokes' Theorem. be a point we can on the opposite side. Then and . in the preamble to Problem This follows from the formula in part (b) since the third column of the determinant defining is zero. where the gives another manifold-with-boundary is opposite to that of the induced orientation. between this and the hint. The curves of Figure 4-5(b) are given by . (Note the discrepancy in sign . by the last paragraph. You may easily convince yourself that calculating case.Exercises: Chapter 5.uky. So. Removing a ball centered at from the interior with boundary . So by Stokes' Theorem and ..) On the other hand. If define and by the above integral is hopeless in this without explicit calculations. If is a compact two-dimensional manifold-with-boundary in and define Let be a point on the same side of as the outward normal and sufficiently close to as desired. The following problem shows how to find 33. suppose that dimension 3. Show that if and both lie in the -plane. . c. a. and and we can take . 7 de 9 16-03-2010 17:10 .edu/~ken/ma570/homework/hw. where This follows by direct substitution using the expression for 31. Suppose of orientation on Problem 5-31 (b). Show that by choosing make as close to Following the hint.ms. Section 4 http://www. if The rest of the proof will be valid only in the case where there is a 3-dimensional compact such that where is a two-dimensional oriented manifold-with-boundary manifold. ms.uky. (If does not intersect itself such an always exists. then . Prove that where The definition of . even if is knotted. Let need to assume that for all be the values where intersects . c. we will is finite. the last paragraph has at has a component in the outward normal is . we will show the first result. The proofs are analogous. if it is negative. page 138. should be . Let . see [6]. if the tangent vector to direction of is is positive. Let be the number of intersections where normal. Choose small enough so that (where by we mean .edu/~ken/ma570/homework/hw. Parameterize with where is not in . Subtracting gives term can be made as small as we like by making sufficiently small. then it Note that this result differs from the problem statement by a sign.. So. show that the number of other intersections.) Suppose that whenever intersects at . .. To complete the proof. Start with 8 de 9 16-03-2010 17:10 . Suppose for some compact oriented two-dimensional manifold-with-boundary .Exercises: Chapter 5. One has By part (a). Section 4 http://www. the tangent vector of is not in . and points in the same direction as the outward . If In the statement. The first b. . what one means is that the component of in the outward normal direction is either in the same or opposite direction as the outward normal. ms.. 41-43. where we have used Problem 3-32 to interchange the order of the limit and the integral. one has these are also the values of . After collecting terms. see also [13].uky. But a straightforward expansion gives: as desired. while if and are show that the curves of Figure 4-6 (c). Volume 2. we see that this is equal to the expression for in Problem 5-32 (b).. and . It remains to check that the remaining terms are equal. and use this result to if and are the curves of Figure 4-6 (b). Section 4 http://www. one has and so .edu/~ken/ma570/homework/hw. d. pp. One then substitutes into: . 409-411. So. we have by Stokes' Theorem that Comparing the two expressions. By inspection in Figure 4-6.) By part (c). we see that two of the terms in the first expression match up with corresponding terms in the second expression.Exercises: Chapter 5. 9 de 9 16-03-2010 17:10 . Show that the integer of (b) equals the integral of Problem 5-32(b). So. Similar results hold for and . The proofs outlined here are from [4] pp. On the other hand. in parts (b) and (c) of the figure. (These results were known to Gauss [7]. let . Then . boundary in -manifold with The generalization: Let be a compact -dimensional manifoldwith-boundary and the unit outward normal on . we have for . Then As in the proof of the divergence theorem. Applying the generalized divergence theorem to the set and . find the volume of in terms of the -dimensional volume of 1 de 2 16-03-2010 17:11 . By Problem 5-25. it follows that 35.. By Stokes' Theorem.edu/~ken/ma570/homework/hw. Let be a differentiable vector fieldd on . Section 5 http://www. Exercises: Chapter 5.. Section 5 34.uky.Exercises: Chapter 5. on . So. Generalize the divergence theorem to the case of an .ms. this says be a compact three. The buoyant force on of the fluid displaced by . So one has the result if we define the buoyant force to be . due to the fluid. Now is the weight of the fluid displaced by .. . Since a fluid exerts equal pressures in all directions.) if is even and since the outward . define the buoyaant force on Prove the following theorem. The vector dimensional manifold-with-boundary with may be thought of as the downward pressure of a fluid of density . Define field in on by is times the volume of and let is odd.ms. (This would make sense otherwise the buoyant force would be negative. as .Exercises: Chapter 5. So the right hand side should be the buoyant force. The definition of buoyant force is off by a sign.edu/~ken/ma570/homework/hw. So the surface area of 36.) 2 de 2 16-03-2010 17:11 .uky. In particular. if .. we . The divergence theorem gives is equal to the weight . Section 5 http://www. . Theorem (Archimedes). (This volume is if One has and normal is in the radial direction.