Calculus on Manifolds. Spivak. Solutions

Exercises 1.1 http://www.ms.uky.edu/~ken/ma570/homework/hw1/html/ch1a.htm Exercises 1.1 1. Prove that One has 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? . . Taking the square root of both sides gives the result. 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 is equivalent to , then substituting shows that the inequality and clearly equality holds if a is non-positive. Similarly, one has equality if for some real . Conversely, if equality holds, then so . By Theorem 1-1 (2), it follows that and are linearly dependent. If for some real back into the equality shows that must be non-positive or must be 0. The case where is treated similarly. 4. Prove that . which is just the triangle inequality. On the other hand, if and . , and , then substituting If , then the inequality to be proved is just then the result follows from the first case by swapping the roles of 5. The quantity is called the distance between . and , . Prove and interpret geometrically the ``triangle inequality": The inequality follows from Theorem 1-1(3): Geometrically, if , , and of the other two sides. 6. Let 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 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 for some ? What if and are continuous? and are continuous, then the No, you could, for example, vary assertion is true. In fact, suppose that for each since and at discrete points without changing the values of the integrals. If , there is an with . Then the inequality holds true in an open neighborhood of since the integrand is always non-negative and is positive on some subinterval of for all . Since the quadratic has no solutions, it must be that its are continuous. So . Expanding out gives discriminant is negative. 3. Show that Theorem 1-1 (2) is a special case of (a). Let , , and for all in 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 of 3 12/6/2011 3:03 PM Exercises 1.1 http://www.ms.uky.edu/~ken/ma570/homework/hw1/html/ch1a.htm 1. Show that If . is norm preserving if and only if is inner product preserving. 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 is 1-1, and that is of the 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 product preserving. 8. If and in are both non-zero, then the angle between and , denoted , is defined to be is 1-1 and for , one has which , since is norm preserving. Thus is norm preserving, and hence also inner , makes sense by Theorem 1-1 (2). The linear transformation is angle preserving if . 1. Prove that if is norm preserving, then is angle preserving. Assume is norm preserving. By Problem 1-7, is inner product preserving. So . 2. If there is a basis are equal. of and numbers such that , prove that is angle preserving if and only if all The assertion is false. For example, if , , , . Now, , and , then , but showing that T is not angle preserving. To correct the situation, add the condition that the means that: Suppose all the be pairwise orthogonal, i.e. because all the cross terms are zero. are equal in absolute value. Then one has because all the are equal and cancel out. So, this condition suffices to make for some and and that be angle preserving. for all . Using bilinearity, this Now suppose that . Then since . So, this condition suffices to make ? not be angle preserving. 3. What are all angle preserving The angle preserving are precisely those which can be expressed in the form V is norm preserving, and the operation is functional composition. where U is angle preserving of the kind in part (b), Clearly, any of this form is angle preserving as the composition of two angle preserving linear transformations is angle preserving. For the converse, suppose that is angle preserving. Let be an orthogonal basis of . Define to be the linear transformation such that for each . Since the are pairwise orthogonal and is angle preserving, the are also pairwise orthogonal. In particular, because the cross terms all cancel out. This proves that Then clearly and is norm preserving. Now define to be the linear transformation maps each . to is angle preserving because it is the composition of two angle preserving maps. Further, 2 of 3 12/6/2011 3:03 PM Exercises 1.1 http://www.ms.uky.edu/~ken/ma570/homework/hw1/html/ch1a.htm a scalar multiple of itself; so 9. If , let . The transformation is a map of the type in part (b). This completes the characterization. have the matrix . Show that is angle preserving and that if , then is 1-1 by Cramer's Rule because the determinant of its matrix is 1. Further, is norm preserving since by the Pythagorean Theorem. By Problem 8(a), it follows that If , then one has is angle preserving. . Further, since is norm preserving, 10. If Let . By the definition of angle, it follows that such that and . for . One has . . is a linear transformation, show that there is a number be the maximum of the absolute values of the entries in the matrix of 11. For and , show that and . Note that and denote points in . This is a perfectly straightforward computation in terms of the coordinates of . If , define is . Define for a unique . using only the definitions of inner by product and norm. 12. Let denote the dual space of the vector space . Show that is a 1-1 linear transformation and conclude that every is a linear tranformation and (b) is a linear transformation. One needs to verify the trivial results that (a) proofs are omitted. Together these imply that Since for , . These follow from bilinearity; the has no non-zero vectors in its kernel and so is 1-1. Since the dual space has dimension n, it follows that is also onto. This proves the last assertion. 13. If , then and are called perpendicular (or orthogonal) if . By bilinearity of the inner product, one has for perpendicular and : . If and are perpendicular, prove that 3 of 3 12/6/2011 3:03 PM htm Exercises: Chapter 1. The intersection of the open intervals is the set containing only intersection of even countably many open sets is not necessarily open. This proves that is open. then is open. . ). So is open. The assertion about finitely many sets Let and be open. and the boundary is the set . we have follows by induction. and the boundary is the is the empty set is the empty set . Section 2 http://www. and so the . and boundary of the sets: The interior of set . then there are open rectangles (resp. Since is open. ) (resp. 16. the exterior is the empty set . and be their union. Prove that the intersection of two (and hence finitely many) open sets is open. and containing and contained in rectangle (Why?). the exterior is .uky. then let . there is an open rectangle containing . If . then there is an . Find the interior. In each case.edu/~ken/ma570/homework/hw2/html/ch1b. so is open. the exterior is .Exercises: Chapter 1. Prove that If . Since the intersection of two open rectangles is an open . If . Let with be a collection of open sets. If and so . be the open rectangle centered at with sides of length . Give a counterexample for infinitely many open sets.ms. Section 2 14. 15. exterior. 1 of 3 12/6/2011 3:08 PM . The interior of The interior of is the set . and the boundary is the set . the proofs are straightforward and omitted. Prove that the union of any (even infinite) number of open sets is open. . Then there is a which contains . Construct a set each vertical line but the boundary of is contains points in each quarter of the square contains at most one point on each horizontal and . Now contains a non-empty open subinterval of and this is necessarily disjoint from . etc. Section 2 http://www. it suffices its complement is open. Hint: It suffices to ensure that and also in each sixteenth. and amongst such the other one is the earliest possible. Then make of all the quarters. Since is compact. show that .htm such that 17. etc. is not in . let . 20. and contains all rational numbers in . there is a finite . To do the construction. the interior of is itself since it is a union of open sets. etc. For example. b) of integers with positive. in increasing order of . then simply and amongst those with same value of eliminate those pairs for which there is an earlier pair with the same value of . which is a contradiction. Since is open. with an obvious lexicographical order make amongst the quarters. Then the collection is closed. it suffices to show that every point in the square is in the boundary of . of the unit sqare.Exercises: Chapter 1. Suppose 2 of 3 12/6/2011 3:08 PM . 1]. there is an open interval containing and disjoint from . it suffices to show that nothing in is in the exterior of . 18. If is the union of open intervals such that each rational number in contained in some . neither has yet been the point used. Let be chosen so that . first make a list of all the rational numbers in the interval [0. Similarly. But every non-empty open subinterval of contains rational numbers.uky. Suppose interior and the exterior. be the open cover consisting of rectangles for all positive integers . To show this. also the exterior of clearly contains as . Let be a rational number in contained in . choose such that is in the portion. Prove the converse of Corollary 1-7: A compact subset of Suppose subcover To show that is compact. show that the boundary of is . traverse the list : for each portion of the square. it follows that is in the boundary of . one could by listing first the quarters. 19. Let be an open interval containing and disjoint from . Suppose . Now. could be made by listing a list all pairs (a. If . which is a contradiction. sixteenths.edu/~ken/ma570/homework/hw2/html/ch1b. Then there is some portion of the square in which is entirely contained within the rectangle and containing . sixteenths. and such that the latter occurring (in ) of them is earliest possible. If it is . Let is closed and bounded. then and so is bounded. then the sixteenths. Since the boundary is the complement of the union of the is in the exterior of . is Clearly. choose any open rectangle containing . If is a closed set that contains every rational number . 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). in increasing lexicographical order. . etc. To show that this works.ms. both and are in the list . non-negative. be as in Problem 1-21 (b) applied with is compact. is bounded and 3 of 3 12/6/2011 3:08 PM . closed.htm where be a finite subcover.Exercises: Chapter 1. such that is contained in is the x-axis and is compact. .e.ms. Clearly. i. If is closed and . Let . So the complement of is open. and so there is an open rectangle . c. such that means that no b. show that there is a compact set . choose to be as in part (a). Then. Let is an open neighborhood of which is disjoint from closed. Let . we know that satisfies the assertion. and for all and For each . Finally. Then . by the triangle be a finite subcover. is the graph of the exponential function. This was chosen so that is entirely contained within the open rectangle. . a. If is open and the interior of and Let if and are required both to be closed with neither compact.uky. and let inequality.edu/~ken/ma570/homework/hw2/html/ch1b. this . If is closed. Section 2 http://www. so and . Let is . Give a counterexample in A counterexample: 22. prove that there is a number such that for all Such an is in the exterior of containing and disjoint from . prove that there is a . It is straightforward to verify that is also true. which shows the assertion. 21. Let . is an open cover of . can be is compact. Then is an open cover of . if and only if each is. Then is continuous and are at 0 and . For define is not continuous . contains an interval around which is in .htm Exercises: Chapter 1. and . if and nowhere else. Show that every straight line through . . If satisfies such that . Prove that and for each Suppose that for every with . Section 3 23. So. if . Let a. Let . Prove that is continuous at . is identically zero in a neighborhood of zero by part (a). Then. and . In particular. one has satisfies . Define by at . for each i. there is an Let . then . Let . suppose that . but For each . Let the line be . On the other hand. Let . 25. then is continuous. then the line intersects the graph of Let . every is clearly continuous at 0. then the whole line is disjoint from at and . Choose for each .Exercises: Chapter 1.ms. So 24. So T is continuous at 26. for each i. for all . So. Prove that a linear transformation By Problem 1-10. cannot be continuous at because every open 1 of 2 12/6/2011 3:09 PM . Then. if . b. a positive . is in . . then such that if and only if This is an immediate consequence of Problem 1-23 and the definition of continuity. Show that each is continuous at 0. show that . it follows by the intermediate value theorem that for all with . . On the other hand. Conversely. If . is chosen as in the definition of and satisfies . the line cannot intersect anywhere to the left of . by if and if . Since the only roots of . Section 3 http://www.uky.edu/~ken/ma570/homework/hw3/html/ch1c. Section 3 http://www. 27. takes on a maximum and a By Theorem 1-9. This shows that 29. 30.uky. Now assume that the have been re-ordered so that they are in increasing order. Let be an increasing function. Clearly these are the minimum and maximum values of . So. which is unbounded. and hence is closed and bounded. are distinct. This . show that 2 of 2 12/6/2011 3:09 PM . If is compact. one has: proves that is continuous.Exercises: Chapter 1. Prove that . The function on the right is an increasing function of .htm rectangle containing . One has .ms. contains points of and for all those points . it follows that and . Let . Let (resp. choose to be a boundary point of which is not in . one has is open by considering the function with The function is continuous. If . with where we have used Problem 1-4 in the simplification. If . is compact. prove that every continuous function minimum value. this is unbounded. Then for any . let . In fact. and they are taken on since they are in . Since 28. To show it is continuous at . let and choose . one has is not closed. and hence are in since it is closed. then is open by Theorem 1-8. Then and are boundary points of . is bounded above by the quantity on the right for any . . . show that there is a continuous function As suggested. If .edu/~ken/ma570/homework/hw3/html/ch1c. 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 . in particular. ) be the greatest lower bound (respectively least upper bound) of . let by Problem 1-4. is continuous at . and let Clearly. Now add up all the inequalities with this value of . a 1 x 2 matrix. but I will often confound with . is the Jacobian. but this follows immediately from Problem 1-10. this is equivalent to their being a function such that for all . if . Section 1 1.ms. A function is differentiable at . then If is independent of the second variable.Exercises Chapter 2. Define when a function is independent of the first variable and find for such . 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 . An argument similar to that of the previous problem shows that . So. Section 1 http://www. you can let . . 4. So. Show that such that .uky. even though one is a linear transformation and the other is a matrix. Conversely. only if there is a function The first assertion is trivial: If is independent of the second variable. Prove that if If is differentiable at show that 2. Let be a continuous real-valued function on the unit circle and . i. What is is independent of the second variable if and in terms of be defined by .e. 3. it would be more proper to say that .htm Exercises Chapter 2. then it is continuous at . we have ? is said to be independent of the second variable if for each for all . then . then because: Note: Actually. define by such that 1 of 4 12/6/2011 3:11 PM . we need only . Just as before.edu/~ken/ma570/homework/hw4/html/ch2a. and so we see that this just says that is identically zero. Then it is trivial to show that obtained from this is as in the is not differentiable at 0. if and only if and are. so that is not differentiable at Define by for all satisfies all the properties of Problem 2-4 and that the function statement of this problem. Let In fact. Show that Suppose otherwise. Just as in the proof of Problem 2-4. 7. Prove that is differentiable at . Show that is differentiable at 0. one would then have: in spite of the fact that the limit is clearly 1. b. .uky. Similarly. Show that . 0) unless is differentiable at . Then one must have: and so . If and is defined by when and . say. one can show that. Let case be a function such that by the squeeze principle using .ms. 6.Exercises Chapter 2. . is linear One has and hence differentiable. On the other hand. then would be the zero map. if were differentiable at 0. using the definition of derivative. one gets : . is not differentiable at (0. we get for fixed . is a function of the kind considered in Problem 2-4. . by approaching zero along the 45 degree line in the first quadrant. 8.edu/~ken/ma570/homework/hw4/html/ch2a. More generally. Show that .htm a. show that is differentiable. In both cases. and in this 2 of 4 12/6/2011 3:11 PM . Section 1 http://www. But with. Thus . Let be defined by . Let be defined by . But for all 5. if are equal up to order at a. if the squeeze principle to conclude that 9. So.uky. are differentiable at . by the squeeze principle. Section 1 http://www. then the function works by the If is differentiable at definition of derivative. 3 of 4 12/6/2011 3:11 PM . To make the converse true. equal up to first order.htm Suppose that . Indeed. this . But clearly changing the value of at changes whether or not is differentiable at .ms.edu/~ken/ma570/homework/hw4/html/ch2a. If and be continuous at : If there is a of the . 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 . Since is continuous. must be differentiable at with .Exercises Chapter 2. then . show that are equal up to order at a. and the function defined by with exist. Two functions is differentiable at with the desired derivative. . add the assumption that specified form with Multiplying this by means that differentiable at b. The converse is not true. we see that . But then the condition is equivalent to the assertion that is . then use the inequality derived from Problem 1-1: and the On the other hand. you can change the value of at without changing whether or not and are equal up to first order. Then one has the inequality: . 4 of 4 12/6/2011 3:11 PM .1 times to the limit to see that the value of the limit is hand.edu/~ken/ma570/homework/hw4/html/ch2a.Exercises Chapter 2.uky.ms.htm Apply L'Hôpital's Rule n . Section 1 http://www. On the other Subtracting these two results gives shows that and are equal up to order at . one has: . Use the theorems of this section to find a. c. we get: is the function of part (c). One has . Using Theorem 2-3 (3) and part (a).edu/~ken/ma570/homework/hw5/html/ch2b. b. . then and we know the derivative of from part (a). then .Exercises: Chapter 2. one has: . i. e. So one gets: 1 of 6 12/6/2011 3:11 PM . and so by the chain rule: for the following: d.ms. If . one has: . then . The chain rule gives: f. If . . If Using the chain rule.htm Exercises: Chapter 2. Section 2 http://www.uky. We have and so by the chain rule. Section 2 10.e. then . Find a. 11. g. So one gets: . Section 2 http://www. One has and and (a) above. then . The chain rule gives: i. Using parts (h).uky. Using the last part: j. . then . If . where have derivatives as given in parts (d) 12. and so: . and so: . is as in part (a).edu/~ken/ma570/homework/hw5/html/ch2b. and .ms. . . we 2 of 6 12/6/2011 3:11 PM . If b. A function have is bilinear if for .Exercises: Chapter 2. one gets . If c. (c). . for the following (where . h. and (a). is continuous): .htm . Section 2 http://www. Find and by By Problem 2-12 and the fact that (Note that is an matrix.edu/~ken/ma570/homework/hw5/html/ch2b. Then we have by an obvious induction using bilinearity.htm a. .uky. So b. 3 of 6 12/6/2011 3:11 PM . and is defined by . its transpose . This follows by applying (b) to the bilinear function 13. it was verified in the proof of Theorem 2-3 (5). . then Let have a 1 in the place only. in this case. is bilinear.ms. show that . . It follows that there is an depending only on such that: . If show that are differentiable. . Since . Show that the formula for in Theorem 2-3 is a special case of (b). we see that it suffices to show the result in the case where and the bilinear function is the product function. Prove that One has by bilinearity and part (a). c. Define a. . But. one has . is differentiable and for all . If .Exercises: Chapter 2. b.) is an matrix. one can apply the chain rule to get the assertion. which we consider as a member of Since c. Prove that if is bilinear. A function is called multilinear if for each choice of the function is a linear transformation. a. Regard an matrix as a point in the as a member of .edu/~ken/ma570/homework/hw5/html/ch2b. Just as in the bilinear case. the remainder looks like a sum of terms as in part (a) except that there may be more than two type arguments.ms. Prove that . with defined by . d. If is multilinear and . Section 2 http://www. 14. This shows the result.uky. This can be argued similarly to Problem 2-12. Prove that are distinct. b. Let be Euclidean spaces of various dimensions.htm Use part (b) applied to to get . we have This is an immediate consequence of Problem 2-12 (b). 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 where is at least two and the zero. a. Exhibit a differentiable function is not differentiable. one could let . Just apply the definition expanding the numerator out using multilinearity. this limit is -fold product is differentiable and by considering each row 4 of 6 12/6/2011 3:11 PM . This shows the result. Then such that the function |f| defined by |f|(t) = |f(t)| is not differentiable at 0. 15.Exercises: Chapter 2. Trivially. show that for . If are differentiable and .Exercises: Chapter 2. Without writing all the details. Show that is differentiable and find 5 of 6 12/6/2011 3:11 PM . show that This follows by the chain rule and part (a).edu/~ken/ma570/homework/hw5/html/ch2b. let are the solutions of the .htm This is an immediate consequence of Problem 2-14 (b) and the multilinearity of the determinant function. c. b. Section 2 http://www.uky. this makes the formulas simpler because the formula for derivative of a for all and be the functions such that for are differentiable. We can take transposes since the determinant of the transpose is the same as the determinant of the original matrix. 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 .ms. If equations: . edu/~ken/ma570/homework/hw5/html/ch2b. Section 2 http://www.htm 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 .uky. Show This follows immediately by the chain rule applied to 6 of 6 12/6/2011 3:11 PM .ms.Exercises: Chapter 2. . . 16. Suppose that is differentiable and has a differentiable inverse . .edu/~ken/ma570/homework/hw6/html/ch2c. Section 3 17. e. b. . . . c. .uky.Exercises: Chapter 2. i. h. Section 3 http://www. . f. 1 of 6 12/6/2011 3:12 PM . d.htm Exercises: Chapter 2.ms. Find the partial derivatives of the following functions: a. g. Exercises: Chapter 2. c. If find Since . .ms. 2 of 6 12/6/2011 3:12 PM . .htm 18. one has . b. . 19. b. d. in terms of the derivatives of and if 20. Find the partial derivatives of the following functions (where a. . . . Section 3 http://www. Find the partial derivatives of a. c.uky. is continuous): .edu/~ken/ma570/homework/hw6/html/ch2c. Define by a.ms. 23. If and . Let a. 21. Show that True since the first term depends only on b. Find a function One could let Find one such that One could let 22. to . Both assertions are immediate consequences of this result. show that is a constant.Exercises: Chapter 2. Find a function such that but 3 of 6 12/6/2011 3:12 PM . . then it is constant on that interval. show that . . and . be defined so that . and from the proof of Problem 2-22. Let be continuous. . Then the line segment from to are all contained in . How should One could let c. e.uky. from to . If and . . one knows that if a function of one variable has zero derivative on a closed interval . Suppose and are arbitrary points of . Section 3 http://www. By the mean value theorem. By . is independent of the second variable. So b. . ? . show that . If and such that . it follows that and . . is not independent of the second variable.edu/~ken/ma570/homework/hw6/html/ch2c.htm d. . . Define by a. Show that is a function. one has . Show that Using part (a).edu/~ken/ma570/homework/hw6/html/ch2c. Consider the function where is a polynomial. and defined by for all .Exercises: Chapter 2. 25. one has (because into one formula and b. Show that One has since for all and for all when . one has ). The assertions follow immediately by substituting in the other. Section 3 http://www. Define by a. By induction on .ms. define and for all . 24.uky. Then for it follows that all . . Then. and . 4 of 6 12/6/2011 3:12 PM . In particular. function is for .htm One could let . and and so for all and . Further. using Problem 2-25 on the side closest to the origin. show that there is a non-negative function 5 of 6 12/6/2011 3:12 PM . If . suppose by induction on . For points other than 1 and -1. 1) and 0 elsewhere.ms.uky.Exercises: Chapter 2. At each of the exceptional points. This follows from part (a). Let c. Show that there is a for .edu/~ken/ma570/homework/hw6/html/ch2c. Show that is a function which is positive on (-1. it suffices to show that induction using L'Hôpital's rule: for each integer . consider the derivative from the left and from the right. let be as in Problem 2-25 and Now use Problem 2-25 to prove that e. Section 3 http://www. We have . Show that is a function which is positive on and zero elsewhere. f. that . To show that this limit is zero. define by works. If is open and is compact. the result is obvious. But this is an easy . b. d. function such that for and Following the hint.htm Now. As suggested in the hint. Define by Show that the maximum of maximum of on on . It is easy to verify that this new satisfies the required conditions. we have positive on . a finite number of them also cover . g.ms. This proves the assertion. Section 3 http://www. Since we have a subcover of . say the rectangles corresponding to form a subcover. We know that for all .Exercises: Chapter 2. one knows is attains its minimum (Problem 1-29). Since the set of these rectangles is an open cover of the compact set . is either the maximum of or the This is obvious because 6 of 6 12/6/2011 3:12 PM . replace with that where is the function of part (b).edu/~ken/ma570/homework/hw6/html/ch2c. The choice 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. let be the open rectangle centered at with sides of length . Let be the function defined for this rectangle as in part (c). Let be as in part (d).htm such that . . let . Finally. 26. Show that we can choose such an so that and for .uky. for and outside of some closed set contained in Let be the distance between and the complement of . Since is compact. and choose For each . .Exercises: Chapter 2. 1 of 4 12/6/2011 3:13 PM . c. is denoted a. b.ms. Find expressions for the partial derivatives of the following functions: a. . Section 4 http://www. 29. For . in the direction . This is obvious from the definitions. . Section 4 28.uky. at . Let .edu/~ken/ma570/homework/hw7/html/ch2d. d. the limit if it exists. Show that and called the directional derivative of .htm Exercises: Chapter 2. a. exists for all . one has since .uky. and therefore One has which shows the result whenever . show that . then The last assertion follows from the additivity of the function 30. Section 4 http://www. 32.0). Show that exists for all and all is not true for all With the notation of Problem 2-4.htm b. . c. Let be defined as in Problem 1-26. but if . 2 of 4 12/6/2011 3:13 PM .Exercises: Chapter 2.ms. Let be defined by for all . Show that . is differentiable at 0 but is differentiable at . Let be defined as in Problem 2-4. part (a) of that problem says that suppose . At is not continuous at 0. Then . . If is differentiable at . But . Show that Clearly. although is not even continuous at (0. By Problem 1-26 (a). The case when is trivially true. . Now . 31.edu/~ken/ma570/homework/hw7/html/ch2d. Show that exists for all . However. Section 4 http://www. But the second term takes on all values between -1 and 1 in every open neighborhood of .edu/~ken/ma570/homework/hw7/html/ch2d. The first term has limit 0 as approaches 0. 0) is the zero linear transformation because . one has . On the other hand. If is also Applying Theorem 2-9 to and so show the result. b. Let be defined by Show that is differentiable at (0. In the case. just as in part (a). If is differentiable and gives .0) but that is not continuous at . it suffices to note that follows from the definition of . . A function is homogeneous of degree differentiable. It follows from the differentiability of .) Further (a). The derivative at (0. Show that the continuity of at up to a sign.uky. that and . 35. Substituting .htm For .Exercises: Chapter 2. in these two formulas . show that if for all and . (The argument given above also shows that they are defined and 0 at the partials are equal to 33. for where is as in part are defined for . does not even exist. and so they cannot be continuous at 0. may be eliminated from the hypothesis of Theorem 2-8. prove that there exist such that 3 of 4 12/6/2011 3:13 PM . So. 34.ms. Proceed as in the proof of Theorem 2-8 for all This is all that is needed in the rest of the proof. Exercises: Chapter 2.edu/~ken/ma570/homework/hw7/html/ch2d. . Theorem 2-9 gives 4 of 4 12/6/2011 3:13 PM . let . So. we have the result with . Then .uky. Section 4 http://www.htm Following the hint. On the other hand.ms. Then there is an open neighborhood defined by and hence of with satisfies for all for all . By replacing with a vector of variables.edu/~ken/ma570/homework/hw8/html/ch2e. Furthermore Since was arbitrary. we see that By applying the previous results to the set 37. It follows that is differentiable. we can apply Problem 2-36. then we replace by one. By Theorem 2-11. is differentiable. For the general case of a map is constant in a non-empty open set reducing the value of the function where is an open subset of with with and drop out . So. is open for any open set For every . Section 5 http://www. Again. Replace with on an open subset of . Note that we have made rectangle. Generalize this result to tthe case of a continuously differentiable function . Assuming that are 1-1. by restricting to an appropriate and get a 1-1 function defined on on a rectangle in one less dimension and mapping into a space of dimension one less.uky. suppose we have (the case where is analogous). b. Let is open. if for some defined by and its inverse will look like . On the other hand. be a continuously differentiable function. a. then consider . Just as in part (a). The function .ms. Now is open but each horizontal line intersects at most once since is 1-1. 1 of 2 12/6/2011 3:14 PM . Section 5 36. Since clearly . if with . in place of . Let be an open set and for all Show also that . we can simply fix the value of . Show that We will show the result is true even if is only defined in a non-empty open subset of Following the hint. Show that a continusously differentiable 1-1 function such that is an open set and .htm Exercises: Chapter 2. the proof of part (a) generalizes to the case where is a function defined on an open subset of where . this will be invertible . is not 1-1. there is an open set . this shows that is differentiable at . it follows that is differentiable. . By repeating this process. The inverse function is clearly of the form and so for all . there is an with and an open subset such that and is open. we know that is not constant in any open set.Exercises: Chapter 2. This is a contradiction since is non-empty and open. . By taking n larger and larger. a. show that is 1-1 on all of . Suppose one has between and such that . Since both factors on the right are non-zero.edu/~ken/ma570/homework/hw8/html/ch2e. this is impossible. Show that for to show that continuity of the derivative cannot be eliminated from the hypothesis of Theorem 2-11. By the mean value theorem. for all for all 39. .uky. but by is not 1-1. By the intermediate value theorem. 2 of 2 12/6/2011 3:14 PM . which we have already taken care of. one has So satisfies the conditions of Theorem 2-11 at except that it is not continuously differentiable for . Define all Clearly. 38. there is a between and where . Use the function defined by .Exercises: Chapter 2. Clearly. is differentiable for . there is a .htm one eventually gets to the case where is equal to 1. If satisfies for all for some . At . we see that is not 1-1 on any neighborhood of 0. Section 5 http://www.ms. b. The function is not 1-1 since . at 0 since Now and it is straightforward to verify that for all sufficiently large positive integers . let be this by . One has . one gets for . . then for some we have and . Show that if This follows immediately from part (a). the Implicit function theorem says that there is a function defined in an open neighborhood of and such that and is differentiable.htm Exercises: Chapter 2. Let . there is exactly one solution of for .Exercises: Chapter 2. If for all was meant to be continuously differentiable. 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.edu/~ken/ma570/homework/hw9/html/ch2f.uky. Find 1 of 2 12/6/2011 3:14 PM . the uniqueness condition guarantees that function provided by the implicit function theorem applied to particular. The determinant condition guarantees that for each . In this problem. Use the implicit function theorem to re-do Problem 2-15(c). call that solution . b. For each there is a unique with defined . By the uniqueness of for all in the domain of the solutions in the last paragraph. for each . show that is differentiable and Just as in the last problem. In is differentiable and differentiating this last relation gives . it must be that .ms. is the same as the . An explicit formula can be obtained by using Cramer's rule. Define by for all . Now. Section 6 40. 41. Section 6 http://www. . Let Suppose that for each be differentiable. . Solving for gives the result. the various functions all glue together into a single function defined on all of and differentiable everywhere. By differentiating the relation . it is assumed that a. c. Exercises: Chapter 2. if by the second derivative . One has of part (a) is true and . The derivative of this function is negative throughout the interval. Also. The derivative of this function is .htm Note that is defined only when and are positive. so the maximum occurs at . Now on the interval. By the result of the previous paragraph the maximum over the entire interval must therefore occur at . For fixed if We will find where the maximum of the minimum's are located in each of the three cases. and the value of the maximum is . The derivative precisely when if and there is a unique is achieved at . and so for fixed . and so the maximum must occur at the left hand end point. The maximum value is . The maximum value is . 2 of 2 12/6/2011 3:14 PM . This function has no zeros in the interval because has derivative which is always negative in the interval and the value of the function is positive at the right end point.edu/~ken/ma570/homework/hw9/html/ch2f. So we need to go back and show that the earlier parts of the problem generalize to this case. at . but just for those in the for where . Now actually.ms.uky. Section 6 http://www. and at . In view of occurs at the last paragraph. Suppose . the minimum of occurs at test. Then . This is a decreasing function and so the maximum occurs at the left hand endpoint. that means that the maximum over the entire interval . . Further . we are not looking for the minimum over all interval . Then we need to maximize . Then . there are no difficulties in doing this. Suppose . Suppose . precisely when and so the hypothesis (since both and are positive). . Let be a refinement of where . because the terms of the sum can differ only on points where and differ.Exercises: Chapter 3. be integrable and let . if property. In fact. Taking differences those subrectangles which contain at least one of the gives 3. 2. is such that every subrectangle of the number of points where and bounds for the values for all .ms. except at finitely many points. then the volume of which is less than as soon as . has the same is any partition. have values which differ. there is a partition of in which every subrectangle has volume less than . where . then any common refinement of this partition and each subrectangle is precisely If and is a partition of . Apply Theorem 3-3 to the partition . Then the hypotheses of Theorem 3-3 are satisfied by and and where is any upper bound for the volume of the subrectangles of . and (resp.edu/~ken/ma570/homework/hw10/html/ch3a. Show that is integrable . then any point is an element of at most of the subrectangles of .htm Exercises: Chapter 3. 1 of 5 12/6/2011 3:15 PM . Let be defined by Show that is integrable and that . Section 1 1. and so is integrable. Let be integrable. the real proof is of course an induction on m. The intuitive iddea of the proof is that the worst case is when the point is in a `corner'. Let and be a partition as in Theorem 3-3 applied to has volume less than and . For this partition. if by dividing each side into equal sized subintervals and . ) are upper (resp. Let and For any you partition Furthermore. In fact. lower) .uky. Section 1 http://www. .ms.uky. and are just positively argument shows the result for weighted sums of the . Since and for each subrectangle of .htm a. and the result for can be obtained by summing (with weights) the inequalities for the . b. Then by part (a) and Lemma 3-1. and so it is at most equal to the greatest lower bound of these values. For any constant . For any partition of and any subrectangle and and of . Let be a (resp.Exercises: Chapter 3. A similar . Let be a partition as in Theorem 3-3 applied to and .edu/~ken/ma570/homework/hw10/html/ch3a. Since . Section 1 http://www. ) be a partition as in Theorem 3-3 applied to (resp. . For each . show that and therefore . By the squeeze principle. By Theorem 3-3. We will show the result in the case where . Adding these inequalities shows that . we have 2 of 5 12/6/2011 3:15 PM . show that . Let common refinement of and . the other case being proved in a similar manner. A similar argument shows the result for . one concludes that c. Further is integrable. one has and since greatest lower bounds are is a lower bound for lower bounds. . . ) and . Show that is integrable and . Let be a common refinement of and . Suppose that all the Theorem 3-3 applied to are integrable where and is any subrectangle of . and that in this .edu/~ken/ma570/homework/hw10/html/ch3a. Let be the (for where is the number of rectangles in of the partition of A obtained by taking the union of all the subsequences defining the partitions of the each dimension). Then there are refinements subrectangles of which are contained in whose rectangles are the set of all . Section 1 http://www. and . by the squeeze principle.uky. Show that is integrable if and only if for each the function . By Theorem 3-3. it follows that is integrable. Let subrectangle case . its and be a partition of . Then there is a partition of whose subrectangles are precisely the subrectangles of which are contained in . Suppose that is integrable and . applied to integral is 4. Let be a partition as in . One has 3 of 5 12/6/2011 3:15 PM . the function is integrable.ms.htm By Theorem 3-3.Exercises: Chapter 3. Then . which consists of restricted to . is integrble. Let be a partition of as in Theorem 3-3 applied to and . the function partition in which is integrable and is the only rectangle. . 6.edu/~ken/ma570/homework/hw10/html/ch3a. .uky. Then this implies that 4 of 5 12/6/2011 3:15 PM . if . and. show that is integrable and . This proves the result. For any rectangle contained in . If .Exercises: Chapter 3. we have . then . On the other hand. then . If is integrable. Let is integrable. it has the desired value. the function 5. Consider the function and .htm By Theorem 3-3. Let be a partition as in Theorem 3-3 applied to and . Using the trivial since be integrable and suppose By Problem 3-3. Section 1 http://www.ms. Show that . we have . by the squeeze principle. uky. the contribution to from these rectangles is also at most . Since there are at most such pairs . It follows that . Since . one can show that that is integrable. . Choose a positive integer so that .htm So. of course. so their contribution to is at most . is integrable and the squeeze principle implies that its integral is 0. 5 of 5 12/6/2011 3:15 PM . it follows that 7. For the remaining rectangles .Exercises: Chapter 3.edu/~ken/ma570/homework/hw10/html/ch3a. . But then. Section 1 http://www. Since . the value of and their total volume is. it follows integrable. since . so if Problem 3-5 implies that (c). is integrable.ms. Let Show that Let is integrable and . Let be any partition of such that every point with lies in a rectangle of of height (in the direction) at most . is integrable by Theorem 3-3. no larger than 1. Further. such a exists and the total volume of all the rectangles containing points of this type is at most . But then by Problem 3-3. be defined by by Problem 3-3 Similarly. By Theorem 3-3. uky. it is of measure zero. Indeed. Let . then is a subset of for some since the union of the is . which is at most equal to the sum of the volumes of the . Then the union of the cover the is also of content 0. 9. say where contains all the and and hence also . then contains only boundary points of . contains all the natural numbers and the total . of measure 0 such that the boundry of The set of rational numbers in the interval is of measure 0 (cf Proof of Problem 3-9 (b)). Choose a partition which refines all of the partitions where Note that is a rectangle of the cover . Let be any rectangle in with non-empty interior. where . Prove that is not of content 0 if for . So the boundary of b. Give an example of a bounded set measure 0. if has non-empty interior. So. Since the intersection of any two rectangles of a partition is contained in their boundaries. but its boundary is not of measure 0 (by Theorem 3-6 and Problem 3-8). a. Let where . On the other . But then is .Exercises: Chapter 3. Suppose for are closed rectangles which form a cover for . where cannot have content 0. Give an example of a closed set of measure 0 which does not have content 0. does not have boundary of and have total volume less than . Then bounded.htm Exercises: Chapter 3. show that the boundary of Suppose a finite set of open rectangles cover of and have total volume less than also has content 0.edu/~ken/ma570/homework/hw11/html/ch3b. Section 2 8. 10. and hence not of content 0 by part (a). . a. Show that an unbounded set Suppose Let . Section 2 http://www. So is not of content 0 as it cannot be covered with rectangles of total area less than the volume of . 1 of 2 12/6/2011 3:16 PM . . The sum of the volumes of the rectangles of is the volume of . b. The set of natural numbers is unbounded. if for volume of all the intervals is .ms. If C is a set of content 0. By replacing the with . contrary to hypothesis. one can assume that for all . if contains an interior point not in for some . then the union of the open intervals hand. are rectangles. there is a rectangle B of the type in part (a) such that has non-zero volume. Show that the set of all rectangles where each rational can be arranged into a sequence (i. In particular. closed and bounded. and cartesian products of countable sets are countable. for each in . be an increasing function. But then there is a finite collection of open intervals which cover the set and have total volume less than . 1].Exercises: Chapter 3. and hence compact. the set of these are countable. But then the sum of the lengths of the intervals in this finite subcover would be less than 1.edu/~ken/ma570/homework/hw11/html/ch3b.ms. there is a finite subcover of . If . and hence so are the set of corresponding 's. it must be countable too. Since the set these open intervals together with the set of form an open cover of [0. 12. 2 of 2 12/6/2011 3:16 PM . Let be the set of Problem 1-18. so is the set of all -tuples of rational numbers. this set of corresponding 's cover . the union of the interiors of the rectangles (where is allowed to range throughout ) is a cover of . Hence the set of discontinuities of is a countable union of finite sets. then it The set would be of content 0 by Theorem 3-6. 13.htm 11. Show that is a set of if finite for every Using the hint. we know by Problem 1-30 that the set of where . In fact. contrary to Theorem 3-5.e. Section 2 http://www. and hence has measure 0 by Theorem 3-4. Let measure 0. contains and is contained in some in .uky. If is any set and is an open cover of . Since the set of these intervals is just a subset of this set. we can even assume that is in the interior of the rectangle . show that the boundary of does not have measure 0. a. Following the hint. show that there is a sequence of members of which also cover . and each are Since the set of rational numbers is countable. If it were also of measure 0. b. By part (a). form a countable set). exists. 19. one gets an open rectangular cover of with sets. so be the set of rational numbers in . the sum of whose volumes is at most . If does not exist by Theorem 3-9. is a bounded set of measure 0 and Using the hint. Let be a which intersects . Since this is true of all . we outside of . Then let be a rectangle of of positive volume. Since has content 0. so theee first set is also of measure 0. So has content 0. so is . Show that if except on a set of measure 0. Then is not of measure 0 by Problem 3-8. Since this holds for all partitions Following the hint.uky. Let be the open set of Problem 3-11. by Problem 3-5. one has for some open rectangles the sum of whose volumes can be made as small as desired. 15. let be a positive integer and partition of such that . then . By replacing the closed rectangles with slightly larger open rectangles. Show that if and has content 0. are The set of where is not continuous is contained in the union of the sets where and not continuous. then it is bounded by Problem 3-9 (a). 16.edu/~ken/ma570/homework/hw12/html/ch3c.Exercises: Chapter 3. Show that if are integrable. Then if have . So 17. Further. But then there is a point of exists. and so is Jordan measurable by Theorem 3-9. one has if the integral . 18. Now apply Theorem 3-4 to conclude that has measure 0. which is not of which can be made as small as desired. then f is 1 of 2 12/6/2011 3:16 PM . it follows that . one has . . let be a partition of where is a closed rectangle containing . But then the boundary of is contained in the closure of . Let . So is a rectangle of . Then the boundary of measure 0.htm Exercises: Chapter 3. show that . If is non-negative and . But then is integrable by Theorem 3-8. Give an example of a bounded set Let of measure 0 such that does not exist. show that has measure 0. so of . for some closed rectangle and is Jordan-measurable If has content 0. which is contained in the union of the closures of the (since this union is closed). is . These last two sets are of measure 0 by Theorem 3-8. Section 3 http://www. so it is a subset of an closed rectangle . But then the boundary of must be of content 0. Section 3 14. and so .ms. The set of where is not continuous is which is not of measure 0. Let choose a finite set for of open rectangles the sum of whose volumes is less than and form a cover of the boundary of .ms. Apply Problem 3-21 with as the Jordan measurable set. show that there is a partition of such that all subrectangles intersecting and is Jordan measurable if and only if for every . Suppose for every . where consists of consists of allsubrectangles contained in . Then by replacing the rectangles with slightly larger ones. hence is Jordan measurable by Theorem 3-9. . 22. show that there is a compact Jordan measurable set be a closed rectangle containing . If is a Jordan measurable set and such that Let Let . Further 2 of 2 12/6/2011 3:16 PM . if it is of measure 0. then taking the union of this set with the set of measure 0 consisting of the points where and differ gives a set of measure 0 which contains the set of points where is not continuous. This is an immediate consequence of Problem 3-12 and Theorem 3-8. Define .uky. If the set where is not continuous is not of measure 0. there is a partition as in the statement. If is a closed rectangle. On the other hand. Show that an increeasing function is integrable on . one can obtain the same result except now one will have in place of and the will be open rectangles. then is not integrable by Theorem 3-8. 21. So this set is also of measure 0. Then its boundary is of content 0 by Theorem 3-9. which is a contradiction.htm not integrable on . Then and clearly is .edu/~ken/ma570/homework/hw12/html/ch3c. This satisfies the condition in the statement of the problem. be the partition as in Problem 3-21. 20. Section 3 http://www. This shows that the boundary of is of content 0. Jordan measurable by Theorem 3-9.Exercises: Chapter 3. Let be a partition of such that every such that the subrectangle of is either contained within each or does not intersect it. and Suppose is Jordan measurable. Exercises: Chapter 3, Section 4 http://www.ms.uky.edu/~ken/ma570/homework/hw13/html/ch3d.htm Exercises: Chapter 3, Section 4 23. Let be a set of content 0. Let be the set of all such that is not of content 0. Show that Following the hint, Theorem. We have or is a set of measure 0. by Problem 3-15 and Fubini's is equivalent to the condition that either is also of measure 0. is integrable with . Now . Both of these having integral 0 implies by Problem 3-18 that the sets where their integrand is non-zero are of measure 0, and so 24. Let be the union of all where is a rational number in written in lowest terms. Use to show that the word ``measure" in Problem 3-23 cannot be replaced with ``content". The set is the set of rational numbers in which is of measure 0, but not of content 0, because the . Let where integral of its characteristic function does not exist. To see that the set has content 0, let be such that . Then the set can be covered by the rectangles and for each in lowest terms with , the rectangle . The sum of the areas of these rectangles is less than 25. Show by induction on for each . that . is not a set of measure 0 (or content 0) if This follows from Problem 3-8 and Theorem 3-6, but that is not an induction. Fubini's Theorem and induction on not of measure 0. 26. Let be integrable and non-negative, and let . Show that . One has and so by Fubini, is Jordan measurable and has area show that and so does not have content 0, and hence is where 27. If is an upper bound on the image of . is continuous, show that 1 of 6 12/6/2011 3:17 PM Exercises: Chapter 3, Section 4 http://www.ms.uky.edu/~ken/ma570/homework/hw13/html/ch3d.htm where the upper bounds need to be determined. By Fubini, the left hand iterated integral is just where Applying Fubini again, shows that this integral is equal to 28. Use Fubini's Theorem to give an easy proof that . if these are continuous. Following the hint, if is not zero for some point , then we may assume (by replacing with if necessary that it is positive at . But then continuity implies that it is positive on a rectangle containing . But then its integral over is also positive. On the other hand, using Fubini on gives: Similarly, one has Subtracting gives: which is a contradiction. obtained by revolving a 29. Use Fubini's Theorem to derive an expression for the volume of a set in Jordan measurable set in the -plane about the -axis. To avoid overlap, it is convenient to keep the set in the positive half plane. To do this, let be the original 2 of 6 12/6/2011 3:17 PM Exercises: Chapter 3, Section 4 http://www.ms.uky.edu/~ken/ma570/homework/hw13/html/ch3d.htm Jordan measurable set in the measurable if is. -plane, and replace it with . Theorem 3-9 can be used to show that is Jordan The problem appears to be premature since we really want to be able to do a change of variables to cylindrical coordinates. Assuming that we know how to do that, the result becomes . 30. Let be the set in Problem 1-17. Show that but that does not exist. as that The problem has a typo in it; the author should not have switched the order of the arguments of trivializes the assertion. The iterated integrals are zero because the inside integral is the zero function. The last integral cannot exist by Theorem 3-9 and Problem 1-17. 31. If and is continuous, define by What is Let , for in the interior of ? be in the interior of , fix . We have by Fubini's Theorem. 32. Let be continuous and suppose . Prove Leibnitz' Rule: Using the hint, we have is continuous. Define . . One has 3 of 6 12/6/2011 3:17 PM edu/~ken/ma570/homework/hw13/html/ch3d. Find One has b. define . is continuous and is continuous. As in Problem 2-21. find . let Show that One has .Exercises: Chapter 3. Let be continuously differentiable and suppose . and .ms.htm 33. Section 4 http://www. 4 of 6 12/6/2011 3:17 PM . If We have and . and so by the chain rule one has 34.uky. where the second assertion used Problem 3-32. If a. is a cylinder with a parallelogram base in the second case. and is the same rectangle 5 of 6 12/6/2011 3:17 PM .ms. a. 1. and 1 respectively.edu/~ken/ma570/homework/hw13/html/ch3d. Section 4 http://www. show that the volume of is is . then is in the first case.Exercises: Chapter 3.htm 35. Let be a linear transformation of one of the following types: If is a rectangle.uky. If the original rectangle . In the three cases. . This shows that the compact portion of the hyperplane is of volume 0. for any linear transformation .ms. (Cavalieri's principle). Suppose that each and measurable and have the same area.uky. Prove that is the volume of If is non-singular. this shows the result in this case too. the result follows in this case. Since the determinant is also 0. the second and third case and get multiplied by b. then subspace. In the second case. Since is multiplicative. 36. 6 of 6 12/6/2011 3:17 PM . then it is a composition of linear transformations of the types in part (a) of the problem.Exercises: Chapter 3. So the volumes do not change in in the first case. is a proper subspace of and is a compact set in this proper is contained in a hyperplane. the into made up of a composition of hyperplane is the image of a linear transformation from maps of the first two types. This shows the result.edu/~ken/ma570/homework/hw13/html/ch3d. In particular. If is singular. This is an immediate consequence of Fubini's Theorem since the inside integrals are equal. Show that and have the same volume. Let and be Jordan measurable subsets of .htm except that the intervals in the the parallelogram base is in the and and places are swapped in the third case. Section 4 http://www. Let are Jordan and define similarly. By choosing the coordinate properly. directions and has corners . But . b. define and . satisfies and such that . .edu/~ken/ma570/homework/hw14/html/ch3e. Find two partitions of unity 1 of 2 12/6/2011 3:18 PM . the integral in the extended sense is same as the that in the old sense. Consider a partition of unity subordinate to the cover . Suppose that . Let and . and the limit is just 38. One has . Show that satisfies does not exist. Clearly. let it be . By summing the with the same in condition (4) of Theorem .Exercises: Chapter 3. a. The assertion that only know the values of the integral of intervals -. It follows that the sum in the middle does not converge as where and so does not exist. As in part (a). Section 5 http://www. Then is bounded on each interval and so by Theorem 3-12. From the hypothesis. Let be a closed set contained in and outside . Now 3-11.uky. Show that exists.so it could be that let us assume that is of constant sign and bounded on each set .htm Exercises: Chapter 3. Take a partition of unity subordinate to the cover for there is only one where . but don't know how behaves on other may not even exist for all To correct the situation. one can assume that there is only one function for each exists if and only converges.ms. If not necessrily true. exists if a natural number. we can assume Suppose that for all . we on the sets . Consider the convergence of . So the sum converges if and only if exists. but as in condition (4) of Theorem 3-11. the integral in the old sense is monotone on each interval of . Suppose that and only if For is a non-negative continuous function. Section 5 37. By forming open covers each set of which consists of intervals added to each of these partial sums. further by re-ordering the terms of this can be done so that there are sequences of partial sums which converge monotonically to the limit for the sum of terms value. So.uky. have terms of the same sign and are each divergent. one gets covers of . one can take a partition of unity subordinate to this cover. 2 of 2 12/6/2011 3:18 PM . Section 5 http://www. 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. By using arrangements with different limiting values. .edu/~ken/ma570/homework/hw14/html/ch3e. one gets the result.ms. Finally. one can make the sum approach any number we want.Exercises: Chapter 3. Because is zero outside .htm and The sums and converge absolutely to different values. Combining results. Combining . . we get Theorem 3-13. where . is a diagonal matrix. Let .edu/~ken/ma570/homework/hw15/html/ch3f. Then . Use Theorem 3-14 to prove Theorem 3-13 without the assumption that Let in place of the of . successively smaller open neighborhoods of . Let Then . 1 of 4 12/6/2011 3:18 PM . So we can define on and . Show that we can write matrix. we know that . then replace have the same form as the and with . is open and Theorem 3-13 applies with be a partition of unity subordinate to an admissible cover is a partion of unity subordinate to the cover is absolutely convergent. One then can verify that results gives and so Now. for and . .htm Exercises: Chapter 3. Section 6 http://www. By Theorem 3-14. If and . 40. Let . and is a linear transformation.uky.Exercises: Chapter 3.ms. Define for . So. We use the same idea as in the proof of Theorem 3-13. if Then the . and . Then . and so also converges since the terms are identical. Now in its statement. prove that in some open set containing is of the form if and only if we can write . Then . inverses of is a diagonal be a point where . Section 6 39. contrary to hypothesis. Since is linear. Suppose and so the other hand. If . the converse is false. Show .htm On the other hand. for all and .Exercises: Chapter 3. then let in the domain of . so diagonal matrix. For example.ms. it suffices to show that . then implies contrary to hypothesis. . it follows that the same value. So. But then .edu/~ken/ma570/homework/hw15/html/ch3f. and show that is the set of Problem 2-23. is 1-1. consider the function . to show that the function imply (or is 1-1. Define a. follow from the last paragraph of the solution of part (a). Then . -axis and the ray from (0.y).uky. Section 6 http://www. On and let . where be the angle between the . 41. for all is not a .) Find P'(x. Show that that Since and implies that . compute by . .0) through . .e. 2 of 4 12/6/2011 3:18 PM . i. if positive b. Then and has So. One has is 1-1. But then . If . Suppose ). show that (Here denotes the inverse of the function The function is called the polar coordinate system on The formulas for and . If . . show that and For the first assertion. Then .pital's Rule allows one to by checking separately for the limit from the left and the limit from . . . Section 6 http://www.edu/~ken/ma570/homework/hw15/html/ch3f. So the first identity holds. show that Assume that and d. Apply Theorem 3-13 to the map . apply part (c) with Applying (c) gives The second assertion follows from Fubini's Theorem. One has . Prove that . Let be the region between the circles of radii which make angles of and with the -axis. c. Clearly. The second identity is a special case of the first. e. show that and and the half-lines through 0 is integrable and If . This is trivial from the formulas . Further. 3 of 4 12/6/2011 3:18 PM . If and by .uky.Exercises: Chapter 3. . the right. If . L'[email protected] One has except in case calculate when . For example. also. So Since part (d) implies that the squeeze principle implies that But using part (d) again. (since the square 4 of 4 12/6/2011 3:18 PM .Exercises: Chapter 3. also exists and is . we get root function is continuous).edu/~ken/ma570/homework/hw15/html/ch3f.htm and conclude that One has and the integrands are everywhere positive.ms.uky. Section 6 http://www. factor were not in the definition of b. Section 1 http://www. . then the right hand side would have been . Consider the linear map for all : defined by .htm Exercises: Chapter 4. What would the right hand side be if the factor did not appear in the definition of ? The result is false if the are not distinct. By multilinearity. Then . One has 1 of 7 12/6/2011 3:19 PM . are all distinct.uky.ms. Section 1 1. One has using Theorem 4-1(3): because all the summands except that corresponding to the identity permutation are zero. Show that is the determinant of thee minor of obtained by selecting columns Assume as in part (a) that the . Let a. If the . Show that be the usual basis of and let be the dual basis.edu/~ken/ma570/homework/hw16/html/ch4a. A computation similar to that of part (a) shows that if some for all . Assume therefore that the are distinct. it follows that we need only verify the result when the are in the subspace generated by the for . in that case. the value is zero.Exercises: Chapter 4. one has for . and . and . .edu/~ken/ma570/homework/hw16/html/ch4a. So 3. If is the volume element determined by 2 of 7 12/6/2011 3:19 PM . Section 1 http://www.htm This shows the result. Then by Theorem 4-6. show that . then must be multiplication by some constant Let . If is a linear transformation and . Show that .Exercises: Chapter 4.ms.uky. 2. . Then we have by blinearity: . such are linearly independent. Taking absolute values and substituting gives the are just the entries of result. is a basis for The function is a continuous function.e. and so is the volume element with respect to is of dimension 1. we have . Since the are linearly independent. If that One has and so is the volume element of determined by and . If is continuous and each . it follows that the image of consists of numbers all of the same sign. i. for some . the right hand sides . a.Exercises: Chapter 4.uky. show that is the usual orientation of . it is the vector for every . whose image does not contain 0 since is a basis for every t. Section 1 http://www. Substitution shows that works. and let where . If . By the intermediate value theorem. 6. we have 5. If that b. By Theorem 4-6. . show that and such that by the definition of and . and the fact that because as desired. show that is an isomorphism such . and .ms. Combining. By the definition. the definition of cross product with completing the basis shows 3 of 7 12/6/2011 3:19 PM . 4. So all the have the same orientation. be an orthonormal basis for V with respect to . . what is ? is the cross product of a single vector. Further.htm where Let .edu/~ken/ma570/homework/hw16/html/ch4a. the determinant is positive. and are an orthonormal basis of determined by and . define a ``cross product" such that : in terms of for all . If is the volume element determined by some inner product be the volume element determined by some inner product be an orthornormal basis (with respect to such that for . and orientation . 4 of 7 12/6/2011 3:19 PM .g. The cross product is the 9.uky. This is the volume element of is a volume element. which shows the result. Let . 7. e. with respect to .Exercises: Chapter 4. So in fact. Let scalar and shows that 8. All of these follow immediately from the definition. .edu/~ken/ma570/homework/hw16/html/ch4a. Then ) such that .ms. Expanding out the determinant shows that: . Section 1 http://www.htm that the cross product is not zero. To show that that for all . . Show that every non-zero and orientation for . note b. and let . Deduce the following properties of the cross product in a. . . There is a . By Problem .edu/~ken/ma570/homework/hw16/html/ch4a. or is zero. . For the second assertion. But this can 5 of 7 12/6/2011 3:19 PM .ms.uky. This is easily verified by and since The second assertion follows from the definition since the determinant of a square matrix with two identical rows is zero. Suppose that . The result is true if either 1-8. the first identity is just substitution using part (b). d. . For the first assertion. one has: So. one needs to show that for all be easily verified by expanding everything out using the formula in part (b). Section 1 http://www. . one has and .htm c. where .Exercises: Chapter 4. and and are both non-zero. one has: since the matrix from Problem 4-3 has the form result in the case where is not zero.edu/~ken/ma570/homework/hw16/html/ch4a.uky. When it is zero. See the proof of part (c). the too. show that .Exercises: Chapter 4. Using the definition of cross product and Problem 4-3. 10. If . This proves the are linearly dependent. where . and the bilinearity of inner product imply that 6 of 7 12/6/2011 3:19 PM .htm The third assertion follows from the second: e. Section 1 http://www. .ms. for each . 12.edu/~ken/ma570/homework/hw16/html/ch4a.ms. one has: But One has .htm 11. by .Exercises: Chapter 4. Use Problem 2-14 to derive a formula for Since the cross product is multilinear. If is an orthogonal basis and is the matrix of with respect to this basis.uky. show that . If to is an inner product on . which shows the result. one can apply Theorem 2-14 (b) and the chain rule to get: 7 of 7 12/6/2011 3:19 PM . If . Using the orthonormality of the basis. define when are differentiable. Section 1 http://www. a linear transformation is called self-adjoint (with respect ) if for all . Define . . Let be a differentiable curve in of at as . Section 2 http://www. 15.edu/~ken/ma570/homework/hw17/html/ch4b. a differentiable function . at is to the graph of at lies on the tangent line to the graph of at is The tangent vector of . Here is the interpretation: .uky. show that and The second assertion follows from: b. Let vector of and define by at . If and . show that . If .htm Exercises: Chapter 4. One has by the definition and the product rule: 14. a. that is.Exercises: Chapter 4. the tangent vector show that the tangent vector to This is an immediate consequence of Problem 4-13 (a). The notation does not fully elucidate the meaning of the assertion. Section 2 13.ms. The end point of the tangent vector of which is certainly on the tangent line 1 of 4 12/6/2011 3:20 PM . If at is . Show that the end point of the tangent . Since where . i. where is the . Section 2 http://www. define a vector field by . gives at . be a curve such that at are perpendicular.Exercises: Chapter 4. this is clearly when 19. define the forms 2 of 4 12/6/2011 3:20 PM . in the direction of . By Problem 2-29. prove that is the direction in which is changing fastest at The direction in which . for all . i. define a vector field by is of the form for some a. define .uky. Show that One has . Given such an .e.htm at 16. If . If For obvious reasons we also write and conclude that . by 18.e. Let vector to . . . which assigns to each an element . If . . . . Show that every vector field on A vector field is just a function . If is a vector field on is changing fastest is the direction given by a unit vector such thatt is largest possible.edu/~ken/ma570/homework/hw17/html/ch4b. Then b.ms. Show that and the tangent Differentiating tangent vector to 17. Prove that The first equation is just Theorem 4-7.htm a. so 3 of 4 12/6/2011 3:20 PM . by part (a) and Theorem 4-10 (3).uky. For the second equation.edu/~ken/ma570/homework/hw17/html/ch4b. Use (a) to prove that One has . one has: For the third assertion: b. Section 2 http://www.ms.Exercises: Chapter 4. one has trivially because is constant when and (or ). is exact. c. i. 4 of 4 12/6/2011 3:20 PM . the assertion is immediate from the definition of in Problem 2-41.Exercises: Chapter 4. Similarly. show that for for some vector . then . 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.e.htm Also. show that the same is true of on such that is closed. By the Theorem 4-11. Section 2 http://www.edu/~ken/ma570/homework/hw17/html/ch4b. 20. if . If is a vector field on a star-shaped open set . we have . By part (a). . Further. So Theorem 4-11. By . Then . Similarly. if some function field on . . i. If every closed form on Suppose that the form there is a form on and so is also exact. Prove that on the set where is defined. In case . if i. For example. Let be a differentiable function with a differentiable inverse is exact. But then and so 21. so the second assertion is also true. as desired.ms. .uky. as desired. So . then .e. show that by part (a) and Theorem . and .e. it must then be exact. 4-10 (3). and so for some is closed. Except when . let also denote the function can be written . Define is easily seen to be such that . and the integers.uky. An -chain is a function such that for all but finitely many .Exercises: Chapter 4. and . and . Show that and are -chains if are. If . the functions The second assertion is obvious since 23.edu/~ken/ma570/homework/hw18/html/ch4c. Define and by and . For and an integer. is the function of Problem 3-41 extended so that it is 0 on the so that is an integer because . Show that there is a singular 2-cube such that Define where 24. and and are -chains if and are. is a singular 1-cube in for some 2-chain . show that there is an integer Given positive . One has and boundaries. let -axis. by and . Section 3 22. such that and for some integers for Show that every -chain singular -cubes Since . Section 3 http://www. If and . define the singular 1-cube . by are positive real numbers. On the other . The boundary of with .htm Exercises: Chapter 4. Let where 1 of 1 12/6/2011 3:20 PM .ms. . Let be the set of all singular -cubes. So desired. as . Independence of parametrization). and Theorem 3-13 augmented by Problem 3-39: 26.htm Exercises: Chapter 4. and use Stokes Theorem to conclude that (recall the definition of in Problem 4-23).uky. Let function such that show that be a singular and -cube and for . Suppose . Theorem 4-9.Exercises: Chapter 4. then Stokes Theorem gives 1 of 10 12/6/2011 3:21 PM . for any 2-chain If . Section 4 http://www. If is a a 1-1 -form. Section 4 25.ms. Show that in One has .edu/~ken/ma570/homework/hw19/html/ch4d. Using the definition of the integral. the chain rule. we have and all with . Note however that no curve is the boundary of any two chain -.edu/~ken/ma570/homework/hw19/html/ch4d.as the sum of the coefficients of a boundary is always 0. When . If of Problem 4-24 is unique. 27. So . Letting be sufficiently large. one has where and and are 2-chains.Exercises: Chapter 4. Suppose that 2 of 10 as above has no complex root. . one gets . b. and that by if is large enough. Show that the integer around 0. Let . This integer is called the winding number of and . Show that is simply with . We assume . If . which is a contradiction. So . 28.uky. is the curve . and the singular 2-cube . prove the Fundamental Theorem of Algebra: Every polynomial with has a root in . it is the curve . we see by part (a) 12/6/2011 3:21 PM .htm because is closed. it is the curve . This would make the boundary these changes have been made. Define thee singular 1-cube . Section 4 http://www. if for all or . Using Problem 4-26. The problem statement is flawed: the author wants to be defined to be . Recall that the set of complex numbers let be by a. we see that cannot be zero since it is the sum of a number of length and one which is smaller in absolute value. and when Then. Since where . When . . the letting . Using Stokes Theorem.ms. we get contradicts the result of the last paragraph. Let an be a singular 1-cube with such that . Now consider the 2-chain . define the singular 2-cube . If is a 1-form on with with . By the proof of the last denote this common value. Now. So . such that . and so is a 2-chain with . Section 4 http://www. if we let and 30. prove that for some and . there is a 2-chain and problem. Henceforth. This . we have . Further. The differential for which there is a is of the type considered in the last . So . when . integrating two singular 1-cubes and with and is independent of path. . applying Stokes' Theorem. By Stokes' Theorem. By Problem 4-24. implies be this value and . Note that . then prepend a curve from 3 of 10 12/6/2011 3:21 PM . In fact. From the result of the last paragraph. So the boundary of . then and so is such that for some function Following the hint. So there is a unique For positive and such that . let . Again. show that theere is a unique number . we get the curve . and in particular. By Stokes' Theorem. On the other hand.edu/~ken/ma570/homework/hw19/html/ch4d. it follows that . by problem. we get the constant curve with value . 29. when .htm and Stokes' Theorem that . if you have . and when or defined by . and so .Exercises: Chapter 4. unique. and so . .uky. we have assumed that values in has no complex root.ms. If is a 1-form on . we get the curve is . uky. This is a contradiction. and is exact. by treating each component separately. for all and with . Then in a closed rectangle of positive volume centered at . Give a counter-example on Let similarly for Suppose .htm (1. If and shows that there is no independence of is a 1-form on a subset of . where are points. For every in the set. Stokes' theorem and and some choice of . show that there is a chain to prove One has . let be any curve from to . The example: path in b. 31. Conclude that if is merely closed. and so . that is.edu/~ken/ma570/homework/hw19/html/ch4d. and set 4 of 10 12/6/2011 3:21 PM .Exercises: Chapter 4. The two integrals are both 0. Then . and .0) to and postpend a path from to (1. Further. is exact. and so the integrals over and are equal.ms. . Now the result follows from Problem 4-32 below. Suppose for some such that . we assume that the subset is pathwise connected. By Stokes' Theorem. and hence closed. be defined by where is the curve with constant value and is exact. 32. Let be singular 1-cubes in such that and with and . Then by Stokes' Theorem. Show that and are if there is a singular 2-cube degenerate. we assume that the subset is open. Let be a chain such that because . Suppose . for closed forms. Fix a point in the subset. 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 . Then since the integrand is continuous and of the same sign throughout the region of integration.0) to get two paths as in the last paragraph. show that Although it is not stated. a. Use this fact. Section 4 http://www. If . we have (since is closed). we would have: . Exercises: Chapter 4. it follows that . and quotient of analytic functions are analytic. Show that is analytic and sum. One has The first pair of terms is because . is closed since . multiplication. but 5 of 10 12/6/2011 3:21 PM . then . calculated with a path that ends in a segment with Similarly. We want to check differentiability of . This establishes the assertion. product. and so does not have a limit as because .) If is differeentiable at every point in an open set and is continuous on .htm .ms. . Because of independence of path. On the other hand. (This quotient involves two complex numbers and this definition is completely different from the one in Chapter 2. and so the last integral is also . if is is in the interior of the subset. . if you use the identities: . if the exists. and division operations are continuous (except when the quotient is zero).uky. Show that the a. continuity of implies that the integrand is . then because is well defined. The assertion that being analytic is preserved under these operations as well as the formulas for the derivatives are then obvious. So is differentiable at . 33. Now. subtraction. Note that because and are continuously differentible. It is straightforward to check that the complex addition. is not (where ).edu/~ken/ma570/homework/hw19/html/ch4d. then is called analytic on . we can assume that constant. Clearly. (A first course in complex variables. similarly the second pair of terms is Finally. define to be differentiable at .) If limit . Section 4 http://www. What complex number is this? Comparing and exist if and only if and gives . . complex number if and only if and . we must have: Comparing the real and imaginary parts gives the Cauchy-Riemann equations. Part (b) shows that an analytic function . Let If the matrix of be a linear transformation (where with respect to the basis is is considered as a vector space over . Section 4 http://www. So. c. If is analytic on . From the last paragraph. show that and satisfy the Cauchy-Riemann} equations: (The converse is also true. . the complex number is .edu/~ken/ma570/homework/hw19/html/ch4d.htm b. has a derivative which is multiplication by a complex number. show that is multiplication by a ).uky. and .) and are continuously differentiable. d. and . considered as a function .ms.Exercises: Chapter 4. this is more difficult to Following the hint. Define where and 6 of 10 12/6/2011 3:21 PM . if prove. it follows that . for every closed for some 2-chain in . then . the 1-form is closed. f. By parts (b) and (d). Prove the Cauchy Integral Theorem: If curve (singular 1-cube with is analytic in ) such that .htm and Show that One has for if and only if that satisfies the Cauchy-Riemann equations.ms.edu/~ken/ma570/homework/hw19/html/ch4d. Clearly this is zero if and only if the Cauchy-Riemann equations hold true for e.Exercises: Chapter 4. Section 4 http://www. then (or . Conclude that in classical notation) equals .uky. Show that if for some function One has . 7 of 10 12/6/2011 3:21 PM . By Stokes' Theorem. then The first assertion follows from part (e) applied to the singular 2-cube . Section 4 http://www. use the fact that to show that is analytic in This then gives g.ms. with .uky. If is analytic on if for .edu/~ken/ma570/homework/hw19/html/ch4d. . .htm if is defined by .Exercises: Chapter 4. Use (f) to evaluate and and conclude: is a closed curve in Cauchy Integral Formula: If is analytic on with winding number around 0. By a trivial modification of Problem 4-24 (to use for Further. 8 of 10 12/6/2011 3:21 PM . defined by ) and Stokes' Theorem. is called a homotopy between the closed curve and the closed curve . if for each the closed curves and do not intersect.htm if is chosen so that for all with . the pair is called a homotopy between the non-intersecting closed curves and . one gets the same singular 2-cube ).uky. define by If each is a closed curve. The Cauchy integral formula follows from this and the result of the last paragraph. similarly. when . The present problem. Using part (f). and Problem 5-33 prove this for (b) but the proof for (c) requires different techniques. one gets the same singular 2-cube . If are nonintersecting closed curves. If and . It is intuitively obvious that there is no such homotopy with the pair of curves shown in Figure 4-6 (a). define by If by is a homotopy of nonintersecting closed curves define Show that When . and the pair of (b) or (c). a. 9 of 10 12/6/2011 3:21 PM . Suppose and are homotopies of closed curves.edu/~ken/ma570/homework/hw19/html/ch4d. When (respectively ). 34. It follows that .Exercises: Chapter 4. So which agrees with the assertion only up to a sign.ms. one gets the singular 2-cube (respectively . we conclude that . Section 4 http://www. show that By Stokes' Theorem and part (a). 10 of 10 12/6/2011 3:21 PM . If is a closed 2-form on .edu/~ken/ma570/homework/hw19/html/ch4d.htm b. one has .Exercises: Chapter 4.uky.ms. Section 4 http://www. each of its points satisfies condition with . Show that is an -dimensional manifold with boundary. Find a counter-example to Theorem 5-2 if condition (3) is omitted. The generalization to manifolds is proved in the same way. say with the function . then the same is a -dimensional which satisfy condition .htm Exercises: Chapter 5. Let be an open set such that boundary is an -dimensional manifold. Then condition holds except for part 3. each such for each such . This proves the assertion. which is absurd since the points of U in the boundary of all map to points with last coordinate 0. Prove a similar assertion for an open subset of an n-dimensional manifold. 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 . prove that 1.Exercises: Chapter 5. Let such that be as in .) Since is open. except you need to restrict attention of condition . It follows that h restricted to an appropriately small open or condition . Section 1 http://www. consider defined by Let (3) since . If there is no open neighborhood of such that . Section 1 is a -dimensional manifold with boundary. Further. . subset of either satisfies condition b. then mind the following example: if is a manifold with boundary. also is map which shows that condition is satisfied is a manifold of dimension .ms. . Then satisfies with . If manifold and is a =dimensional manifold. and because those points which don't . a. By working in the set 1 of 3 12/6/2011 3:22 PM . Let .edu/~ken/ma570/homework/hw20/html/ch5a. But then the line segments from to must contain a point o the boundary of . (It is well to bear in . Following the hint. it follows that is a manifold of dimension . but . The boundary of condition is the set of points . then there is a sequence of points of such that the sequence converges to .uky. So satisfy is in . . must satisfy 2. one gets back into to a coordinate system around . In works for every point in particular. .ms. Let be the projection on the last coordinates. it is not obvious what one means by ``rotate". If is defined in some open set by . If . -dimensional manifold. Example: the torus (Figure 5-4). If . Conversely. a. Then the function -dimensional (vector) subspace of is a satisfies all the desired conditions. . . In the case where n > 3. Consider the case where n = 3. the map . suppose is as in condition for some point in the graph. so the graph is a manifold of dimension n. and choose .edu/~ken/ma570/homework/hw20/html/ch5a. the graph of is -dimensional manifold if and only if If is differentiable. is a -dimensional manifold . 8. so that all the together form be a basis for the subspace. it is easy to see that the Jacobian has the proper rank. Then apply the Implicit function theorem to .Exercises: Chapter 5. One can verify 6. defined by is an is easily verified to be a coordinate system around all points of the graph of f. Let and is obtained by revolving around the axis since the graph is the set of points which map to zero by .htm the case where one is contained within 4. then is defined by . Show that the graph of is differentiable. show that has measure 0. one has condition and . Define a map . If is a -dimensional manifold in .uky. is a -dimensional manifold and . Section 1 http://www. and the same argument applies. then such that and a differentiable function when . and be as in condition applied to be defined . by satisfies the condition . The Jacobian is Since either or is non-zero. Prove a partial converse to Theorem 5-1: If there is an open set and Let by 5. Prove that a Let a basis for that containing has rank . Let be the For each holding for some function 2 of 3 12/6/2011 3:22 PM . The differentiable function obtained from this theorem must be none other than 7. show that is a -dimensional manifold. to be closed. where we can choose to be a ball with center at rational coordinates and rational radius. If is is a closed -dimensional manifold with boundary in . Section 1 http://www. Now each maps points of in to points with the last coordinates 0. the definition of Jordan measurable is satisfied. Take a thin plate including the image of . is in the boundary of . show that is .uky.Exercises: Chapter 5. . it follows that bounded. is Jordan- is a counter-example if we do not require -dimensional manifold with boundary in . By choosing the thickness of the plate sufficiently small.htm domain of one of these functions. By part (a). Give a counter-example if is not closed. If is a compact measurable. the boundary of manifold contained in . So if But then is in the interior of because the dimension of The open unit interval in c.edu/~ken/ma570/homework/hw20/html/ch5a. By part (b). Finally. Then is a countable cover of . show that the boundary of Clearly. boundary of . b. we can guarantee that this value is no more than for the element of the cover. By Problem 5-1. then since is closed. since 3 of 3 12/6/2011 3:22 PM .ms. This shows the result. its inverse image has volume which can be bounded by where is the volume of the plate (by the change of variables formula). every element of by the condition . If is in the . it must satisfy condition is n. is an -dimensional is is of measure 0. Since is in .e. we know that implies that where . Section 2 9. . .Exercises: Chapter 5. (2) if there is . But analogous to the author's observation of p.edu/~ken/ma570/homework/hw21/html/ch5b. 119. where the tangent vector of 10. such that (1) For each . Then let be as in condition is a coordinate . For . it follows that consists of the tangent vectors at of curves be a coordinate system around is a rectangle centered at . Conversely. 1 of 5 12/6/2011 3:23 PM . If . Show that Let assume that curve vary. If is an orientation preserving for all as desired. 11. In order for this to be well defined. let . show that the (the orientation so defined is the usual orientation of two definitions of given above agree. Clearly. and with .uky.ms. one can be the as and ranges through out . Let be such that . Show that there which is a coordinate system around is a unique orientation of such that is orientation-preserving for all Define the orientation to be the for every . and this is only orientation which -dimensional manifold-with-boundary in . Then in and in with . Then we have i. define as the usual orientation of . suppose that for the point system about is a curve in with . with a subset. then . Section 2 http://www. the definition makes could satisfy this condition. Suppose is a collection of coordinate systems for . We know by the proof of Theorem 5-2. that . by replace . we must show that we get the same orientation if we use use and .htm Exercises: Chapter 5. and clearly it cannot be extended to the point in a differentiable manner. by definition. is a sum of finitely 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. one might have a vector field defined on by . This is a vector field of outward pointing unit vectors. . But now we see that the resulting is a differentiable extension of to all of . For . one can assume that the 1. Then to which extends the restriction of and be a partition of unity subordinate to . Since is closed. to . In fact. many differentiable vector fields . and is perpendicular to is the usual orientation of is the induced orientation on and . we have now assured that in a neighborhood of any point. But then . show that there is an open set for . define differentiable vector field on Let choose a with be the projection on the first . . . If . Let a. For example. let be the essentially unique diffeomorphism such that 2 of 5 12/6/2011 3:23 PM . If is a differentiable vector field on on with .htm Let and where be a coordinate system about with .edu/~ken/ma570/homework/hw21/html/ch5b. let b. Note that . For is a . is the dimension of .ms. there is a diffeomorphism where coordinates. Let . Section 2 http://www. This assures that there are at most finitely many which intersect any given bounded set.uky. We can then replace with the union of all these finite subcovers for all . a. and so we can choose a finite subcover of . where satisfying condition . is compact. is closed. . Then is a differentiable extension of . show that we can choose are open rectangles with sides at most In the construction of part (a). and so. If be as in Theorem 5-1. 12. Let is the unit normal in the second sense. 13. Define non-zero only for elements of Finally. and a differentiable vector field Let For every . Show that the orientations ince can be defined consistently.htm and . Now which is 1-1 because is a diffeomorphism. we have a defined in an open neighborhood of . Let be . and . Since it is 1-1. show that the components of the outward normal at . Then . This shows that is is an orientable -dimensional manifold. this follows from Problem 5-10 with c. If as desired. we can assume locally that the orientation imposed by is the given orientation . Problem 5-4 does the problem locally. So for each .uky. Each is non-zero only inside some . that the are distinct for distinct .Exercises: Chapter 5. If . and a differentiable Choose an orientation for . are some multiple of We have we get perpendicular to 14. Let we can assume by replacing the with sums of the 3 of 5 12/6/2011 3:23 PM . Further.ms. let and so . As the hint says. Let . is a coordinate system about every point of . b. . it maps its domain onto a space of dimension and so the vectors. Section 2 http://www. and . we can assume that takes on non-negative values. as in the proof of the implicit function theorem. being a basis.edu/~ken/ma570/homework/hw21/html/ch5b. Let and Also. show that there is an open set so that and has rank 1 for . By replacing with its square. Show that are linearly independent. Define vectors by The notation will be changed. is 1-1 so that the . . be defined by defined. using Problem 5-13. so that is orientable. by . We have changed the order of the arguments to correct an apparent typographical error in the problem statement. . and so by considering the components. and be a partion of unity subordinate to . must map to linearly independent vectors. uky. which is precisely the condition to be proved. is in the subspace generated by the . and be as in Problem 5-4 in a neighborhood of we have a coordinate systems of the form . The following problem gives a nice theoretical use for Lagrangian multipliers. is of rank and is of dimension . One can attempt to find we must solve equations by solving the system of equations. Let minimum) of be as in Theorem 5-1. In particular. Let be the set of end-points of normal vectors (in both directions) of length and suppose is small enough so that is also an -dimensional manifold. Show that is orientable (even if is not). which is often very simple if we leave the equation for last. Then Let be a coordinate system in a neighborhood of the extremum at and so . But we also have for all near . This is Lagrange's method. In particular. such that . If on occurs at is equivalent to a single ring is differentiable and the maximum (or . . Now the image of is just the tangent space . 16. 4 of 5 12/6/2011 3:23 PM . What is if is the M"{o}bius strip? Let . Let be an -dimensional manifold in .edu/~ken/ma570/homework/hw21/html/ch5b.Exercises: Chapter 5. Then and of the form (respectively ) In the case of the M"{o}bius strip. In particular. the . 15. if in unknowns . This is an orientation for . and so the row of is perpendicular to the tangent space . and so the rows of are also perpendicular to . and so the rows of generate the entire subspace of vectors perpendicular to . Choose an orientation on each piece so that adding gives the usual orientation on . and the useful but irrelevant is called a Lagrangian multiplier. But. and so .htm be defined by . Let be as in Problem 5-13.ms. Then satisfies the desired conditions. . show that there are The maximum on on is sometimes called the maximum of subject to the constraints . Section 2 http://www. . this is true at . Since is compact. Now. so that the manifold is . the case has already been shown. By considering the maximum of One has on show that there is and with . takes on a for which the Lagrangian multiplier equations are true. Apply Problem 5-16 with is a Lagrangian multiplier precisely when maximum on .uky. has a basis of eigenvectors with eigenvalues respectively. 89 for the definition). Suppose It is true for dimension . . Then apply part (a) to find the eigenvector with eigenvalue .Exercises: Chapter 5. it is clear that as a map of has a basis of eigenvectors. is of dimension . . So.htm 17.edu/~ken/ma570/homework/hw21/html/ch5b. Then . Since as a map of is self-adjoint and is also self-adjoint (cf p. In this case. and so the maximum has a . This shows the result. Show that Proceed by induction on . and so . a. show that . This shows that . c. All the together is the basis of eigenvectors for . Let be self-adjoint with matrix . Section 2 http://www. If . 5 of 5 12/6/2011 3:23 PM . show that and is self-adjoint. If Suppose .ms. b. so that . is the same as be the open interval and .uky. prove that . An absolute -form on -tensor on defined. In order for this to work. Section 3 http://www. 20. with the usual orientation. has the usual orientation. With the set of positive real numbers. Show that Theorem 5-5 holds for noncompact . One has with . if we let One can also let is not required to be compact. Make the definition the same as done in the section. Section 3 18. Show that Theorem 5-5 is false if For example. is a function such that of the form is an absolute for . In our case. An absolute can be is empty. 1 of 2 12/6/2011 3:23 PM . that constants. Give a counter-example if One has with 21. subset of provided that vanishes outside of a compact The compactness was used to guarantee that the sums in the proof were finite.Exercises: Chapter 5. nor that the singular -cubes be orientation preserving. as that -tensor on is a function . even if is not orientable. one has but . . it also works under this assumption because all but finitely many summands are zero if vanishes outside of a compact subset of . The singular -cubes We can assume in the situation of Chapter 3 that with can be taken to be linear maps where and are scalar . except don't require the manifold be orientable. If is a -form on a compact is not compact. 19.edu/~ken/ma570/homework/hw22/html/ch5c. . a. this is automatic because Theorem 4-9 gives . b. So. If is an -dimensional manifold (or manifold-with-boundary) in . as defined in this section.ms. as defined in show that Chapter 3. and there the crucial step was to replace with its absolute value so that Theorem 3-13 could be applied. one has -dimensional manifold . we need to have the argument of Theorem 5-4 work.htm Exercises: Chapter 5. . the two integrals give the same value. Show that . for example.htm 22.Exercises: Chapter 5. By Stokes' Theorem. and are compact. as stated. So. . If is an -dimensional manifold-with-boundary and -dimensional manifold with boundary. the result. and have the orientations induced by the .uky. and . Then is an -dimensional manifoldFollowing the hint. prove that is an where is an usual orieentations of -form on and . . but. is not correct. it 2 of 2 12/6/2011 3:23 PM . let with-boundary and its boundary is the union of and . the orientation of are opposite in the two cases. Section 3 http://www. we have So the result is equivalent to would be true if were closed.edu/~ken/ma570/homework/hw22/html/ch5c.ms. Because the outward directed normals at points of are in opposite directions for and . To see this. is orientation preserving. This is the form which matches the proposed solution since Furthermore.Exercises: Chapter 5.) Let be an orientation-preserving -cube. show that Consider the 1-form defined by . By Theorem 4-9. 1 of 9 12/6/2011 3:24 PM . Section 4 23. Section 4 http://www.edu/~ken/ma570/homework/hw23/html/ch5d.uky. show that where if and only if where .e. But then by orthonormality. If is an -dimensional manifold in and . i. Since this is also equal to . it is the volume element. (Note that this depends on the numerical factor in the definition of . If is an oriented one-dimensional manifold in and is orientation-preserving. 25. Then 24. with the usual orientation. so must have the usual orientation. choose an orthonormal basis .e. is the volume as defined in Chapter 3. . . we have and so applying this to be an gives Since Now. so that the volume of .htm Exercises: Chapter 5. it follows that the value is precisely 1.ms. and so it is the volume element in the sense of this section. is orientation preserving. i. as defined in this section. Now. Generalize Theorem 5-6 to the case of an oriented The generalization is defined by -dimensional manifold in . as desired. it must be that . Let orthonormal basis with the usual orientation. . if the are an is the orthonormal basis with orientation volume element . for some scalar and so for all . Letting .ms. was the orientation used to determine the outward normal). So Expanding in terms of cofactors of the last row gives: As in the 2-dimensional case.edu/~ken/ma570/homework/hw23/html/ch5d.Exercises: Chapter 5. Section 4 http://www. As in the 2-dimensional case.htm where is the unit outward normal at (where .uky. we get 2 of 9 12/6/2011 3:24 PM . Exercises: Chapter 5, Section 4 http://www.ms.uky.edu/~ken/ma570/homework/hw23/html/ch5d.htm 26. a. If surface is non-negative and the graph of , show that the area of is in the -plane is revolved around the -axis in to yield a One can use singular 2-cubes of the form calculate out to , , and . b. Compute the area of . . The quantities . So the surface area is , , and Apply part (a) with . So and . One has . is a and 27. If is a norm preserving linear transformation and the same volume as . Although it is not stated, it is assumed that is orientable. -dimensional manifold in , show that has 3 of 9 12/6/2011 3:24 PM Exercises: Chapter 5, Section 4 http://www.ms.uky.edu/~ken/ma570/homework/hw23/html/ch5d.htm By Problem 1-7, is inner product preserving and so it maps orthonormal bases to orthonormal bases. Further, if is a singular -cube which is a coordinate system for in a neighborhood of , then is a singular -cube which is a coordinate system for in a neighborhood of . Depending on the sign of , the new -cube is either orientation preserving or reversing. In particular, the 28. is the volume element of if is the volume element of (which is also orientable). Since the volume is just the integral of the volume element and the integral is calculated via -cubes, it follows that the volumes of and are equal. a. If is a -dimensional manifold, show that an absolute can be defined as . -tensor |dV| can be defined, even if is not orientable, so that the volume of This was already done in Problem 5-21. b. If is defined by show that is a M@ouml;bius strip and find its area. To see that it is a M@ouml;bius strip, note that in cylindrical coordinates, the equations are: . In particular, for fixed , we have fixed and the path is a line segment traversed from to . Calculating the length of the line, one gets that is a line segment of length 2. Again, for fixed , the line segment in the -plane has slope . Note that this varies from down to as ranges from 0 to , i.e. the line segment starts vertically at and reduces in slope until it becomes vertical again at . This corresponds to twisting the paper 180 degrees as it goes around the ring, which is the M@ouml;bius strip. To find the area, one can actually, just use the formulas for an orientable surface, since one can just remove the line at . In thatt case one can verify, preferably with machine help, that , , and . So the area is . Numerical evaluation of the integral , which is just slightly larger than , the area of a circular ring of yields the approximation radius 2 and height 2. 29. If there is a nowhere-zero Suppose -form on a -dimensional manifold . If is a singular , show that is orientable , we have is the nowhere-zero -form on -cube, then for every because the space some p. Then if it were negative at another point is of dimension 1. Choose a -cube so that the value is positive for into , then because this is a continuous function from , the intermediate value theorem would guarantee a point positive for all For every , choose a . -cube of this type with -cubes, say and where the function were zero, which is absurd. So the value is in its image. Define . This is well defined. (where is Indeed, if we had two such , then the ) are positive for . But then the values are of the same sign regardless of which non-zero k-form in used. So, both maps define the same orientation on . 30. a. If length is differentiable and . is defined by , show that has This is an immediate consequence of Problem 5-23. 4 of 9 12/6/2011 3:24 PM Exercises: Chapter 5, Section 4 http://www.ms.uky.edu/~ken/ma570/homework/hw23/html/ch5d.htm b. Show that this length is the least upper bound of lengths of inscribed broken lines. We will need to be continuously differentiable, not just differentiable. , then by the mean value theorem, Following the hint, if for some . Summing over gives a Riemann sum for . Taking the limit as the mesh approaches 0, shows that these approach the integral. Starting from any partition, and taking successively finer partitions of the interval with the mesh approaching zero, 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. 31. Consider the 2-form defined on by a. Show that is closed. This is a straightforward calculation using the definition and Theorem 4-10. For example, . The other two terms give similar results, and the sum is zero. b. Show that For let . Show that . Conclude that on restricted to the tangent space of is not exact. Nevertheless, we denote . is by times since, the volume element and that as we shall see, is the analogue of the 1-form As in the proof of Theorem 5-6 (or Problem 5-25), the value of using cofactors of the third row. can be evaluated by expanding The second assertion follows from taken to be 5-26. If , then Stokes' Theorem would imply that , we conclude that c. If is not exact. for some and the fact that the outward directed normal can be by Problem be an appropriate choice of orientation. One has . Since the value is is a tangent vector such that in . , show that for all . If a two-dimensional manifold the origin, show that is part of a generalized cone, that is, is the union of segments of rays through 5 of 9 12/6/2011 3:24 PM htm If . By part (c). This follows from the formula in part (b) since the third column of the determinant defining The curves of Figure 4-5(b) are given by easily convince yourself that calculating is zero. the line through . The union of those rays through 0 which intersect . Show that if and both lie in the -plane. Let . or equivalently as . . in the preamble to Problem 31. one has on the generalized cone. b. the integral over the part of the boundary making up part of a generalized cone is zero. we have of part (b). The following problem shows 6 of 9 12/6/2011 3:24 PM .G) is a homotopy of nonintersecting closed curves. then This follows immediately from Problem 4-34 (b) and Problem 5-31 (a).ms. Prove that the solid angle subtended by We take the orientations induced from the usual orientation of Following the hint. The orientation of the part of the boundary on the orientation of the same set as a part of solid angle subtended by 32. and . If show that . Section 4 http://www. (Actually. By Problem 5-9.edu/~ken/ma570/homework/hw23/html/ch5d. The solid is defined as the area of is . line) is in d. and by (cf. Define the linking number l(f. But then .) Note that this is essentially the same situation as in Problem 5-22. Applying Stokes' Theorem gives because is closed by part (a). for any point and the origin lies on the surface and so its tangent line (the same . see the remarks at the end of the next section. where This follows by direct substitution using the expression for c. is a solid cone times the area of . angle subtended by for . Problem 4-34 is opposite to that of and the last integral is the by the interpretation of be nonintersecting closed curves. Let . then using part (b). But then is identically 0 for all points be a compact two-dimensionaal manifold-with-boundary such that every ray through 0 intersects at most once (Figure 5-10). choose small enough so that there is a three dimensional manifold-with-boundary N (as in Figure 5-10) such that is the union of and .Exercises: Chapter 5.g) of } a. So. Show that if (F.uky. will be a manifold-with-corners. You may by the above integral is hopeless in this case. and a part of a generalized cone. Parameterize values where with intersects where in the outward normal direction is either in the same or is not in for all (where by . see [6]. we will need to assume that small enough so that 7 of 9 12/6/2011 3:24 PM . Removing a ball centered at from the interior of gives another manifoldwith-boundary with boundary . Subtracting gives and . The rest of the proof will be valid only in the case where there is a 3-dimensional compact oriented manifoldwith-boundary such that where is a two-dimensional manifold. Let we mean be the .edu/~ken/ma570/homework/hw23/html/ch5d. sufficiently close to Following the hint. as close Show that by choosing to as desired. and the number of other intersections. a.) Suppose that whenever intersects of is not in . Let is finite. page 138.htm how to find without explicit calculations. Suppose . Let be the number of intersections where points in the same . even if is knotted. One has . Choose . To complete the proof. So by Stokes' Theorem and Problem 5-31 (b). (Note the .uky. suppose that where is a compact manifold-with-boundary of dimension 3. orientation. show that . Then and we can take . discrepancy in sign between this and the hint. If define 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 we can make be a point on the opposite side. if . then by Stokes' Theorem. the tangent vector direction as the outward normal. by the last paragraph. for some compact oriented two-dimensional manifold-with-boundary . what one means is that the component of opposite direction as the outward normal. Section 4 http://www. (If does not always exists.Exercises: Chapter 5. and . where the orientation on is opposite to that of the induced .ms. Suppose intersect itself such an at sufficiently small.) On the other hand. If In the statement. 33. The first term can be made as small as we like by making b. So. the last paragraph has Note that this result differs from the problem statement by a sign. So. Prove that where The definition of should be . we will show the first result. Section 4 http://www. The proofs are analogous. Start with where we have used Problem 3-32 to interchange the order of the limit and the integral.edu/~ken/ma570/homework/hw23/html/ch5d.htm By part (a). if the tangent vector to at is .ms.Exercises: Chapter 5. then it is is positive.uky. c. has a component in the outward normal direction of . . if it is negative. On the other hand. we have by Stokes' Theorem that 8 of 9 12/6/2011 3:24 PM . then . and use this result to show that if and are the curves of Figure 4-6 (b).htm Comparing the two expressions. pp. we see that this is equal to the expression for By inspection in Figure 4-6. and in parts (b) and (c) of the figure. one has .ms.uky. After collecting terms. 41-43. . these are also the values of 9 of 9 12/6/2011 3:24 PM . while if and are the curves of Figure 4-6 (c). One then substitutes into: in Problem 5-32 (b). So. The proofs outlined here are from [4] pp. Show that the integer of (b) equals the integral of Problem 5-32(b). 409-411. see also [13]. . (These results were known to Gauss [7]. It remains to check that the remaining terms are equal.edu/~ken/ma570/homework/hw23/html/ch5d. Volume 2. we see that two of the terms in the first expression match up with corresponding terms in the second expression. Similar results hold for and . But a straightforward expansion gives: as desired. d.Exercises: Chapter 5.) By part (c). one has and so . So. Section 4 http://www. So.) in terms of the if and -dimensional volume of is even and . the unit be a compact -dimensional manifold-with-boundary and be a differentiable vector fieldd on .edu/~ken/ma570/homework/hw24/html/ch5e. we have . this says the 1 of 2 12/6/2011 3:25 PM . So and is odd. since the outward normal is in the radial . By for By Stokes' Theorem. In particular. it follows that 35. if . Section 5 http://www.ms.htm Exercises: Chapter 5. (This volume is if One has direction. on . Generalize the divergence theorem to the case of an The generalization: Let outward normal on . Section 5 34. Applying the generalized divergence theorem to the set find the volume of .Exercises: Chapter 5. Then Problem 5-25. Then As in the proof of the divergence theorem. Let -manifold with boundary in . let .uky. . we define the buoyaant force on the following theorem.uky. as . be a compact three-dimensional manifoldmay be thought of as the downward with-boundary with pressure of a fluid of density in . due to the fluid.) result if we define the buoyant force to be is equal to the weight of the fluid displaced by . The buoyant force on The definition of buoyant force is off by a sign. Prove directions. (This would make sense otherwise the 2 of 2 12/6/2011 3:25 PM .Exercises: Chapter 5. .htm surface area of 36.edu/~ken/ma570/homework/hw24/html/ch5e. Theorem (Archimedes). Now is the . The divergence theorem gives weight of the fluid displaced by buoyant force would be negative.ms. So one has the . The vector field . Section 5 http://www. So the right hand side should be the buoyant force. Define on by is times the volume of and let . Since a fluid exerts equal pressures in all .