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# Calculus on manifolds pdf

Calculus on Manifolds: A Modern Approach to Classical Theorems of Advanced Calculus . Print/export. Create a book · Download as PDF · Printable version. Calculus on Manifolds. A Solution Manual for Spivak (). Jianfei Shen. School of Economics, The University of New South Wales. Sydney, Australia. Page 1. Page 2. Page 3. Page 4. Page 5. Page 6. Page 7. Page 8. Page 9. Page Page Page Page Page Page Page Page Page Author: TAMARA BLACKINGTON Language: English, Spanish, German Country: Cuba Genre: Fiction & Literature Pages: 688 Published (Last): 17.12.2015 ISBN: 619-1-14979-310-3 ePub File Size: 20.39 MB PDF File Size: 18.27 MB Distribution: Free* [*Regsitration Required] Downloads: 27618 Uploaded by: SHERLYN

Calculus on Manifolds. A MODERN APPROACH TO CLASSICAL THEOREMS. OF ADVANCED CALCULUS. ADDISON-WESLEY PUBLISHING COMPANY. "advanced calculus" in which the subtlety of the concept.a and methods makes Thia study of manifolds, which could be justified solely on the caite.info of their. Calculus on Manifolds. A MODERN APPROACH TO CLASSICAL THEOREMS. OF ADVANCED CALCULUS. O K. ADDISON-WESLEY PUBLISHING COMPANY.

Now and it is straightforward to verify that for all sufficiently large positive integers. Apply Theorem to the map by. Now on the interval, and so the maximum must occur at the left hand end point. Suppose is compact. Now, when , we get the constant curve with value ; when , we get the curve ; and when or , we get the curve. On the other hand, if , then the line intersects the graph of at and and nowhere else. But then, so if integrable.

Chapter 2, Section 4 Find expressions for the partial derivatives of the following functions: Chapter 2, Section 4 Page 2 of 4. If is differentiable at , show that and therefore. The case when is trivially true. The last assertion follows from the additivity of the function. Let be defined as in Problem Show that exists for all , but if , then is not true for all and all.

With the notation of Problem , part a of that problem says that exists for all. Now suppose. Then , But. Show that exists for all , although is not even continuous at 0,0. Chapter 2, Section 4 Page 3 of 4.

Clearly, is differentiable at. At , one has since. For , one has. The first term has limit 0 as approaches 0. But the second term takes on all values between -1 and 1 in every open neighborhood of. So, does not even exist.

Show that is differentiable at 0,0 but that is not continuous at. The derivative at 0, 0 is the zero linear transformation because , just as in part a.

However, for where is as in part a. It follows from the differentiability of , that and are defined for. The argument given above also shows that they are defined and 0 at. Further the partials are equal to up to a sign, and so they cannot be continuous at 0. Show that the continuity of at may be eliminated from the hypothesis of Theorem Proceed as in the proof of Theorem for all.

In the case, it suffices to. Chapter 2, Section 4 Page 4 of 4. This is all that is needed in the rest of the proof.

A function is homogeneous of degree if for all and. If is also differentiable, show that. Applying Theorem to gives. On the other hand, and so. Substituting in these two formulas show the result. If is differentiable and , prove that there exist such that. Following the hint, let.

On the other hand, Theorem gives. So, we have the result with. Chapter 2, Section 5 Page 1 of 3. Chapter 2, Section 5 Let be an open set and a continusously differentiable function such that for all.

Show that is an open set and is differentiable. Show also that is open for any open set. For every , there is an with. By Theorem , there is an open set and an open subset such that and. Since clearly , this shows that is open. Furthermore is differentiable. It follows that is differentiable at. Since was arbitrary, it follows that is differentiable. By applying the previous results to the set in place of , we see that is open. Let be a continuously differentiable function.

Show that is not 1- 1. We will show the result is true even if is only defined in a non-empty open subset of. Following the hint, we know that is not constant in any open set. So, suppose we have the case where is analogous. Then there is an open neighborhood of with for all. The function defined by satisfies for all. Assuming that and hence are , we can apply Problem The inverse function is clearly of the form and so for all. Now is open but each horizontal line intersects at most once since is This is a contradiction since is non- empty and open.

Generalize this result to tthe case of a continuously differentiable function with. By replacing with a vector of variables, the proof of part a generalizes to the case where is a function defined on an open subset of where.

Chapter 2, Section 5 Page 2 of 3. For the general case of a map where is an open subset of with , if is constant in a non-empty open set , then we replace with and drop out reducing the value of by one. On the other hand, if for some , then consider the function defined by. Just as in part a , this will be invertible on an open subset of and its inverse will look like.

Replace with. Note that we have made. Again, by restricting to an appropriate rectangle, we can simply fix the value of and get a function defined on on a rectangle in one less dimension and mapping into a space of dimension one less.

By repeating this process, one eventually gets to the case where is equal to 1, which we have already taken care of. If satisfies for all , show that is on all of. Suppose one has for some. By the mean value theorem, there is a between and such that. Since both factors on the right are non-zero, this is impossible. Show that for all but is not Clearly, for all. The function is not since for all. Chapter 2, Section 5 Page 3 of 3. So satisfies the conditions of Theorem at except that it is not continuously differentiable at 0 since for.

Now and it is straightforward to verify that for all sufficiently large positive integers. By the intermediate value theorem, there is a between and where. By taking n larger and larger, we see that is not on any neighborhood of 0. Chapter 2, Section 6 Page 1 of 3.

Chapter 2, Section 6 Use the implicit function theorem to re-do Problem c. Define by for. One has for all. The determinant condition guarantees that for each , there is exactly one solution of , call that solution. Now, for each , the Implicit function theorem says that there is a function defined in an open neighborhood of and such that and is differentiable.

By the uniqueness of the solutions in the last paragraph, it must be that for all in the domain of. In particular, the various functions all glue together into a single function defined on all of and differentiable everywhere. By differentiating the relation , one gets for. Note that this is of the same form as the set of equations for except that the right hand side functions have changed. An explicit formula can be obtained by using Cramer's rule. Let be differentiable. For each defined by.

Suppose that for each there is a unique with ; let be this. In this problem, it is assumed that was meant to be continuously differentiable. Just as in the last problem, the uniqueness condition guarantees that is the same as the function provided by the implicit function theorem applied to. In particular, is differentiable and differentiating this last relation gives. Solving for gives the result.

Chapter 2, Section 6 Page 2 of 3. Show that if , then for some we have and. Note that is defined only when and are positive. So we need to go back and show that the earlier parts of the problem generalize to this case; there are no difficulties in doing this. One has precisely when and so the hypothesis of part a is true and. Also, since both and are positive , and so for fixed , the minimum of occurs at by the second derivative test.

Now actually, we are not looking for the minimum over all , but just for those in the interval. The derivative for. Further precisely when and there is a unique where.

For fixed , is achieved at if , at if , and at if. We will find where the maximum of the minimum's are located in each of the three cases.

Then we need to maximize. The derivative of this function is negative throughout the interval; so the maximum occurs at. The maximum value is. The derivative of this function is. 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.

Now on the interval, and so the maximum must occur at the left hand end point. In view of the last paragraph, that means that the maximum over the entire interval. Chapter 2, Section 6 Page 3 of 3. This is a decreasing function and so the maximum occurs at the left hand endpoint. By the result of the previous paragraph the maximum over the entire interval must therefore occur at , , and the value of the maximum is.

Chapter 3, Section 1 Page 1 of 6. Chapter 3, Section 1 1. Apply Theorem to the partition where. For this partition,. Let be integrable and let except at finitely many points. Show that is integrable and. For any , there is a partition of in which every subrectangle has volume less than. In fact, if you partition by dividing each side into equal sized subintervals and , then the volume of each subrectangle is precisely which is less than as soon as. Furthermore, if is any partition, then any common refinement of this partition and has the same property. If and is a partition of , then any point is an element of at most of the subrectangles of. Let and be a partition as in Theorem applied to and. Let be a refinement of such that every subrectangle of has volume less than where , is the number of points where and have values which differ, and resp.

Then the hypotheses of Theorem are satisfied by and , and so is integrable. In fact, and where is any upper bound for the volume of the subrectangles of , because the terms of the sum can differ only on those subrectangles which contain at least one of the points where and differ.

Taking differences gives. Chapter 3, Section 1 Page 2 of 6. Let be integrable. For any partition of and any subrectangle of , show that and and therefore and. For each , one has and since greatest lower bounds are lower bounds.

Adding these inequalities shows that is a lower bound for , and so it is at most equal to the greatest lower bound of these values. A similar argument shows the result for.

Since , , and are just positively weighted sums of the , , and the result for can be obtained by summing with weights the inequalities for the. Let be a common refinement of and. Then by part a and Lemma ,. Chapter 3, Section 1 Page 3 of 6. We will show the result in the case where ; the other case being proved in a similar manner. Let be a partition as in Theorem applied to and. Since and for each subrectangle of , we have. By Theorem , applied to and , the function is integrable; by the squeeze principle, its integral is.

## 34158129 Calculus on Manifolds Spivak M PDF

Let and be a partition of. Show that is integrable if and only if for. Chapter 3, Section 1 Page 4 of 6. Suppose that is integrable and. Let be a partition of as in Theorem applied to and.

Then there is a partition of whose subrectangles are precisely the subrectangles of which are contained in. By Theorem , it follows that is integrable.

Suppose that all the are integrable where is any subrectangle of. Let be a partition as in Theorem applied to and where is the number of rectangles in. Let be the partition of A obtained by taking the union of all the subsequences defining the partitions of the for each dimension. Then there are refinements of the whose rectangles are the set of all subrectangles of which are contained in. By Theorem , the function is integrable, and, by the squeeze principle, it has the desired value.

Chapter 3, Section 1 Page 5 of 6. By Problem , the function is integrable and.

## Spivak Calculus of Manifolds Solutions

Using the trivial partition in which is the only rectangle, we have since. This proves the result. Consider the function. For any rectangle contained in , we have and. On the other hand, if , then. Then this implies that.

Similarly, one can show that is integrable. But then by Problem , it follows that is integrable. But then, so if integrable. Further, since , Problem implies that. Since by Problem c , it follows that. Chapter 3, Section 1 Page 6 of 6. Choose a positive integer so that. Let be any partition of such that every point with lies in a rectangle of of height in the direction at most. Since there are at most such pairs , such a exists and the total volume of all the rectangles containing points of this type is at most.

Since , the contribution to from these rectangles is also at most. For the remaining rectangles , the value of and their total volume is, of course, no larger than 1; so their contribution to is at most. It follows that. By Theorem , is integrable and the squeeze principle implies that its integral is 0. Chapter 3, Section 2 Page 1 of 3. Chapter 3, Section 2 8. Prove that is not of content 0 if for.

Suppose for are closed rectangles which form a cover for. By replacing the with , one can assume that for all. 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.

Since the intersection of any two rectangles of a partition is contained in their boundaries, if contains an interior point not in for some , then contains only boundary points of.

So, if has non-empty interior, then is a subset of for some since the union of the is. The sum of the volumes of the rectangles of is the volume of , which is at most equal to the sum of the volumes of the.

So is not of content 0 as it cannot be covered with rectangles of total area less than the volume of. Suppose where are rectangles, say. Let where and. Then contains all the and hence also. But then is bounded, contrary to hypothesis.

The set of natural numbers is unbounded, and hence not of content 0 by part a. On the other hand, it is of measure zero. Indeed, if , then the union of the open intervals for.

If C is a set of content 0, show that the boundary of also has content 0. Suppose a finite set of open rectangles ,.

Let where. Then the union of the cover the boundary of and have total volume less than. So the boundary. Chapter 3, Section 2 Page 2 of 3. Give an example of a bounded set of measure 0 such that the boundry of does not have measure 0. The set of rational numbers in the interval is of measure 0 cf Proof of Problem b , but its boundary is not of measure 0 by Theorem and Problem Let be the set of Problem If , show that the boundary of does not have measure 0.

The set closed and bounded, and hence compact. If it were also of measure 0, then it would be of content 0 by Theorem But then there is a finite collection of open intervals which cover the set and have total volume less than.

Since the set these open intervals together with the set of form an open cover of [0, 1], there is a finite subcover of. But then the sum of the lengths of the intervals in this finite subcover would be less than 1, contrary to Theorem Show that is a set of measure 0. Using the hint, we know by Problem that the set of where if finite for every. Hence the set of discontinuities of is a countable union of finite sets, and hence has measure 0 by Theorem Show that the set of all rectangles where each and each are rational can be arranged into a sequence i.

Since the set of rational numbers is countable, and cartesian products of countable sets are countable, so is the set of all -tuples of rational numbers.

Since the set of these intervals is just a subset of this set, it must be countable too. If is any set and is an open cover of , show that there is a sequence of members of which also cover.

Following the hint, for each in , there is a rectangle B of the type in part a such that has non-zero volume, contains and is contained in some in. In fact, we can even assume that is in the interior of the rectangle. Chapter 3, Section 2 Page 3 of 3. By part a , the set of these are countable, and hence so are the set of corresponding 's; this set of corresponding 's cover.

Chapter 3, Section 3 Page 1 of 3. Chapter 3, Section 3 Show that if are integrable, so is. The set of where is not continuous is contained in the union of the sets where and are not continuous. These last two sets are of measure 0 by Theorem ; so theee first set is also of measure 0. But then is integrable by Theorem 3- 8. Show that if has content 0, then for some closed rectangle and is Jordan-measurable and. If has content 0, then it is bounded by Problem a ; so it is a subset of an closed rectangle.

Since has content 0, one has for some open rectangles the sum of whose volumes can be made as small as desired. But then the boundary of is contained in the closure of , which is contained in the union of the closures of the since this union is closed. But then the boundary of must be of content 0, and so is Jordan measurable by Theorem Further, by Problem , one has which can be made as small as desired; so. Let be the set of rational numbers in. Then the boundary of is , which is not of measure 0.

So does not exist by Theorem If is a bounded set of measure 0 and exists, show that. Using the hint, let be a partition of where is a closed rectangle containing.

Then let be a rectangle of of positive volume. Then is not of measure 0 by Problem , and so. But then there is a point of outside of ; so. Since this is true of all , one has. Since this holds for all partitions of , it follows that if the integral exists. If is non-negative and , show that has measure 0. Following the hint, let be a positive integer and.

Let be a partition of such that. Then if is a. Chapter 3, Section 3 Page 2 of 3. By replacing the closed rectangles with slightly larger open rectangles, one gets an open rectangular cover of with sets, the sum of whose volumes is at most. So has content 0. Now apply Theorem to conclude that has measure 0.

Let be the open set of Problem Show that if except on a set of measure 0, then f is not integrable on. The set of where is not continuous is which is not of measure 0.

If the set where is not continuous is not of measure 0, then is not integrable by Theorem On the other hand, 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. So this set is also of measure 0, which is a contradiction. If is a closed rectangle, show that is Jordan measurable if and only if for every there is a partition of such that , where consists of all subrectangles intersecting and consists of allsubrectangles contained in.

Suppose is Jordan measurable. Then its boundary is of content 0 by Theorem Let and choose a finite set for of open rectangles the sum of whose volumes is less than and such that the form a cover of the boundary of.

Let be a partition of such that every subrectangle of is either contained within each or does not intersect it. This satisfies the condition in the statement of the problem. Suppose for every , there is a partition as in the statement. Then by replacing the rectangles with slightly larger ones, one can obtain the same result except now one will have in place of and the will be open rectangles.

This shows that the boundary of is of content 0; hence is Jordan measurable by Theorem If is a Jordan measurable set and , show that there is a compact Jordan measurable set such that. Chapter 3, Section 3 Page 3 of 3. Let be the partition as in Problem Then and clearly is Jordan measurable by Theorem Chapter 3, Section 4 Page 1 of 7. Chapter 3, Section 4 Let be a set of content 0. Let be the set of all such that is not of content 0. Following the hint, is integrable with by Problem and Fubini's Theorem.

Now is equivalent to the condition that either or. Both of these having integral 0 implies by Problem that the sets where their integrand is non-zero are of measure 0, and so is also of measure 0. Let be the union of all where is a rational number in written in lowest terms.

The set is the set of rational numbers in which is of measure 0, but not of content 0, because the integral of its characteristic function does not exist. To see that the set has content 0, let. Let be such that. Then the set can be covered by the rectangles and for each in lowest terms with , the rectangle where. The sum of the areas of these rectangles is less than. Show by induction on that is not a set of measure 0 or content 0 if for each.

Fubini's Theorem and induction on show that and so does not have content 0, and hence is not of measure 0. Let be integrable and non-negative, and let. Show that is Jordan measurable and has area. Chapter 3, Section 4 Page 2 of 7.

Applying Fubini again, shows that this integral is equal to. 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. Chapter 3, Section 4 Page 3 of 7. Use Fubini's Theorem to derive an expression for the volume of a set in obtained by revolving a 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 Jordan measurable set in the -plane, and replace it with. Theorem can be used to show that is Jordan measurable if is. 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. The problem has a typo in it; the author should not have switched the order of the arguments of as that trivializes the assertion.

The iterated integrals are zero because the inside integral is the zero function. The last integral cannot exist by Theorem and Problem If and is continuous, define by. Chapter 3, Section 4 Page 4 of 7. Let be continuous and suppose is continuous.

Prove Leibnitz' Rule: Using the hint, we have. Chapter 3, Section 4 Page 5 of 7. If is continuous and is continuous, define. One has and where the second assertion used Problem Let be continuously differentiable and suppose.

As in Problem , let. Chapter 3, Section 4 Page 6 of 7. Let be a linear transformation of one of the following types:. In the three cases, is , 1, and 1 respectively. If the original rectangle , then is.

## Calculus on Manifolds (book) - Wikipedia

Chapter 3, Section 4 Page 7 of 7. In the second case, the parallelogram base is in the and directions and has corners. So the volumes do not change in the second and third case and get multiplied by in the first case.

Prove that is the volume of for any linear transformation. If is non-singular, then it is a composition of linear transformations of the types in part a of the problem. Since is multiplicative, the result follows in this case. If is singular, then is a proper subspace of and is a compact set in this proper subspace. In particular, is contained in a hyperplane.

By choosing the coordinate properly, the hyperplane is the image of a linear transformation from into made up of a composition of maps of the first two types. This shows that the compact portion of the hyperplane is of volume 0.

Since the determinant is also 0, this shows the result in this case too. Cavalieri's principle. Let and be Jordan measurable subsets of. Let and define similarly. Suppose that each and are Jordan measurable and have the same area. Show that and have the same volume. This is an immediate consequence of Fubini's Theorem since the inside integrals are equal. Chapter 3, Section 5 Page 1 of 2. Chapter 3, Section 5 Suppose that is a non-negative continuous function.

Show that exists if and only if exists. For a natural number, define and. Consider a partition of unity subordinate to the cover. By summing the with the same in condition 4 of Theorem , one can assume that there is only one function for each , let it be. Now exists if and only converges. So the sum converges if and. Let Suppose that satisfies and for all. Show that does not exist, but.

Take a partition of unity subordinate to the cover where for. As in part a , we can assume there is only one as in condition 4 of Theorem Consider the convergence of. One has where. It follows that the sum in the middle does not converge as and so does not exist. The assertion that. If not necessrily true. From the hypothesis, we only know the values of the integral of on the sets , but don't know how behaves on other intervals -- so it could be that may not even exist for all To correct the situation, let us assume that is of constant sign and bounded on each set.

Then is bounded on each interval and so by Theorem , the integral in the extended sense is same as the that in the old sense. Clearly, the integral in the old sense is. Chapter 3, Section 5 Page 2 of 2. Let be a closed set contained in. Suppose that satisfies and outside. Find two partitions of unity and such that and converge absolutely to different values. The sums and have terms of the same sign and are each divergent. So, by re-ordering the terms of , one can make the sum approach any number we want; further this can be done so that there are sequences of partial sums which converge monotonically to the limit value.

By forming open covers each set of which consists of intervals for the sum of terms added to each of these partial sums, one gets covers of.

Finally, one can take a partition of unity subordinate to this cover. By using arrangements with different limiting values, one gets the result. Chapter 3, Section 6 Page 1 of 5. Chapter 3, Section 6 Use Theorem to prove Theorem without the assumption that. Let Then is open and Theorem applies with in place of the in its statement. Let be a partition of unity subordinate to an admissible cover of.

Then is a partion of unity subordinate to the cover. Now is absolutely convergent, and so also converges since the terms are identical. By Theorem , we know that. Combining results, we get Theorem If and , prove that in some open set containing we can write , where is of the form , and is a linear transformation.

Show that we can write if and only if is a diagonal matrix. We use the same idea as in the proof of Theorem Let be a point where. Let , and. Define for ,. So we can define on successively smaller open neighborhoods of , inverses of and.

One then can verify that. Combining results gives. Chapter 3, Section 6 Page 2 of 5. Now, if is a diagonal matrix, then replace with. Then the have the same form as the and. On the other hand, the converse is false. For example, consider the function. Since is linear, ; so is not a diagonal matrix. Show that is , compute , and show that for all. Show that is the set of Problem Since , to show that the function is , it suffices to show that and imply.

Then implies that or. If , it follows that. But then and has the same value, contrary to hypothesis. So, is 1- 1. Suppose , i. If , then implies and so. But then contrary to hypothesis. On the other hand, if , then let and let be the angle between the positive -axis and the ray from 0,0 through. Chapter 3, Section 6 Page 3 of 5. Here denotes the inverse of the function. Find P' x,y. The function is called the polar coordinate system on. Further, L'H ocirc;pital's Rule allows one to calculate when by checking separately for the limit from the left and the limit from the right.

Let be the region between the circles of radii and and the half- lines through 0 which make angles of and with the -axis. If is integrable and , show that. Assume that and. Apply Theorem to the map by. One has and. So the first identity holds. The second identity is a special case of the first. Chapter 3, Section 6 Page 4 of 5. Since part d implies that , the squeeze. Chapter 3, Section 6 Page 5 of 5. But using part d again, we get also exists and is since the square root function is continuous.

Chapter 4, Section 1 Page 1 of 8. Chapter 4, Section 1 1. Let be the usual basis of and let be the dual basis. 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; in that case, the value is zero.

Assume therefore that the are distinct. One has using Theorem If the factor were not in the definition of , then the right hand side would have been.

Show that is the determinant of thee minor of. A computation similar to that of part a shows that if some for all. By multilinearity, it follows that we need only verify the result when the are in the subspace generated by the for.

Consider the linear map defined by. One has for all:. Chapter 4, Section 1 Page 2 of 8. If is a linear transformation and , then must be multiplication by some constant. Then by Theorem , one has for ,. Chapter 4, Section 1 Page 3 of 8. If is the volume element determined by and , and , show that. Let be an orthonormal basis for V with respect to , and let where.

Then we have by blinearity: By Theorem ,. Taking absolute values and substituting gives the result. If is the volume element of determined by and , and is an isomorphism such that and such that , show that.

One has by the definition of and the fact that is the volume element with respect to and. Further, for some because is of dimension 1. Combining, we have , and so as desired. If is continuous and each is a basis for , show that.

The function is a continuous function, whose image does not contain 0 since is a basis for every t. By the intermediate value theorem, it follows that the image of consists of numbers all of the same sign. So all the have the same orientation. If , what is? Substitution shows that works. If are linearly independent, show that.

Chapter 4, Section 1 Page 4 of 8. Since the are linearly independent, the definition of cross product with completing the basis shows that the cross product is not zero. So in fact, the determinant is positive, which shows the result. Show that every non-zero is the volume element determined by some inner product and orientation for.

Let be the volume element determined by some inner product and orientation , and let be an orthornormal basis with respect to such that. There is a scalar such that. Let , , , and for. Then are an orthonormal basis of with respect to , and.

This shows that is the volume element of determined by and. The cross product is the such that for all. Chapter 4, Section 1 Page 5 of 8. The result is true if either or is zero. Suppose that and are both non- zero. By Problem , and since , the first identity is just. This is easily verified by substitution using part b. The second assertion follows from the definition since the determinant of a square matrix with two identical rows is zero. Chapter 4, Section 1 Page 6 of 8. So, one needs to show that for all.

But this can be easily verified by expanding everything out using the formula in part b. Chapter 4, Section 1 Page 7 of 8. This proves the result in the case where is not zero. When it is zero, the are linearly dependent, and the bilinearity of inner product imply that too. If is an inner product on , a linear transformation is called self- adjoint with respect to if for all. If is an orthogonal basis and is the matrix of with respect to this basis, show that.

One has for each. Using the orthonormality of the basis, one has: But , which shows the result. Chapter 4, Section 1 Page 8 of 8. If , define by. Use Problem to derive a formula for when are differentiable. Since the cross product is multilinear, one can apply Theorem b and the chain rule to get:. Chapter 4, Section 2 Page 1 of 5. Chapter 4, Section 2 If and , show that and.

The notation does not fully elucidate the meaning of the assertion. Here is the interpretation:. Let be a differentiable curve in , that is, a differentiable function.

Define the tangent vector of at as. If , show that the tangent vector to at is. Chapter 4, Section 2 Page 2 of 5. Let and define by. Show that the end point of the tangent vector of at lies on the tangent line to the graph of at. The tangent vector of at is. In fact, several of the concepts introduced in Calculus on Manifolds reappear in the first volume of this classic work in more sophisticated settings. The cover of Calculus on Manifolds features snippets of a July 2, letter from Lord Kelvin to Sir George Stokes containing the first disclosure of the classical Stokes' theorem i.

From Wikipedia, the free encyclopedia. Using this language, Cartan stated the generalized Stokes' theorem in its modern form, publishing the simple, elegant formula shown here in Katz Mathematics Magazine 52 3: Retrieved Archived from the original on Retrieved from " https: Hidden categories: Articles with short description Pages to import images to Wikidata. Namespaces Article Talk. Views Read Edit View history.