# Calculus on manifolds solutions

My options to some of Spivak"s exercises (I skip the few I uncovered non-interesting). I guarantee no correctness. Hopefully any type of self-learners out there deserve to advantage from this. Please execute send comments, corrections and also conversation to me by e-mail.

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Spivak"s Notation Spivak denotes points in a Euclidean room as row vectors in message, yet provided in matrix equations -- for instance, peak of Page 4 -- they have the right to magically turn into column vectors. This deserve to be confutilizing and while analysis the book one need to always remember to think of vectors as points in an abstract room, not as of "rows" or "columns" of some kind. In addition, direct transformation matrices have to constantly be considered extitmatrices, even if they are (n imes 1) or (1 imes n) matrices. To make this clear, also though Spivak will signify a suggest in (mathbfR^3) as ((a,b,c)), a straight transormation (fcolonmathbfR^3 ightarrow mathbfR) is still a (1 imes 3) matrix. This is particulary confutilizing in cases like Problem 2-13, wright here even the author feels compelled to provide an explanatory note.

My Notation I will certainly signify matrices and direct transformations via resources letters; the i-th row of matrix (A) by (A_i,:), and also its j-th column by (A_:,j).

## 1- Functions on Euclidean Void

1-1 ((sum_i |x^i|)^2 = sum_i |x^i|^2 + r), wbelow (r) is a non-negative number. This is trivially greater or equal to (|x|^2 = sum_i |x^i|^2).

1-10 This outcome comes up many times in the book. Let (T) signify the (n imes m) matrix of the direct transdevelopment, and set (k=Th), an n-dimensional vector with (T_i,: cdot h) as its i-th entry. By Thm 1-1(4), (|k_i|leq |T_i,:|cdot |h|) which is smaller sized or equal to (leq m|max(T_i,j)|cdot |h|). Since this is independent of i, we have the right to collection (M = nm|max(T_i,j)|) by Problem 1-1.

1-12 Let (u,vin mathbfR^n) be 2 different vectors, for example (u^i eq v^i). Then (phi_u(e_i) = u^i) and also (phi_v(e_i)=v^i) so (T) is 1-1. It is linear by bilinearity of the dot product. The last part follows from the reality that eincredibly injective direct transofrmation is also surjective.

1-13 By Thm 1-2(5), (|x+y|^2 = sum_i (x^i+y^i)^2 = sum_i (x^i)^2+(y^i)^2+x^iy^i = |x|^2+|y|^2+langle x,y angle = |x|^2+|y|^2).

1-14 By Spivak"s meaning, a collection is open up iff it is any amount of open rectangles. Therefore, any kind of amount of open sets is trivially also open up. Because taking the interarea of two open up rectangles likewise offers an open up rectangle, finite intersections are open. For the limitless counter-instance, take (igcup_i=1^infty (-1/i,1/i) = \), a singleton closed collection.

1-19 Let (B = Acap<0,1>). Then (B) is an interarea of 2 closed sets and so is itself closed, and is in enhancement bounded. By Corollary 1-7, (B) is compact. Now assume tbelow is some (x in <0,1>, x otin B); then (igcup_i=1^infty (-1,x-1/i) cup igcup_j=1^infty (x+1/j,2)) is an open cover of (B). However, it has actually no finite subcover, which contradicts the compactness of (B).

1-20 Assume (Ain mathbfR^n) is not bounded. Then (igcup_k=1^infty (-k,k)^n=(-k,k) imes(-k,k) imescdots(-k,k)) is an open cover of (A) that has actually no finite subcover, a contradiction. Now assume (A) is not closed -- that is, tright here is a suggest (x otin A) on (A)s boundary. Let (U_k(x)) be an open square of side length (k) focused on (x). Then (igcup_kinmathbfR), that is the sum of complements of all such open squares, is an open cover of (A) (the just point it does not contain is (x)). However, for any finite subcover tbelow is a area of (x) the subcover doesn"t contain, and this community has actually a non-empty interarea via (A) considering that (x) is on the boundary.

1-21 (a) Assume the premise is not true, and also for any type of (d) take an open square focused on (x), via side of size (2d). Because each such square has some suggest (yin A),(x) is on the boundary of (A). However, (A) is closed and also for this reason includes its own boundary, which contradicts the assumption that (x otin A).(b) Note that by (a), tright here is already for each (yin B) a number (d_y) such that (|y-x| geq d ) for all (xin A); we just need to prove that tbelow is a smallest such (d_y>0). Now, considering that (A) is closed, its compliment (A") is open up. Because (B subcollection A"), for any type of (yin B) we have the right to discover an open up rectangle (U_y) disjoint through (A). Due to the fact that all such open up rectangles form a cover of the compact collection (B), tbelow is their finite subset (U_y_1, U_y_2,...,U_y_N\) likewise covers. Now choose (d) to be the fifty percent of the largest side size of these rectangles. (c) Let (A) be the collection ( y leq log(x)\) and also let (B) be its reflection along the y-axis. The 2 sets are closed but not bounded and also the distance in between them takes on arbitrarily low worths.

1-22 Complement of (U) is closed and disjoint from (C), so we can find a nonzero distance (d) in between the sets by Problem 1-21. Now cover each suggest in (C) with an open bevery one of radius (d/2) and also take a finite subcover (hatD). By construction, eincredibly allude in this subcover is at least (d/2) distance away from the enhance of (U). Its clocertain (D) is hence likewise consisted of in (U), and is a finite union of closed sets. It is also bounded and also contains (C) by building and construction.

1-26 (a) Take any type of line (ax-by=0, bin,1\). If (a leq 0) or (b=0) then the totality line is in (mathbfR^2-A). Now think about the situation where (a>0,b=1). Then the line intersects the parable (y=x^2) at (x=0 ext and x=a). Therefore tright here is an interval on the line, equivalent to (xin (-1,a)), exterior of the collection (A). (b) (f(0,0)=0), but the sequence ((f(x_i)), x_i = (1/i, frac(1/i)^22)) is of consistent value 1 while (x_i ightarrow (0,0)) -- for this reason (f) is not constant at (0) (a less complicated means ). For any type of sequence ((t_i) ightarrow 0), ((g_h(t_i))) consists of the worths of (f) alengthy a particular line. However before, we"ve shown in (a) over that eincredibly such sequence converges to zero, for this reason (g_h) is constant at zero.

1-29 This is a critical truth. Assume (f) has actually no minimum or maximum. Then (f(A)) is boundless. However before, (f) is consistent so (f(A)) have to be compact and also thus closed and bounded, a contradiction.

## 2 - Differentiation

2-1 ( 0 = lim_h ightarrowhead 0 frach geq lim_h ightarrowhead 0frach-lim_h ightarrowhead 0frach geq lim_h ightarrow 0 frach - M_a), wbelow (M_a) is some nonnegative scalar by Problem 1-10. From this it adheres to that (0=lim_h ightarrowhead 0|h|M_a geq lim_h ightarrow 0|f(a+h)-f(a)|) and by the "squeeze theorem", (f(a+h)) converges to (f(a)).

2-4 (a) Note that (h(t) = |tx|g(x/|x|)=t|x|g(x/|x|)) if (t>0), (h(0)=0), (h(t) = -|tx|g(x/|x|)=t|x|g(x/|x|)) if (t2-5 Let (ain mathbfR^2) and also define (g(a/|a|) = fraca^1fraca^2). This fulfills all the needs from Problem 2-4, and (f(x,y) = |(x,y)|gleft(frac(x,y) ight)) as desired.

2-6 We can present that if the derivative exists it have to be (Df(0,0)=0) simply like in Problem 2-4. However, take for example (lim_t ightarrowhead 0fracsqrt=fracsqrt22), a contradiction.

2-7 We"ll show that the derivative at (0) is zero, that is (lim_h ightarrowhead 0 frac = 0). This adheres to as (f(0) = 0) and then (lim_h ightarrow 0frac leq lim_h ightarrowhead 0frac^2h = 0), so by the "squeeze theorem" the limit equates to zero.

2-8 The proof is not very illuminating and also adheres to from later sections anyway.2-9 (a) Let (f) be differentiable at (a), that is for some (f"(a)). We should define (g) such that (g(a+h) = f(a)+f"(a)h). Usingfunctions well sufficient. For the converse, assume for some (a_0,a_1). Then (lim_h ightarrow 0fracf(a+h)-a_0-a_1 hh = 0,)which implies that Now if (a_0 eq f(a)), the LHS above goes to infinity, so the over equation is satisfied if and only if (a_0 = f(a), f"(a) = a_1.)

2-10 In each case, create the feature carefully in term of projections and attribute compositions, for example:(a) (f = exp circ (ln circ pi^1 cdot pi^2)) so by theorems in this area, (f"(a,b,c) = exp"(lnacdot b) = a^b(fracba, lna, 0)=(ba^b-1, lnacdot a^b,0)).(b) By Thm 2-3(3) the first row of (f") is the same as in (a), so (c) (f = sin circ (pi^1 cdot sin circ pi^2)), henceeginalign*f"(a,b) &= sin"(a sinb)\ &= sin"(asinb)\ &= (sinbcos(asinb), acosbcos(asinb)).endalign*(d) Let (g(y,z) = sin(y sinz)). Then (f = sin circ ), and also we already recognize (g") from trouble 2-10(c) over. Thuseginalign*f"(a,b,c) &= cos(asin(bsinc))<(1, 0, 0)g(b,c)+ acdot g"(b,c)cdot eginpmatrix 0 & 1 & 0\ 0 & 0 & 1endpmatrix>\ &=cos(asin(bsinc))(g(b,c), g"(b,c)^1, g"(b,c)^2).endalign*(e) Similarily to the above, let (g(y,z) = y^z) as in Problem 2-10(a), and (f = exp circ ln circ pi^1 cdot g cdot (pi^2, pi^3)). The remainder of the calculations is automatic.(f) Use (g) as over and note that (f = g circ (pi^1, pi^2+pi^3)).(g) Use (g) as above and also (f = g circ (pi^1+pi^2, pi^3)).(h) (f = sin circ (pi^1 cdot pi^2)), and also (f"(a,b) = cos(ab)<(0,a)+(b,0)> = (bcos(ab),acos(ab))).(i) My variation of the book is quite unreadable here, yet looks favor (f = g circ (sin circ pi^1 cdot pi^2, cos circ )), wright here (g) is identified prefer in Problem 2-10(e).(j) This adheres to directly from Thm 2-3(3) and also the troubles above. We get

2-11 Define (Int(x) = int_a^x g), and also from the Fundamental Theorem of Calculus (e.g. Theorem 5.1 in Apostol), (DInt(x) = g(x)). Then we have(a) (f = Int circ (pi^1+pi^2)), so (f"(x,y) = Int"(x,y)cdot (1,1) = (g(x+y), g(x+y))).(b) (f = Int circ (pi^1cdot pi^2)), so (f"(x,y) = g(xy)(y,x)=(yg(xy),xg(xy))).(c) Let (h(x,y,z) = sin(xsin(ysin(z)))). Note that we currently recognize (h") from Problem 2-10(d). Then (f = Int circ h - Int circ (pi^2cdot pi^1)), by fundamental properties of the integral (view e.g. various theorems 1.* in Apostol). Hence (f"(x,y,z) = g(h(x,y,z))h"(x,y,z)-(yg(xy), xg(xy), 0)).

2-12 (a) Keep in mind that for any type of vector (a=(a^1,...,a^N)) we have (|a|leq|(a^i_1, a^i_2,...,a^i_k)|) for any kind of choice of (i). We have the right to therefore bound the sequence in question by a sequence via limit zero as follows:eginalign*frac &leq frac1|f(h,k)| = frac1kigl|sum_j=1^m k^j f(h,e_j)igr|\ &leq frac1igl|sum_j=1^m |k| f(h,e_j)igr|\ &leq fracsum_j=1^m |f(h,e_j)|,endalign*So (lim_(h,k) ightarrowhead 0frac leq lim_(h,k) ightarrow 0sum_j=1^m|f(h,e_j)|=sum_j=1^m|f(0,e_j)| = 0).(b) We have the right to ssuggest inspect that the equation defining the derivative holds:eginalign*lim_(h,k) ightarrow 0frac &= lim_(h,k) ightarrow 0frac \&= lim_(h,k) ightarrowhead 0frac\&= 0.endalign*(c) It"s simple to inspect that (p) is bistraight. Then, from (b) over used to (f=p) we have actually (Dp(a,b)(x,y) = p(a,y)+p(x,b) = ay+bx), as in Thm 2-3.

2-13 It"s easy to verify that the inner product IP is bidirect, so we deserve to use Problem 2-12.(a)(D(IP)(a,b)(x,y) = IP(a,y)+IP(x,b) = langle a,y angle + langle b, x angle) (as the inner product is symmetric). From this and the interpretation of matrix-vector multiplication we get ((IP)"(a,b) = (b, a)).(b) (h = IP circ (f, g)), and we deserve to use the Chain Rule to acquire eginalign*h"(a) &= IP"(f(a),g(a))cdot eginpmatrixf"(a)\g"(a)endpmatrix\ &= (g(a), f(a))cdot eginpmatrixf"(a)\g"(a)endpmatrix\ &= langle g(a), f"(a)^T angle + langle f(a), g"(a)^T angle,endalign*wbelow the last line follows from the meaning of the inner product.(c) Since (|x|^2 = langle x,x angle) for any type of (x), the constant-norm assumption gives us which offers by (b) aboveeginalign*0 &= langle f"(t)^T, f(t) angle + langle f(t), f"(t)^T angle\ &= 2langle f"(t)^T, f(t) angle.endalign*(d) We deserve to simply use (f(t)=t) (as (|t|) is not differentiable at zero). This doesn"t really fit via the remainder of the problem though...

2-16 We have actually (Dfcirc f^-1 = D(id)), and also from the Chain Rule (f"(f^-1(a))(f^-1)"(a)=mathbf1), wright here (mathbf1) is the identification matrix. This straight means the ehigh quality in question.

2-17,2-18 These are simply a series of one-dimensional differentiations using the definition of a partial derivative. The most crucial understanding is exactly how much less complicated it is to percreate these calculations than to straight use the chain preeminence prefer in Problem 2-10.

2-19 (f(1,y) = 1) which is consistent in (y), so (D_2f(1,y)=0).2-20 (a) (D_1f(x,y) = h(y)Dg(x)), (D_2f(x,y) = g(x)Dh(y)). (b) Using the Chain Rule, (D_1f(x,y) = h(y)g(x)^h(y)-1Dg(x)), (D_2f(x,y) = g(x)^h(y)ln(g(x))Dh(y)). (e) Aget making use of the Chain Rule, (D_1f(x,y) = D_2f(x,y) = Dg(x+y)).

2-21 (a) The initially term of the amount is independent of (y), and the derivative (with respect to y) of the second term amounts to (g_2(x,y)) by the Fundapsychological Theorem of Calculus. (b) This question is not rather clear. The function < f(x,y) = int_0^x g_1(t,y), dt + int_0^y g_2(0,t)!dt> is of comparable for and also provides the desired result, but so does for example for any kind of function (h). (c) Let for the initially problem, and also let for the second.

See more: Calculus Concepts And Contexts 4Th Edition, Calculus: Concepts And Contexts, 4Th Edition

2-22 Let (D_2f = 0). Then, by one-variable calculus (in certain the Mean Value Theorem, watch e.g. Apostol) (f(x,y_1) = f(x,y_2)) for all (y_1,y_2). That is, (f) is independent of the second variable. If in addition (D_1f=0), then (f) is constant in both variables by simmilar thinking.

2-23 (a) Let ((x_1,y_1)) and ((x_2,y_2)) be two points in (A). Because (D_1f=0), the worth of (f) is consistent along the lines (l_1 = xin mathbfR\) and (l_2 = xin mathbfR\). In particular, we have (f(-1,y_1) = f(x_1,y_1), f(-1,y_2) = f(x_2,y_2)). Due to the fact that (D_2f=0) we need to likewise have actually (f(-1,y_1)=f(-1,y_2)), thus (f) is constant on (A). (b) The line ((xgeq 0, y=0)) is not in the domajor, so the Median Value Theorem does not necessarily hold for intervals of develop (, , xgeq 0, y_10). For example, the attribute Let first (x eq 0). Then by the Chain Rule(f^(1)(x) = 2x^-3e^-x^-2). It is straightforward by induction that any (f^(i)(x)) will be a sum of product of power functions and also the exponential (e^-x^-2), all of which are differentiable and also therefore by the Product Rule and Chain Rule (f) has actually derivatives of all orders at (x eq 0). At (x=0), let eginalign*f"(0) = lim_h ightarrow 0fracf(h)h &= lim_h ightarrowhead 0frac1/he^h^-2\ &= lim_h ightarrowhead 0frac-h^-2-2h^-3e^h^-2\ &= 0.5 lim_h ightarrowhead 0 h e^-1/h^2 = 0,endalign*wbelow we offered L"Hospital"s ascendancy to acquire the second line. Calculating (f""(0)) requires making use of L"Hostpital"s dominion twice:eginalign*f""(0) = lim_h ightarrow 0fracf"(h)h &= frac2h^-3e^-h^-2h\ &=4lim_h ightarrow 0e^-h^-2 = 0.endalign*Due to the fact that we"ve prcooktop that (f^(i)) is differentiable, it should also be constant by Problem 2-1, so (fin C^infty) as wanted.

2-26 (a) Let (g(x)) be the function described in Problem 2-25 above. Then on the ((-1,1)) interval, (f(x) = g(x-1)g(x+1)). Due to the fact that (g) is infinitely differentiable on that interval, by the Product Rule and also induction (f) is likewise infinitely differentiable on ((-1,1)). For (x in (-infty,-1) cup (1,infty)) (f) is constant, so we just need to check that the derivative exists at (1,-1). For instance, for (x=1) we haveeginalign*lim_h ightarrowhead 1 fracg(h+1)g(h-1)h &= g(2)lim_h ightarrowhead 0 g(h)\ &= e^-1/40,endalign*by Problem 2-25. That (f) is positive complies with from the definition of an exponential. (b) Let (h(x) = figl(frac2epsilon(x-1)igr)). Then (h) is positive on ((0,epsilon)) and also zero elsewhere, and by the Chain Rule is in (C^infty). Now think about (H(x) = int_0^x h/int_0^epsilon h). It"s basic to check out that (H(x)=0) if (xleq 0) and (H(x)=1) if (xgeqepsilon). In addition, (H^(i)(x) = h^(i-1)(x)) for (xgeq 1) by the Fundamental Theorem of Calculus, so (Hin C^infty).(c)(g in C^infty) by the Product and Chain rules, and also it"s simple to view that it"s positive just on the stated square and also zero in other places.(d)Let (U_x) be an open up rectangle focused on (xin C), such that (U_x subsetneq A). Then we can pick a finite subcover (U_x_1, ..., U_x_m) of compact (A). Let (f) be a amount of (m) attributes of develop offered in Problem 2-26(c) over, each positive on (U_x_m) and zero in other places. (gin C^infty) bereason it"s a amount of infinitely differentiable features (note that we needed (A) to be compact, as we can"t apply the "sum rule" of differentiation to uncountable sums). In addition, (g geq 0) on (A) and also is zero exterior of (igcup_i arU_x_i) (where we aget provided the compactness of (A)).(e) We only have to display that (f(x) geq epsilon) for some (epsilon > 0) for any type of (xin A). This is true for any kind of continuous feature by Problem 1-29. The hint completes the difficulty.

2-28 In each case, let (a) be the value of the component attributes as in the book"s examples. Then (a)eginalign*D_1F(x,y) &= D_1f(a)k(y)g"(x)+D_2f(a)g"(x)\D_2F(x,y) &= D_2f(a)g(x)k"(y)+D_2f(a)k"(y).endalign*(b) eginalign*D_1F(x,y,z) &= D_1f(a)g"(x+y)\D_2F(x,y,z) &= D_1f(a)g"(x+y) + D_2f(a)h"(y+z)\D_3F(x,y,z) &= D_2f(a)h"(y+z).endalign*(c) eginalign*D_1F(x,y,z) &= D_1f(a)yx^y-1+D_3f(a)x^zlnx\D_2F(x,y,z) &= D_1f(a)x^ylny+D_2f(a)zy^z-1\D_3F(x,y,z) &= D_2f(a)y^zlnz+D_3f(a)xz^x-1.endalign*(d) eginalign*D_1F(x,y) &= D_1f(a)+D_2f(a)g"(x)+D_3f(a)D_1h(x,y)\D_2F(x,y) &= D_3f(a)D_2h(x,y).endalign*

2-29 (a) Follows straight from the meaning of a partial derivative. (b) Let (D_txf(a) = lim_s ightarrow 0fracf(a+s:tx)-f(a)s). We have the right to substitute (u=st) to get (c) Due to the fact that (f) is differentiable, we have <0 = lim_h ightarrowhead 0fracf(a+h)-f(a)-Df(a)hh> for any type of sequence (h ightarrowhead 0). In particular, for any nonzero (xin mathbfR^n), eginalign* 0 &= lim_t ightarrow 0fracf(a+tx)-f(a)-tDf(a)xx\ &= frac1xlim_t ightarrowhead 0fracf(a+tx)-f(a)t-Df(a)x, endalign* which indicates that (D_xf(a) = Df(a)x) for any kind of nonzero (x), and the same holds for (x=0) trivially. Linearity of the directional derivative currently follows from linearity of (Df(a)).2-31 This follows straight from Problem 1-26, via (0=Dg_h(0) = D_hf(0,0)). Of course, in this instance (f) is not consistently differentiable, having actually no derivative at the boundary of (A).

2-32 (a) To see that (f) is differentiable, compute the derivave directy:wbelow the last ehigh quality follows as (|sin1/h|leq 1) for any type of (hin mathbfR). For any type of (x eq 0), It"s basic to watch that (f") has actually no limit as (x) goes to zero, thus the derivative is not continuous.(b) This is totally analoguous to (a). Keep in mind additionally that we deserve to rewrite (f) for (zinmathbfR_+) as (f(z) = |z|^2sinfrac1).

2-33 Note that continuity is used in the proof twice. First, using the mean-value theorem requires continuity of (g_i(x):=f(xa^1,...,a^i-1,x,a^i+1,...,a^n)) for each (i), which is guaranteed by the visibility of partial derivatives. Secondly, the last equality of the proof is based upon the continuity of the derivaitves, that isbereason once However, using the intend value theorem only to dimensions (igeq 2), we arrive at eginalign*lim_h ightarrow 0frac &leq lim_h ightarrowhead 0frac1|f(c_1)-f(a)-D_1f(a)cdot h^1|+lim_h ightarrowhead 0sum_i=2^n|D_if(c_i)-D_if(a)|\ &leq lim_h ightarrow 0iggl|fracf(c_1)-f(a)h^1-D_1f(a)iggr|\ &leq lim_h ightarrow 0|D_1f(a)-D_1f(a)|\ &= 0.endalign*

2-34 For any (xin mathbfR^n), let (hcolon mathbfR ightarrow mathbfR^n, h(t)=tx). Then (gcolon mathbfR ightarrow mathbfR,g = fcirc h). From the Chain Rule, eginalign* g"(1) &= f"(x)cdot h"(1)\ &= sum_i=1^nx^i D_if(x).endalign*From homogeneity of (f) we likewise have actually (g(t)=t^mf(x)) which from the Product Rule offers straight (g"(1) = mf(x)).

2-35 Like in Problem 2-34, the derivative of (h_x(t)) is (h_x"(t) = sum_i=1^nx^iD_if(tx)). Because noe-dimensional integration is a direct operation (see e.g. Apostol), we haveeginalign*f(x) &=int_0^1h_x"(t)dt\ &= int_0^1 sum_i=1^n x^iD_if(tx)dt\ &= sum_i=0^nx^iint_0^1D_if(tx)dt.endalign*Now, take (g_i(x) = int_0^1D_if(tx)dt).

2-36 By the Inverse Function Theorem there exists for any (yin f(A)) an open up set (W_y subcollection f(A)) such that (y in W_y), which proves that (f(A)) is open. In enhancement, and (f) has actually a constant inverse from some open set (V_f^-1(y) i f^-1(y)) onto (W_y), which is differentiable. We only should present that if (W_y_1, W_y_2) overlap, the values of the inverse functions agree. Assume (f) has actually a neighborhood inverse (gcolon W_y_1 ightarrowhead V_1) and a regional inverse (hcolon W_y_2 ightarrow V_2), and also that for some (yin W_y_1cap W_y_2) (a=g(y), b=h(y), a eq b). Due to the fact that both (g,h) are neighborhood inverses, this implies (f(a)=f(b)=y), which contradicts the injectivity of (f). Therefore, (f) has a "global" inverse (f^-1) on (A) offered by the IFT locally. Since the regional inverses agree on all the open up subsets of (f(A)) their derivatives given by the IFT must also agree, therefore (f^-1) is differentiable.

2-37 (a) Assume for contradiction that (f) is injective. Then if (D_2f(x,y)=0), there need to exist an open up collection (A i (x,y)) such that (D_1f(x,y) eq 0 ext if (x,y)in A) - otherwise, ((x,y)) would certainly be a maximum which contradicts the injectivity of (f). Hence, we deserve to assume without loss of generality that Now consider the function (g(x,y) = (f(x,y),y)) defined on (A). Its derivative has a nonzero determinant, as . Therefore, (g) is of the create discussed in Problem 2-36 and it has actually a (differentiable) inverse on the open set (g(A) = (f(x,y),y), (x,y)in A\). However before, this is impossible: for some ((x,y)in A) let (f(x,y)=b). Then (g(x,y) = (b,y)) so the feature (g^-1) is not identified on any allude ((b,z), z eq y) as this would certainly imply (f(x,z)=b) which contradicts that (f) is injective.(b) Similarily as over, we must have (D_1f(a^1,...,a^n) eq 0) for all (a:=(a^1,...,a^n)) in some open collection (A). Then, consider the attribute (g(a^1,...,a^n) = (f(a),a^n-m,...,a^n).) Its derivative is <eginpmatrixD_1f(a) & D_2f(a) & D_3f(a) & cdots & D_nf(a)\ 0 & 1 & 0 & cdots & 0\ vdots & & & & \ 0 & cdots & & & 1 endpmatrix,>which has nonzero determinant. The remainder of the proof is entirely analoguous to that in (a) above. Note that one have the right to apply this to the instance of (m2-38 (a) Due to the fact that (f) is differentiable (it has a nonzero derivative everywhere), the Typical Value Theorem uses. Assume (f(x_1) = f(x_2), x_1 eq x_2). Then by MVT, for some suggest (ain (x_1,x_2)) we have (f"(a)=0), a contradiction. (b) The derivative matrix is <eginpmatrixe^xcosx & -e^xsiny\e^xsiny & e^xcosyendpmatrix,>whose determinant is (detf"(x,y)=e^2x) for any ((x,y)). However, (f(x,y) = f(x,y+2pi)) for any kind of ((x,y)), by periodicity of the trigonometric functions.

2-39 Checking that (Df) is not repetitively differentiable at (0) proceeds similarily to Problem 2.32. It"s basic to watch that (f"(0)=1/2), so we could apply the IFT at (0) had actually it not forced a consistent derivative. However before, for any type of (x > 0) we have (f"(x) = 1/2-cos1/x+2xsin1/x). The derivative function is constant at (x>0) and also for small (x) approximates (1/2-cos1/x) arbitrarily well: it has actually worth zero at points with negative and also positive second derivatives. Thus (f) has actually both local maxima and minima arbitrarily cshed to zero, and also cannot be inverted about the beginning.

2-41 NOTE: I"ll assume the attribute is in (C^infty) (sufficient to assume twice-consistently differentiability). (a) Consider the function (D_2f:mathbfR^2 ightarrow mathbfR). It is continuously differentiable through a zero at (x,c(x)), and also (D_2(D_2f)(x,c(x)) eq 0) by presumption. Therefore the Implicit Function Theorem applies directly and also guarantees that (c) is differentiable. To compute its derivative, differentiate <0=D_2f(x,c(x))> on both sides. Using the Chain Rule giveseginalign*0 &= D_1(D_2f)(x,c(x))\ &= D_2,1f(x,c(x))+D_2,2f(x,c(x))c"(x)\ &indicates c"(x)=-fracD_2,1f(x,c(x))D_2,2f(x,c(x)).endalign*(b) Due to the fact that (c"(x)=0), it complies with from (a) above that (D_2,1f(x,y)=0) for (y=c(x)), that is (y) such that (D_2f(x,y)=0).(c) Do the correct differentiations and also boundary bookkeeping, remembering that Theorem 2-6 uses thanks to outcomes in (a,b) above.3-1 Consider the partition (P=((0,1/2,1),(0,1))). Then (L(f,P) = U(f,P) = 0.5) which means (with Corollary 3-2) that (supL(f,P) = infU(f,P)) -- for this reason the attribute is integrable and also (int_<0,1> imes <0,1> = 0.5).

3-2 Let (x_1,cdots,x_N) be all the points such that (f(x_i) eq g(x_0)) and also let (delta = max). Let (P) be a partition and note that thanks to Lemma 3-1 we have the right to assume that (P) is fine sufficient to encshed each (x_i) in a separate rectangle (S_i). Finally, let (epsilon = maxv(S_i)). Then we have actually eginalign*|L(f,P)-L(g,P)| &leq Ndeltaepsilon\|U(f,P)-U(g,P)| &leq Ndeltaepsilon\endalign*Hence by refining (P) so that the (x_i) are enclosed in smaller rectangles we have the right to get the reduced and top sums of (g) arbitrarily cshed to the reduced and upper sums of (f). This, together with integrcapability of (f), indicates that (g) is integrable on (A) and also has actually the exact same integral.

3-3 Easy, just follow the meanings.

3-4 If (f|_S)) is integrable for all (S), let (Delta(f,P) = max_SU(f.) If (N) is the number of subrectangles in (P) we have (U(f,P)-L(f,P) leq NDelta(f,P).) Due to the fact that by refining (P) we can obtain (Delta) to be arbitrarily tiny, (f) is integrable. For the other direction, assume (f) is integrable but for some subrectangle (S), (infU(f-sup_S,S) = delta > 0.) Then for any kind of partition (P) we need to have (U(f,P) - L(f,P) geq delta) -- otherwise we can refine the partition (P) right into (P") by adding the limits of (S). But this would indicate that (infU(f-supL(f 3-6 Let (P) be a partition. Then for any type of subrectangle (S), This mirrors that (U(f,P) - L(f,P) geq U(|f|,P) - L(|f|,P)) and also so (|f|) is integrable. Furthermore,eginalign*left|int_Af ight| &= left|infU(f,P) ight|\ &= infleft\ &leq infv(S)\ &= inff ight\ &= int_Aleft|f ight|.endalign*

3-7 First note that for any type of partition (P), any kind of (m_S(f) = 0). Therefore, it is enough to show that for little enough rectangles, (M_S(f)v(S)) gets arbitrarily cshed to zero. For a organic number (q), take into consideration the partition

Clearly if (x in

, p3-10 (a) The boundary of (C) belongs to its closure, which by definition indicates that the boundary belongs to every closed collection containing (C). If (S_1,cdots,S_N) is any type of cover of (C) by closed sets, it is likewise a closed set (being a finite sum of closed sets), and it contains (C), therefore it likewise covers the boundary. (b) Let (C) be the set of all rational numbers included in the interval (<0,1>). We"ve checked out that this set has actually measure zero. However before, its boundary is the totality interval, obviously does not have actually measure zero.

3-11 Assume we deserve to cover the boundary by a cover of closed intervals (, ,cdots) such that (sum_i l_i-k_i leq epsilon) for any (epsilon). But by Problem 1-18, the boundary of (A) is (<0,1>-A), so (A = <0,1>- extboundaryA). Hence, we would certainly have (sum_i (b_i-a_i) = 1-epsilon), a contradiction.

3-12 The outcome of the Hint complies with conveniently from Problem 1-30. We have the right to use it to display that the points at which (f) is disconsistent have the right to be ordered in a sequence. By Theorem 1-10, (f) is disconsistent at (a) iff its oscillation at (a) is nonzero. Hence for each (iin 1,2,cdots) we deserve to consequently take the finite set of (x) such that (o(f,x) > 1/n). Because finite sets have actually measure zero, Theorem 3-4 means the outcome.

3-13 (a) Treat each rectangle as a finite sequence ((a_1,b_1,a_2,b_2,cdots, a_n,b_n)). Since we deserve to order the rational numbers, there is a bijection between the rectangles and the set of vectors (k=(k_1,k_2,cdots,k_2n)) through positive integer coefficients -- for this reason we only need to ararray this set. Now for any kind of (N_jin 1,2cdots) there is only a finite variety of such vectors via (sum_i k_i leq N_j). By Theorem 3-4, the set of such vectors hence has actually measure zero.(b) The Hint solves the difficulty.

3-14 Let (B_f) and also (B_g) be the sets of points at which (f) and also (g) are discontinuous. Clearly (fdot g) deserve to just be disconsistent at (x) such that either (g) is diconsistent at (x) or (f) is disconsistent at (g(x)). Hence (B_fdot gsubset B_f cup B_g) which is a collection of measure zero.

3-15 By Problem 3-9(a), (C) is a bounded set and therefore is a subset of some closed rectangle (A). By Problem 3-10(a), the boundary of (C) has actually content zero and also therefore measure zero, thus (C) is Jordan-measurable. Because (C) has content zero, for any type of partition (P) we deserve to create a subpartition in which the elements of (C) are enclosed in a finite number of subrectangles of joint volume much less than (epsilon); this mirrors that (int_Achi_C = 0.)

3-16 The subset of the rationals had in the interval (<0,1>) (we"ve currently watched that its boundary is the totality interval, which we understand does not have meacertain zero).

3-17 Let (P) be a partition of (A). For any type of (epsilon), encshed (C) in closed rectangles (S_1,cdots,S_icdots) such that (amount v(S_i) 1/n}) is enclosed in rectangles of arbitrarily tiny volume; otherwise the set would contribute at least (epsilon 1/n) to the largest lowe bound on (U). Therefore, (xcolon f(x)>1/n\) has content zero. Now, alert that (C = xcolon f(x) eq 0 = igcup_i xcolon f(x) > 1/i\) -- that is, (C) is a countable set-sum of sets of measure zero, for this reason is itself of meacertain zero.

3-19 By Problem 3-11, (U) has actually a boundary of meacertain not zero. But < extBoundary(U) - V subcollection extBoundary(U-V),>and for this reason if (V) has meacertain zero, ( extBoundary(U-V)) has a subset of meacertain not zero, and also thus cannot itself have actually meacertain zero.

3-20 This adheres to straight from Problem 3-12.

3-21 By interpretation of the boundary, (xin extBoundary(C)) iff it deserve to be enclosed in an arbitrarily tiny rectangle (S_Xin S_1.) From this, and also from the fact that (S_2subcollection S_1), the outcome of the trouble follows easily.

3-22 Partition the underlying collection as in Problem 3-21, so that the boundary of (A) is covered by rectangles in (S_1-S_2) of volume less than (epsilon). Then (S_2) is a closed collection (consisting of a finite amount of closed sets), and it"s bounded as a subcollection of bounded (A). Thus, (C) is compact and also (int_A-C1 3-23 Due to the fact that (C) has content zero, the characteristic attribute on it is integrable through integral zero per Problem 3-15. Fubini"s Theorem then tells us that which by linearity of integration (Problem 3-3) impliesNow, the integrand also is non-negative, so by Problem 3-18 the collection at which it is non-zero has measure zero. We will display this set amounts to (A"). Take any type of such (x); then we have mathcalLint_Bchi_C(x,y)dy.>But this suggests that (y: (x,y) in C\) is not of content zero (otherwise both the lower and also top amount would certainly need to be equal), and also consequently (xin A").

3-24 First, display that (C) has content zero. Take any (epsilon > 0), and also let (S_1 = <0,1> imes<0,epsilon/2>). This rectangle covers all the elements ((p/q,y), yleq 1/q)of (C) through (1/q leq epsilon/2). The staying elements are consisted of in a finite number (say (N)) of segments of a line of develop (p/q imes 1/q), which can easily be spanned by (N) rectangles of volume much less than (epsilon/2). Now, for any (p/q) the collection (<0,1/q>) is not of content zero (Theorem 3-5), for this reason in this case (A") is the set of all rational numbers in (<0,1>) which has measure, however not content, zero.

3-25 Using Fubini"s Theorem and induction it"s simple to check out that imes cdots imes 1 > 0>for any (n) -- that is, the set has nonzero volume. However before, by Problem 3-15 sets of content zero have zero volume.

3-26 It"s straightforward to see that the boundary of (A_f) equals (<0,1> imes cup imes <0,f(0)> cup 1\times <0,f(1)> cup extgraph(f)). The first 3 elements of this collection amount are easily viewed to have content zero. We thus only should prove that ((x,f(x))colon xin \) has content zero. This is almost instant from the meaning of integrability: (f) is integrable so we can partition () right into intervals () such that (f)-m_(f))>is arbitrarily small. In other words, we have the right to cover the graph of (f) via rectangles ( imes (f),M_(f)>) of arbitrarily small volume. Now we have the right to use Fubini"s Theorem. First, note that (f) must attain a maximum on the compact interval (). Call this maximum (B). Theneginalign*int_A_f 1 &= int_ imes<0,B> chi_A_f\&=int_int_<0,f(x)>1,dx\&=int_f(x),dx.endalign*

3-27 Let (C = (x,y)in ^2colon xleq y\). By Problem 3-26, (C) is Jordan-measurable. Due to the fact that (f) is consistent it is integrable, Fubini"s Theorem offers us directly< int_Cf = int_a^b int_a^y f(x,y)dx dy = int_a^b int_x^b f(x,y)dy dx.>

3-28 For any kind of (a), either (D_1,2f(a) = D_2,1f(a)) or we deserve to assume (D_1,2f(a) - D_2,1f(a) > 0) (if the difference is smaller then zero, we use specifically the same thinking to arrive at a contradiction). Now, because the derivatives are consistent, (D_1,2f(a) - D_2,1f(a) > 0) implies that there is an open up set (O) containing a rectangle (A = imes i a) such that (D_1,2f - D_2,1f > 0) on (A). Now, we have the right to use Fubini"s Theorem to geteginalign*int_A D_1,2f - D_2,1f &= int_ int_D_1,2f(x,y) - D_2,1f(x,y),dy,dx\&=int_igl(D_1f(x,d)-D_1f(x,c)igr),dx -int_igr(D_2f(b,y)-D_2f(a,y)igl),dy\&=f(b,d)-f(b,c)-f(a,d)+f(a,c)-(f(b,d)-f(a,c)-f(a,d)+f(a,c))\&=0,endalign*a contradiction. Note that over, we"ve supplied the theorem to write dvery own the iterated integrals in the first line, and to switch the order of integration of the second integral in the second line.

3-30 ERRATUM: The trouble as stated is trivial. We"ll show instead that int_<0,1>chi_C(x,y),dx,dy = int_<0,1>int_<0,1>chi_C(x,y),dy,dx = 0.>First, note that (int_<0,1>^2chi_C) does not exist, considering that its boundary is the entirety unit square. However before, chi_C(x,y)dx = 0>for any kind of (y), as by definition of (C) tbelow is just one suggest on the line (<0,1> imes y\). By similar thinking, chi_C(x,y)dy =0.>From these two facts, the outcome adheres to directly as (int_<0,1> 0 = 0).

3-31 Define (g_x^-i(x^i) := g_x^1,cdots,x^i-1,x^i+1,cdots,x^n(x^i) = f(x^1,cdots,x^n).). Also, let (A_-i(x)) suggest the collection ( imes cdots imes imes imes cdots imes ). Due to the fact that (f) is constant, by Fubini"s Theorem we have the right to openly exchange the order of integration to getint_A_-i(x) f.>This yieldseginalign*D_iF(x) &= D_iint_ int_A_-i(x) f\ &= int_A_-i(x) f(t^1,cdots,t^i-1,x^i,t^i+1,cdots,t^n) dt^-i,endalign*

3-32 Because (int_c^yD_2f(x,y),dy = f(x,y)-f(x,c)), we have Now, eginalign*F(y) &= int_a^bleft(int_c^yD_2f(x,y),dy + f(x,c) ight),dx\ &= int_a^bint_c^yD_2f(x,y),dy,dx+int_a^bf(x,c),dx\ &= int_c^yint_a^bD_2f(x,y),dx,dy+int_a^bf(x,c),dx,endalign*wright here we supplied linearity of the integral to acquire the second equality, and also Fubini"s Theorem to acquire the final etop quality. From this and the Fundamental Theorem of Calculus it follows straight thatKeep in mind that the usage of Fubini"s Theorem here didn"t call for that (D_2f) be constant. Certainly, the exchange of the derivative ordering forced only that both functions (g(x) = D_2f(x,y)) and also (g(y) = D_2f(x,y)) be integrable, so that correct reduced and also top integrals from the theorem are equal.

3-33 (a) From the Fundapsychological Theorem of Calculus, (D_1F(x,y) = f(x,y)). By Problem 3-32, (D_2F(x,y) = int_a^xD_2f(t,y),dt.) (b) We have the right to compose (G(x) = F(g(x),x).) Using the Chain Rule, we haveeginalign*DG(x) &= D_1F(g(x),x)Dg(x)+D_2F(g(x),x)D id\ &= f(g(x),x)g"(x)+int_a^g(x)D_2f(t,x),dt.endalign*

3-34 Keeping in mind Leibnitz"s dominion, computeeginalign*D_1f(x,y) &= g_1(x,0) + int_0^y D_1g_2(x,t),dt\ &= g_1(x,0)+int_0^yD_2g_1(x,t),dt\ &= g_1(x,y).endalign*

3-36 This complies with directly from Fubini"s Theorem.

3-37 (a) Suppose first that (int_(0,1)f) exists. That is, there is an admissible open up cover (mathcalO) of ((0,1)) and a partition of unity (Phi) subordinate to (mathcalO) such that converges, assuming (f) is bounded in some open up collection around each suggest of ((0,1)). By Theorems 3-11 and also 3-12 we can select for any type of sequence (epsilon ightarrowhead 0.) Furthermore, by Theorem 3-11(4), we deserve to arselection (phiinPhi) in a sequence such that whenever before (ileq j), (phi_i) is zero exterior ((epsilon_i,1-epsilon_i)) and (phi_j) is zero exterior ((epsilon_j,1-epsilon_j)) with ((epsilon_jleq epsilon_i)). Then each partial sum (sum_i=1^kint_(0,1)phi_i f) amounts to (int_epsilon_k^1-epsilon_kf) for some (epsilon_k, ) through (epsilon_k ightarrowhead 0.) Hence, the extfinished integral exists if and just if the limit exists.(b) This question is damaged -- view errata e.g. at http://www.jirka.org/spivak-errata.html.

3-38 Define (mathcalA_n=int_A_nf.) Because the series ((mathcalA_n)) converges conditionally it deserve to be rearranged to converge to arbitrary actual values. Let ((alpha_n)_n=1^infty) be one such replan that converges to (alpha), and also ((eta_n)_n=1^infty) an additional rearrangement, converging to (eta.) We deserve to now pick (i_1) such that (|sum_n=1^i_1alpha_n - alpha| leq 1/4). Next, pick (i_2) such that (|sum_n=1^i_2alpha_n - alpha| leq 1/8). In this means, we can pick an index sequence ((i_1,i_2,cdots)) that partitions the original 2 sequences in a way such that (|sum_n=i_k^i_k+1alpha_n| leq 1/2^k). If (a_k=sum_n=i_k^i_k+1alpha_n,) the sequence ((a_i)_i=1^infty) converges absolutely to (alpha.) Of course we deserve to partition ((eta_i)) similarily to acquire a sequence of sums that converges absolutely to (eta.) Now take into consideration the adhering to partition of unity:where with some abuse of notation we compose (xin alpha_i) if (alpha_i = int_A_nf) and also (xin A_n.) Then by building, (sum_phiinPhi|intphicdot f| = alpha.) We have the right to construct a partition of unity (Phi) such that the sum converges to (eta) in the same means.

3-39 Let (B) be the set of critical points (i.e. (det g"(x) = 0) if (xin B.). By Sard"s Theorem and also Problem 3-4, eginalign*int_g(A)f &= int_g(A)-g(B)f + int_g(B) f\ &= int_g(A)-g(B)f.endalign*Since the determinant is nonzero on (g(A)-g(B)), we deserve to apply Theorem 3-13 as follows:eginalign*int_g(A)chi_g(A)-g(B)cdot f &= int_A (chi_g(A)-g(B)cdot f circ g)|det g"|\ &= int_g(A) (chi_g(A)-g(B)circ g)cdot (fcirc g) cdot |det g"|\ &= int_g(A) (fcirc g) cdot |det g"|,endalign*wbelow the last etop quality complies with as (chi_g(A)-g(B) = 0) if and also only if (|det g"| = 0) by building of (g(B).)

3-41 (a) If (rcos heta = scosphi) and also (rsin heta = ssinphi), we can square and add together the 2 equations to see that (r=s.) But then (sin heta=sinphi) and also sine is (1-1) on ((0,2pi)), so (f) is injective. We deserve to compute the derivative directly:It"s equally easy to check that 0,>and the last part of the subproblem likewise adheres to quickly (the half-line (y=0, x>0) is excluded bereason the second dimension of the doprimary does not contain an integer multiple of (2pi)).(b) Checking that this is the inverse is easy. For the derivative in case (x eq 0), compute directlyFor (x=0) use L"Hospital"s rule to acquire the exact same outcome. (c) The initially part adheres to straight from Theorem 3-13 and the reality that (|det P"(x,y)| = sqrtx^2+y^2 = r(x,y).) The collection (B_r) is simply the ball of radius (r), which is the collection in between the circle of radius (0) and the circle of radius (r), so the outcome also follows. (d) First, note that if (h(x,y) = e^-(x^2+y^2)) and (g(r, heta) = e^-r^2) then (h(x,y) = g(r(x,y), heta(x,y)).) Then it follows from (c) thateginalign*int_B_r e^-(x^2+y^2) &= int_0^r,int_0^2pir e^-r^2 d heta dr\ &= 2pi int_0^r re^-r^2 dr\ &= 2pi igl(-frac12e^-r^2igr)_0^r\ &= pi(1-e^-r^2),endalign*wbelow we supplied Fubini"s Theorem to arrive at the second ehigh quality. For the second equality, use Fubini"s Theorem again:eginalign*int_C_re^-(x^2+y^2)dx,dy &= int_-r^rint_-r^r e^-(x^2+y^2)dx,dy\ &= int_-r^rint_-r^r e^-x^2e^-y^2dx,dy\ &= left(int_-r^r e^-x^2dx ight)^2.endalign* (e) (C_r) is the square of side (r), which includes (B_r) and is contained in (B_rsqrt2.) By (d), we recognize that (lim_R ightarrowinfty int_B_R e^-(x^2+y^2)dx,dy = pi), which for this reason means that (lim_R ightarrowinfty int_C_R e^-(x^2+y^2)dx,dy = pi.) From this and the second equation in (d) it ultimately adheres to that (int_-r^r e^-x^2,dx = sqrtpi.)