Analysis on Manifolds 8. The Inverse Function Theorem

christangdt留下评论

这一节和下一节讨论的是两个比较特殊的多元微分情景，即反函数和隐函数，本节的主要结论是只要导数Df(x)是非奇异的，那么f有一个可微的反函数g。作者将过程分成了三个定理逐步证明。

Exercises

Exercise 1. Let f:\mathbf{R}^2\to\mathbf{R}^2 be defined by the equation f(x,y)=(x^2-y^2,2xy).
( a ) Show that f is one-to-one on the set A consisting of all (x,y) with x>0.
( b ) What is the set B=f(A)?
( c ) If g is the inverse function, find Dg(0,1).
Solution:
( a ) Suppose f(x,y)=f(a,b), in which x>0,a>0, then we know x^2-y^2=a^2-b^2 and xy=ab, also since |f(x,y)|=|f(a,b)|, we have (x^2-y^2 )^2+4x^2 y^2=(a^2-b^2 )^2+4a^2 b^2, thus x^2+y^2=a^2+b^2, together with x^2-y^2=a^2-b^2 we have x^2=a^2, and thus x=a, together with xy=ab we get y=b, so f is one-to-one.
( b ) Since A={(x,y):x>0}, we write x=a \cos\theta,y=a \sin\theta, in which a>0,\theta\in(-\pi/2,\pi/2), then f(a,\theta)=(a^2 \cos2\theta,a^2\sin2\theta ), in which 2\theta\in(-\pi,\pi), thus we can never have \cos2\theta=-1, also \cos2\theta and \sin2\theta cannot be 0 simultaneously, we get B=\mathbf{R}^2-{(x,0):x\leq0}.
( c ) We have Dg(0,1)=[Df(g(0,1))]^{-1}, in which g(0,1)=(x,y), which satisfies f(x,y)=(0,1), solve it we get x=y=1/\sqrt{2}, as

\displaystyle{Df(x,y)=\begin{bmatrix}2x&2y\\-2y&2x\end{bmatrix},\quad Df\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)=\begin{bmatrix}\sqrt{2}&\sqrt{2}\\-\sqrt{2}&\sqrt{2}\end{bmatrix},\quad \left[Df\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)\right]^{-1}=\frac{1}{4}\begin{bmatrix}\sqrt{2}&-\sqrt{2}\\{\sqrt{2}}&\sqrt{2}\end{bmatrix}}

we know that

\displaystyle{Dg(0,1)=\frac{1}{4}\begin{bmatrix}\sqrt{2}&-\sqrt{2}\\{\sqrt{2}}&\sqrt{2}\end{bmatrix}}

Exercise 2. Let f:\mathbf{R}^2\to\mathbf{R}^2 be defined by the equation f(x,y)=(e^x\cos{y},e^x\sin{y}).
( a ) Show that f is one-to-one on the set A consisting of all (x,y) with 0<y<2\pi.
( b ) What is the set B=f(A)?
( c ) If g is the inverse function, find Dg(0,1).
Solution:
( a ) Suppose f(x,y)=f(a,b), in which 0<y<2\pi,0<b<2\pi, then since |f(x,y)|=|f(a,b)|, we have e^{2x}=e^{2a}, thus x=a, so we have \cos{y}=\cos{b} and \sin{y}=\sin{b}, thus y=b.
( b ) Since A={(x,y):0<y<2\pi}, we know \cos{y}\neq 1, and when \sin{y}=0,\cos{y}=-1, thus B=\mathbf{R}^2-{(0,y):y\leq 0}.
( c ) We have Dg(0,1)=[Df(g(0,1))]^{-1}, in which g(0,1)=(x,y), which satisfies f(x,y)=(0,1), solve it we get x=0,y=\pi/2, as

\displaystyle{Df(x,y)=\begin{bmatrix}e^x\cos{y}&e^x\sin{y}\\-e^x\sin{y}&e^x\cos{y}\end{bmatrix},\quad Df\left(0,\frac{\pi}{2}\right)=\begin{bmatrix}0&1\\-1&0\end{bmatrix}}

we have

\displaystyle{Dg(0,1)=\left[Df\left(0,\frac{\pi}{2}\right)\right]^{-1}=\begin{bmatrix}0&-1\\1&0\end{bmatrix}}

Exercise 3. Let f:\mathbf{R}^n\to\mathbf{R}^n be given by the equation f(\mathbf{x})=|\mathbf{x}|^2\cdot\mathbf{x}. Show that f is of class C^{\infty} and that f carries the unit ball B(\mathbf{0};1) onto itself in a one-to-one fashion. Show, however, that the inverse function is not differentiable at \mathbf{0}.
Solution: That f is of class C^{\infty} is easy to verify. Next we show f carries B(\mathbf{0};1) onto itself in a one-to-one fashion.
Suppose f(\mathbf{x})=f(\mathbf{y}) with \mathbf{x},\mathbf{y}\in B(\mathbf{0};1), write \mathbf{x}=(x_1,\dots,x_n),\mathbf{y}=(y_1,\dots,y_n), then we have

\displaystyle{(x_1^2+\cdots +x_n^2)x_j=(y_1^2+\cdots +y_n^2)y_j,\quad j=1,\dots,n}

When all x_j and y_j are 0, the result holds naturally, if any x_k\neq 0, then the corresponding y_k\neq 0 and we have

\displaystyle{\frac{x_k}{y_k}=\frac{y_1^2+\cdots +y_n^2}{x_1^2+\cdots +x_n^2}}

Let x_k/y_k=C, then we have x_k=Cy_k and x_j=Cy_j for j=1,\dots,n, thus

\displaystyle{(C^2y_1^2+\cdots +C^2y_n^2)\cdot (Cy_1,\dots,Cy_n)=(y_1^2+\cdots +y_n^2)\cdot (y_1,\dots,y_n)}

which means C^3=1 and C=1, so \mathbf{x}=\mathbf{y}.
Conversely, for any \mathbf{y}\in B(\mathbf{0};1), we can write \mathbf{y}=(y_1,\dots,y_n), then if \mathbf{y}\neq \mathbf{0} we let \mathbf{x}=(x_1,\dots,x_n) with

\displaystyle{x_j=\frac{y_j}{y_1^2+\cdots+y_n^2},\quad j=1,\dots,n}

we have f(\mathbf{x})=\mathbf{y}. If \mathbf{y}=\mathbf{0} we just notice that f(\mathbf{0})=\mathbf{0}.
Finnaly, we show the inverse function is not differentiable at \mathbf{0}. Assume so, then Df(\mathbf{0}) shall be non-singular. But it is easy to calculate D_jf_i(\mathbf{0})=0 for i,j=1,\dots,n, thus Df(\mathbf{0}) is the square matrix with all entries 0, and singular, this is a contradiction.

Exercise 4. Let g:\mathbf{R}^2\to\mathbf{R}^2 be given by the equation g(x,y)=(2ye^{2x},xe^y). Let f:\mathbf{R}^2\to\mathbf{R}^3 be given by the equation f(x,y)=(2x-y^2,2x+y,xy+y^3).
( a ) Show that there is a neighborhood of (0,1) that g carries in a one-to-one fashion onto a neighborhood of (2,0).
( b ) Find D(f\circ g^{-1}) at (2,0).
Solution:
( a ) By Lemma 8.1, it is enough to to show that Dg(0,1) is non-singular, since

\displaystyle{Dg(x,y)=\begin{bmatrix}4ye^{2x}&2e^{2x}\\e^y&xe^y\end{bmatrix}}

we have

\displaystyle{Dg(0,1)=\begin{bmatrix}4&2\\e&0\end{bmatrix},\quad \det{Dg(0,1)}=-2e}

( b ) First, g^{-1}(2,0)=(0,1) and f(0,1)=(-1,1,1). By the chain rule we have

\displaystyle{D(f\circ g^{-1})(2,0)=Df(0,1)\cdot Dg^{-1}(2,0)=Df(0,1)\cdot [Dg(0,1)]^{-1}}

We calculate Df(0,1):

\displaystyle{Df(x,y)=\begin{bmatrix}2&-2y\\2&1\\y&x+3y^2\end{bmatrix},Df(0,1)=\begin{bmatrix}2&-2\\2&1\\1&3\end{bmatrix}}

also we have

\displaystyle{[Dg(0,1)]^{-1}=\frac{1}{2e}\begin{bmatrix}0&2\\e&-4\end{bmatrix}}

so

\displaystyle{D(f\circ g^{-1})(2,0)=\frac{1}{2e}\begin{bmatrix}2&-2\\2&1\\1&3\end{bmatrix}\begin{bmatrix}0&2\\e&-4\end{bmatrix}=\frac{1}{2e}\begin{bmatrix}-2e&-4\\e&0\\3e&-10\end{bmatrix}}

Exercise 5. Let A be open in \mathbf{R}^n; let f:A\to\mathbf{R}^n be of class C^r; assume Df(x) is non-singular for \mathbf{x}\in A. Show that even if f is not one-to-one on A, the set B=f(A) is open in \mathbf{R}^n.
Solution: Choose any \mathbf{b}\in\ B, since B=f(A), there exists \mathbf{a}\in A such that f(\mathbf{a})=\mathbf{b}, since Df(\mathbf{a}) is non-singular, by Theorem 8.3, there is a neighborhood U of \mathbf{a} such that V=f(U) is open in \mathbf{R}^n and f is an isomorphism on U, the inverse function g=f^{-1} is of class C^r.
Since A is open, we can find a neighborhood U' of \mathbf{a} such that U'\subset A, the set U\cap U' is a neighborhood of \mathbf{a} and is contained in A, which means f(U\cap U') is contained in B. Since g is continuous and f(U\cap U')=g^{-1}(U\cap U'), we see that f(U\cap U') is open, the conclusion follows as \mathbf{b}\in\ f(U\cap U').

Analysis on Manifolds 7. The Chain Rule

christangdt留下评论

本节主要说明链式法则，即多维空间中，可导函数的复合也是可导的。Theorem 7.1是本节的主要定理。Corollary 7.2说明C^r类函数的复合也是C^r类。Theorem 7.3是多维空间的均值定理。Theorem 7.4说明多维空间下反函数定理也成立，但是反函数的导数变为原函数导数的逆矩阵，由此可知，反函数可导需要原函数导数矩阵非奇异。

Exercises

Exercise 1. Let f:\mathbf{R}^3\to\mathbf{R}^2 satisfy the condition f(\mathbf{0})=(1,2) and

\displaystyle{Df(\mathbf{0})=\begin{bmatrix}1&2&3\\0&0&1\end{bmatrix}}

Let g:\mathbf{R}^2\to\mathbf{R}^2 be defined by the equation

\displaystyle{g(x,y)=(x+2y+1,3xy)}

Find D(g\circ f)(\mathbf{0}).

Solution: We first have

\displaystyle{Dg(x,y)=\begin{bmatrix}1&2\\3y&3x\end{bmatrix},Dg(1,2)=\begin{bmatrix}1&2\\6&3\end{bmatrix}}

then by the chain rule

\displaystyle{\begin{aligned}D(g\circ f)(\mathbf{0})&=Dg(f(\mathbf{0})){\cdot}Df(\mathbf{0})=Dg(1,2){\cdot}Df(\mathbf{0})\\&=\begin{bmatrix}1&2\\6&3\end{bmatrix}\begin{bmatrix}1&2&3\\0&0&1\end{bmatrix}=\begin{bmatrix}1&2&5\\6&12&21\end{bmatrix}\end{aligned}}

Exercise 2. Let f:\mathbf{R}^2\to\mathbf{R}^3 and g:\mathbf{R}^3\to\mathbf{R}^2 be given by the equations

\displaystyle{f(x)=(e^{2x_1+x_2},3x_2-\cos{x_1},x_1^2+x_2+2)}
\displaystyle{g(y)=(3y_1+2y_2+y_3^2,y_1^2-y_3+1)}

( a ) If F(x)=g(f(x)), find DF(\mathbf{0}).
( b ) If G(y)=f(g(y)), find DG(\mathbf{0}).

Solution: We first have f(0,0)=(1,-1,2),g(0,0,0)=(0,1) and

\displaystyle{Df(x_1,x_2 )=\begin{bmatrix}2e^{2x_1+x_2}&e^{2x_1+x_2}\\{\sin}x_1&3\\2x_1&1\end{bmatrix},Dg(y_1,y_2,y_3 )=\begin{bmatrix}3&2&2y_3\\2y_1&0&-1\end{bmatrix}}

( a ) By the chain rule we have

\displaystyle{\begin{aligned}DF(\mathbf{0})&=Dg(f(\mathbf{0})){\cdot}Df(\mathbf{0})=Dg(1,-1,2){\cdot}Df(0,0)\\&=\begin{bmatrix}3&2&4\\2&0&-1\end{bmatrix}\begin{bmatrix}2&1\\0&3\\0&1\end{bmatrix}=\begin{bmatrix}6&13\\4&1\end{bmatrix}\end{aligned}}

( b ) By the chain rule we have

\displaystyle{\begin{aligned}DG(\mathbf{0})&=Df(g(\mathbf{0})){\cdot}Dg(\mathbf{0})=Df(0,1){\cdot}Dg(0,0,0)\\&=\begin{bmatrix}2e&e\\0&3\\0&1\end{bmatrix}\begin{bmatrix}3&2&0\\0&0&-1\end{bmatrix}=\begin{bmatrix}6e&4e&-e\\0&0&-3\\0&0&-1\end{bmatrix}\end{aligned}}

Exercise 3. Let f:\mathbf{R}^3\to\mathbf{R} and g:\mathbf{R}^2\to\mathbf{R} be differentiable. Let F:\mathbf{R}^2\to\mathbf{R} be defined by the equation

\displaystyle{F(x,y)=f(x,y,g(x,y))}

( a ) Find DF in terms of the partials of f and g.
( b ) If F(x,y)=0 for all (x,y), find D_1g and D_2g in terms of the partials of f.

Solution:
( a ) We have DF(x,y)=\begin{bmatrix}D_1 F(x,y)&D_2 F(x,y) \end{bmatrix}, and

\displaystyle{\begin{aligned}D_1 F(x,y)&=\lim_{t\to 0}\frac{f(x+t,y,g(x+t,y))-f(x,y,g(x,y))}{t}\\&=\lim_{t\to 0}\frac{f(x+t,y,g(x+t,y))-f(x,y,g(x+t,y))}{t}+\lim_{t\to 0}\frac{f(x,y,g(x+t,y))-f(x,y,g(x,y))}{t}\\&=D_1 f(x,y,g(x,y))+D_3 f(x,y,g(x,y)){\cdot}D_1 g(x,y)\end{aligned}}
\displaystyle{D_2 F(x,y)=D_2 f(x,y,g(x,y))+D_3 f(x,y,g(x,y)){\cdot}D_2 g(x,y)}

( b ) In this case D_1 F(x,y)=D_2 F(x,y)=0, thus

\displaystyle{D_1 g(x,y)=[D_3 f(x,y,g(x,y))]^{-1}{\cdot}(-D_1 f(x,y,g(x,y)))}
\displaystyle{D_2 g(x,y)=[D_3 f(x,y,g(x,y))]^{-1}{\cdot}(-D_2 f(x,y,g(x,y)))}

Exercise 4. Let g:\mathbf{R}^2\to\mathbf{R}^2 be defined by the equation g(x,y)=(x,y+x^2). Let f:\mathbf{R}^2\to\mathbf{R} be the function defined in Example 3 of Section 5. Let h=f\circ g. Show that the directional derivatives of f and g exist everywhere, but there is a \mathbf{u}\neq\mathbf{0} for which h'(\mathbf{0};\mathbf{u}) does not exist.

Solution: The directional derivatives of f exists everywhere is clear from the text. Since Dg(x,y) exists everywhere and Dg(x,y)=\begin{bmatrix}1&2x\\0&1\end{bmatrix}, the directional derivatives of g exists everywhere since

\displaystyle{g'(\mathbf{a};\mathbf{u})=Dg(\mathbf{a}){\cdot}\mathbf{u}}

To show \exists\mathbf{u}\neq 0, s.t. h'(\mathbf{0};\mathbf{u}) does not exist, first notice h(\mathbf{0})=f(g(\mathbf{0}))=f(0,0)=0, let \mathbf{u}=(1,0)^T, then when t\to 0,t\neq 0, we have

\displaystyle{\frac{h(\mathbf{0}+t\mathbf{u})-h(\mathbf{0})}{t}=\frac{h(t,0)}{t}=\frac{f(t,t^2)}{t}=\frac{1}{t} \frac{t^2 t^2}{(t^4+t^4)}=\frac{1}{2t}}

thus the limit

\displaystyle{h'(\mathbf{0};\mathbf{u})=\lim_{t\to 0}\frac{h(\mathbf{0}+t\mathbf{u})-h(\mathbf{0})}{t}=\lim_{t\to 0}\frac{1}{2t}}

does not exist.

Analysis on Manifolds 6.Continuously Differentiable Functions

christangdt留下评论

这一节说明偏导数如果全部存在且连续，则导数存在。Theorem 6.1回顾了一元函数的中值定理；Theorem 6.2是本节的主定理；Theorem 6.3说明C^2函数的二阶偏导可以交换次序。本节的讨论看出，连续性，特别是每个分量上的连续性是可导的重要必要条件，在每个分量上连续的函数是性质很好的。作者也提到在实践中大部分时候只需要处理C^1类函数。

Exercises

Exercise 1. Show that the function f(x,y)=|xy| is differentiable at \mathbf{0}, but is not of class C^1 in any neighborhood of \mathbf{0}.

Solution: We let B=\begin{bmatrix}0&0\end{bmatrix}, then for any \mathbf{h}=(a,b)\in \mathbf{R}^2,h\to 0, we have

\displaystyle{\frac{f(\mathbf{0}+\mathbf{h})-f(\mathbf{0})-\mathbf{B}\cdot\mathbf{h}}{|\mathbf{h}|}=\frac{|ab|}{\max{(|a|,|b|)}}\to 0}

thus f is differentiable at \mathbf{0}, and f'(0)=\begin{bmatrix}0&0\end{bmatrix}.
Let A be a neighborhood of \mathbf{0}, then \exists a>0, s.t. B(\mathbf{0},a)\subset A, let {\delta}=a/2, then for any {\epsilon}>0, the two points x_1=(-{\epsilon}/4,{\delta}) and x_2=({\epsilon}/4,{\delta}) satisfy the property that |x_1-x_2 |<{\epsilon}, notice that when x>0,y>0, f(x,y)=xy and D_1 f(x,y)=y, when x<0,y>0,f(x,y)=-xy and D_1 f(x,y)=-y, so we have D_1 f(x_1 )=-{\delta} and D_1 f(x_2 )={\delta}, thus |D_1 f(x_1 )-D_1 f(x_2 )|=2{\delta}>{\delta}, which means D_1 f is not continuous in A. Thus f is not of class C^1 in A, and the conclusion follows since A is an arbitrary neighborhood of \mathbf{0}.

Exercise 2. Define f:\mathbf{R}\to\mathbf{R} by setting f(0)=0, and

\displaystyle{f(t)=t^2\sin{(1/t)}\quad\text{if}\quad{t\neq 0}.}

( a ) Show f is differentiable at 0, and calculate f'(0).
( b ) Calculate f'(t) if t\neq 0.
( c ) Show f' is not continuous at 0.
( d ) Conclude that f is differentiable on \mathbf{R} but not of class C^1 on \mathbf{R}.

Solution:
( a ) f is a real valued function, so by the existence of the limit

\displaystyle{\lim_{h\to 0}\frac{f(0+h)-f(0)}{h}=\lim_{h\to 0}h\sin{\left(\frac{1}{h}\right)}=0}

we see f is differentiable at 0 and f'(0)=0.
( b ) If t\neq 0, we can directly have

\displaystyle{f'(t)=2t\sin\left(\frac{1}{t}\right)+t^2\cos\left(\frac{1}{t}\right)\left(-\frac{1}{t^2}\right)=2t\sin\left(\frac{1}{t}\right)-\cos\left(\frac{1}{t}\right)}

( c ) We already have f'(0)=0, but \lim_{t\to 0,t\neq 0}f'(t) does not exist.
( d ) Since for any x\in \mathbf{R},f' (x) exists, f is differentiable on \mathbf{R}, but f' is not continuous at 0, thus not of class C^1 on \mathbf{R}.

Exercise 3. Show that the proof of Theorem 6.2 goes through if we assure merely that the partials D_jf exist in a neighborhood of \mathbf{a} and are continuous at \mathbf{a}.

Solution: If D_j f exists in a neighborhood of \mathbf{a} and are continuous at \mathbf{a}, then we can choose a neighborhood of \mathbf{a} and name it A, the proofs goes through in A.

Exercise 4. Show that if A\subset\mathbf{R}^m and f:A\to\mathbf{R}, and if the partials D_jf exist and are bounded in a neighborhood of \mathbf{a}, then f is continuous at \mathbf{a}.

Solution: Let A be the neighborhood such that D_j f exists and are bounded, then we can have M>0 s.t.

\displaystyle{|D_jf(\mathbf{a})|\leq M,\quad 1\leq j\leq m,\forall\mathbf{a}\in A\subset\mathbf{R}^m}

For {\forall}{\epsilon}>0, let {\delta}={\epsilon}/Mm, then use similar techniques from the proof of Theorem 6.2, if h\in \mathbf{R}^m,0<|h|<{\delta}, then there exits q_j\in A s.t.

\displaystyle{|f(\mathbf{a}+\mathbf{h})-f(\mathbf{a})|=\left|\sum_{j=1}^m D_j f(\mathbf{q}_j ) h_j \right|\leq Mm|h_j |\leq Mm|\mathbf{h}|<Mm{\delta}={\epsilon}}

thus f is continuous at \mathbf{a}.

Exercise 5. Let f:\mathbf{R}^2\to\mathbf{R}^2 be defined by the equation

\displaystyle{f(r,\theta)=(r\cos{\theta},r\sin{\theta})}

It is called the polar coordinate transformation.
( a ) Calculate Df and \det Df.
( b ) Sketch the image under f of the set S=[1,2]\times[0,\pi].

Solution:
( a ) Let B=\begin{bmatrix}\cos{\theta}&-r\sin{\theta}\\{\sin\theta}&r\cos\theta\end{bmatrix}, let \mathbf{h}=(h_1,h_2), then if \mathbf{h}\to\mathbf{0}, we have

\displaystyle{\begin{aligned}&\quad\frac{f(r+h_1,\theta+h_2)-f(r,\theta)-B\cdot\mathbf{h}}{|\mathbf{h}|}\\&=\frac{((r+h_1)\cos(\theta+h_2),(r+h_1)\sin(\theta+h_2))-(r\cos\theta,r\sin\theta)}{|\mathbf{h}|}\\&\quad-\frac{h_1\cos\theta-rh_2\sin\theta,h_1\sin\theta+rh_2\cos\theta}{|\mathbf{h}|}=(A,B)\end{aligned}}

in which

\displaystyle{A=\frac{(r+h_1)\cos(\theta+h_2)-r\cos\theta-h_1\cos\theta+rh_2\sin\theta}{|\mathbf{h}|}}
\displaystyle{B=\frac{(r+h_1)\sin(\theta+h_2)-r\sin\theta-h_1\sin\theta-rh_2\cos\theta}{|\mathbf{h}|}}

Use the fact that |h_1|\leq |\mathbf{h}| and |h_2|\leq |\mathbf{h}|, we have

\displaystyle{\begin{aligned}|A|&=\left|\frac{r(\cos(\theta+h_2)-\cos\theta+h_2\sin\theta)-h_1(\cos(\theta+h_2)-\cos\theta)}{|\mathbf{h}|}\right|\\&\leq|r|\left|\frac{\cos\theta(\cos{h_2}-1)+\sin\theta(h_2-\sin{h_2})}{h_2}\right|+|\cos(\theta+h_2)-\cos\theta|\end{aligned}}

from the L’Hospital rule we have

\displaystyle{\lim_{h_2\to 0}\frac{\cos{h_2}-1}{h_2}=\lim_{h_2\to 0}\frac{-\sin{h_2}}{1}=0,\quad \lim_{h_2\to 0}\frac{h_2-\sin{h_2}}{h_2}=\lim_{h_2\to 0}\frac{1-\cos{h_2}}{1}=0}

thus A\to 0, similarly B\to 0, so

\displaystyle{\frac{f(r+h_1,\theta+h_2)-f(r,\theta)-B\cdot\mathbf{h}}{|\mathbf{h}|}\to\mathbf{0}\quad\text{as}\quad\mathbf{h}\to\mathbf{0}}

Thus

\displaystyle{Df(r,\theta)=\begin{bmatrix}\cos\theta&-r\sin\theta\\{\sin\theta}&r\cos\theta\end{bmatrix},\det Df(r,\theta)=\det\begin{bmatrix}\cos\theta&-r\sin\theta\\{\sin\theta}&r\cos\theta\end{bmatrix}=r}

( b ) The graph is as follows:

Exercise 6. Repeat Exercise 5 for the function f:\mathbf{R}^2\to\mathbf{R}^2 given by

\displaystyle{f(x,y)=(x^2-y^2,2xy)}

Take S to be the set

\displaystyle{S={(x,y)|x^2+y^2\leq a^2\quad\text{and}\quad x\geq 0\quad\text{and}\quad y\geq 0}.}

We remark that if one identifies the complex numbers \mathbf{C} with \mathbf{R^2} in the usual way, then f is just the function f(z)=z^2.

Solution:
( a ) It is easy to see D_1f and D_2f are continuous, thus

\displaystyle{Df(x,y)=\begin{bmatrix}2x&-2y\\2y&2x\end{bmatrix},\quad\det Df(x,y)=4(x^2+y^2)}

( b ) Let x=a\cos{t},y=a\sin{t}, then f(a,t)=(a^2\cos{2t},a^2\sin{2t}), and t\in[0,\pi/2], the graph is a semi-circle on the upper half of the \mathbf{R}^2 with radius \mathbf{a}.

Exercise 7. Repeat Exercise 5 for the function f:\mathbf{R}^2\to\mathbf{R}^2 given by

\displaystyle{f(x,y)=(e^x\cos{y},e^x\sin{y})}

Take S to be the set S=[0,1]\times[0,\pi].
We remark that if one identifies \mathbf{C} with \mathbf{R^2} as usual, then f is the function f(z)=e^z.

Solution:
( a ) It is easy to see D_1f and D_2f are continuous, thus

\displaystyle{Df(x,y)=\begin{bmatrix}e^x\cos{y}&-e^x\sin{y}\\e^x\sin{y}&e^x\cos{y}\end{bmatrix},\quad\det Df(x,y)=e^{2x}}

(b) The graph is the same as Exercise 6.5, but the inner circle has radius 1, the outer circle has radius e.

Exercise 8. Repeat Exercise 5 for the function f:\mathbf{R}^3\to\mathbf{R}^3 given by

\displaystyle{f(\rho,\phi,\theta)=(\rho\cos{\theta}\sin{\phi},\rho\sin{\theta}\sin{\phi},\rho\cos{\phi}).}

It is called the spherical coordinate transformation. Take S to be the set
\displaystyle{S=[1,2]\times[0,\pi/2]\times[0,\pi/2].}

Solution:
( a ) It is easy to see D_1f,D_2f and D_3f are continuous, thus

\displaystyle{Df(\rho,\phi,\theta)=\begin{bmatrix}\cos\theta\sin\phi&\rho\cos\theta\cos\phi&-\rho\sin\theta\sin\phi\\{\sin\theta}\sin\phi&\rho\sin\theta\cos\phi&\rho\cos\theta\sin\phi\\{\cos\phi}&-\rho\sin\phi&0\end{bmatrix}}
\displaystyle{\begin{aligned}\det Df(\rho,\phi,\theta)&=\cos\phi\det\begin{bmatrix}\rho\cos\theta\cos\phi&-\rho\sin\theta\sin\phi\\{\rho}\sin\theta\cos\phi&\rho\cos\theta\sin\phi\end{bmatrix}+\rho\sin\phi\det\begin{bmatrix}\cos\theta\sin\phi&-\rho\sin\theta\sin\phi\\{\sin}\theta\sin\phi&\rho\cos\theta\sin\phi\end{bmatrix}\\&=\rho^2(\cos^2\phi\sin\phi+\sin^3\phi)=\rho^2\sin\phi\end{aligned}}

( b ) The graph is as follows:

Exercise 9. Let g:\mathbf{R}\to\mathbf{R} be a function of class C^2. Show that

\displaystyle{\lim_{h\to 0}\frac{g(a+h)-2g(a)+g(a-h)}{h^2}=g''(a)}

Solution: We can find p\in(a,a+h),q\in(a-h,a), such that

\displaystyle{g(a+h)-g(a)=g'(p)h,\quad g(a)-g(a-h)=g'(q)h}

thus

\displaystyle{\lim_{h\to 0}\frac{g(a+h)-2g(a)+g(a-h)}{h^2}=\lim_{h\to 0}\frac{g'(p)h-g'(q)h}{h^2}=\lim_{h\to 0}\frac{g'(p)-g'(q)}{h}}

then we can find \xi\in(q,p), s.t. g'(p)-g'(q)=g''(\xi), thus

\displaystyle{\lim_{h\to 0}\frac{g'(p)-g'(q)}{h}=g''(\xi)}

since g'' is continuous, when h\to 0,\xi\to a, we have g''(\xi)\to g''(a) as h\to 0.

Exercise 10. Define f:\mathbf{R}^2\to\mathbf{R} by setting f(0)=0, and

\displaystyle{f(x,y)=xy(x^2-y^2)/(x^2+y^2)\quad\text{if}\quad (x,y)\neq\mathbf{0}.}

( a ) Show D_1f and D_2f exist at \mathbf{0}.
( b ) Calculate D_1f and D_2f at (x,y)\neq\mathbf{0}.
( c ) Show f is of class C^1 on \mathbf{R}^2.
( d ) Show that D_2D_1f and D_1D_2f exist at \mathbf{0}, but are not equal there.

Solution:
( a ) By definition we have

\displaystyle{D_1f(\mathbf{0})=\lim_{t\to 0}\frac{f(\mathbf{0}+t\mathbf{e_1})-f(\mathbf{0})}{t}=\lim_{t\to 0}\frac{f(t,0)}{t}=\lim_{t\to 0}\frac{0}{t}=0}
\displaystyle{D_2f(\mathbf{0})=0}

( b ) If (x,y)\neq\mathbf{0}, then

\displaystyle{\begin{aligned}D_1f(x,y)&=\frac{\partial{f}}{\partial{x}}=\frac{(3x^2y-y^3)(x^2+y^2)-2x^2y(x^2-y^2)}{(x^2+y^2)^2}\\&=\frac{3x^4y+3x^2y^3-x^2y^3-y^5-2x^4y+2x^2y^3}{(x^2+y^2)^2}\\&=\frac{x^4y+4x^2y^3-y^5}{(x^2+y^2)^2}\end{aligned}}
\displaystyle{D_2f(x,y)=-\frac{y^4x+4y^2x^3-x^5}{(x^2+y^2)^2}=\frac{x^5-y^4x-4y^2x^3}{(x^2+y^2)^2}}

( c ) When x,y>0, we have

\displaystyle{\begin{aligned}D_1f(x,y)&=y\left(\frac{x^4+4x^2y^2-y^4}{(x^2+y^2)^2}\right)=y\left(1+\frac{2x^2y^2-2y^4}{(x^2+y^2)^2}\right)\\&{\leq}y\left(1+\frac{2x^2y^2+2y^4}{(x^2+y^2)^2}\right)\leq{y}\left(1+\frac{2x^2y^2}{2x^2y^2}+\frac{2y^4}{y^4}\right)\\&=y(1+1+2)=4y\end{aligned}}
\displaystyle{D_2f(x,y)=x\left(\frac{x^4-y^4-4x^2y^2}{(x^2+y^2)^2}\right)\leq{x}\left(\frac{x^4+y^4+4x^2y^2}{(x^2+y^2)^2}\right)\leq(1+1+2)x=4x}

thus |D_1 f(x,y)|\leq 4|y|,D_2 f(x,y)\leq 4|x|, so both D_1 f and D_2 f are continuous at \mathbf{0}, since the continuity at points other than \mathbf{0} is obvious, we have f being of class C^1 on \mathbf{R}^2.
( d ) A simple calculation shows

\displaystyle{D_2D_1f(\mathbf{0})=\lim_{t\to 0}\frac{D_1f(\mathbf{0}+t\mathbf{e}_2)-D_1f(\mathbf{0})}{t}=\lim_{t\to 0}\frac{D_1f(0,t)}{t}=\lim_{t\to 0}\frac{-t^5}{t^5}=-1}
\displaystyle{D_1D_2f(\mathbf{0})=\lim_{t\to 0}\frac{D_2f(\mathbf{0}+t\mathbf{e}_1)-D_2f(\mathbf{0})}{t}=\lim_{t\to 0}\frac{D_2f(t,0)}{t}=\lim_{t\to 0}\frac{t^5}{t^5}=1}