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Solution to Principles of Mathematical Analysis Chapter 9 Part A


Chapter 9 Functions of Several Variables

Exercise 1

(By analambanomenos) Let $\mathbf x,\mathbf y$ be in the span of $S$ and let $c$ be a scalar; we need to show that $\mathbf x+\mathbf y$ and $c\mathbf x$ are in the span of $S$. We have
\begin{align*}
\mathbf x &= c_1\mathbf x_1+\cdots+c_m\mathbf x_m \\
\mathbf y &= d_1\mathbf y_1+\cdots+d_n\mathbf y_n
\end{align*} for some $\mathbf x_1,\ldots,\mathbf x_m,\mathbf y_1,\ldots,\mathbf y_n$ in $S$ and some scalars $c_1,\ldots,c_m,d_1,\ldots,d_n$. Hence
\begin{align*}
\mathbf x+\mathbf y &= c_1\mathbf x_1+\cdots+c_m\mathbf x_m+d_1\mathbf y_1+\cdots+d_n\mathbf y_n \\
c\mathbf x &= cc_1\mathbf x_1+\cdots+cc_m\mathbf x_m
\end{align*}shows that $\mathbf x+\mathbf y$ and $c\mathbf x$ are in the span of $S$.


Exercise 2

(By analambanomenos) Let $A\in L(X,Y)$ and $B\in L(Y,Z)$. Let $\mathbf x,\mathbf y\in X$ and let $c$ be a scalar. Then
\begin{align*}
(BA)(\mathbf x+\mathbf y) &= B\big(A(\mathbf x+\mathbf y)\big) \\
&= B\big(A(\mathbf x)+A(\mathbf y)\big) \\
&= B\big(A(\mathbf x)\big) + B\big(A(\mathbf y)\big) \\
&= (BA)(\mathbf x)+(BA)(\mathbf y) \\
(BA)(c\mathbf x) &= B\big(A(c\mathbf x)\big) \\
&= B\big(cA(\mathbf x)\big) \\
&= cB\big(A(\mathbf x)\big) \\
&= c(BA)(\mathbf x)
\end{align*}which shows that $BA$ is linear.

Let $A\in L(X)$ be invertible. Let $\mathbf x,\mathbf y\in X$ and let $c$ be a scalar. Then there are $\mathbf x’,\mathbf y’\in X$ such that
\[
A(\mathbf x’)=\mathbf x, \quad A(\mathbf y’)=\mathbf y, \quad A^{-1}(\mathbf x)=\mathbf x’, \quad A^{-1}(\mathbf y)=\mathbf y’
\]Then
\begin{align*}
A^{-1}(\mathbf x+\mathbf y) &= A^{-1}\big(A(\mathbf x’)+A(\mathbf y’)\big) \\
&= A^{-1}\big(A(\mathbf x’+\mathbf y’)\big) \\
&= \mathbf x’+\mathbf y’ \\
&= A^{-1}(\mathbf x)+A^{-1}(\mathbf y) \\
A^{-1}(c\mathbf x) &= A^{-1}\big(cA(\mathbf x’)\big) \\
&= A^{-1}\big(A(c\mathbf x’)\big) \\
&= c\mathbf x’ \\
&= cA^{-1}(\mathbf x)
\end{align*}which shows that $A^{-1}$ is linear. Since $A^{-1}$ is 1-1 and maps $X$ onto itself, it is also invertible.


Exercise 3

(By analambanomenos) Suppose $A(\mathbf x_1)=A(\mathbf x_2)$. Then \[
A(\mathbf x_1-\mathbf x_2)=A(\mathbf x_1)-A(\mathbf x_2)=\mathbf 0 \text{ implies that }\mathbf x_1-\mathbf x_2=\mathbf 0 \] so that $\mathbf x_1=\mathbf x_2$. Hence $A$ is 1-1.


Exercise 4

(By analambanomenos) Let $\mathbf x,\mathbf y\in\mathscr N(A)$ and let $c$ be a scalar. Then
\begin{align*}
A(\mathbf x+\mathbf y) &= A(\mathbf x)+A(\mathbf y) = \mathbf 0+\mathbf 0 = \mathbf 0 \text{ so that }\mathbf x+\mathbf y\in\mathscr N(A) \\
A(c\mathbf x) &= cA(\mathbf x) = c\mathbf 0 = \mathbf 0 \text{ so that }c\mathbf x\in\mathscr N(A)
\end{align*} which shows that $\mathscr N(A)$ is a vector space.

Let $\mathbf x,\mathbf y\in\mathscr R(A)$, so that $\mathbf x=A(\mathbf x’)$ and $\mathbf y=A(\mathbf y’)$, and let $c$ be a scalar. Then
\begin{align*}
\mathbf x+\mathbf y &= A(\mathbf x’)+A(\mathbf y’) = A(\mathbf x’+\mathbf y’) \text{ so that }\mathbf x+\mathbf y\in\mathscr R(A) \\
c\mathbf x &= cA(\mathbf x’) = A(c\mathbf x’) \text{ so that }c\mathbf x\in\mathscr R(A)
\end{align*}which shows that $\mathscr R(A)$ is a vector space.


Exercise 5

(By analambanomenos) I’m going to show this by induction on $n$. For the case $n=1$, let $A\mathbf e_1=k$, and let $\mathbf y=k\mathbf e_1$. If $\mathbf x\in \R1$, then $\mathbf x=x\mathbf e_1$ for some scalar $x$, so $A\mathbf x=xA\mathbf e_1=ck=\mathbf x\cdot\mathbf y$.

Suppose the assertion is true for the case $n$ and let $A\in L(\R{n+1},\R1)$. Restricting $A$ to the subspace $\R n$ (spanned by $\mathbf e_1,\ldots,\mathbf e_n$) yields an element of $L(\R n,\R1)$, so by the induction assumption there is a $\mathbf y’\in\R n$ such that $A\mathbf x’=\mathbf x’\cdot\mathbf y’$ for all $\mathbf x’\in\R n$. If $A\mathbf e_{n_1}=k$, let $\mathbf y=\mathbf y’+k\mathbf e_{n+1}$. If $\mathbf x\in\R{n+1}$, then $\mathbf x=\mathbf x’+x\mathbf e_{n+1}$ for some $\mathbf x’\in\R n$ and some scalar $x$. Then, since the scalar product of $\mathbf e_{n+1}$ with every element of $\R n$ is 0, we have
\begin{align*}
A\mathbf x &= A\mathbf x’+xA\mathbf e_{n+1} \\
&= \mathbf x’\cdot \mathbf y’ + xk \\
&= (\mathbf x’+x\mathbf e_{n+1})\cdot\mathbf y’ + (\mathbf x’+x\mathbf e_{n+1})\cdot (k\mathbf e_{n+1}) \\
&= \mathbf x\cdot\mathbf y
\end{align*}If $|{\mathbf x}|\le1$, then $|{A\mathbf x}|=|{\mathbf x\cdot\mathbf y}|\le|{\mathbf x}|\,|{\mathbf y}|\le|{\mathbf y}|$, so that $\|{A}\|\le|{\mathbf y}|$. Let $\tilde{\mathbf y}=\mathbf y/|{\mathbf y}|$. Then $|{\tilde{\mathbf y}}|=1$ and
\[|{A\tilde{\mathbf y}}|=\frac{|{\mathbf y\cdot\mathbf y}|}{|{\mathbf y}|}=\frac{|{\mathbf y}|^2}{|{\mathbf y}|}=|{\mathbf y}|.
\]Hence $\|{A}\|=|{\mathbf y}|$.


Exercise 6

(By analambanomenos) The usual rules of differentiaion show that the partial derivatives of $\R2$ at $(x,y)\ne(0,0)$ are\[D_1f(x,y)=\frac{y(y^2-x^2)}{(x^2+y^2)^2} \qquad D_2f(x,y)=\frac{x(x^2-y^2)}{(x^2+y^2)^2}\]And since $f$ is equal to 0 everywhere along the $x$ and $y$ axes, they also exist and are equal to 0 at $(0,0)$.

For $x\ne 0$, $f(x,x)=x^2/(x^2+x^2)=1/2$. Since $f(0,0)=0$, $f$ is not continuous along the line $y=x$ at $(0,0)$.


Exercise 7

(By analambanomenos) Suppose that $|D_jf|<M$ in $E$, for $j=1,\ldots,n$. Following the hint to mimic the proof of Theorem 8.21, fix $\mathbf x\in E$ and let $\varepsilon>0$. Since $E$ is open, there is an open ball $S\subset E$, with center at $\mathbf x$ and radius $r<(Mn)^{-1}$. Suppose $\mathbf h=\sum h_j\mathbf e_j$, $|\mathbf h|<r$, put $\mathbf v_0=\mathbf 0$, and $\mathbf v_k=h_1\mathbf e_1+\cdots+h_k\mathbf e_k$, for $1\le k\le n$. Then
\begin{equation}\label{9.7.1}f(\mathbf x+\mathbf h)-f(\mathbf x)=\sum_{j=1}^n\big(f(\mathbf x+\mathbf v_j)-f(\mathbf x+\mathbf v_{j-1})\big). \end{equation}Since $| \mathbf v_k|<r$ for $1\le k\le n$ and since $S$ is convex, the segments with endpoints $\mathbf x+\mathbf v_{j-1}$ and $\mathbf x+\mathbf v_j$ lie in $S$. Since $\mathbf v_j=\mathbf v_{j-1}+h_j\mathbf e_j$, the mean value theorem, Theorem 5.10, shows that the $j$th summand in \eqref{9.7.1} is equal to
\[
h_j(D_jf)(\mathbf x+\mathbf v_{j-1}+\theta_jh_j\mathbf e_j)
\] for some $\theta_j\in(0,1)$. By \eqref{9.7.1}, it follows that
\begin{align*}
\big|f(\mathbf x+\mathbf h)-f(\mathbf x)\big| &\le \sum_{j=1}^n\big|f(\mathbf x+\mathbf v_j)-f(\mathbf x+\mathbf v_{j-1})\big| \\
&= \sum_{j=1}^n\big|h_j(D_jf)(\mathbf x+\mathbf v_{j-1}+\theta_jh_j\mathbf e_j)\big| \\
&< \sum_{j=1}^n|h_j|M<(Mn)r<\varepsilon
\end{align*}which shows that $f$ is continuous at $\mathbf x$. Since $\mathbf x\in E$ was arbitrary, we have $f$ continuous on $E$.


Exercise 8

(By analambanomenos) You can use Theorem 9.17 to express $f’$ as a sum of the partial derivatives and easily reduce the problem to the the single-variable case, Theorem 5.8. However, I thought I’d use the new definition of derivative (commonly called a Fr\’echet derivative, by the way) instead.

By Exercise 5, there is a $\mathbf y\in\R n$ such that $f’(\mathbf x)=\mathbf y\cdot\mathbf x$. Let $\mathbf h=h\mathbf y/|\mathbf y|$, and take the limit in the definition of derivative as $h$ approaches 0 through positive numbers. Then we get
\begin{align*}
0 &= \lim_{\mathbf h\rightarrow \mathbf 0}\frac{\big|f(\mathbf x+\mathbf h)-f(\mathbf x)-f’(\mathbf x)\mathbf h\big|}{|\mathbf h|} \\
&= \lim_{h\rightarrow 0+}\frac{\big|f(\mathbf x+\mathbf h)-f(\mathbf x)-\mathbf y\cdot(h\mathbf y)/|\mathbf y|\big|}{h|\mathbf y|/|\mathbf y|} \\
&= \lim_{h\rightarrow 0+}\frac{\big|f(\mathbf x+\mathbf h)-f(\mathbf x)-h|\mathbf y|\big|}{h} & \hbox{(since $\mathbf y\cdot\mathbf y=|\mathbf y|^2$)} \\
&= \lim_{h\rightarrow 0+}\frac{f(\mathbf x)-f(\mathbf x+\mathbf h)+h|\mathbf y|}{h} & \hbox{(since $f(\mathbf x+\mathbf h)-f(\mathbf x)$ and $-h|\mathbf y|$ are $\le0$)} \\
&= |\mathbf y| + \lim_{h\rightarrow 0+}\frac{f(\mathbf x)-f(\mathbf x+\mathbf h)}{h}
\end{align*}Since the term in the limit is non-negative for small $h$, this forces $|\mathbf y|\le 0$, or $\mathbf y=\mathbf 0$. Hence $f’(\mathbf x)=0$.


Exercise 9

(By analambanomenos) Fix $\mathbf x\in E$. Since $E$ is open, there is a open ball $S\subset E$ containing $\mathbf x$. By the Corollary to Theorem 9.19, $\mathbf f$ is constant on $S$. Hence the set $E’$ of all points $\mathbf z\in E$ such that $\mathbf f(\mathbf z)=\mathbf f(\mathbf x)$ is an open subset of $E$. Similarly, the set $E-E’$ is also open in $E$ (being the union of open sets on which $\mathbf f$ has a constant value not equal to $\mathbf f(\mathbf x)$). By Exercise 2.19(b), $E’$ and $E-E’$ are two separated sets whose union is $E$. Since $E$ is connected, we must have $E’=E$, so $\mathbf f$ is constant on $E$.


Exercise 10

(By analambanomenos) Let $E$ satisfy the weaker condition: if $(x,x_2,\ldots,x_n)\in E$ and $(y,x_2,\ldots,x_n)\in E$, then for all $x<z<y$ we have $(z,x_2,\ldots,x_n)\in E$. For such points, define $f_{x_2,\ldots,x_n}(z)=f(z,x_2,\ldots,x_n)$, then $f_{x_2,\ldots,x_n}’(z)=(D_1f)(z,x_2,\ldots,x_n)=0$, so by the Mean Value Theorem $f_{x_2,\ldots,x_n}$ must be constant on $[x,y]$. Hence $f(z,x_2,\ldots,x_n)$ must equal this constant value for all $z$ such that $(z,x_2,\ldots,x_n)\in E$, so that $f(\mathbf x)$ depends only on $x_2,\ldots,x_n$.

To see a counterexample, let $E\subset\R2$ be the square $(x,y)$, $-1<x<1$, $-1<y<1$ less the positive $y$-axis, $(0,y)$, $0<y<1$. Define
\[
f(x,y)=
\begin{cases}
0 & -1<x<1,\;-1<y<0 \\
-y & -1<x<0,\;0\le y<1 \\
y & 0<x<1,\;0\le y<1
\end{cases}
\]Then $f$ is continuous, $(D_1f)(x,y)=0$ for all $(x,y)\in E$, but the value of $f$ does not depend only on $y$.


Exercise 11

(By analambanomenos) We have \begin{align*}
\nabla(fg) &= \sum_{i=1}^n\big(D_i(fg)\big)\mathbf e_i \\
&= \sum_{i=1}^n\big(f(D_ig)+g(D_if)\big)\mathbf e_i \\
&= f\sum_{i=1}^n(D_ig)\mathbf e_i + g\sum_{i=1}^n(D_if)\mathbf e_i \\
&= f\nabla g+g\nabla f \\[3ex]
0 &= \nabla(1) \\
&= \nabla(f\cdot f^{-1}) \\
&= f\nabla(f^{-1}) + f^{-1}\nabla f \\
f\nabla(f^{-1}) &= -f^{-1}\nabla f \\
\nabla(f^{-1}) &= -f^{-2}\nabla f
\end{align*}


Exercise 12

(By analambanomenos) The range of $\mathbf f$ is a torus with inner radius $b-a$ and outer radius $b+a$, centered at the origin, whose plane is perpendicular to the $z$-axis. To see this, the central circle of such a torus is the set of points $(b\cos t,b\sin t,0)$. Then the circle of radius $a$ which intersects the half-plane that goes through the $z$-axis and the point $(b\cos t,b\sin t,0)$ is the circle centered at that point and perpendicular to the $x$-$y$-plane, with radius vector $a(\cos t,\sin t,0)$, which is described by \[(b\cos t,b\sin t,0)+a(\cos t,\sin t, 0)\cos s+ a(0,0,1)\sin s = \big(f_1(s,t),f_2(s,t),f_3(s,t)\big).\]Note that $\mathbf f$ is a one-to-one mapping of the square $S$ given by $(x,y)$, $0\le x<2\pi$, $0\le y<2\pi$ onto the torus.

(a) Taking the partial derivatives of $f_1$ with respect to $s$ and $t$ to get $\nabla f_1$, we want to solve\[(\nabla f_1)(s,t)=\big(-a\sin s\cos t,-(b+a\cos s)\sin t\big)=(0,0).\]Restricing ourselves to the square $S$, the first component equals 0 when either $s=0$, $s=\pi$, $t=\pi/2$, or $t=3\pi/2$. Since $b+a\cos s>0$ for all $s$, the second component equals 0 only when $t=0$ or $t=\pi$. Hence $\nabla f_1$ equals $\mathbf 0$ in $S$ only at the four points\[
(0,0),\quad(0,\pi),\quad(\pi,0),\quad(\pi,\pi),
\]which map onto the four points\[
(b+a,0,0),\quad(-b-a,0,0),\quad(b-a,0,0),\quad(a-b,0,0),
\]respectively.

(b) Taking the partial derivatives of $f_3$ with respect to $s$ and $t$ to get $\nabla f_3$, we want to solve\[
(\nabla f_3)(s,t)=(a\cos s,0)=(0,0).
\]which occurs in $S$ at the points $(s,t)$, where $s=\pi/2$ or $s=3\pi/2$, and $0\le t<2\pi$. The points $(\pi/2,t)$, $0\le t<2\pi$, are mapped by $\mathbf f$ to the circle $(b\cos t, b\sin t, a)$, and the points $(3\pi/2,t)$, $0\le t<2\pi$, are mapped to the circle $(b\cos t, b\sin t, -a)$.

(c) At $(b+a,0,0)$, corresponding to $(s,t)=(0,0)$, where $\cos s=\cos t=1$, $f_1$ attains its maximum value, $b+a$, and at $(-b-a,0,0)$, corresponding to $(s,t)=(0,\pi)$, where $\cos s=1$ and $\cos t=-1$, $f_1$ attains its minimum value, $-b-a$. At $(b-a,0,0)$, corresponding to $(s,t)=(\pi,0)$, where $\cos s=-1$ and $\cos t=1$, $f_1$ increases if you fix $t=0$ and vary $s$, and decreases if you fix $s=\pi$ and vary $t$. Similarly, at $(a-b,0,0$, corresponding to $(s,t)=(0,\pi)$, where $\cos s=1$ and $\cos t=-1$, $f_1$ decreases if you fix $t=\pi$ and vary $s$, and increases if you fix $s=\pi$ and vary $t$.

The points $(b\cos t,b\sin t,a)$, $0\le t<2\pi$, correspond to the points $(s,t)=(\pi/2,t)$ where $f_3$ attains its maximum value of $a$, and $(b\cos t,b\sin t,-a)$, $0\le t<2\pi$, correspond to the points $(s,t)=(3\pi/2,t)$ where $f_3$ attains its minimum value of $-a$.

(d) To show that the “irrational winding of the torus” (it even has its own Wikipedia page) is dense, I first show a general proposition about continuous functions. Suppose $F$ is a continuous mapping from $X$ to $Y$ (both metric, or even just topological spaces). Then if $D$ is a dense subset of $X$, then $F(D)$ is a dense subset of $F(Y)$. For if not, then there would be a nonempty open set $G$ in $F(Y)$ disjoint from $F(D)$. Then $F^{-1}(D)$ would be an open subset of $X$ disjoint from $D$, contradicting the density of $D$.

Hence, to show that the image of $\mathbf g$ is dense in the torus $K$, it suffices to show that its preimage under the mapping $\mathbf f$ is dense in $\R2$. Call this set $K’$, it is the set of points of the form\[
(s+m2\pi,\lambda s+n2\pi),\qquad\hbox{$s$ real, $m$ and $n$ integers}.
\]First I want to show that the intersection of $K’$ with the $s$-axis is dense on the axis. If $(s,0)$ is a point in this set, then from the above, we that $\lambda s$ is a multiple of $2\pi$, so $s=(n\lambda^{-1}+m)2\pi$ for some integers $m$ and $n$. To see that this set is dense, I first show another general proposition.

If $\alpha$ is an irrational number, then the set of points $D$ in $[0,1]$ which are equivalent to integral multiples of $\alpha$ modulo 1 are dense in that interval. That is, let $n$ be an integer, then $n\alpha$ is in the interval $(m,m+1)$ for some integer $m$, so that $x_n=n\alpha-m$ is between 0 and 1. These points are all distinct, since if $x_{n_1}=n_1\alpha-m_1=x_{n_2}=n_2\alpha-m_2$, for some integers $n_1\ne n_2$, then $\alpha=(m_1-m_2)/(n_2-n_1)$ would be rational. Hence $D$ is an infinite set.

Another simple fact is that if $x_{n_1}<x_{n_2}$ are two points in $D$ then $0<x_{n_2}-x_{n_1}<1$, so that $x_{n_2}-x_{n_1} = (n_2-n_1)\alpha-(m_2-m_1)$ is also in $D$. Also, if $0<jx_n<1$ for some positive integer $j$, then $jx_{n}=jn\alpha-jm$ is also in $D$.

Let $\varepsilon>0$ and let $k$ be an integer large enough so that $k^{-1}<\varepsilon$. Divide $[0,1]$ into $k$ intervals \[
I_j=\big[jk^{-1},(j+1)k^{-1}\big]\qquad j=0,\ldots,(k-1)
\](This isn’t strictly a partition, but I am ignoring the points on the boundaries, which are rational numbers which cannot be in $D$.) Since $D$ is infinite, one of these intervals contains at least two points in $D$, say $x_{n_1}<x_{n_2}$. Then $x_{n_2}-x_{n_1}$ is a point $x_n\in D$ such that $0<x_n<k^{-1}$. Hence each of the intervals $I_j$, $j=0,\ldots,(k-1)$, contains a point $jx_n$ of $D$. If $x\in[0,1]$, then $x$ is in one of the $I_j$, so there is a point $x_n\in D$ such that $|x-x_n|<k^{-1}<\varepsilon$. Hence $D$ is dense in $[0,1]$.

Going back to the intersection of $K’$ with the $s$-axis, it is not hard to see that the intersection of $K’$ with the interval $[0,2\pi]$ is the image of the set $D$ above, with $\alpha=\lambda^{-1}$, under the mapping $s\mapsto 2\pi s$, hence $K’$ is dense in the interval $[0,2\pi]$ in the $s$-axis. Since the other points of $K’$ on the $s$-axis are translates of this dense set by multiples of $2\pi$, we see that $K’$ is dense in the entire axis.

For other lines in $\R2$ parallel to the $s$-axis, $(s,t)\in K’$ if $(s-\lambda^{-1}t,0)\in K’$, that is, the intersection of $K’$ with this line is the image of the dense interesection of $K’$ with the $s$-axis under the translation $(s,0)\mapsto(s+\lambda^{-1}t,t)$. Hence $K’$ is also dense in this line, and so is dense in all of $\R2$.

If $\mathbf g(t_1)=\mathbf g(t_2)$, then $\mathbf f(t_1,\lambda t_1)=\mathbf f(t_2,\lambda t_2)$ which would only happen if $(t_1-t_2)=n2\pi$ and $\lambda(t_1-t_2)=m2\pi$ for some integers $m$ and $n$. But that would imply $\lambda=m/n$ which contradicts the irrationality of $\lambda$.

By Theorem 9.17 we have
\begin{align*}
\mathbf f’(s,t)\mathbf e_1 &= \big((D_1f_1)(s,t),(D_1f_2)(s,t),(D_1f_3)(s,t)\big) \\
&= \big(-a\sin s\cos t,-a\sin s\sin t,a\cos s\big) \\
\mathbf f’(s,t)\mathbf e_2 &= \big((D_2f_1)(s,t),(D_2f_2)(s,t),(D_2f_3)(s,t)\big) \\
&= \big(-(b+a\cos s)\sin t,(b+a\cos s)\cos t,0\big).
\end{align*} If we let ${\boldsymbol\gamma}(t)=(t,\lambda t)$, so that $\boldsymbol\gamma’(t)=\mathbf e_1+\lambda\mathbf e_2$, then we have by the chain rule
\begin{align*}
\mathbf g’(t) &= \mathbf f’\big(\boldsymbol\gamma(t)\big)\boldsymbol\gamma’(t) \\
&= \mathbf f’(t,\lambda t)(\mathbf e_1+\lambda\mathbf e_2) \\
&= \big(-a\sin t\cos\lambda-\lambda(b+a\cos t)\sin\lambda,-a\sin t\sin\lambda t+\lambda(b+a\cos t)\cos\lambda t,a\cos t\big) \\
\big|\mathbf g’(t)\big|^2 &= a^2\sin^2t\cos^2\lambda t+2\lambda a(b+a\cos t)\sin t\cos\lambda t\sin\lambda t+\lambda^2(b+a\cos t)^2\sin^2\lambda t\;+ \\
&\phantom{=}\;\; a^2\sin^2t\sin^2\lambda t-2\lambda a(b+a\cos t)\sin t\sin\lambda t\cos\lambda t+\lambda^2(b+a\cos t)^2\cos^2\lambda t\;+ \\
&\phantom{=}\;\; a^2\cos^2t \\
&= a^2+\lambda^2(b+a\cos t)^2.
\end{align*}

Baby Rudin 数学分析原理完整第九章习题解答

Linearity

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