\input homework_macros.tex

%\distiller
\printer

\def\dd#1#2{\frac{d#1}{d#2}}
\toks0={
8.9  p. 677:   5,  7, 11, 17, 33.
}
\def\Li{\displaystyle\lim_{n\to\infty}}
\def\Sum#1{\displaystyle\sum_{#1}^\infty}
\def\Int{\displaystyle\int}

\SN{\jobname}
\np$f(x)=\sin x$, $a=\frac\pi4$.
$f(x)=\sin x$, $f^{(1)}(x)=\cos x$,
$f^{(2)}(x)=-\sin x$, $f^{(3)}(x)=-\cos x$
$f^{(4)}(x)=\sin x$.
Hence
$f(\frac\pi4)=\frac{\sqrt2}{2}$, 
$f^{(1)}(\frac\pi4)=\frac{\sqrt2}{2}$,
$f^{(2)}(\frac\pi4)=-\frac{\sqrt2}{2}$, 
$f^{(3)}(\frac\pi4)=-\frac{\sqrt2}{2}$
$f^{(4)}(\frac\pi4)=\frac{\sqrt2}{2}$.

Hence

\noindent$P_0(x)=\frac{\sqrt2}{2}$;
$P_1(x)=\frac{\sqrt2}{2}+\frac{\sqrt2}{2}(x-\frac\pi4)$;
$P_2(x)=\frac{\sqrt2}{2}+\frac{\sqrt2}{2}(x-\frac\pi4)-
\frac{\sqrt2}{4}(x-\frac\pi4)^2$;
\par\noindent
$P_3(x)=\frac{\sqrt2}{2}+\frac{\sqrt2}{2}(x-\frac\pi4)-
\frac{\sqrt2}{4}(x-\frac\pi4)^2-\frac{\sqrt2}{12}(x-\frac\pi4)^3$.

\np$f(x)=\sqrt x$, $a=4$.
$f(x)=x^{\frac12}$; $f^{(1)}(x)=\frac12 x^{-\frac12}$;
$f^{(2)}(x)=-\frac14 x^{-\frac32}$;
$f^{(3)}(x)=\frac38 x^{-\frac52}$;
$f^{(4)}(x)=-\frac{15}{16} x^{-\frac72}$.

\noindent$f(4)=2$; $f^{(1)}(4)=\frac14$;
$f^{(2)}(4)=-\frac{1}{32}$;
$f^{(3)}(4)=\frac{3}{256}$;
$f^{(4)}(4)=-\frac{15}{2048}$.

Hence

\noindent$P_0(x)=2$;
$P_1(x)=2+\frac14(x-4)$;
$P_2(x)=2+\frac14(x-4)-\frac1{64}(x-4)^2$;
\par\noindent
$P_3(x)=2+\frac14(x-4)-\frac1{64}(x-4)^2+\frac{1}{512}(x-4)^3$.


\np$\Frac1{1+x}$.
The series $\Sum{n=0}(-1)^nx^n=\Frac1{1+x}$ so this series is the
Maclaurin series.
With more work, we can evaluate the Maclaurin series from the
definition.
First check that $f(x)=(1+x)^{-1}$ and
$f^{(n)}(x)=(-1)(-2)\cdots(-n)(1+x)^{-n-1}=(-1)^nn!(1+x)^{-n-1}$
so again the Maclaurin series is
$\Sum{n=0}\Frac{f^{(n)}(0)}{n!}x^n=
\Sum{n=0}\Frac{(-1)^nn!}{n!}x^n=\Sum{n=0}(-1)^nx^n$.
Notice that this calculation of the Maclaurin series does less than
the first argument since the first argument also shows that the
Maclaurin series converges to the function on the interval $(-1,1)$.

\def\cosh{{\rm cosh}\ }
\def\sinh{{\rm sinh}\ }

\np Given that $e^x$ has the power series $\Sum{n=0}\Frac{x^n}{n!}$,
it follows that
$e^{-x}=\Sum{n=0}\Frac{(-1)^nx^n}{n!}$ so
$\cosh x=\Frac{e^x+e^{-x}}2$ can be found by adding the two
power series to get
$\cosh x=\Sum{n=0}\Frac{x^{2n}}{(2n)!}$.
Hence this is the Maclaurin series and the series represents
the function everywhere since the same is true for $e^x$.
We can also compute the Maclaurin series via the definition as follows.
$f(x)=\cosh x$; $f^{(1)}(x)=\sinh x$;
$f^{(2)}(x)=\cosh x$.
Hence $f^{(2n+1)}(x)=\sinh x$ and $f^{(2n)}(x)=\cosh x$.
It follows that $f^{(2n)}(0)=1$ and $f^{(2n+1)}(0)=0$,
so the Maclaurin series for $\cosh x$ is
$\Sum{n=0}\Frac{x^{2n}}{(2n)!}$.

\np
The linearization at $x=a$ is always $f(a)+f^\prime(a)(x-a)$ 
and the quadratic approximation is always
$f(a)+f^\prime(a)(x-a)+\Frac{f^{\prime\prime}(a)}{2!}(x-a)^2$.
For $f(x)=\ln(\cos x)$ and $a=0$ we can compute as follows:
$f^\prime(x)=\Frac{\dd{\cos x}x}{\cos x}=\Frac{-\sin x}{\cos x}=-\tan x$;
$f^{\prime\prime}(x)=\dd{-\tan x}x=-\sec^2x$.
Hence at $0$, $f(0)=0$, $f^\prime(0)=0$ and $f^{\prime\prime}(0)=-1$.
Hence the linearization is $0$ and the quadratic approximation is
$-x^2/2$.


\end