\input homework_macros.tex

%\distiller
\printer

\def\smfrac#1#2{{\scriptscriptstyle#1}\over{\scriptscriptstyle#2}}
\def\dd#1#2{\frac{d#1}{d#2}}
\toks0={
8.11 p. 697:   1,  5, 21, 23, 43, 53.
}
\def\Li{\displaystyle\lim_{n\to\infty}}
\def\Sum#1{\displaystyle\sum_{#1}^\infty}
\def\Int{\displaystyle\int}

\SN{\jobname}
\np
Write the first four terms of the binomial series for
$(1+x)^{\smfrac12}=
\Sum{n=0}{{\smfrac12}\choose n}x^n$.
The first four terms are
$1+\frac12 x+\Frac{\frac12(\frac12-1)}{2!}x^2+
\Frac{\frac12(\frac12-1)(\frac12-2)}{3!}x^3
$, or
$1+\frac12 x-\frac{1}{8}x^2+\frac{1}{16}x^3$.

\np
Write the first four terms of the binomial series for
$(1+\frac x2)^{-2}=\Sum{n=0}{{-2}\choose n}\Bigl(\frac x2\Bigr)^n$.
The first four terms are
$1-2\ \Frac x2+\Frac{-2(-2-1)}{2!}\Bigl(\Frac x2\Bigr)^2 +
\Frac{-2(-2-1)(-2-2)}{3!}\Bigl(\Frac x2\Bigr)^3
$, or
$1-x+\frac34 x^2-\frac12 x^3$.

\np
Solve $y^\prime-xy=0$ with $y(0)=1$.
We are looking for a power series solution,
$y=\Sum{n=0}a_n x^n$.
Then $y^\prime=\Sum{n=1}na_n x^{n-1}=\Sum{n=0}(n+1)a_{n+1}x^n$ and
$xy=\Sum{n=0}a_nx^{n+1}$
so $y^\prime -x y=a_1+ \Sum{n=1} \bigl((n+1)a_{n+1}-a_{n-1}\bigr)x^n$.
The only way a power series can vanish is if all its coefficients
vanish, so $a_1=0$ and the recursion relation
$(n+1)a_{n+1}-a_{n-1}=0$ holds.
The initial value condition $y(0)=1$ shows $a_0=1$.
The recursion relation can be rewritten
$(k+2)a_{k+2}=a_k$, or
$a_{k+2}=\frac{a_k}{k+2}$.
Since $a_1=0$, $a_{2k+1}=0$ for all integers $k\geq 0$.
Since $a_0=1$, $a_{2k}=\Frac{1}{2^k k!}$ for all integers $k\geq 0$.
Hence $y=\Sum{n=0}\Frac{x^{2n}}{2^nn!}$ is the solution.
One can see easily that
$e^{\smfrac{x^2}2}=\Sum{n=0}\Frac{x^{2n}}{2^nn!}$ as well.
$y=e^{\smfrac{x^2}2}$ is the answer you will get if you apply
your technique for solving first order linear equations to this
problem.

\np
Solve$(1-x)y^\prime-y=0$ with $y(0)=2$.
Same yoga as the last problem.
We are looking for a power series solution,
$y=\Sum{n=0}a_n x^n$.
Then $y^\prime=\Sum{n=1}na_n x^{n-1}=\Sum{n=0}(n+1)a_{n+1}x^n$.
Then $(1-x)y^\prime=y^\prime-xy^\prime=
\Sum{n=0}(n+1)a_{n+1}x^n-\Sum{n=0}(n+1)a_{n+1}x^{n+1}=
a_1+\Sum{n=1} \bigl((n+1)a_{n+1}-na_n\bigr)x^n$ and
$(1-x)y^\prime-y=
a_1-a_0+\Sum{n=1} \bigl((n+1)a_{n+1}-na_n-a_n\bigr)x^n
=a_1-a_0+\Sum{n=1} (n+1)(a_{n+1}-a_n)x^n$.
In order for this power series to vanish,
$a_1-a_0=0$ and $a_{n+1}-a_n=0$ for all $n\geq1$.
Hence $a_0=a_1=a_2=\cdots=a_n=\cdots$.
From the initial value condition $y(0)=2$ we see $a_0=2$,
so $y=\Sum{n=0}2x^n$ is our solution.
Note that this series is a geometric series so
$y=\Frac2{1-x}$ is another expression for the solution.



\np
Find a polynomial that will approximate the function 
$F(x)=\Int_0^x\sin t^2\> dt$ on the interval $[0,1]$ with an error
of magnitude less than $10^{-3}$.
First write down an exact power series solution:
$\sin x=\Sum{n=0}\Frac{(-1)^nx^{2n+1}}{(2n+1)!}$ so
$\sin t^2=\Sum{n=0}\Frac{(-1)^n(t^2)^{2n+1}}{(2n+1)!}=
\Sum{n=0}\Frac{(-1)^n(t)^{4n+2}}{(2n+1)!}$ and
$F(x)=\Int_0^x\sin t^2\> dt=
\Sum{n=0}\Frac{(-1)^n\Frac{(t)^{4n+3}}{4n+3}\Big\vert^x_0}{(2n+1)!}=
\Sum{n=0}\Frac{(-1)^n x^{4n+3}}{(4n+3)\cdot(2n+1)!}$.

On the interval $[0,1]$ the series is alternating because $x^{4n+3}\geq0$
and one can see that for a fixed $x$ in this interval
$\Frac{x^{4n+3}}{(4n+3)\cdot(2n+1)!}$ decreases to $0$.
Hence the estimate associated to the Alternating Series Test applies.

This says that the polynomial of degree $m$,
$\displaystyle\sum^m_{n=0}\Frac{(-1)^n x^{4n+3}}{(4n+3)\cdot(2n+1)!}$
has the property that
$\Big\vert F(x)-
\displaystyle\sum^m_{n=0}\Frac{(-1)^n x^{4n+3}}{(4n+3)\cdot(2n+1)!}
\Big\vert\leq\Frac{x^{4m+7}}{(4m+7)\cdot(2m+3)!}
$ for every $x$ in the interval $[0,1]$.
Since $x^{4m+7}$ increases as $x$ does,
$\Big\vert F(x)-
\displaystyle\sum^m_{n=0}\Frac{(-1)^n x^{4n+3}}{(4n+3)\cdot(2n+1)!}
\Big\vert\leq\Frac{1}{(4m+7)\cdot(2m+3)!}$
and we may choose any $m$ so that
$\Frac{1}{(4m+7)\cdot(2m+3)!}<10^{-3}$.
When $m=1$, $(4m+7)\cdot(2m+3)!=11\cdot 5!=11\cdot 120=1320>1000$.
Hence the polynomial
$\Frac{x^3}3-\Frac{x^7}{42}$
approximates $F(x)$ to within $10^{-3}$ on the interval $[0,1]$.
The answer in the back of the book is not completely correct.
If one is approximating  $F(x)$ one would like to see a polynomial
in $x$, not $t$.
The book is not wrong to include an extra term.
Once the polynomial of degree $m$ works, so does the polynomial
of degree $m+1$.


\def\Lim{\displaystyle\lim}
\np
Find $\Lim_{x\to \infty}x^2(e^{^{-\smfrac1{x^2}}}-1)$.
Since $e^x=\Sum{n=0}\Frac{x^n}{n!}$
$e^{^{-\smfrac1{x^2}}}=\Sum{n=0}\Frac{(-1)^n}{n!\ x^{2n}}$
so $e^{^{-\smfrac1{x^2}}}-1=\Sum{n=1}\Frac{(-1)^n}{n!\ x^{2n}}$
and hence
$x^2\bigl(e^{^{-\smfrac1{x^2}}}-1\bigr)=
\Sum{n=1}\Frac{(-1)^n}{n!\ x^{2n-2}}=
\Sum{n=0}\Frac{(-1)^{n-1}}{(n+1)!\ x^{2n}}$.
As $x\to\infty$, $\Frac{1}{x^2}$ goes to $0$ so only
the degree $0$ term is left in the limit and that term
is $-1$.

Hence $\Lim_{x\to \infty}x^2(e^{^{-\smfrac1{x^2}}}-1)=-1$.
\end