\input homework_macros.tex

%\distiller
\printer

\def\dd#1#2{\frac{d#1}{d#2}}
\toks0={
8.8  p. 671:   3,  7, 13, 19, 29, 33, 41.
}
\def\Li{\displaystyle\lim_{n\to\infty}}
\def\Sum#1{\displaystyle\sum_{#1}^\infty}
\def\Int{\displaystyle\int}

\SN{\jobname}
\np
$\Sum{n=0} (-1)^n(4x+1)^n$ is not in the usual form, we need
$(x-b)^n$, so we factor out a $4^n$:
$\Sum{n=0} (-1)^n(4x+1)^n=\Sum{n=0} (-1)^n4^n(x+\frac14)^n=
\Sum{n=0} (-4)^n\bigl(x-(-\frac14)\bigr)^n$.
To calculate the radius of convergence, use either the radius or the
root test.
For the ratio test, compute
$\Li \Frac{\vert (-4)^{n+1}\vert R^{n+1}}{\vert (-4)^{n}\vert R^n}=4R$
and $4R=1$ or $R=\frac14$.
For the root test, compute
$\Li\root n \of{\vert (-4)^{n}\vert R^n}=4R$ so again $R=\frac14$.
At the two endpoints, we need to evaluate the the convergence
properties of 
$\Sum{n=0} (-4)^n(\frac14)^n$ and $\Sum{n=0} (-4)^n(-\frac14)^n$.
\itemitem{$\bullet$} $\Sum{n=0} (-4)^n(\frac14)^n=\Sum{n=0} (-1)^n$ 
diverges since $\Li (-1)^n$ does not exists and henceis not $0$
\itemitem{$\bullet$} $\Sum{n=0} (-4)^n(-\frac14)^n=\Sum{n=0} 1$ diverges
since $\Li 1\neq 0$

Putting this all together, the series converges absolutely for
$x$ in the interval $(-\frac12,0)$ and diverges for all $x$ not
in this open interval.

\np$\Sum{n=0}\Frac{n\>x^n}{n+2}$
To compute the radius of convergence, use the ratio test:
$\Li \Frac{\frac {n+1}{n+3} R^{n+1}}{\frac {n}{n+2} R^n}=
\Li\Frac{(n+1)(n+2)R^{n+1}}{n(n+3) R^n}=
\Li\Frac{(n+1)(n+2)}{n(n+3)} R=
\Li\Frac{(1+\frac1n)(1+\frac2n)}{1+\frac3n} R=R$
so the radius of convergence is $1$.
At the end points, we must study the series
$\Sum{n=0}\Frac{n}{n+2}$ and $\Sum{n=0}\Frac{n\>(-1)^n}{n+2}$.
Since $\Li\Frac{n}{n+2}=1$ both series diverge since the limit of
the terms of either sequence do not go to $0$.
Hence the series converges absolutely for all $x$ in the
interval $(-1,1)$ and it diverges for all other $x$.

\np$\Sum{n=0}\Frac{x^{2n+1}}{n!}$.
Use the ratio test and compute
$\Li\Frac{\frac1{(n+1)!}R^{n+1}}{\frac1{n!}R^{n}}=
\Li\Frac R n=0$ for any $R$ so the radius of convergence is infinity.
Hence the series converges absolutely everywhere.

\np$\Sum{n=0}\Frac{\sqrt n\> x^n}{3^n}$
Use the ratio test and compute
$\Li\Frac{\frac{\sqrt{n+1\>}}{3^{n+1}} R^{n+1}}
{\frac{\sqrt{n}}{3^{n}} R^{n}}=\Li\Frac{\sqrt{n+1\>}}{\sqrt n}\Frac R3=
\Li\sqrt{\frac{n+1}n}\Frac R3=\Frac R3$ 
so the radius of convergence is $3$.
At the two end points we need to study the two series
$\Sum{n=0}\Frac{\sqrt n\> 3^n}{3^n}=\Sum{n=0}\sqrt n$
and $\Sum{n=0}\Frac{\sqrt n\> (-3)^n}{3^n}=\Sum{n=0}\sqrt n(-1)^n$.
Since $\Li\sqrt n=\infty$, both the above series diverge since
the limit of the terms in the series do not go to $0$.
Hence the series converges absolutely for $x$ in $(-3,3)$ and
diverges for all other $x$.

\np$\Sum{n=1}\Frac{(4x-5)^{2n+1}}{n^{3/2}}$.
As in \#3, we first need to rewrite the series in standard form:
$\Sum{n=1}\Frac{(4x-5)^{2n+1}}{n^{3/2}}=
\Sum{n=1}\Frac{4^{2n+1}}{n^{3/2}}\Bigl(x-\frac54\Bigr)^{2n+1}$.
Compute the radius of convergence using the root test
(the ratio test can also be made to work).
$\Li\root n\of{\Frac{4^{2n+1}}{n^{3/2}}R^{2n+1}} =
\Li \Frac{4^{2+\frac1n}}{\Bigl(n^{\frac{1}{n}}\Bigr)^{3/2}} R^{2+\frac1n}$.
Since $\Li n^{\frac 1n}=1$,
$\Li\root n\of{\Frac{4^{2n+1}}{n^{3/2}}R^{2n+1}} = 4^2R^2$ so
$R^2=\frac1{4^2}$ so $R=\frac14$.
Understanding the endpoints involves understanding the series
$\Sum{n=1}\Frac{4^{2n+1}}{n^{3/2}}\Bigl(\frac14\Bigr)^{2n+1}=
\Sum{n=1}\Frac{1}{n^{3/2}}$
and
$\Sum{n=1}\Frac{4^{2n+1}}{n^{3/2}}\Bigl(-\frac14\Bigr)^{2n+1}=
\Sum{n=1}\Frac{-1}{n^{3/2}}$.
Since one of the series is the negative of the other, either both
converge absolutely or both diverge.
Since $\Sum{n=1}\Frac{1}{n^{3/2}}$ is a $p${--}series with
$p=\frac32$ and since $\frac32>1$, this $p${--}series converges.
Hence the series converges absolutely for all $x$ in the interval
$[1,\frac 32]$.

\np$\Sum{n=0}\Frac{(x-1)^{2n}}{4^n}$.
To compute the radius of convergence, use the root test
$\Li\root n\of{\Frac{R^{2n}}{4^n}}=\Frac{R^2}{4}$ so the
radius of convergence is $2$.
The two endpoint series are
$\Sum{n=0}\Frac{(2)^{2n}}{4^n}=\Sum{n=0}1$ and
$\Sum{n=0}\Frac{(-2)^{2n}}{4^n}=\Sum{n=0}1$
so both series diverge as the terms do not go to $0$.
To actually sum the series note that it is a geometric series
$\Sum{n=0}\Frac{(x-1)^{2n}}{4^n}=
\Sum{n=0}\Bigl(\Frac{(x-1)^{2}}{4}\Bigr)^n=
\Frac{1}{1-{\frac{(x-1)^2}{4}}}=
\Frac{4}{4-(x^2-2x+1)}=\Frac{4}{3+2x-x^2}$

\np$\sin x=x-\Frac{x^3}{3!}+\Frac{x^5}{5!}-\Frac{x^7}{7!}
+\Frac{x^9}{9!}-\Frac{x^{11}}{11!}+\cdots$
\par\noindent a) Since $\dd{\sin x}x=\cos x$ and we can
differentiate the series term by term, so
$\cos x=1-\Frac{3x^2}{3!}+\Frac{5x^4}{5!}-\Frac{7x^6}{7!}
+\Frac{9x^8}{9!}-\Frac{11x^{10}}{11!}+\cdots$ or
$\cos x=
1-\Frac{x^2}{2!}+\Frac{x^4}{4!}-\Frac{x^6}{6!}
+\Frac{x^8}{8!}-\Frac{x^{10}}{10!}+\cdots$

\par\noindent b)
$\sin(2x)=2x-\Frac{(2x)^3}{3!}+\Frac{(2x)^5}{5!}-\Frac{(2x)^7}{7!}
+\Frac{(2x)^9}{9!}-\Frac{(2x)^{11}}{11!}+\cdots$ or
$\sin(2x)=2x-\Frac{8x^3}{3!}+\Frac{32x^5}{5!}-\Frac{128x^7}{7!}
+\Frac{512x^9}{9!}-\Frac{2048x^{11}}{11!}+\cdots$

\par\indent c)
Since $\sin(2x)=2\cos x\sin x$ we can use the product theorem
to compute $$\eqalign{
\cos x \sin x=\ &
(1-\Frac{x^2}{2!}+\Frac{x^4}{4!}-\Frac{x^6}{6!}
+\Frac{x^8}{8!}-\Frac{x^{10}}{10!}+\cdots)\cdot(
x-\Frac{x^3}{3!}+\Frac{x^5}{5!}-\Frac{x^7}{7!}
+\Frac{x^9}{9!}-\Frac{x^{11}}{11!}+\cdots)\cr
=\ &x-\Frac{x^3}{2!}+\Frac{x^5}{4!}-\Frac{x^7}{6!}
+\Frac{x^9}{8!}-\Frac{x^{11}}{10!}+\cdots\cr
&\hphantom{x-}-\Frac{x^3}{3!}+\Frac{x^5}{3!\cdot2!}-\Frac{x^7}{3!\cdot4!}+
\Frac{x^{9}}{3!\cdot6!}-\Frac{x^{11}}{3!\cdot8!}+\cdots\cr
&\hphantom{x-\Frac{x^3}{3!}--}
\Frac{x^5}{5!}-\Frac{x^7}{5!\cdot2!}+\Frac{x^9}{5!\cdot4!}-
\Frac{x^{11}}{5!\cdot6!}+\cdots\cr
&\hphantom{x-\Frac{x^3}{3!}+\Frac{x^5}{5!}-}
-\Frac{x^7}{7!}+\Frac{x^9}{7!\cdot2!}-\Frac{x^{11}}{7!\cdot4!}+
\cdots\cr
&\hphantom{x-\Frac{x^3}{3!}+\Frac{x^5}{5!}-\Frac{x^7}{7!}---}
\Frac{x^9}{9!}-\Frac{x^{11}}{9!\cdot2!}+\cdots\cr
&\hphantom{x-\Frac{x^3}{3!}+\Frac{x^5}{5!}-\Frac{x^7}{7!}+\Frac{x^9}{9!}--}
-\Frac{x^{11}}{11!}-\cdots\cr
&\vdots\cr
=\ & x-\Biggl({3\choose 1}+{3\choose 3}\Biggr)\Frac{x^3}{3!}+
\Biggl({5\choose 1}+{5\choose 3}+{5\choose 5}\Biggr)\Frac{x^5}{5!}
-\Biggl({7\choose 1}+{7\choose 3}+{7\choose 5}+
{7\choose 7}\Biggr)\Frac{x^7}{7!}\cr
&\Biggl({9\choose 1}+{9\choose 3}+{9\choose 5}+{9\choose 7}+
{9\choose 9}\Biggr)\Frac{x^9}{9!}\cr
&-\Biggl({11\choose 1}+{11\choose 3}+{11\choose 5}+
{11\choose 7}+{11\choose 9}+{11\choose 11}\Biggr)
\Frac{x^{11}}{11!}+\cdots\cr
=\ &
x - \Frac{4x^3}{3!}+\Frac{16x^5}{5!}-\Frac{64x^7}{7!}
+\Frac{256x^9}{9!}-\Frac{1024x^{11}}{11!}+\cdots\cr
}$$

One can now easily verify that as far as we have calculated, the
series for $\sin(2x)$ and that for $2\cos x\sin x$ agree.

\end