\magnification1200\footline{}
\let\ds=\displaystyle

\def\frac#1#2{{{#1}\over{#2}}}
\def\ll{\bigskip\noindent\vrule width\hsize height .1pt depth 0pt
\medskip}
\def\LL{\bigskip\noindent\vrule width\hsize height .1pt depth 0pt
\vskip-8pt
\noindent\vrule width\hsize height .1pt depth 0pt
\medskip}

\long\def\Quiz#1\eq#2\ans#3\ea{%
\noindent\vbox{#1\ll\par
\centerline{\bf Solution}\par#3\LL
}\vfil\null\par}

\long\def\quiz#1\eq{
\noindent\vbox{#1\ll}
}
\long\def\ans#1\ea{
\noindent\vbox{\centerline{\bf Solution}\par#1\LL}
}

\Quiz
\centerline{\bf Math.~126 Quiz \#1}
\centerline{January  23, 2001}
\vskip10pt
\noindent
Consider the region below the curve $y=\cos x$
between $-\frac\pi2$ and $\frac\pi2$ and above
the $x${--}axis.
Assume the density is constant, $\delta$.
\item{a)} Find the mass of
this region.
\item{b)}
Write a definite integral whose value will be the moment
about the $y${--}axis of this region.
Give a short reason why the moment about the $y${--}axis
of this region is $0$ even though you do not yet know 
how to do the integral.
\item{c)}
Write a definite integral whose value will be the moment
about the $x${--}axis of this region.
You should realize that we have talked about evaluating this
integral, but don't do it today.
\item{d)}
Write a definite integral whose value will be the volume obtained
by rotating this region about the $x${--}axis.
Use the disk method from last semester.

\bigskip\noindent{\bf Remark for after the quiz.}
The proportionality relation between the moment about the $x${--}axis and
the volume of the solid obtained by rotation about the $x${--}axis
is true in general and goes back to Pappus of Alexandria around 300AD.
\eq
\ans
\item{(a)}$\ds Mass=\delta \int_{-\frac\pi2}^{\frac{\pi}2} \cos x\ dx=
\delta(-\sin x)\Big\vert_{-\frac\pi2}^{\frac{\pi}2}=
\delta\bigl( 1-(-1)\bigr)=2\delta
$

\item{(b)}$\ds Moment_y=\delta \int_{-\frac\pi2}^{\frac{\pi}2} x\cos x\ dx=
0$ since the region is symmetric about the $y${--}axis.

\item{(c)}$\ds Moment_x=
\delta \int_{-\frac\pi2}^{\frac{\pi}2} \frac12\cos^2 x\ dx
$
\item{}The integral can be evaluated as follows:$\ds \delta 
\int_{-\frac\pi2}^{\frac{\pi}2} \frac14\bigl(1+\cos(2x)\bigr)
=\frac{\delta}4 
x-\frac12\sin 2x\Big\vert_{-\frac\pi2}^{\frac{\pi}2}=\frac{\pi\delta}4
$
\item{(d)}$\ds Volume=\pi\int_{-\frac\pi2}^{\frac\pi2}
\cos^2 x\ dx
$
\ea
%%%%%%%%%%%%%%%%%%%%%%
\Quiz
\centerline{\bf Math.~126 Quiz \#2}
\centerline{January  30, 2001}
\vskip10pt
\noindent

It is true that 
for positive integers $2$, $3$, $4$, \dots\ the following holds.

$$1/2 + 1/3 + \cdots + 1/n  < \ln n  < 1 + 1/2 + 1/3 + \cdots + 1/n
\leqno(\ast)$$

\item{1.}Explain why.

\item{2.}Given that $3^5=243$ and that
$5.0 < 1/2 + 1/3 + \cdots+ 1/243$
use $(\ast)$ with $n=3^5$ and the laws of logarithms to argue $e<3$.
\eq
\ans
\noindent
\def\F#1/#2{\frac{#1}{#2}}
\item{1.}By definition $\ln n=\ds\int_1^n\frac{dx}{x}$.
Recall $\F1/2 + \F1/3 + \cdots + \F1/n$ is the right{--}hand
Riemann sum for the partition of $[1,n]$ into $n$ pieces of
length $1$:
$f^\prime(x)=\frac{-1}{x^2}<0$.
$1+\F1/2 + \F1/3 + \cdots + \F1/{n-1}$ is the left{--}hand
Riemann sum for the partition of $[1,n]$ into $n$ pieces of
length $1$.

\item{}The graph of $f(x)=\frac1x$ is decreasing
since $f^\prime(x)=\frac{-1}{x^2}<0$, so the right{--}hand
Riemann sum is less than the integral and the left hand
Riemann sum is greater.
Hence \par
$\F1/2 + \F1/3 + \cdots + \F1/n<\ln n<
1+\F1/2 + \F1/3 + \cdots + \F1/{n-1}<
1+\F1/2 + \F1/3 + \cdots + \F1/n$.
\bigskip
\item{2.}
Using the facts given, we see
$5.0<\F1/2 + \F1/3 + \cdots+ \F1/{3^5}<\ln 3^5$.
Hence $5.0<\ln 3^5=5\cdot\ln 3$ and therefore
$1<\ln 3$.
Since $\ln e=1$ we see $\ln e< \ln 3$
and since $\ln$ is an increasing function
($\frac{d\> \ln x}{dx}=\frac{1}{x}>0$)
$e<3$.
\ea
%%%%%%%%%%%%%%%%%%%%%%
\Quiz
\centerline{\bf Math.~126 Quiz \#3}
\centerline{February  6, 2001}
\vskip10pt
\noindent

\item{1.}Solve $6^{5k}=4$. Leave your answer as a quotient
involving numbers and natural logs of numbers.
\vfill

\item{2.}Compute ${{\textstyle dy}\over{\textstyle dx}}$
where $y=x^{e^x}\ .$
\eq
\ans
\vskip2pt
\item{1.}$\ln 6^{5k}=\ln 4$ so $5k\ln 6=\ln4$ or
$\ds k=\frac{\ln4}{5\ln 6}$
\item{2.} Rewrite $y=x^{e^x}=e^{e^x \ln x}$ so
$\ds\frac{dy}{dx}=\bigl(e^{e^x \ln x}\bigr)\frac{d\>{e^x \ln x}}{dx}$.

\noindent
$\frac{d\>{e^x \ln x}}{dx}=
e^x\ln x+e^x\frac1x$, so

$\ds\frac{dy}{dx}=x^{e^x}\bigl(
e^x\ln x+\frac{e^x}x\bigr)$
\ea
%%%%%%%%%%%%%%%%%%%%%%
\Quiz
\centerline{\bf Math.~126 Quiz \#4}
\centerline{February  13, 2001}
\vskip10pt
\noindent
Solve the initial value problem
$$x{{d y}\over{d x}} =
x^2{\cos x} -y$$
$$y(\pi)=1$$
\eq
\ans
First put it in standard form: $\ds {{d y}\over{d x}} +{{y}\over{x}}=
x\cos x$

\noindent
Then $\ds P={{1}\over{x}}$ and
$\ds Q=x\cos x$.
Compute $\ds \int P dx=\ln x+C$ so we may take 
$v=e^{\ln x}=x$.
Then $\ds \int vQ dx=\int {\cos x}\> dx=\sin x +C$.

\noindent
Hence $\ds y={{1}\over{v}}\int vQ dx=
{{\sin x}\over{x}}+ {{C}\over{x}}$
so $\ds y(\pi)={{\sin \pi}\over{\pi}}+ {{C}\over{\pi}}=
{{C}\over{\pi}}=1$.
Hence $C=\pi$ and
$$y={{\sin x}\over{x}}+ {{\pi}\over{x}}$$
\ea
%%%%%%%%%%%%%%%%%%%%%%
\Quiz
\centerline{\bf Math.~126 Quiz \#5}
\centerline{February  27, 2001}
\vskip10pt
\noindent

\noindent Evaluate the integral
$$\int \sin \bigl(\sqrt x\>\bigr)\> dx \ .$$

\noindent{\bf Hint:} First do a substitution and
then an integration by parts.
\eq
\ans
Substitute $w=\sqrt x$: then 
$\ds dw =\frac{dx}{2\sqrt x}$, so
$2\sqrt x\ dw=dx$ and $2w dw=dx$.
Hence \hfill\par
$\ds\int \sin \bigl(\sqrt x\>\bigr)\> dx=2\int w\sin(w)\>dw$

Parts:
$\ds \matrix{
u=w &du=dw\cr
dv=\sin(w) dw&v=-\cos(w)\cr
}$ so

$$\eqalign{%
2\int w\sin(w)\>dw=&
-2w\cos w+2\int \cos w\> dw\cr
=&
-2w\cos w+2\sin w + C \cr
}$$
Finally
$$\int \sin \bigl(\sqrt x\>\bigr)\> dx =
-2\sqrt{x}\cos{\sqrt{x}}+2\sin{\sqrt x}+C
\ .$$
\ea
%%%%%%%%%%%%%%%%%%%%%%
\Quiz
\centerline{\bf Math.~126 Quiz \#6}
\centerline{March  6, 2001}
\def\frac#1#2{{{\textstyle#1}\over{\textstyle#2}}}
Expand $$\frac{2x^3-2x^2+3x+1}{(x^2-x-2)(x^2+1)}$$
as a sum of partial fractions.
\vskip-20pt
\eq
\ans
$$\frac{2x^3-2x^2+3x+1}{(x^2-x-2)(x^2+1)}=\frac A{(x-2)}+\frac B{(x+1)}+
\frac{Cx+D}{(x^2+1)}$$
$$2x^3-2x^2+3x+1=A(x+1)(x^2+1)+B(x-2)(x^2+1)+(Cx+D)(x+1)(x-2)$$

\noindent{\bf Equate coefficients:}
$$\eqalign{x^3:&
2=A+B+C\cr
x^2:&
-2=A-2B+D-C\cr
x^1:&
3=A+B-2C-D\cr
x^0:&
1=A-2B-2D\cr
}$$
From $x^3$: $A=2-B-C$ so the other equations become
\vskip-10pt
$$\eqalign{
x^2:\quad & -2=2-B-C-2B+D-C\cr
& -4=-3B-2C+D\cr
x^1:\quad & 3=2-B-C+B-2C-D\cr
& 1=-3C-D\cr
x^0:\quad & 1=2-B-C-2B-2D\cr
& -1=-3B-2D\cr
}$$
From $x^1$: $D=-3C-1$ so
\vskip-10pt
$$\eqalign{
x^2:\quad & -4=-3B-2C+-3C-1\cr
&-3=-3B-5C\cr
x^0:\quad & -1=-3B-2(-3C-1)\cr
&-3=-3B+6C\cr
}$$
It follows that $C=0$ and $B=1$,
whence $D=-1$ and $A=1$.

\noindent{\bf Plug in:}\par
$x=2:
16-8+6+1=A(3)(5)
$
or
$15=15A$ or $A=1\ .$


$x=-1:
-2-2-3+1=B(-3)(2)$
or
$-6=(-6)B$ or $B=1\ .$

$x=0:
1=A+(-2)B+(-2)D
$
or
$1=-1+(-2)D$ or $D=-1\ .$

$x=1:
2-2+3+1=A(2)(2)+B(-1)(2)+(C+D)(2)(-1)
$
or
$4=4-2-2(C-1)$ or 
$2=2-2C$
or
$C=0\ .$
\vskip-20pt
\ea
%%%%%%%%%%%%%%%%%%%%%%
\Quiz
\centerline{\bf Math 126, Quiz \#7}
\centerline{March 27, 2001}
\vskip10pt
\noindent

\def\frac#1#2{{{\textstyle#1}\over{\textstyle#2}}}
\def\Strut{\vrule width0pt height9pt depth0pt}

\noindent
Which improper integrals below converge and which diverge?
A brief indication of your reasoning should be given.
\itemitem{a)}$\displaystyle\int_0^\infty e^{-x^3}\ dx$

\itemitem{b)}$\displaystyle\int_0^\infty \frac{1}{\root 3 \of{\Strut x^2+1}}\ dx$
\eq
\ans
\item{a)}\vskip-10pt
$\ds\int_0^\infty e^{-x^3}\ dx$ converges if and only if
$\ds\int_1^\infty e^{-x^3}\ dx$ converges.

On the interval $[1,\infty)$, $x\geq1$ so $x^2\geq 1$ and
$x^3\geq x$ so
$e^x\leq e^{x^3}$ so $e^{-x^3}\leq e^{-x}$.
Now
$\ds\int_1^\infty e^{-x}\>dx$ converges since
$\ds\lim_{t\to\infty}\int_1^t e^{-x}\>dx=
\lim_{t\to\infty} -e^{-x}\
\Big\vert_1^t=e^{-1}-\lim_{t\to\infty}e^{-t}=e^{-1}-0$.
By the first comparison test for improper integrals, 
$\ds\int_1^\infty e^{-x^3}\ dx$ converges and hence
so does $\ds\int_0^\infty e^{-x^3}\ dx$.
\item{b)}
Roughly speaking $\frac{1}{\root 3 \of{\Strut x^2+1}}$ behaves
near $\infty$ like $x^{-2/3}$.
More precisely,\hfil\par\noindent
$\ds\lim_{x\to\infty}
\frac{\frac{1}{\root 3 \of{\Strut x^2+1}}}{x^{-2/3}}=
\ds\lim_{x\to\infty}
\frac{x^{2/3}}{\root 3 \of{\Strut x^2+1}}=
\ds\lim_{x\to\infty}
\frac{1}{\root 3 \of{\Strut 1+\frac{1}{x^2}}}=1$.
Since $0<1<\infty$ we would like to use the limit comparison test,
comparing the integral for $\frac{1}{\root 3 \of{\Strut x^2+1}}$
with the one for $x^{-2/3}$.
Annoyingly, $x^{-2/3}$ has an additional singularity at $0$ so we
proceed as follows.

$\ds\int_0^\infty \frac{1}{\root 3 \of{\Strut x^2+1}}\ dx$ converges
if and only if 
$\ds\int_1^\infty \frac{1}{\root 3 \of{\Strut x^2+1}}\ dx$ converges
and by the limit comparison test for improper integrals,
$\ds\int_1^\infty \frac{1}{\root 3 \of{\Strut x^2+1}}\ dx$ converges
if and only if
$\ds\int_1^\infty x^{-2/3}\ dx$ converges.

But $\ds\int_1^\infty x^{-2/3}\ dx=
\lim_{t\to\infty} \frac{x^{1/3}}{1/3}\ \Big\vert_1^t
\lim_{t\to\infty} 3-3t^{1/3}=3-\infty
$
so $\ds\int_1^\infty x^{-2/3}\ dx$ diverges and hence so does
$\ds\int_0^\infty \frac{1}{\root 3 \of{\Strut x^2+1}}\ dx$.
\ea

\Quiz
\centerline{\bf Math.~126 Quiz \#8}
\centerline{April 3, 2001}
\vskip10pt
\noindent

\def\frac#1#2{{{\textstyle#1}\over{\textstyle#2}}}
\def\isum#1{{\displaystyle{\sum_{n={#1}}^{\infty}}}\>}
\def\sL#1{\displaystyle\lim_{#1\to\infty}}
\def\frac#1#2{{{#1}\over{#2}}}
\noindent
For each of the series below, do two things.
First compute $\sL{n} a_n$ and then use this
calculation to say if you are {\bf certain} that 
$\isum{1} a_n$ diverges or if the limit calculation
does not suffice to say if the series converges
or diverges.
Just circle {\bf diverges} or {\bf can not tell}
after the series for the second part of each question
\def\ans{\vskip10pt
\itemitem{a.} $\sL{n} a_n=$
\vskip30pt
\itemitem{b.}$\isum{1} a_n$\hskip10pt
{\bf diverges} \hskip 40pt {\bf can not tell}
}
\bigskip
\item{A.}$a_n=\frac1{\sqrt n}$:
\ans
\vfill
\item{B.}$a_1=0$ and $a_n=a_{n-1}+1$ for all $n\geq2$:
\eq
\ans
\item{A.}
$\sL{n}\frac{1}{\sqrt{n}}=\sqrt{\sL{n}\frac{1}{{n}}}=\sqrt 0=0$.
Since the terms go to $0$, we can not tell if the series diverges
or converges.
We know from our later work that this is a $p${--}series with
$p=\frac12\leq 1$ so it does diverge, but not because of this calculation.

\item{B.}
Since $a_n>a_{n-1}$ the series is increasing so either $\sL{n} a_n=\infty$
or it exists.
In either case, $\sL{n}a_{n}=\sL{n}a_{n+1}=\sL{n} a_n+1$, which implies
$\sL{n} a_n=\infty$.
A second way to do this calcualtion is to observe that
$a_n=n-1$ and hence $\sL{n} a_n=\infty$.
In any case $\sL{n} a_n\neq 0$ so we are sure that this series diverges.
\ea

\Quiz
\centerline{\bf Math.~126 Quiz \#9}
\centerline{April 10, 2001}
\vskip10pt
\noindent
\def\frac#1#2{{{#1}\over{#2}}}

Show that $\displaystyle\sum_{n=1}^\infty\
\frac{(-1)^{n+1}}{\sqrt{n^2+1}}$ converges
by {\sl checking\/} the hypotheses of the
Alternating Series Test.

Then show the calcualtions needed to find an $m$
such that 
$$\displaystyle0<\sum_{n=1}^\infty\
\frac{(-1)^{n+1}}{\sqrt{n^2+1}}-
\sum_{n=1}^m\
\frac{(-1)^{n+1}}{\sqrt{n^2+1}}<0.01$$
\eq\ans
For the Alternating Series Test,
$a_n=\frac{(-1)^{n+1}}{\sqrt{n^2+1}}=(-1)^{n+1} u_n$
with $u_n=\frac{1}{\sqrt{n^2+1}}>0$, so the series is
alternating.
The limit $\displaystyle\lim_{n\to\infty}\frac{1}{\sqrt{n^2+1}}=0$
so the terms go to $0$.
Finally we check that $u_{n+1}<u_{n}$: let
$f(x)=\frac{1}{\sqrt{x^2+1}}$ so $u_n=f(n)$.
Compute $f^\prime(x)=-\frac32 (x^2+1)^{-3/2}(2x)$
and $f^\prime(x)>0$ for $x>0$. 
Hence $f$ is decreasing on the interval $[0,\infty)$
so $u_{n+1}=f(n+1)>f(n)=u_n$.

Since the difference indicated is positive, $m$ must be odd
and for odd $m$
$$\displaystyle0<\sum_{n=1}^\infty\
\frac{(-1)^{n+1}}{\sqrt{n^2+1}}-
\sum_{n=1}^m\
\frac{(-1)^{n+1}}{\sqrt{n^2+1}}< u_{m+1}$$
Hence any $m$ odd such that $u_{m+1}\leq 0.01=\frac{1}{100}$
will suffice:
$\frac{1}{\sqrt{m^2+1}}\leq\frac1{100}$ or
$\frac{1}{m^2+1}\leq\frac1{10^4}$, or
$10^4\leq m^2+1$.
The smallest integer satisfying this inequality is $m=100$ so
$m=101$ is the smallest integer 
satisfying both our requirements.
\ea
\Quiz
\centerline{\bf Math.~126 Quiz \#10}
\centerline{April 17, 2001}
\vskip10pt
\noindent

\def\frac#1#2{{{#1}\over{#2}}}
\def\isum#1{\displaystyle{\sum_{n={#1}}^{\infty}}\>}

\noindent
Compute the interval of convergence of each of the
following power series.
For each series, indicate where the convergence is
conditional and where it is absolute.

\itemitem{a.}$\isum0 \frac{x^{2n}}{7^n}$


\itemitem{b.}$\isum0 \frac{x^{2n}}{(n+1)7^n}$
\eq
\ans
First compute the radii of convergence.
For a. we calculate 
$\displaystyle\lim_{n\to\infty}\root{n}\of{\frac{x^{2n}}{7^n}}=
\lim_{n\to\infty}\frac{x^{2}}{7}$ so $\frac{R^2}{7}=1$ or $R=\sqrt7\,$.
For b. we calculate 
$\displaystyle\lim_{n\to\infty}\root{n}\of{\frac{x^{2n}}{(n+1) 7^n}}=
\lim_{n\to\infty}\frac{1}{\root n\of{n+1}}\cdot\frac{x^{2}}{7}=
\frac{x^{2}}{7}$
so again$R=\sqrt7\,$.

At the endpoints in a. the series to consider are
$\isum0 \frac{(\pm\sqrt{7\,}^{2n}}{7^n}=\isum0 1$ and
both of these series diverge.
At the endpoints in part b. the series to consider are
$\isum0 \frac{(\pm\sqrt{7\,}^{2n}}{(n+1)7^n}=\isum0 \frac{1}{n+1}$.
Both of these series are the harmonic series and thus diverge.
Therefore, in both cases, the interval of convergence is
$(-\sqrt 7, \sqrt 7)$.

By our theory, the convergence is absolute on the open interval (always
the case for a power series) so in both cases the convergence is 
absolute on $(-\sqrt 7, \sqrt 7)$ and divergent elsewhere.
Neither series conditionally converges anywhere.
\ea

\Quiz
\centerline{\bf Math.~126 Quiz \#11}
\centerline{May 1, 2001}
\vskip10pt
\noindent

\def\frac#1#2{{{\textstyle#1}\over{\textstyle#2}}}
\def\isum#1{{\displaystyle{\sum_{n={#1}}^{\infty}}}\>}

\noindent

The MacLaurin series for $\sin x$ is
$$\sin x = \isum1 \frac{(-1)^nx^{2n+1}}{(2n+1)!}$$

\itemitem{a.} What is the radius of convergence of this series?
\itemitem{}(Just an answer is sufficient - no reason need be given.)

\itemitem{b.}
Write down the MacLaurin series for $\frac{\sin x}{x}$.
Give a reason why it is the MacLaurin series.

\itemitem{c.} Write down the MacLaurin series for the
function
$\displaystyle\int_0^x\frac{\sin t^2}{t^2}\> dt$.
\eq
\ans
\item{a.}
The radius of convergence of the MacLaurin series for $\sin x$ is $\infty$.
If you want to check this:
$$\eqalign{
\ds\lim_{n\to\infty}
\frac{\frac{x^{2(n+1)+1}}{(2(n+1)+1)!}}{\frac{x^{2n+1}}{(2n+1)!}}=&
\lim_{n\to\infty}\frac{\frac{x^{2n+3}}{(2n+3)!}}{\frac{x^{2n+1}}{(2n+1)!}}=
\lim_{n\to\infty}x^2\cdot\frac{(2n+1)!}{(2n+3)!}\cr=&
\lim_{n\to\infty}\frac{x^2}{(2n+3)(2n+2)}=0\cr
}$$

\item{b.}
$\ds\frac{\sin x}{x} = \isum1 \frac{(-1)^nx^{2n}}{(2n+1)!}$
is a valid equation since we are just dividing an equality
(the MacLaurin series for $\sin x$) by $x$.
Hence we have a power series centered at $0$ for $\frac{\sin x}{x}$ 
and this {\sl must\/} be the MacLaurin series because a power series
is its own Taylor series.

\noindent{\bf Remark:}
To be completely precise, we should have written the function
$\frac{\sin x}{x}$ as
$$f(x)=\cases{\frac{\sin x}{x}&$x\neq0$\cr 1&$x=0$\cr}$$
We know from 125 that $f(x)$ is continuous at $0$.
The power series certainly represents $f(x)$ if $x\neq 0$ 
and equality at $x=0$ follows by evaluating the power series
at $0$.
Hence the power series represents $f(x)$ everywhere and is
therefore its MacLaurin series.
Notice that as a side benefit we have shown that $f(x)$ is infinitely
differentiable (this being obvious except at $x=0$).

\item{c.}
We are actually using the function $f(x)$ from the remark but
we will write it as $\frac{\sin x}{x}$.
Using term{--}for{--}term integration,
$$\displaystyle\int_0^x\frac{\sin t^2}{t^2}\> dt=
\isum1 \frac{(-1)^nx^{2n+1}}{(2n+1)\cdot(2n+1)!}
$$
is a valid equation and hence the series on the right is the
MacLaurin series for the function on the left.

\ea
\end
