\magnification1200\footline{}

\def\frac#1#2{{{#1}\over{#2}}}
\def\equiz{\par\noindent\vrule width\hsize height 0pt depth .1 pt
\par}
\def\leaderfill{\leaders\hbox to .5em{\hss.\hss}\hfil}

\def\dquiz{\par\noindent\hbox to \hsize{\leaderfill}
\par\medskip}
\long\def\stp\centerline#1 \centerline#2{%
\par\bigskip\noindent{#1} \qquad{ #2}\par\medskip\noindent}

%%%%%%%%%%%%%%%
%################
\def\frac#1#2{{{#1}\over{#2}}}
\def\sL#1{\displaystyle\lim_{#1\to\infty}}
\def\isum#1{{\displaystyle{\sum_{n={#1}}^{\infty}}}\>}
\stp
\centerline{\bf Math.~126 Quiz \#11}
\centerline{April 30, 2002}
\vskip10pt
\noindent

The MacLaurin series for $\cos x$ is
$$\cos x = \isum0 \frac{(-1)^{n}x^{2n}}{(2n)!}$$

\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 $\displaystyle\frac{\cos x -1}{x^2}$.
Give a reason why it is the MacLaurin series.

\itemitem{c.} Write down the MacLaurin series for the
function
$\displaystyle\int_0^x\ \cos t^2\> dt$.

\dquiz
For (a), the radius of convergence is $\infty$, 
a result you should memorize.
To see why (which you were NOT asked to do), compute
$\displaystyle\lim_{n\to\infty}
\frac{\frac{x^{2n+2}}{(2n+2)!}}{\frac{x^{2n}}{(2n)!}}=
\lim_{n\to\infty}\frac{(2n)!}{(2n+2)!)}\cdot \vert x\vert^2=
$\par\noindent
$\lim_{n\to\infty}\frac{\vert x\vert^2}{(2n+1)(2n+2)}=0$.

For (b), $\displaystyle\cos x-1= \isum1 \frac{(-1)^{n}x^{2n}}{(2n)!}$
and hence $\displaystyle\frac{\cos x-1}{x^2}=
\isum1 \frac{(-1)^{n}x^{2n-2}}{(2n)!}$. 
This is equal to $\displaystyle\isum0 \frac{(-1)^{n-1}x^{2n}}{(2n+2)!}$.
Why is this the MacLaurin series for 
$\displaystyle\frac{\cos x-1}{x^2}$?
The best answer is that by the calculations we just did, the series 
represents the function.
The MacLaurin series for the power series is itself (that's a theorem)
and since the series and the function are the same, the series is 
also the MacLaurin series for the function.

For (c), $\displaystyle\cos t^2= \isum0 \frac{(-1)^{n}t^{4n}}{(2n)!}$,
so
$\displaystyle\int_0^x \cos t^2\ dt=
\int_0^x \isum0 \frac{(-1)^{n}t^{4n}}{(2n)!}\ dt=$\par\noindent$
\displaystyle\isum0 \frac{(-1)^{n}x^{4n+1}}{(4n+1)(2n)!}$.
\equiz

%################
\def\sL#1{\displaystyle\lim_{#1\to\infty}}
\def\isum#1{{\displaystyle{\sum_{n={#1}}^{\infty}}}\>}
\stp
\centerline{\bf Math.~126 Quiz \#10}
\centerline{April 16, 2002}
\vskip10pt
\noindent

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

Then show the calcualtions needed to find an $m$
such that 
$$\displaystyle0<\sum_{n=1}^\infty\
\frac{(-1)^{n+1}}{\sqrt{n^2+n}}-
\sum_{n=1}^m\
\frac{(-1)^{n+1}}{\sqrt{n^2+n}}<0.01$$
\dquiz
\def\next{\vskip.2in \noindent}
\def\ul{\underline}
To see that the series converges using the Alternating Series Test, we
need to show that $\lim_{n\rightarrow \infty} b_n = 0$ and that
$b_{n+1}\leq b_n$.  First, note that $$b_n=\frac{1}{\sqrt{n^2+n}}.$$
Since $$\lim_{n\rightarrow \infty} \sqrt{n^2+n} =\infty,$$ we have that
$$\lim_{n\rightarrow \infty} \frac{1}{\sqrt{n^2+n}} =0.$$
\next
To check that $b_{n+1}\leq b_n$, ask:  Is
$$\frac{1}{\sqrt{(n+1)^2+n+1}} \leq \frac{1}{\sqrt{n^2+n}}?$$
After clearing fractions and squaring, we see that the inequality holds
if $$n^2+n\leq (n+1)^2 +n+1$$  holds.  This second inequality is clearly
true.
(Note:  another way is to show that 
$\frac{d}{dn}b_n=-\frac{1}{2}(n^2+n)^{-\frac{3}{2}}(2n+1)$
and this is clearly $<0$ for $n\geq 1$.)
\next
\next
Recall that $|R_m|\leq b_{m+1}$.  Thus, for the error calculation, we need
to solve$$b_{m+1}<0.01\ .$$
( Also, since $$0<\sum_{n=1}^\infty\
           \frac{(-1)^{n+1}}{\sqrt{n^2+n}}-
           \sum_{n=1}^m\
           \frac{(-1)^{n+1}}{\sqrt{n^2+n}},$$
 note that $m$ must be odd since the remainder has the same sign as
$b_{m+1}$.)
\equiz
%################
\stp
\centerline{\bf Math.~126 Quiz \#9}
\centerline{April 9, 2002}
\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.}$\displaystyle a_n=\frac1{n\>!}$:
\medskip
\item{}Recall $n\>!=n(n-1)\cdots2\cdot1$
\ans
\vfill
\item{B.}$\displaystyle a_n=\frac{n}{1+\sqrt{n}}$:
\ans
\dquiz
For A we note $0< \frac1{n\>!}=\frac1n\cdot\frac1{n-1}
\cdots\frac12\cdot\frac11 <\frac1n$ and since
$\displaystyle\lim_{n\to\infty}\frac{1}{n}=0$, the 
Squeeze Theorem promises 
$\displaystyle\lim_{n\to\infty}\frac{1}{n\>!}=0$.
This tells us nothing about whether the series
$\isum{1} \frac{1}{n\>!}$ converges.
(Remark: It is correct that the series does converge, in fact to
$e-1$, but this calculation does not show it.)

For B we write
$\displaystyle\frac{n}{1+\sqrt{n}}=
\frac{\sqrt{n}}{\frac{1}{\sqrt{n}}+1}
$ so
$\displaystyle\lim_{n\to\infty}\frac{n}{1+\sqrt{n}}=
\lim_{n\to\infty}\frac{\sqrt{n}}{\frac{1}{\sqrt{n}}+1}=
\frac{\infty}{0+1}=\infty$.
The Divergence Test shows that the series
$\isum{1}\frac{n}{1+\sqrt{n}}$ diverges.
\equiz
%################
\stp
\centerline{\bf Math.~126 Quiz \#8}
\centerline{April 2, 2002}

\noindent
Write an equation for the hyperbola which has
foci at $(3,3)$ and $(3,11)$ and which has asymptotes
$y=x+4$ and $y=10-x$.
What are the coordinates of the vertices?
\dquiz
The center is halfway between the foci so has coordinates $(3,7)$.
Thus $c=4$ since it is the distance from the center to a focus.

The vertices of the hyperbola are on the line $x=3$ since
they lie on the line between the foci, so the equation has the form
$$\frac{(y-7)^2}{a^2}-\frac{(x-3)^2}{b^2}=1$$
where $16=c^2=a^2+b^2$.

The asymptotes have slope $\pm 1$ from the slope{--}intercept 
form of the equation. 
The slopes are $\pm\frac{a}{b}$ for any hyperbola so $a=b$
and $16=2a^2$ so $a^2=b^2=8$.

The vertices are a distance $a$ from the center.
Since $a^2=8$, $a=2\sqrt 2$ and the vertices are at
$(3,7\pm 2\sqrt2\ )$.
\equiz
%################
\stp
\centerline{\bf Math.~126 Quiz \#7}
\centerline{March 26, 2002}

\noindent
\item{1.}
Write down a definite integral which gives the length of
the curve
$x(t)=t^3+4t+1$, $y(t)=t^4+t^2$ for $0\leq t\leq 5$.
Do {\bf NOT} attempt to evaluate either this integral or the integrals you
will write down in parts 2 and 3.
\vfill
\item{2.}Write down a definite integral which gives the surface
area of the surface of revolution obtained by rotating
the parameterized curve in 1 around the $x${--}axis.
\vfill
\item{3.}Write down a definite integral which gives the surface
area of the surface of revolution obtained by rotating
the parameterized curve in 1 around the $y${--}axis.
\dquiz
We have the curve $x(t)=t^3 +4t +1$,  
$y(t)=t^4+t^2$ for $0\leq t\leq 5$.

Note $\displaystyle ds=\sqrt{\biggl(\frac{dx}{dt}\biggr)^2+
\biggl(\frac{dy}{dt}\biggr)^2}\ dt= \sqrt{(3t^2+4)^2+(4t^3+2t)^2}\ dt$
\bigskip
\item{1.}  The definite integral which gives the length of this curve is
$$L=\int_0^5 \sqrt{(3t^2+4)^2+(4t^3+2t)^2}\ dt\ .$$

\item{2.}  The definite integral which gives the surface area of the surface of
revolution obtained be rotating the parameterized curve around the x-axis
is $$S=\int_0^5 2\pi (t^4+t^2)\sqrt{(3t^2+4)^2+(4t^3+2t)^2}\ dt$$
since $t^4+t^2$ is the distance from the point $\bigl(x(t),y(t)\bigr)$
on the curve to the $x${--}axis.


\item{3.}  The definite integral which gives the surface area of the surface of
revolution obtained be rotating the parameterized curve around the y-axis
is $$S=\int_0^5 2\pi (t^3+4t+1)\sqrt{(3t^2+4)^2+(4t^3+2t)^2}\ dt$$
since $t^3 +4t +1$ is the distance from the point $\bigl(x(t),y(t)\bigr)$
on the curve to the $y${--}axis.
\equiz
%################
\stp
\centerline{\bf Math.~126 Quiz \#6}
\centerline{March 5, 2002}

\noindent
Solve the initial value problem
$$x\frac{d y}{d x} =
x^2{\cos x} +y$$
$$y(\pi)=1$$
\dquiz
First, bring the equation to the canonical form:
$$\frac{dy}{dx}+\biggl(-\frac{1}{x}\biggr)y=x\cos x\leqno(\ast)$$
Next compute the integrating factor:
$$I(x)=e^{\int -\frac{1}{x}dx}=e^{-\ln x}=\frac{1}{x}$$
Multiply the equation $(\ast)$ by the integrating factor, $I(x)$, 
to obtain:
$$\biggl(\frac{1}{x}y\biggr)^\prime=\cos x$$
Integrate this:
$$\frac{1}{x}y=\int\cos x\> dx=\sin x +C$$
and thus
$$y=x(\sin x + C)$$
To determine $C$, compute $y(\pi)$ two ways:
$$1=y(\pi)=\pi(\sin(\pi)+C)=\pi C$$
so $C=\frac1\pi$ and the solution is:
$$y=x\Bigl(\sin x+\frac1\pi\Bigr)$$

\equiz
%################
\stp
\centerline{\bf Math.~126 Quiz \#5}
\centerline{February  26, 2002}

\noindent
Which improper integrals below converge and which diverge?
Indicate
your reasoning and be careful.\vskip10pt
\itemitem{a)}$\displaystyle\int_0^\infty 
\frac{1}{\sqrt{x^3}}\ dx$
\itemitem{b)}$\displaystyle\int_1^\infty \frac{\sin^2x}{x^2}\ dx$
\dquiz
$$\eqalign{
\hbox to 0pt{\hss{\rm (a)}
\hskip 1in}\int_0^\infty\frac{1}{\sqrt{x^3}}\ dx=
&\int_0^1
\frac{1}{\sqrt{x^3}}\ dx+\int_1^\infty 
\frac{1}{\sqrt{x^3}}\ dx\cr
=&\lim_{t\to 0+}\int_t^1\frac{1}{\sqrt{x^3}}\ dx+
\lim_{t\to\infty}\int_1^t\frac{1}{\sqrt{x^3}}\ dx\cr
=&\lim_{t\to 0+}\frac{-2}{\sqrt{x}}\ \bigg\vert^1_t+
\lim_{t\to\infty}\frac{-2}{\sqrt{x}}\ \bigg\vert^t_1\cr
=&-2+\lim_{t\to 0+}\frac{2}{\sqrt{t}}+
\lim_{t\to \infty}\frac{-2}{\sqrt{t}}+2\cr
=&+\infty\cr
}$$
so $\displaystyle\int_0^\infty 
\frac{1}{\sqrt{x^3}}\ dx$ diverges.
\bigskip
\noindent{b)} Use the Comparison Theorem. Note 
$\displaystyle0\leq \frac{\sin^2 x}{x^2}\leq \frac{1}{x^2}$
since $-1\leq\sin x\leq 1$.
Evaluate
$\displaystyle \int_1^\infty\frac{1}{x^2}\>dx=
\lim_{t\to\infty}\int_1^t\frac{1}{x^2}\>dx=
\lim_{t\to\infty}\frac{-1}{x}\bigg\vert^t_1=
\lim_{t\to\infty}\frac{-1}{t}+1=1+0$.
Since $\displaystyle \int_1^\infty\frac{1}{x^2}\>dx$
converges and since 
$\displaystyle0\leq \frac{\sin^2 x}{x^2}\leq \frac{1}{x^2}$,
$\displaystyle\int_1^\infty \frac{\sin^2x}{x^2}\ dx$ converges.
\medskip
\equiz

%################
\stp
\centerline{\bf Math.~126 Quiz \#4}
\centerline{February  19, 2002}

Expand
$$\frac{2x^3+5x^2+5x+5}{(x+1)^2(x^2+2)}$$
using the method of partial fractions.
\par\noindent{\bf Remark:}
Indicate your setup clearly since half the points are
for the correct setup.
\dquiz
\noindent{\bf Setup:}
$$\frac{2x^3+5x^2+5x+5}
{(x+1)^2(x^2+2)}
=
\frac{A}{x+1}+\frac{B}{(x+1)^2}+\frac{Cx+D}{x^2+2}$$

\noindent{\bf Solve:}
Clear denominators by multiplying through by
$(x+1)^2(x^2+2)$:
$$\eqalign{
(1)\hskip20pt2x^3+5x^2&+5x+5=A(x+1)(x^2+2)+B(x^2+2)+(Cx+D)(x+1)^2\cr
(2)\hskip 53pt=&A(x^3+x^2+2x+2)+B(x^2+2)+C(x^3+2x^2+x)+D(x^2+2x+1)\cr
}$$

Set $x=-1$ in (1). This give $3=3B$ so $B=1$.

Setting $B=1$ and collecting like terms in (2) 
gives the system of equations
$$\matrix{
\hskip .5in&2&=&A+C\hfill&\hskip 20pt&(3)\cr
&5&=&A+2C+D+1\hfill\cr
&5&=&2A+C+2D\hfill&&(4)\cr
&5&=&2A+D+2\hfill&&(5)\cr
}$$

Solve:
$C=2-A$ from (3); $D=3-2A$ from (5); so (4) becomes
$$5=2A+(2-A)+2(3-2A)\ ;$$ so $3A=3$; $A=1$.
Then $C=2-A=1$ and $D=3-2A=1$.
Hence
$$\frac{2x^3+5x^2+5x+5}
{(x+1)^2(x^2+2)}
=
\frac{1}{x+1}+\frac{1}{(x+1)^2}+\frac{x+1}{x^2+2}$$

\equiz

%################
\stp
\centerline{\bf Math.~126 Quiz \#3}
\centerline{February  12, 2002}
\noindent Evaluate the integral
$$\int \sec^2 \bigl(\sqrt x\>\bigr)\> dx \ .$$

\noindent{\bf Hint:} First do a substitution and
then an integration by parts.
\dquiz
\noindent First, apply the substitution $t=\sqrt x$, so $x=t^2$ and $dx=2t\> dt$
\par\noindent(or $dt=\frac{dx}{2\sqrt x}$, $dx=2\sqrt{x}\> dt=2t\> dt$).

Then $\int \sec^2(\sqrt x\>)\>dx=\int \sec^2 t\>\cdot\> 2t\> dt=
2\int t\> \sec^2 t\> dt$.

To compute $\int t\> \sec^2 t\> dt$ use integration by parts with
$$\matrix{
dv\> =\> \sec^2t\> dt& v\>=\>\tan t\cr
u\> =\> t&du\> = \> dt\cr
}$$
so
$\int t\> \sec^2 t\> dt\> = \> t\>\tan t-\int \tan t\> dt=
t\> \tan t -\ln\vert\sec t\vert+C$.
\par\noindent
Hence $\int \sec^2(\sqrt x\>)\>dx=2\Bigl(
\sqrt{x}\ \tan(\sqrt x\>)\ -\ 
\ln\big\vert\sec(\sqrt{x}\>)\big\vert\Bigr)+C$.

\noindent{\bf Remark:}
Some people computed $\int\tan t$ by substitution instead of from memory.
\par\noindent
$\int \tan t\>dt=\int\frac{\sin t}{\cos t}\> dt=-\int \frac{du}{u}=
-\ln\vert \cos t\vert+C=\ln\vert\sec t\vert+C$ after the substitution
$u=\cos t$, $du=-\sin t\> dt$.
\equiz
%################
\stp
\centerline{\bf Math.~126 Quiz \#2}
\centerline{January  29, 2002}

\itemitem{1.} Solve the equation $3^{2x}=5$ for $x$.
An answer involving $\ln$ of other numbers is fine.

\itemitem{2.}Use logarithmic differentiation to find
$\frac{dy}{dx}$ if 
$$y=\frac{(x^2-1)^{2.3}(x^3+2)^{1.1}}{(x^2+1)^{0.4}}\ .$$

\dquiz
\item{1.}$\ln\bigl(3^{2x}\bigr)=\ln 5$
\item{}$2x \ln 3 =\ln 5$
\item{}$2x=\frac{\ln 5}{\ln 3}$
\item{}$2x=\frac{\ln 5}{2\ln 3}$
\medskip
\item{2.}$\ln y=\ln\Bigl(\frac{(x^2-1)^{2.3}
(x^3+2)^{1.1}}{(x^2+1)^{0.4}}\Bigr)=
\ln\bigl((x^2-1)^{2.3}\bigr)+
\ln\bigl((x^3+2)^{1.1}\bigr)-
\ln\bigl((x^2+1)^{0.4}\bigr)$
\item{}$\ln y=2.3\> \ln(x^2-1)+
1.1\>\ln(x^3+2)-
0.4\>\ln(x^2+1)$
\bigskip
\noindent Differentiating both sides,
\item{}$\frac{d}{dx}\Bigl(\ln y\Bigr)=
\frac{d}{dx}\Bigl(2.3\> \ln(x^2-1)+
1.1\>\ln(x^3+2)-
0.4\>\ln(x^2+1)\Bigr)
$
\noindent or
\def\dd{\displaystyle}
\item{}$\dd\frac 1y\>\frac{dy}{dx}=
\frac{2.3(2x)}{x^2-1}+
\frac{1.1(3x^2)}{x^3+2}-
\frac{0.4(2x)}{x^2+1}=
\frac{(4.6)x}{x^2-1}+
\frac{(3.3)x^2}{x^3+2}-
\frac{(0.8)x}{x^2+1}
$
\noindent
So
$$\dd\frac{dy}{dx}=\biggl(
\frac{(x^2-1)^{2.3}(x^3+2)^{1.1}}{(x^2+1)^{0.4}}\biggr)\biggl(
\frac{(4.6)x}{x^2-1}+
\frac{(3.3)x^2}{x^3+2}-
\frac{(0.8)x}{x^2+1}
\biggr)$$

\equiz

%################
\stp
\centerline{\bf Math.~126 Quiz \#1}
\centerline{January  22, 2002}
The function, $f(x)=x^3-3x^2+x$ for $x$ in the interval $[-2,0]$,
has an inverse function because $f$ is strictly increasing 
on this interval.

\itemitem{a.} What is the domain of the inverse function, $f^{-1}$?
\itemitem{b.} What is the value of the derivative of $f^{-1}$
for $x=-5$? 

\dquiz
\item{a.} Domain $(f^{-1})$ $=$ Range$(f)$.
Since $f$ is increasing on $[-2,0]$ the range of $f$ is
\par$\bigl[f(-2),f(0)\bigr]$ which is $[-22,0]$.

\item{b.}
First note that $f^{-1}(5)=y$ if and only if $-5=f(y)$,
or $-5=x^3-3x^2+x$ or $0=x^3-3x^2+x+5$.
Since $-5\in [-22,0]$ there is precisely one solution to
this equation in the interval $[-2,0]$
and by trial and error you find that $f(-1)=-5$.
In other words, $f^{-1}(-5)=-1$.

Next compute $f^\prime(x)=3x^2-6x+1$.

Then $\bigl(f^{-1}\bigr)^\prime(-5)=\displaystyle
\frac{1}{f^\prime\bigl(f^{-1}(-5)\bigr)}=
\frac{1}{f^\prime(-1)}=\frac{1}{10}$.
\equiz

\end