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\title{Math 166: Honors Calculus II\\
Final Exam Solutions}
\date{May 6, 1999}

\begin{document}

\maketitle

\bigskip

Problems 1 and 2 are definitions and theroems that can be found in
the text.

\nextproblem

3. a)
$y = \arctan(x) \Leftrightarrow \tan(y) = x$. $\frac{dy}{dx} =
\frac{1}{dx/dy} = \frac{1}{\sec^2(y)} = \frac{1}{1+\tan^2(y)} =
\frac{1}{1+x^2}$.

b) If $f(x) = \frac{x+1}{x-1}$ then
$f'(x) = -2/(x-1)^2$. Thus, $f(x)$ is strictly decreasing on the intervals
$x<1$ and $x>1$ and therefore has an inverse on each of those intervals.
To find $f^{-1}(x)$, solve $y = \frac{x+1}{x-1}$ for $x$:
$yx - y = x+1$, $x(y-1) = y+1$ and $x = \frac{y+1}{y-1}$. Therefore,
$f^{-1}(x) = \frac{x+1}{x-1} = f(x)$ (!).

\nextproblem

4. a) Let $A(x) = \int_0^x \exp(1-t^2)\,dt$, so $A'(x) =
\exp(1-x^2)$. Then $\int_{\sin(x)}^{\cos(x)}\exp(1-t^2)\,dt =
A(\cos(x)) - A(\sin(x))$ and by the chain rule $\frac{d}{dx}
\int_{\sin(x)}^{\cos(x)}\exp(1-t^2)\,dt = A'(\cos(x))(-\sin(x)) -
A'(\sin(x))\cos(x) = -\sin(x)\exp(1-\cos^2(x))-\cos(x)\exp(1-\sin^2(x))
= -\sin(x)\exp(\sin^2(x))-\cos(x)\exp(\cos^2(x))$.

b) Let $f(x) = x^{-x}$, $x > 0$.  Since $f'(x) = \exp(-x\log(x))[-\log(x)
- 1]$, we see that $f(x)$ is increasing ($f'(x)>0$) if $\log(x)+1<0$
and decreasing ($f'(x)<0$) if $\log(x)+1>0$. This implies that $f(x)$
has a maximum when $\log(x)+1 = 0$, i.e., at $x = e^{-1}=1/e$ with
maximum value $f(1/e) = e^{1/e}$.

\nextproblem

5. a) Substituting $u = x+4$, $du = dx$ gives $\int
\frac{x}{\sqrt{x+4}}\,dx = \int u^{1/2} - 4u^{-1/2} du =
\frac{2}{3}(x+4)^{3/2} - 8(x+4)^{1/2} + C = \frac{2}{3}(x-16)\sqrt{x+4}
+ C$.

b) $\int\frac{1}{x^2+2x}\,dx = \frac{1}{2}\int\frac{1}{x} -
\frac{1}{x+2}\,dx = \frac{1}{2}(\log|x| - \log|x+2|) + C$.

c) $\int\frac{1}{\sqrt{3-2x-x^2}}\,dx = \int\frac{1}{\sqrt{4-(x+1)^2}}\,dx
=\arcsin((x+1)/2) + C$.

d) Substituting $u = \tan(\theta/2)$, $d\theta = \frac{2du}{1+u^2}$,
$\cos(\theta) = \frac{1-u^2}{1+u^2}$ gives
$\int\frac{1}{2 + \cos(\theta)}\,d\theta
= \int \frac{1}{2 + \frac{1-u^2}{1+u^2}} \frac{2}{1+u^2}\,du
= \int \frac{2}{3 + u^2}\,du
= \frac{2}{\sqrt{3}}\arctan\big(\frac{u}{\sqrt{3}}\big) + C
= \frac{2}{\sqrt{3}}\arctan\big(\frac{\tan(\theta/2)}{\sqrt{3}}\big) + C$

e) $\int_0^\infty e^{-x}\sin(x)\,dx = \lim_{a\to\infty}
\int_0^a e^{-x}\sin(x)\,dx$. Integrating by parts with
$u=\sin(x)$, $dv=e^{-x}dx$ gives $\int_0^a e^{-x}\sin(x)\,dx
= -\sin(a)e^{-a}+\int_0^a\cos(x)e^{-x}\,dx$. Repeating
this with $u=\cos(x)$ gives $\int_0^a e^{-x}\sin(x)\,dx =
-\sin(a)e^{-a}-\cos(a)e^{-a}+1-\int_0^a\sin(x)e^{-x}\,dx$,
so $\int_0^a e^{-x}\sin(x)\,dx =
\frac{1}{2}-\frac{1}{2}e^{-a}(\sin(a)+\cos(a))$. Therefore, in the limit
as $a\to\infty$ we get $\int_0^\infty e^{-x}\sin(x)\,dx = 1/2$.

\nextproblem

6. a) $f(x) = \sum_{k=0}^n \frac{f^{(k)}(a)}{n!}(x-a)^k +
\frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}$ for some $c$ between $a$ and $x$.

b) $\frac{x}{x^2+3} = \frac{x}{3}\Big(\frac{1}{1-(-x^2/3)}\Big)
= \frac{x}{3} \Big(1 +\Big(\frac{-x^2}{3}\Big) +
\Big(\frac{-x^2}{3}\Big)^2 + \cdots + \Big(\frac{-x^{2}}{3}\Big)^n +
\frac{(-x^2/3)^{n+1}}{1+(x^2/3)}\Big)$, so \newline
$T_{2n+1}\Big(\frac{x}{x^2 + 3}\Big) = \frac{x}{3} - \frac{x^3}{3^2}
+ \frac{x^5}{3^3} - \cdots + \frac{(-1)^nx^{2n+1}}{3^{n+1}}$

c) Since $\log(1+u) = u - \frac{1}{2}u^2 + \frac{1}{3}u^3 + o(u^3)$,
$\log(1+x^2) - \log(1+x^3) = [x^2 - \frac{1}{2}x^4 + \frac{1}{3}x^6 +
o(x^7)] - [x^3 - \frac{1}{2}x^6 + o(x^8)] = x^2 - x^3 - \frac{1}{2}x^4 +
\frac{5}{6}x^6 + o(x^7)$. Hence $T_7[\log(1+x^2) - \log(1+x^3)] = x^2 -
x^3 - \frac{1}{2}x^4 + \frac{5}{6}x^6$.

\nextproblem

7. a) $\frac{\cos(x^2) -1}{x^2\sin^2(x)}
= \frac{(1-x^4/2 + o(x^7)) - 1}{x^2(x+o(x^2))^2}
= \frac{-x^4/2 + o(x^7)}{x^2(x^2+o(x^3))}
= \frac{-1/2 + o(x^3)}{1+o(x)} \to -\frac{1}{2}$ as $x\to0$.

b) $\lim_{x\to0^+}
\frac{1}{x}\Big(\frac{1}{\sqrt{1-x}}-\frac{1}{\sqrt{1+x}}\Big)
=\lim_{x\to0^+}\frac{1}{x}
\Big(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1-x}\sqrt{1+x}}\Big)\newline
=\lim_{x\to0^+}\frac{(1+x)-(1-x)}{x\sqrt{1-x^2}(\sqrt{1-x}+\sqrt{1+x})}
=\lim_{x\to0^+}\frac{2}{\sqrt{1-x^2}(\sqrt{1-x}+\sqrt{1+x})}
=\frac{2}{1(1+1)} = 1$

c) $\lim_{x\to\infty}\frac{\sinh(1+x)}{\cosh(x)}
=\lim_{x\to\infty}\frac{\frac{1}{2}(e^{1+x}-e^{-1-x})}{\frac{1}{2}(e^x+e^{-x})}
=\lim_{x\to\infty}\frac{e^{1+x}(1-e^{-2-2x})}{e^x(1+e^{-2x})}
=\lim_{x\to\infty}\frac{e(1-e^{-2-2x})}{1+e^{-2x}} = e$.

d) $\lim_{n\to \infty} \Big(n^{100} + (-1)^n\Big)^{1/n}
=\lim_{n\to\infty} n^{100/n}\Big(1 + \frac{(-1)^n}{n^{100}}\Big)^{1/n}
=\lim_{n\to\infty}\exp[100\log(n)/n]
 \lim_{n\to\infty}\Big(1+\frac{(-1)^n}{n^{100}}\Big)^{1/n}
=\exp[0](1+0)^0 = 1$
  
\nextproblem

8. a) $\sum_{n=1}^\infty \Big(\frac{1}{n} - \frac{1}{n+2}\Big)
= \sum_{n=1}^\infty \Big(\frac{1}{n}-\frac{1}{n+1}
                       +\frac{1}{n+1}-\frac{1}{n+2}\Big)
= \sum_{n=1}^\infty \Big(\frac{1}{n}-\frac{1}{n+1}\Big)
 +\sum_{n=1}^\infty \Big(\frac{1}{n+1}-\frac{1}{n+2}\Big)
= \Big(\frac{1}{1}-\lim_{n\to\infty}\frac{1}{n+1}\Big) +
\Big(\frac{1}{1+1}-\lim_{n\to\infty}\frac{1}{n+2}\Big)
= 1 + \frac{1}{2} = \frac{3}{2}$\quad [telescoping sums].

b) $\sum_{n=1}^\infty \frac{2+(-1)^n}{3^n}
=2\sum_{n=1}^\infty\Big(\frac{1}{3}\Big)^n
+\sum_{n=1}^\infty \Big(\frac{-1}{3}\Big)^n
=2\frac{1/3}{1-(1/3)} + \frac{-1/3}{1-(-1/3)} 
=1 - \frac{1}{4} =\frac{3}{4}$.

c) $2x^3 + 4x^5 + 6x^7 + 8x^9 + \cdots
= x^2(2x + 4x^3 + 6x^5 + 8x^7 + \cdots)
= x^2\frac{d}{dx}(x^2 + x^4 + x^6 + x^8 + \cdots)
= x^2\frac{d}{dx}\Big[\frac{x^2}{1-x^2}\Big]
= x^2\frac{2x(1-x^2)-(-2x)x^2}{(1-x^2)^2}
= \frac{2x^3}{(1-x^2)^2}$.

d) The series $\frac{1}{2} - \frac{1}{4} + \frac{1}{6} - \frac{1}{8} + \cdots$
converges by Leibniz' Rule. Since
$\frac{1}{2}x^2 - \frac{1}{4}x^4 + \frac{1}{6}x^6 - \frac{1}{8}x^8 + \cdots
=\int_0^x(t - t^3 + t^5 - t^7 + \cdots)\,dt
=\int_0^xt(1 - t^2 + t^4 - t^6 + \cdots)\,dt
=\int_0^x\frac{t}{1+t^2}\,dt =\frac{1}{2}\log(1+x^2)$,
Abel's Theorem implies that
$\frac{1}{2} - \frac{1}{4} + \frac{1}{6} - \frac{1}{8} + \cdots =
\lim_{x\to1^-} \frac{1}{2}\log(1+x^2) = \frac{1}{2}\log(2)$
\nextproblem

9. a) Since $s>1$, $\log(a)^{1-s}\to0$ as $a\to\infty$, so
$\int_2^\infty \frac{1}{x\log(x)^s}\,dx
= \frac{1}{1-s}\log(x)^{1-s}\Big|_2^\infty < \infty$.
By the Integral Test, the absolute value series $\sum_{n=1}^\infty
\frac{1}{n\log(n)^s}$ converges.  Therefore the series $\sum_{n=1}^\infty
\frac{(-1)^n}{n\log(n)^s}$ is absolutely convergent.

b)  Since $\lim_{n\to\infty}\Big(1+\frac{1}{n}\Big)^n = e$,
the terms of the series
$\sum_{n=1}^\infty (-1)^n\Big(1+\frac{1}{n}\Big)^n$ do not approach~$0$,
so the series diverges by the Simple Divergence Test.

c) First observe that $\frac{n}{n^2+1}\approx \frac{1}{n}$ since
$\lim_{n\to\infty} \frac{n}{n^2+1}\cdot\frac{n}{1} = 1$. The harmonic
series $\sum_{n=1}^\infty\frac{1}{n}$ diverges (e.g., by the Integral
Test), so the absolute value series $\sum_{n=1}^\infty \frac{n}{n^2+1}$
diverges. However, the terms $\frac{n}{n^2+1}$ strictly decrease to zero,
so the series $\sum_{n=1}^\infty \frac{(-1)^n n}{n^2+1}$ converges by
Leibniz' Rule and is therefore conditionally convergent.

\nextproblem

10. a) Applying the Ratio Test to the absolute value of the terms of the
series $\sum_{n=0}^\infty (1 + (-3)^n)x^n$ yields $\lim_{n\to\infty}
\frac{|1+(-3)^{n+1}||x|^{n+1}}{|1+(-3)^n||x|^n} =\lim_{n\to\infty}
\frac{|1/(-3)^n+(-3)|}{|1/(-3)^n+1|}|x| = \frac{|0+(-3)|}{|0+1|}|x|
=3|x|$. The series thus converges for $|x|<1/3$ and diverges for
$|x|>1/3$.  At the endpoints $x = \pm1/3$, the series $\sum_{n=0}^\infty
\frac{1 + (-3)^n}{(\pm3)^n}$ is divergent since its terms do not
approach~0. Therefore, the interval of convergence is $(-1/3,1/3)$.

b) Applying the Ratio Test to the absolute value of the
terms of the series $\sum_{n=1}^\infty \frac{(x+1)^{2n}}{n3^n}$
yields\newline $\lim_{n\to\infty}
\frac{|x+1|^{2n+2}}{(n+1)3^{n+1}}\cdot\frac{n3^n}{|x+1|^{2n}}
=\lim_{n\to\infty} \Big(\frac{n}{n+1}\Big)\frac{1}{3}|x+1|^2
= \frac{1}{3}|x+1|^2$. The series thus converges if
$\frac{1}{3}|x+1|^2<1$, i.e., $|x+1|<\sqrt{3}$, and diverges
if $\frac{1}{3}|x+1|^2>1$, i.e., $|x+1|>\sqrt{3}$. At the
endpoints $x+1=\pm\sqrt{3}$, the series is $\sum_{n=1}^\infty
\frac{(\pm\sqrt{3})^{2n}}{n3^n} =\sum_{n=1}^\infty\frac{1}{n}$ which
diverges (the harmonic series). Therefore the interval of convergence
is $(-1-\sqrt{3}, -1+\sqrt{3})$.

\end{document}
