Practice Exam: Series and Taylor Series

Practice Exam: Series and Taylor Series
Time: 60 minutes

Problem 1 (15 points) Use the fourth degree Taylor polynomial of $\cos(2x)$ to find the exact value of

$\begin{displaymath}\lim_{x\to0}\frac{1-\cos(2x)}{3x^2}\end{displaymath}$

Explain your reasoning!

Solution: Taking derivatives, if necessary, we obtain that the fourth degree Taylor polynomial for $\cos y$ equals $\displaystyle 1-\frac{y^2}{2}+\frac{y^4}{24}$ . Using the substitution y=2x, the Taylor polynomial for $\cos(2x)$ becomes

$\begin{displaymath}1-2x^2+\frac{16x^4}{24}.\end{displaymath}$

Consequently for $x\approx 0$ ,

$\begin{displaymath}\frac{1-\cos(2x)}{3x^2}\approx\frac{2x^2-\frac{16x^4}{24} }{3x^2}=\frac{2}{3}-\frac{2}{9}x^2.\end{displaymath}$

Thus

$\begin{displaymath}\lim_{x\to0}\frac{1-\cos(2x)}{3x^2}=\lim_{x\to 0}\frac{2}{3}-\frac{2}{9}x^2=\frac{2}{3}.\end{displaymath}$

Problem 2 (15 points) Find the third degree Taylor polynomial of

$\begin{displaymath}f(x)=\sqrt{x}\end{displaymath}$

with center x₀=3.

Solution: We have to find the first three derivatives of f(x) at x₀=3. $f(x)=\sqrt{x}$ , so $f(3)=\sqrt{3}$ . $\displaystyle f^\prime(x)=\frac{1}{2}x^{-1/2}$ , so $\displaystyle f^\prime(3)=\frac{1}{2}3^{-1/2}$ . $\displaystyle f^{\prime\prime}(x)=\left(-\frac{1}{2}\right)\frac{1}{2}x^{-3/2}$ , so $\displaystyle f^{\prime\prime}(3)=\left(-\frac{1}{2}\right)\frac{1}{2}3^{-3/2}$ . Finally, $\displaystyle f^{\prime\prime\prime}(x)=\left(-\frac{3}{2}\right)\left(-\frac{1}{2}\right)\frac{1}{2}x^{-5/2}$ , so $\displaystyle f^{\prime\prime\prime}(3)=\left(-\frac{3}{2}\right)\left(-\frac{1}{2}\right)\frac{1}{2}3^{-5/2}$ .

Thus the third degree Taylor polynomial with center x₀=3 is given by

$\begin{displaymath}\sqrt{3}+\frac{1}{2}(\sqrt{3})^{-1/2} (x-3)-\frac{1}{8}(\sqrt{3})^{-3/2}(x-3)^2+\frac{1}{16}(\sqrt{3})^{-5/2}(x-3)^3.\end{displaymath}$

Problem 3 (15 points) Let $f(x)=\ln(1+x^2)$ . Find the Taylor series of f(x) with center x₀=0 and its radius of convergence.

Solution: This is easiest if you remember that the Taylor series with center y₀=0 for $\ln(y+1)$ has radius of convergence 1 and is given by

$\begin{displaymath}y-\frac{y^2}{2}+\frac{y^3}{3}-\frac{y^4}{4}+\cdots=\sum_{n=1}^\infty (-1)^{n+1}\frac{y^n}{n}.\end{displaymath}$

Using the substitution y=x², one then obtains the Taylor series for f(x):

$\begin{displaymath}x^2-\frac{x^4}{2}+\frac{x^6}{3}-\frac{x^8}{4}+\cdots= \sum_{n=1}^\infty (-1)^{n+1}\frac{x^{2n}}{n}.\end{displaymath}$

Since $\vert y\vert<1\Leftrightarrow \vert x^2\vert<1\Leftrightarrow \vert x\vert<1$ , the Taylor series for f(x) will also have 1 as its radius of convergence.

Alternatively, observe that $\displaystyle f^\prime(x)=\frac{2x}{1+x^2}$ , then write down a geometric series expression for $f^\prime(x)$ and integrate.

Problem 4 (15 points) Find the radius of convergence of the power series

$\begin{displaymath}\sum_{n=0}^\infty (-3)^n\sqrt{n+1}\,(x+1)^{2n+1}\end{displaymath}$

Solution: The absolute value of the general term of the series is

$\begin{displaymath}b_n=3^n\sqrt{n+1}\vert x+1\vert^{2n+1}.\end{displaymath}$

Consequently

$\begin{eqnarray*}\lim_{n\to\infty}\frac{b_{n+1}}{b_n}&=&\lim_{n\to\infty} \frac{... ...n\to\infty}\frac{\sqrt{n+2}}{\sqrt{n+1}}\\ &=&3 \vert x+1\vert^2 \end{eqnarray*}$

The series will converge (diverge), if this quantity is less than 1 (bigger than 1).

$\begin{displaymath}3\vert x+1\vert^2<1 \Leftrightarrow \vert x+1\vert<\frac{1}{\sqrt{3}},\end{displaymath}$

so the radius of convergence is $1/\sqrt{3}$ .

Problem 5 (20 points) Find the exact value of the following series:

$\displaystyle \frac{4}{5}-\frac{16}{25}+\frac{64}{125}-\frac{256}{625}+\cdots$

Solution:

$\begin{eqnarray*}&&\frac{4}{5}-\frac{16}{25}+\frac{64}{125}-\frac{256}{625}+\cdo... ...ac{4}{5}\right)^k=\frac{4}{5}\cdot\frac{1}{1-(-4/5)}=\frac{4}{9} \end{eqnarray*}$
$\displaystyle 1+\frac{1}{2}+\frac{1}{4\cdot 2!}+\frac{1}{8\cdot 3!}+\frac{1}{16\cdot 4!}+\frac{1}{32\cdot 5!}+\cdots$

Solution: $\displaystyle 1+\frac{1}{2}+\frac{1}{4\cdot 2!}+\frac{1}{8\cdot 3!}+\frac{1}{16\cdot 4!}+\frac{1}{32\cdot 5!}+\cdots=e^{1/2}=\sqrt{e}.$

You can only do this problem if you recognize the given series as a special case (x=1/2) of the Taylor expansion $\displaystyle e^x=\sum_{n=0}^\infty \frac{x^n}{n!}$ .

Problem 6 (20 points) An antibiotic decays exponentially in the human body with a half-life of about 2.5 hours. Suppose a patient takes a 250 mg tablet of the antibiotic every 6 hours.

1.
Write an expression for Q₂, Q₃, Q₄, where Q_n is the amount (in mg) of the antibiotic in the body after the $n^{\mbox{th}}$ tablet is taken. Note that Q₁=250 mg.
2.
Write an expression for Q_n, and put it in closed form.
3.
Assume the antibiotic treatment consists of a total of 28 tablets. Give a numerical estimate for the amount of antibiotic in the body immediately after the patient takes the last tablet of the treatment.

Solution: Using an exponential decay model of the form $\displaystyle Q(t)=Q_0 e^{-kt}$ , where t is measured in hours, and the given information that $Q(2.5)=\frac{1}{2}Q_0$ , we can compute $\displaystyle k=\frac{\ln 2}{2.5}$ . Consequently

$\begin{displaymath}Q_1=Q_0+Q(6)=Q_0+Q_0 e^{-\frac{6}{2.5} \ln2 }=Q_0(1+e^{-\frac{6}{2.5} \ln2}).\end{displaymath}$

In a similar vein, we obtain

$\begin{displaymath}Q_2=Q_0+Q_1 e^{-\frac{6}{2.5} \ln2 }=Q_0\left(1+e^{-\frac{6}{2.5} \ln2}+\left( e^{-\frac{6}{2.5} \ln2}\right)^2\right),\end{displaymath}$

and

$\begin{displaymath}Q_3=Q_0+Q_2 e^{-\frac{6}{2.5} \ln2 }=Q_0\left(1+e^{-\frac{6}{... ... \ln2}\right)^2+\left( e^{-\frac{6}{2.5} \ln2}\right)^3\right).\end{displaymath}$

In general,

$\begin{displaymath}Q_n=Q_0\left(1+e^{-\frac{6}{2.5} \ln2}+\left( e^{-\frac{6}{2.... ...{-\frac{6}{2.5} \ln2}\right)^{n+1}}{1-e^{-\frac{6}{2.5} \ln2}}.\end{displaymath}$

In particular, $Q_{28}\approx 308.44.$