19 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:

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 Qn 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 Qn, 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, $Q_{28}\approx 308.44.$

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