3.6 Taylor Series

Which of the functions are constant, linear, and quadratic?

A: linear B: constant C: quadratic

You don't have to actually calculate the entire series $T(x)$ to answer the question.

$T(5)=\arctan^3\left(e^5+7\right)$

If you don't have these memorized, it's good to be able to derive them. For instance, $\log(1+x)$ is the antiderivative of $\dfrac{1}{1+x}$, whose Taylor series can be found by modifying the geometric series $\sum x^n$.

See Example 3.6.16 in the CLP-2 text.

(a) $f^{(20)}(3)=20^2\left(\dfrac{20!}{20!+1}\right)$ (b) $g^{(20)}(3)=10^2\left(\dfrac{20!}{10!+1}\right)$
(c) $h^{(20)}(0)=-\dfrac{20!\cdot 5^{11}}{11}$; $h^{(22)}(0)=0$

The series will bear some resemblance to the Maclaurin series for $\log(1+x)$.

$\displaystyle\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n}(x-1)^n$

The terms $f^{(n)}(\pi)$ are going to be similar to the terms $f^{(n)}(0)$ that we used in the Maclaurin series for sine.

$\displaystyle \sum_{n=0}^\infty \frac{(-1)^{n+1}}{(2n+1)!}(x-\pi)^{2n+1}$

The Taylor series will look similar to a geometric series.

$\displaystyle\frac{1}{10}\sum_{n=0}^\infty\left(\frac{10-x}{10}\right)^n$ with interval of convergence $(0,20)$.

Your answer will depend on $a$.

$\displaystyle \sum_{n=0}^\infty \frac{3^ne^{3a} }{n!}(x-a)^n$, with infinite radius of convergence

You should know the Maclaurin series for $\dfrac{1}{1-x}$. Use it.

$\displaystyle-\sum\limits_{n=0}^\infty 2^nx^n$

You should know the Maclaurin series for $\dfrac{1}{1-x}$. Use it.

$b_n = 3(-1)^n + 2^n$

You should know the Maclaurin series for $e^x$. Use it.

$c_5=\dfrac{3^5}{5!}$

Review Example 3.5.20 in the CLP–2 text.

$\displaystyle\sum_{n=0}^\infty (-1)^n \frac{2^{n+1} x^{n+1}}{n+1}$ for all $|x|<\frac{1}{2}$

You should know the Maclaurin series for $\sin x$. Use it.

$a=1$, $b=-\dfrac{1}{3!}=-\dfrac{1}{6}$.

You should know the Maclaurin series for $e^x$. Use it.

$\displaystyle\int \frac{e^{-x^2}-1}{x} \,\dee{x} = C -\frac{x^2}{2} + \frac{x^4}{8} + \cdots$.

It is not clear from the wording of the question whether or not the arbitrary constant $C$ is to be counted as one of the “first two nonzero terms”.

You should know the Maclaurin series for $\arctan(x)$. Use it.

$\displaystyle\sum\limits_{n=0}^\infty (-1)^n\frac{2^{2n+1} x^{2n+6}}{(2n+1)(2n+6)} +C =\sum\limits_{n=0}^\infty (-1)^n\frac{2^{2n} x^{2n+6}}{(2n+1)(n+3)} +C$

You should know the Maclaurin series for $\dfrac{1}{1-x}$. Use it.

$f(x) = 1+ \displaystyle\sum\limits_{n=0}^\infty (-1)^n \frac{3^n }{3n+2} x^{3n+2}$

Set $(-1)^n\dfrac{x^{2n+1}}{2n+1}=C\dfrac{(-1)^n}{(2n+1)3^n}$, for some constant $C$. What are $x$ and $C$?

$\dfrac{\pi}{2\sqrt{3}}$

There is an important Taylor series, one of the series in Theorem 3.6.5 of the

CLP-2 text, that looks a lot like the given series.

$\dfrac{1}{e}$

There is an important Taylor series, one of the series in Theorem 3.6.5 of the

CLP-2 text, that looks a lot like the given series.

$e^{1/e}$

There is an important Taylor series, one of the series in Theorem 3.6.5 of the

CLP-2 text, that looks a lot like the given series. Be careful about the limits of summation.

$e^{1/\pi}-1$

There is an important Taylor series, one of the series in Theorem 3.6.5 of the

CLP-2 text, that looks a lot like the given series.

$\log(3/2)$

Split the series into a sum of two series. There is an important Taylor series, one of the series in Theorem 3.6.5 of the

CLP-2 text, that looks a lot like each of the two series.

$(e+2)e^e-2$

Try the ratio test.

The sum diverges–see the solution.

Write it as the sum of two Taylor series.

$\dfrac{1+\sqrt{2}}{\sqrt{2}}$

Can you think of a way to eliminate the odd terms from $e^x=\displaystyle \sum_{n=0}^\infty \frac{x^n}{n!}$ ?

(a) See the solution. (b) $\displaystyle\frac12\left(e+\frac{1}{e}\right)$

The series you're adding up are alternating, so it's simple to bound the error using a partial sum.

(a) 50,000 (b) three terms ($n=0$ to $n=2$) (c) six terms ($n=0$ to $n=5$)

The Taylor Series is alternating, so bounding the error in a partial-sum approximation is straightforward.

29

The Taylor Series is not alternating, so use Theorem 3.6.1-b in the CLP-2 text to bound the error in a partial-sum approximation.

$S_{13}$ or higher

The Taylor Series is not alternating, so use Theorem 3.6.1-b in the CLP-2 text to bound the error in a partial-sum approximation.

$S_9$ or higher

Use Theorem 3.6.1-b in the CLP-2 text to bound the error in a partial-sum approximation. This theorem requires you to consider values of $c$ between $x$ and $x=0$; since $x$ could be anything from $-2$ to $1$, you should think about values of $c$ between $-2$ and $1$.

$S_{18}$ or higher

Use Theorem 3.6.1-b in the CLP-2 text to bound the error in a partial-sum approximation.

To bound the derivative over the appropriate range, remember how to find absolute extrema.

The error is in the interval $\displaystyle\left(\dfrac{-5^7}{14\cdot 3^7}\left[1+\frac{1}{3^7} \right]\quad,\quad \dfrac{-5^7}{7\cdot 6^7} \right)\approx\left(-0.199, -0.040\right)$

See Example 3.6.21 in the

CLP-2 text

$-1$

See Example 3.6.21 in the

CLP-2 text

$\dfrac{1}{5!}=\dfrac{1}{120}$

Set $f(x)=\left(1+x+x^2\right)^{2/x}$, and find $\lim\limits_{x\rightarrow 0}\log\left(f(x)\right)$.

$e^2$

Use the substitution $y=\dfrac{1}{x}$, and compare to Question 34.

$\sqrt{e}$

Start by differentiating $\displaystyle\sum_{n=0}^\infty x^n$.

$\displaystyle\frac{2}{(6/7)^3}=\frac{343}{108}$

The series bears a resemblance to the Taylor series for arctangent.

$\displaystyle \sum_{n=0}^\infty (-1)^n\frac{x^{2n+4}}{(2n+1)(2n+2)}=x^3\arctan x - \frac{x^2}{2}\log(1+x^2)$

For simplification purposes, note $\displaystyle(1)(3)(5)(7)\cdots(2n-1)=\dfrac{(2n)!}{2^n \,n!}$ .

(a) the Maclaurin series for $f(x)$ is $\displaystyle \sum_{n=0}^\infty\frac{(2n)!}{2^{2n}\,(n!)^2}x^n$, and its radius of convergence is $R=1$.

(b) the Maclaurin series for $\arcsin x$ is $\displaystyle\sum_{n=0}^\infty\frac{(2n)!}{2^{2n}\,(n!)^2(2n+1)}x^{2n+1}$, and its radius of convergence is $R=1$.

You know the Maclaurin series for $\log(1+y)$. Use it! Remember that you are asked for a series expansion in powers of $x-2$. So you want $y$ to be some constant times $x-2$.

$\log(x) =\displaystyle \log 2 +\sum\limits_{n=1}^\infty \frac{(-1)^{n-1}}{n\,2^n}(x-2)^n$. It converges when $0<x\le 4$.

See Example 3.5.21 in the

CLP-2 text. For parts (b) and (c), review § 3.3.4 in the

CLP-2 text.

(a) $\displaystyle\sum\limits_{n=0}^\infty (-1)^n\ \frac{x^{4n+1}}{4n+1}$ (b) $0.493967$

(c) The approximate value of part (b) is larger than the true value of $I(1/2)$

Look at the signs of successive terms in the series.

$\dfrac{1}{66}$

The magic word is “series”.

Any interval of length 0.0002 that contains $0.03592$ and $0.03600$ is fine.

See Example 3.6.14 in the

CLP-2 text. For parts (b) and (c), review § 3.3.4 in the

CLP-2 text.

(a) $\displaystyle\sum\limits_{n=1}^\infty (-1)^n\ \frac{x^n}{n\ n!}$ (b) $-0.80$ (c) See the solution.

See Example 3.6.14 in the

CLP-2 text. For part (b), review the fundamental theorem of calculus in § 1.3 of the

CLP-2 text. For part (c), review § 3.3.4 in the

CLP-2 text.

(a) $\displaystyle\Si(x) =\sum\limits_{n=0}^\infty (-1)^n\ \frac{x^{2n+1}}{(2n+1)(2n+1)!}$ (b) $x=\pi$ (c) $1.8525$

See Example 3.6.14 in the

CLP-2 text. For parts (b) and (c), review § 3.3.4 in the

CLP-2 text.

(a) $\displaystyle I(x)=\sum\limits_{n=1}^\infty(-1)^n\frac{x^{2n-1}}{(2n)!(2n-1)}$ (b) $\displaystyle I(1)= -\frac{1}{2}+\frac{1}{4!3} \pm \frac{1}{6!5} =-0.486\pm0.001$

(c) $I(1)<-\dfrac{1}{2}+\dfrac{1}{4!3}$

See Example 3.6.14 in the

CLP-2 text. For parts (b) and (c), review § 3.3.4 in the

CLP-2 text.

(a) $\displaystyle I(x)=\sum_{n=1}^\infty (-1)^{n+1}\frac{x^{2n-1}}{(2n)!}= \frac{1}{2!}x-\frac{1}{4!}x^3+\frac{1}{6!}x^5-\frac{1}{8!}x^8+\cdots$

(b) 0.460 (c) $\displaystyle I(1)<\frac{1}{2!}-\frac{1}{4!}+\frac{1}{6!}<0.460$

(a) See the solution. (b) The series converges for all $x$.

Use the Maclaurin series for $e^x$.

See the solution.

For part (c), compare two power series term-by-term.

(a) $\cosh(x)=\displaystyle \sum\limits_{\genfrac{}{}{0pt}{}{n=0}{n\text{ even}}}^\infty \frac{x^n}{n!} = \sum\limits_{n=0}^\infty\frac{x^{2n}}{(2n)!}$ for all $x$.

For Newton's method, recall we approximate a root of the function $g(x)$ in iterations: given an approximation $x_n$, our next approximation is $x_{n+1}=x_n-\dfrac{g(x_n)}{g'(x_n)}$.

To gauge your error, note that from approximation to approximation, the first digits stabilize. Keep refining your approximation until the first two digits stop changing.

(a) $\sqrt[3]{3}\approx 1.26$ (b) 12 terms ($S_{11}$)