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Sequences and Series

3.5 Power Series

27 problems · hints, answers and solutions shown beside each one

1Stage 1Conceptual

Do you understand the idea? Usually little or no calculation.

Q1Stage 1

Suppose f(x)=∑n=0∞(3−x4)nf(x)=\displaystyle\sum_{n=0}^\infty\left(\dfrac{3-x}{4}\right)^n. What is f(1)f(1)?

Hint

f(1)f(1) is the sum of a geometric series.

Answer

2

Full solution
f(1)=∑n=0∞(3−14)n=∑n=0∞(12)n\begin{align*}f(1)&=\displaystyle\sum_{n=0}^\infty\left(\dfrac{3-1}{4}\right)^n\\ &=\displaystyle\sum_{n=0}^\infty\left(\frac12\right)^n\end{align*}

This is a geometric series with r=12r=\frac{1}{2}, so we know that it converges and

=11−12=2\begin{align*}&=\frac{1}{1-\frac12}=2\end{align*}

The question does not ask us to find the interval of convergence of the series defining f(x)f(x). But we will do so anyway, to get a bit more practice. We may rewrite the nthn^{\rm th} term of the series defining f(x)f(x) as

(3−x4)n=arnwith a=1 and r=3−x4\begin{align*} \left(\frac{3-x}{4}\right)^n = a r^n\qquad\text{with }a=1\text{ and }r= \frac{3-x}{4} \end{align*}

That is, for every fixed xx, we have a geometric series with r=3−x4r= \frac{3-x}{4}. So, by (3.2.2) in the CLP-2 text, the series converges if and only if

∣r∣=∣3−x4∣<1  ⟺  −1<3−x4<1  ⟺  −4<3−x<4  ⟺  −4<3−x and 3−x<4  ⟺  −1<x<7\begin{align*} &|r|= \left|\frac{3-x}{4}\right|<1 \\ \iff & -1< \frac{3-x}{4} < 1 \\ \iff & -4< 3-x < 4 \\ \iff & -4< 3-x \ \text{and}\ 3-x < 4 \\ \iff & -1< x < 7 \end{align*}
Q2Stage 1

Suppose f(x)=∑n=1∞(x−5)nn!+2f(x)=\displaystyle\sum_{n=1}^\infty \dfrac{(x-5)^n}{n!+2}. Give a power series representation of f′(x)f'(x).

Hint

Calculate ddx{(x−5)nn!+2}\displaystyle\diff{}{x}\left\{\frac{(x-5)^n}{n!+2}\right\} when nn is a constant.

Answer

f(x)=∑n=1∞n(x−5)n−1n!+2f(x)=\displaystyle\sum_{n=1}^\infty \dfrac{n(x-5)^{n-1}}{n!+2}

Full solution

By Theorem 3.5.13 in the CLP-2 text, we may differentiate our function term-by-term for all xx obeying ∣x−5∣<R|x-5|<R, where RR is the radius of convergence of the power series. The series defining f(x)f(x) is reminiscent of the exponential series ∑n=0∞Xnn!\sum_{n=0}^\infty \frac{X^n}{n!} of Example 3.5.5 in the CLP-2 text. In that example, we showed that ∑n=0∞Xnn!\sum_{n=0}^\infty \frac{X^n}{n!} has radius of convergence ∞\infty. Since

∣(x−5)nn!+2∣≤Xnn!with X=∣x−5∣\begin{align*} \left|\frac{(x-5)^n}{n!+2}\right|\le \frac{X^n}{n!}\quad\text{with}\ X=|x-5| \end{align*}

the comparison test, Theorem 3.3.8 in the CLP-2 text, tells us that ∑n=1∞(x−5)nn!+2\sum_{n=1}^\infty \frac{(x-5)^n}{n!+2} converges for all xx. So we may differentiate our function term-by-term.

f(x)=∑n=1∞(x−5)nn!+2f′(x)=∑n=1∞ddx{(x−5)nn!+2}=∑n=1∞n(x−5)n−1n!+2\begin{align*} f(x)&=\displaystyle\sum_{n=1}^\infty \dfrac{(x-5)^n}{n!+2}\\ f'(x)&=\displaystyle\sum_{n=1}^\infty \diff{}{x}\left\{\dfrac{(x-5)^n}{n!+2}\right\}\\ &=\sum_{n=1}^\infty \frac{n(x-5)^{n-1}}{n!+2} \end{align*}

Keep in mind that xx is our variable, and for each term, nn is constant.

Q3Stage 1

Let f(x)=∑n=a∞An(x−c)nf(x)=\displaystyle \sum_{n=a}^\infty A_n(x-c)^n for some positive constants aa and cc, and some sequence of constants {An}\{A_n\}. For which values of xx does f(x)f(x) definitely converge?

Hint

There is only one.

Answer

only x=cx=c

Full solution

If x=cx=c, then

f(x)=Aa(c−c)a+Aa+1(c−c)a+1+Aa+2(c−c)a+2+⋯=Aa⋅0+Aa+1⋅0+Aa+2⋅0+⋯=0\begin{align*} f(x)&=A_a(c-c)^a+A_{a+1}(c-c)^{a+1}+A_{a+2}(c-c)^{a+2}+\cdots\\ &=A_a \cdot 0+A_{a+1}\cdot 0+A_{a+2}\cdot 0+\cdots\\ &=0 \end{align*}

So, f(x)f(x) converges (to the constant 0) when x=cx=c. (Had we allowed a=0a=0, it would be possible for f(x)f(x) to converge to a nonzero number A0A_0, because we use the convention 00=10^0=1.)

Depending on the sequence {An}\{A_n\}, it's possible that f(x)f(x) diverges for all x≠cx \neq c. For example, suppose An=n!A_n=n!, so f(x)=∑n=0∞n!(x−c)nf(x)=\displaystyle\sum_{n=0}^\infty n!(x-c)^n. If x≠cx \neq c, then the limit lim⁡n→∞∣(n+1)!(x−c)n+1n!(x−c)n∣=lim⁡n→∞(n+1)∣x−c∣\displaystyle\lim_{n\to\infty} \left|\frac{(n+1)!(x-c)^{n+1}}{n!(x-c)^n}\right|=\lim_{n\to\infty}(n+1)|x-c| is infinity, since x−c≠0x-c\neq 0. So, the series diverges.

We've now shown that the series definitely converges at x=cx=c, but at any other point, it may fail to converge.

Q4Stage 1

Let f(x)f(x) be a power series centred at c=5c=5. If f(x)f(x) converges at x=−1x=-1, and diverges at x=11x=11, what is the radius of convergence of f(x)f(x)?

Hint

Use Theorem 3.5.9 in the CLP-2 text.

Answer

R=6R=6

Full solution

According to Theorem 3.5.9 in the CLP-2 text, because f(x)f(x) diverges somewhere, and because it converges at a point other than its centre, f(x)f(x) has a positive radius of convergence RR. That is, f(x)f(x) converges whenever ∣x−5∣<R|x-5|<R, and it diverges whenever ∣x−5∣>R|x-5|>R.

Since we are told that the series diverges at x=11x=11, the statement ∣11−5∣<R|11-5|<R must be false. That is, we must have R≤∣11−5∣=6R\le |11-5|=6.

Since we are told that the series converges at x=−1x=-1, the statement ∣−1−5∣>R|-1-5|>R must be false. That is, we must have R≥∣−1−5∣=6R\ge |-1-5|=6.

Therefore, R=6R=6.

2Stage 2Procedural

Practising the skill itself, until applying it is automatic.

Q5Stage 2Past exam · M105 2012A

(a) Find the radius of convergence of the series

∑k=0∞(−1)k2k+1xk\begin{equation*} \sum_{k=0}^\infty (-1)^k 2^{k+1} x^k \end{equation*}

(b) You are given the formula for the sum of a geometric series, namely:

1+r+r2+⋯=11−r,∣r∣<1\begin{equation*} 1+r+r^2 + \cdots =\frac{1}{1-r},\qquad|r|<1 \end{equation*}

Use this fact to evaluate the series in part (a).

Hint

Review the discussion immediately following Definition 3.5.1 in the

CLP-2 text.

Answer

(a) R=12R = \dfrac{1}{2} (b) 21+2x\dfrac{2}{1+2x} for all ∣x∣<12|x|<\dfrac{1}{2}

Full solution

(a) We give two solutions. For the first solution, we apply the ratio test for the series whose kthk^{\rm th} term is ak=(−1)k2k+1xka_k= (-1)^k 2^{k+1} x^k. Then

lim⁡k→∞∣ak+1ak∣=lim⁡k→∞∣(−1)k+12k+2xk+1(−1)k2k+1xk∣=lim⁡k→∞∣2x∣=∣2x∣\begin{align*} \lim_{k\to\infty} \bigg| \frac{ a_{k+1} }{ a_k } \bigg| &= \lim_{k\to\infty} \bigg| \frac{(-1)^{k+1} 2^{k+2} x^{k+1}} {(-1)^k 2^{k+1} x^k} \bigg| \\ &= \lim_{k\to\infty}|2x| = |2x| \end{align*}

Therefore, by the ratio test, the series converges for all xx obeying ∣2x∣<1|2x|<1, i.e. ∣x∣<12|x|<\frac{1}{2}, and diverges for all xx obeying ∣2x∣>1|2x|>1, i.e. ∣x∣>12|x|>\frac{1}{2}. So the radius of convergence is R=12R = \frac{1}{2}.

For the second solution we apply (3.7) in the CLP-2 text. To do so, we set Ak=(−1)k2k+1A_k = (-1)^k 2^{k+1} and compute

A=lim⁡k→∞∣Ak+1Ak∣=lim⁡k→∞∣(−1)k+12k+2(−1)k2k+1∣=lim⁡k→∞2=2\begin{equation*} A = \lim_{k\to\infty} \bigg| \frac{A_{k+1}}{A_k}\bigg| = \lim_{k\to\infty} \bigg| \frac{(-1)^{k+1} 2^{k+2}}{(-1)^k 2^{k+1}}\bigg| = \lim_{k\to\infty} 2 =2 \end{equation*}

so that R=1A=12R=\frac{1}{A}=\frac{1}{2}, again.

(b) The series is

∑k=0∞(−1)k2k+1xk=2∑k=0∞(−2x)k=2∑k=0∞rk∣r=−2x=2×11−r=21+2x\begin{align*} \sum_{k=0}^\infty (-1)^k 2^{k+1} x^k =2 \sum_{k=0}^\infty (-2x)^k =2\sum_{k=0}^\infty r^k\Big|_{r=-2x} =2\times\frac{1}{1-r} =\frac{2}{1+2x} \end{align*}

for all ∣r∣=∣2x∣<1|r|=|2x|<1, i.e. all ∣x∣<12|x|<\frac{1}{2}.

Q6Stage 2Past exam · M105 2014A

Find the radius of convergence for the power series ∑k=0∞xk10k+1(k+1)!\displaystyle\sum_{k=0}^\infty \frac{x^k}{10^{k+1}(k+1)!}

Hint

Review the discussion immediately following Definition 3.5.1 in the

CLP-2 text.

Answer

R=∞R = \infty

Full solution

We give two solutions. For the first solution, we apply the ratio test for the series whose kthk^{\rm th} term is ak=xk10k+1(k+1)!a_k= \frac{x^k}{ 10^{k+1}(k+1)! }. Then

lim⁡k→∞∣ak+1ak∣=lim⁡k→∞∣xk+110k+2(k+2)!⋅10k+1(k+1)!xk∣=lim⁡k→∞∣10k+110k+2∣⋅∣(k+1)!(k+2)!∣⋅∣xk+1xk∣=lim⁡k→∞110(k+2)∣x∣=0<1\begin{align*} \lim_{k\to\infty} \bigg| \frac{ a_{k+1} }{ a_k } \bigg| &= \lim_{k\to\infty} \bigg| \frac{x^{k+1}}{10^{k+2}(k+2)!} \cdot \frac{10^{k+1}(k+1)!}{x^k} \bigg| \\ &=\lim_{k\to\infty} \left|\frac{10^{k+1}}{10^{k+2}}\right|\cdot\left|\frac{(k+1)!}{(k+2)!}\right|\cdot\left|\frac{x^{k+1}}{x^k}\right|\\ &= \lim_{k\to\infty}\frac{1}{10(k+2)}|x| = 0 < 1 \end{align*}

for all xx. Therefore, by the ratio test, the series converges for all xx and the radius of convergence is R=∞R = \infty.

For the second solution we apply (3.7) in the CLP-2 text. To do so, we set Ak=110k+1(k+1)!A_k = \dfrac{1}{10^{k+1}(k+1)!} and compute

A=lim⁡k→∞∣Ak+1Ak∣=lim⁡k→∞∣10k+110k+2∣⋅∣(k+1)!(k+2)!∣=0\begin{align*} A = \lim_{k\to\infty} \left| \frac{A_{k+1}}{A_k}\right| =\lim_{k\to\infty} \left|\frac{10^{k+1}}{10^{k+2}}\right|\cdot\left|\frac{(k+1)!}{(k+2)!}\right| = 0 \end{align*}

So, by the bullet just after (3.7) in the CLP-2 text, RR is again +∞+\infty.

Q7Stage 2Past exam · 2014A

Find the radius of convergence for the power series ∑n=0∞(x−2)nn2+1\displaystyle\sum_{n=0}^\infty \frac{(x - 2)^n}{n^2+1} .

Hint

Review the discussion immediately following Definition 3.5.1 in the

CLP-2 text.

Answer

11

Full solution

We give two solutions. For the first solution, we apply the ratio test with an=(x−2)nn2+1a_n = \frac{(x - 2)^n}{n^2+1}.

lim⁡n→∞∣an+1an∣=lim⁡n→∞∣(x−2)n+1(n+1)2+1⋅n2+1(x−2)n∣=lim⁡n→∞n2+1(n+1)2+1∣x−2∣=lim⁡n→∞1+1/n2(1+1/n)2+1/n2∣x−2∣=∣x−2∣\begin{align*} \lim_{n\rightarrow\infty}\Big|\frac{a_{n+1}}{a_n}\Big| &=\lim_{n\rightarrow\infty}\bigg|\frac{(x-2)^{n+1}}{(n+1)^2+1}\cdot \frac{n^2+1}{(x-2)^n}\bigg|\\ &=\lim_{n\rightarrow\infty}\frac{n^2+1}{(n+1)^2+1} |x-2|\\ &=\lim_{n\rightarrow\infty}\frac{1+1/n^2}{(1+1/n)^2+1/n^2} |x-2|\\ &=|x-2| \end{align*}

So, the series converges if ∣x−2∣<1|x-2|<1 and diverges if ∣x−2∣>1|x-2|>1. That is, the radius of convergence is 1.

For the second solution we apply (3.7) in the CLP-2 text. To do so, we set An=1n2+1A_n = \frac{1}{n^2+1} and compute

A=lim⁡n→∞∣An+1An∣=lim⁡n→∞n2+1(n+1)2+1=lim⁡n→∞1+1/n2(1+1/n)2+1/n2=1\begin{equation*} A = \lim_{n\to\infty} \bigg| \frac{A_{n+1}}{A_n}\bigg| = \lim_{n\to\infty} \frac{n^2+1}{(n+1)^2+1} = \lim_{n\to\infty} \frac{1+1/n^2}{(1+1/n)^2+1/n^2} =1 \end{equation*}

so that R=1A=1R=\frac{1}{A}=1, again.

Q8Stage 2Past exam · 2013A

Consider the power series ∑n=1∞(−1)n(x+2)nn\displaystyle\sum\limits_{n=1}^\infty \frac{(-1)^n(x+2)^n}{\sqrt{n}}, where xx is a real number. Find the interval of convergence of this series.

Hint

See Example 3.5.11 in the

CLP-2 text.

Answer

The interval of convergence is −1<x+2≤1-1<x+2\le 1 or (−3,−1](-3,-1].

Full solution

We will first find the radius of convergence RR of the given series. Once we have found RR, we will know that the series converges for ∣x+2∣<R|x+2|<R and diverges for ∣x+2∣>R|x+2|>R. Then we will determine what happens at the two end points, x+2=Rx+2=R and x+2=−Rx+2=-R, of the interval.

To determine the radius of convergence, we apply (3.7) in the CLP-2 text. To do so, we set An=(−1)nnA_n= \frac{(-1)^n}{\sqrt{n}} and compute

A=lim⁡n→∞∣An+1An∣=lim⁡n→∞nn+1=lim⁡n→∞11+1/n=1\begin{align*} A = \lim_{n\to\infty} \bigg| \frac{A_{n+1}}{A_n}\bigg| &=\lim_{n\rightarrow\infty} \frac{\sqrt{n}}{\sqrt{n+1}} \\ &=\lim_{n\rightarrow\infty} \frac{1}{\sqrt{1+1/n}} \\ &=1 \end{align*}

so that R=1A=1R=\frac{1}{A}=1.

So the series must converge when ∣x+2∣<1|x+2|<1 and must diverge when ∣x+2∣>1|x+2|>1. When x+2=1x+2=1, the series reduces to

∑n=1∞(−1)nn\begin{align*} \sum_{n=1}^\infty \frac{(-1)^n}{\sqrt{n}} \end{align*}

which converges by the alternating series test. When x+2=−1x+2=-1, the series reduces to

∑n=1∞1n\begin{align*} \sum_{n=1}^\infty \frac{1}{\sqrt{n}} \end{align*}

which diverges by the pp–series test with p=12p=\frac{1}{2}. So the interval of convergence is −1<x+2≤1-1<x+2\le 1 or (−3,−1](-3,-1].

Q9Stage 2Past exam · 2016Q5

Find the radius of convergence and interval of convergence of the series

∑n=0∞(−1)nn+1(x+13)n\begin{align*} \sum_{n=0}^{\infty} \frac{(-1)^n}{n+1} \left(\frac{x+1}{3}\right)^n \end{align*}
Hint

See Example 3.5.11 in the

CLP-2 text.

Answer

The radius of convergence is 33. The interval of convergence is −4<x≤2-4<x\le2, or simply (−4,2](-4,2].

Full solution

We will first find the radius of convergence RR of the given series. Once we have found RR, we will know that the series converges for ∣x+1∣<R|x+1|<R and diverges for ∣x+1∣>R|x+1|>R. Then we will determine what happens at the two end points, x+1=Rx+1=R and x+1=−Rx+1=-R, of the interval.

We could determine the radius of convergence by applying (3.7), from the CLP-2 text, with An=(−1)n(n+1)3nA_n = \frac{(-1)^n}{(n+1)3^n}. As an alternate method, we apply the ratio test for the series whose nthn^{\rm th} term is an=(−1)nn+1(x+13)na_n= \frac{(-1)^n}{n+1} \left(\frac{x+1}{3}\right)^n.

lim⁡n→∞∣an+1an∣=lim⁡n→∞∣(−1)n+1n+2(x+13)n+1(−1)nn+1(x+13)n∣=lim⁡n→∞∣(−1)n+1(−1)n∣⋅∣n+1n+2∣⋅∣(x+1)n+1(x+1)n∣⋅∣3n3n+1∣=lim⁡n→∞(n+1n+2)⋅∣x+13∣=∣x+1∣3\begin{align*} \lim_{n\to\infty} \bigg| \frac{a_{n+1}}{a_n} \bigg| &= \lim_{n\to\infty} \bigg| \frac{ \frac{(-1)^{n+1}}{n+2} \left(\frac{x+1}{3}\right)^{n+1}} { \frac{(-1)^n}{n+1} \left(\frac{x+1}{3}\right)^n} \bigg| \\ &= \lim_{n\to\infty}\left|\frac{(-1)^{n+1}}{(-1)^n} \right|\cdot\left|\frac{n+1}{n+2} \right|\cdot\left|\frac{(x+1)^{n+1}}{(x+1)^n} \right|\cdot\left|\frac{3^n}{3^{n+1}} \right| \\ &= \lim_{n\to\infty} \left(\frac{n+1}{n+2}\right)\cdot \bigg| \frac{x+1}3 \bigg| = \frac{|x+1|}3 \end{align*}

Therefore, by the ratio test, the series converges when ∣x+1∣3<1\frac{|x+1|}3 < 1 and diverges when ∣x+1∣3>1\frac{|x+1|}3 > 1. In particular, it converges when

∣x+1∣<3  ⟺  −3<x+1<3  ⟺  −4<x<2\begin{align*} |x+1| < 3 \iff -3 < x+1 < 3 \iff -4 < x < 2 \end{align*}

and the radius of convergence is R=3R = 3.

Next, we consider the endpoints 22 and −4-4. At x=2x=2, i.e. x+1=3x+1=3, the series is simply ∑n=0∞(−1)nn+1\sum_{n=0}^\infty \frac{(-1)^{n}}{n+1}, which is an alternating series: the signs alternate, and the unsigned terms decrease to zero. Therefore the series converges at x=2x=2 by the alternating series test.

At x=−4x=-4 the series is

∑n=0∞(−1)nn+1(−4+13)n=∑n=0∞(−1)nn+1(−1)n=∑n=0∞1n+1,\sum_{n=0}^\infty \frac{(-1)^{n}}{n+1} \left(\frac{-4+1}{3}\right)^n = \sum_{n=0}^\infty \frac{(-1)^{n}}{n+1} (-1)^n = \sum_{n=0}^\infty \frac1{n+1},

since (−1)n⋅(−1)n=(−1)2n=((−1)2)n=1(-1)^n \cdot (-1)^n = (-1)^{2n} = \left((-1)^2\right)^n = 1. This series diverges, either by comparison or limit comparison with the harmonic series (the pp-series with p=1p=1). (For that matter, it is exactly equal to the standard harmonic series ∑n=1∞1n\sum_{n=1}^\infty \frac{1}{n}, re-indexed to start at n=0n=0.)

In summary, the interval of convergence is −4<x≤2-4<x\le2, or simply (−4,2](-4,2].

Q10Stage 2Past exam · 2015A

Find the interval of convergence for the power series

∑n=1∞(x−2)nn4/5(5n−4).\begin{align*} \sum_{n=1}^\infty \frac{(x-2)^n}{n^{4/5}(5^n-4)}. \end{align*}
Answer

−3≤x<7-3\le x< 7 or [−3,7)[-3,7)

Full solution

We will first find the radius of convergence RR of the given series. Once we have found RR, we will know that the series converges for ∣x−2∣<R|x-2|<R and diverges for ∣x−2∣>R|x-2|>R. Then we will determine what happens at the two end points, x−2=Rx-2=R and x−2=−Rx-2=-R, of the interval.

To determine the radius of convergence, we apply (3.7) in the CLP-2 text. To do so, we set An=1n4/5(5n−4)A_n= \frac{1}{n^{4/5}(5^n-4)} and compute

Alim⁡n→∞∣An+1An∣=lim⁡n→∞n4/5(5n−4)(n+1)4/5(5n+1−4)=lim⁡n→∞(1−4/5n)(1+1/n)4/5(5−4/5n)=15\begin{align*} A & \lim_{n\to\infty} \bigg| \frac{A_{n+1}}{A_n}\bigg| \\ &=\lim_{n\rightarrow\infty}\frac{n^{4/5}(5^n-4)}{(n+1)^{4/5}(5^{n+1}-4)} \\ &=\lim_{n\rightarrow\infty}\frac{(1-4/5^n)}{(1+1/n)^{4/5}(5-4/5^n)} \\ &=\frac{1}{5} \end{align*}

so that R=1A=5R=\frac{1}{A}=5.

Therefore the series converges if ∣x−2∣<5|x-2|<5 and diverges if ∣x−2∣>5|x-2|>5. When x−2=+5x-2=+5, i.e. x=7x=7, the series reduces to ∑n=1∞5nn4/5(5n−4)=∑n=1∞1n4/5(1−4/5n)\sum\limits_{n=1}^\infty \frac{5^n}{n^{4/5}(5^n-4)} =\sum\limits_{n=1}^\infty \frac{1}{n^{4/5}(1-4/5^n)} which diverges by the limit comparison test with bn=1n4/5b_n=\frac{1}{n^{4/5}}. When x−2=−5x-2=-5, i.e. x=−3x=-3, the series reduces to ∑n=1∞(−5)nn4/5(5n−4)=∑n=1∞(−1)nn4/5(1−4/5n)\sum\limits_{n=1}^\infty \frac{(-5)^n}{n^{4/5}(5^n-4)} =\sum\limits_{n=1}^\infty \frac{(-1)^n}{n^{4/5}(1-4/5^n)} which converges by the alternating series test. So the interval of convergence is −3≤x<7-3\le x< 7 or [−3,7)[-3,7).

Q11Stage 2Past exam · 2016Q6

Find all values xx for which the series ∑n=1∞(x+2)nn2\displaystyle\sum_{n=1}^\infty\frac{(x+2)^n}{n^2} converges.

Answer

The given series converges if and only if −3≤x≤−1-3\le x\le -1. Equivalently, the series has interval of convergence [−3,−1][-3,-1].

Full solution

We determine the radius of convergence by applying (3.7) in the CLP-2 text. To do so, we set An=1n2A_n= \frac{1}{n^2} and compute

A=lim⁡n→∞∣An+1An∣=lim⁡n→∞n2(n+1)2=lim⁡n→∞1(1+1/n)2=1\begin{align*} A &= \lim_{n\to\infty} \bigg| \frac{A_{n+1}}{A_n}\bigg| \\ &=\lim_{n\rightarrow\infty}\frac{n^2}{{(n+1)}^2} \\ &=\lim_{n\rightarrow\infty}\frac{1}{{(1+1/n)}^2} \\ &=1 \end{align*}

So we have convergence for

∣x+2∣<1  ⟺  −1<x+2<1  ⟺  −3<x<−1\begin{equation*} |x+2|<1 \iff -1<x+2<1 \iff -3<x<-1 \end{equation*}

and divergence for ∣x+2∣>1|x+2|>1. For ∣x+2∣=1|x+2|=1, i.e. for x+2=±1x+2=\pm 1, i.e. for x=−3,−1x=-3,-1, the series reduces to ∑n=1∞(±1)nn2\sum\limits_{n=1}^\infty\frac{(\pm 1)^n}{n^2}, which converges absolutely, because ∑n=1∞1np\sum\limits_{n=1}^\infty\frac{1}{n^p} converges for p=2>1p=2>1. So the given series converges if and only if −3≤x≤−1-3\le x\le -1.

Q12Stage 2Past exam · 2016Q6

Find the interval of convergence for the following series.

  1. ∑n=1∞4nn(x−1)n\displaystyle \sum_{n=1}^\infty \frac{4^n}{n}(x-1)^n.

  2. ∑n=1∞4nn(2x+1)n\displaystyle \sum_{n=1}^\infty \frac{4^n}{n}(2x+1)^n.

  3. ∑n=1∞4nn(2x+1)2n\displaystyle \sum_{n=1}^\infty \frac{4^n}{n}(2x+1)^{2n}.

Answer

(a) 34≤x<54\frac{3}{4}\le x<\frac{5}{4} or [34,54)\big[\frac{3}{4},\frac{5}{4}\big) (b) −58≤x<−38-\frac{5}{8}\le x<-\frac{3}{8} or [−58,−38)\big[-\frac{5}{8},-\frac{3}{8}\big) (c) −34<x<−14-\frac{3}{4}< x<-\frac{1}{4}, or (−34,−14)\big(-\frac{3}{4},-\frac{1}{4}\big)

Full solution

(a) We could determine the radius of convergence by applying (3.7), from the CLP-2 text, with An=4nnA_n = \frac{4^n}{n}. As an alternate method, we apply the ratio test for the series whose nthn^{\rm th} term is an=4nn(x−1)na_n= \frac{4^n}{n}(x-1)^n. Since

lim⁡n→∞∣an+1an∣=lim⁡n→∞∣4n+1(x−1)n+1/(n+1)4n(x−1)n/n∣=lim⁡n→∞4∣x−1∣nn+1=4∣x−1∣lim⁡n→∞nn+1=4∣x−1∣⋅1.\begin{align*} \lim_{n\to\infty} \bigg| \frac{a_{n+1}}{a_n} \bigg| &= \lim_{n\to\infty} \bigg| \frac{4^{n+1}(x-1)^{n+1}/(n+1)}{4^n(x-1)^n/n} \bigg| \\ &= \lim_{n\to\infty} 4|x-1| \frac{n}{n+1} \\ &= 4|x-1| \lim_{n\to\infty} \frac{n}{n+1} = 4|x-1|\cdot1. \end{align*}

the series converges if

4∣x−1∣<1  ⟺  −1<4(x−1)<1  ⟺  −14<x−1<14  ⟺  34<x<54\begin{equation*} 4|x-1|<1 \iff -1<4(x-1)<1 \iff -\frac{1}{4}<x-1<\frac{1}{4} \iff \frac{3}{4}<x<\frac{5}{4} \end{equation*}

and diverges if 4∣x−1∣>14|x-1|>1. Checking the right endpoint x=54x=\frac{5}{4}, we see that

∑n=1∞4nn(54−1)n=∑n=1∞1n\begin{align*} \sum_{n=1}^\infty \frac{4^n}{n}\bigg( \frac{5}{4}-1 \bigg)^n = \sum_{n=1}^\infty \frac{1}{n} \end{align*}

is the divergent harmonic series. At the left endpoint x=34x=\frac{3}{4},

∑n=1∞4nn(34−1)n=∑n=1∞(−1)nn\begin{align*} \sum_{n=1}^\infty \frac{4^n}{n}\bigg( \frac34-1 \bigg)^n = \sum_{n=1}^\infty \frac{(-1)^n}{n} \end{align*}

converges by the alternating series test. Therefore the interval of convergence of the original series is 34≤x<54\frac{3}{4}\le x<\frac{5}{4}, or [34,54)\big[\frac{3}{4},\frac{5}{4}\big).

(b) We could determine the radius of convergence by applying (3.7), from the CLP-2 text. But to do so, we would first have to rewrite the series as ∑n=1∞8nn(x+12)n\sum_{n=1}^\infty \frac{8^n}{n}\big(x+\frac{1}{2}\big)^n to get the nthn^{\rm th} term into the form An(x−c)nA_n (x-c)^n required by (3.7). Instead, we use the alternate method that consists of applying the ratio test with an=4nn(2x+1)na_n= \frac{4^n}{n}(2x+1)^n. Repeating the computation of part (a), just with x−1x-1 replaced by 2x+12x+1,

lim⁡n→∞∣an+1an∣=lim⁡n→∞∣4n+1(2x+1)n+1/(n+1)4n(2x+1)n/n∣=4∣2x+1∣lim⁡n→∞nn+1=4∣2x+1∣⋅1.\begin{align*} \lim_{n\to\infty} \bigg| \frac{a_{n+1}}{a_n} \bigg| &= \lim_{n\to\infty} \bigg| \frac{4^{n+1}(2x+1)^{n+1}/(n+1)}{4^n(2x+1)^n/n} \bigg| \\ &= 4|2x+1| \lim_{n\to\infty} \frac{n}{n+1} = 4|2x+1|\cdot1. \end{align*}

So the series converges if

4∣2x+1∣<1  ⟺  −1<4(2x+1)<1  ⟺  −14<2x+1<14  ⟺  −54<2x<−34  ⟺  −58<x<−38\begin{align*} 4|2x+1|<1 &\iff -1<4(2x+1)<1 \\ &\iff -\frac{1}{4}<2x+1<\frac{1}{4} \\ &\iff -\frac{5}{4}<2x<-\frac{3}{4} \\ &\iff -\frac{5}{8}<x<-\frac{3}{8} \end{align*}

and diverges if 4∣2x+1∣>14|2x+1|>1. At the right endpoint x=−38x=-\frac{3}{8}, the series becomes

∑n=1∞4nn(−34+1)n=∑n=1∞1n\begin{align*} \sum_{n=1}^\infty \frac{4^n}{n}\bigg( -\frac{3}{4}+1 \bigg)^n = \sum_{n=1}^\infty \frac{1}{n} \end{align*}

which is the divergent harmonic series. At the left endpoint x=−58x=-\frac{5}{8},

∑n=1∞4nn(−54+1)n=∑n=1∞(−1)nn\begin{align*} \sum_{n=1}^\infty \frac{4^n}{n}\bigg( -\frac54+1 \bigg)^n = \sum_{n=1}^\infty \frac{(-1)^n}{n} \end{align*}

converges by the alternating series test. Therefore the interval of convergence of the original series is −58≤x<−38-\frac{5}{8}\le x<-\frac{3}{8}, or [−58,−38)\big[-\frac{5}{8},-\frac{3}{8}\big).

(c) This time we cannot determine the radius of convergence by applying (3.7), from the CLP-2 text, because the nthn^{\rm th} term is not in the form An(x−c)nA_n (x-c)^n required by (3.7). But we may still apply the ratio test with an=4nn(2x+1)2na_n= \frac{4^n}{n}(2x+1)^{2n}. Repeating the computation of part (a), just with x−1x-1 replaced by (2x+1)2(2x+1)^2,

lim⁡n→∞∣an+1an∣=lim⁡n→∞∣4n+1(2x+1)2(n+1)/(n+1)4n(2x+1)2n/n∣=4∣2x+1∣2lim⁡n→∞nn+1=4∣2x+1∣2⋅1.\begin{align*} \lim_{n\to\infty} \bigg| \frac{a_{n+1}}{a_n} \bigg| &= \lim_{n\to\infty} \bigg| \frac{4^{n+1}(2x+1)^{2(n+1)}/(n+1)}{4^n(2x+1)^{2n}/n} \bigg| \\ &= 4|2x+1|^2 \lim_{n\to\infty} \frac{n}{n+1} = 4|2x+1|^2\cdot1. \end{align*}

So the series converges if

4∣2x+1∣2<1  ⟺  −1<2(2x+1)<1  ⟺  −12<2x+1<12  ⟺  −32<2x<−12  ⟺  −34<x<−14\begin{align*} 4|2x+1|^2<1 &\iff -1<2(2x+1)<1 \\ &\iff -\frac{1}{2}<2x+1<\frac{1}{2} \\ &\iff -\frac{3}{2}<2x<-\frac{1}{2} \\ &\iff -\frac{3}{4}<x<-\frac{1}{4} \end{align*}

and diverges if 4∣2x+1∣2>14|2x+1|^2>1. At the right endpoint x=−14x=-\frac{1}{4}, the series becomes

∑n=1∞4nn(−12+1)2n=∑n=1∞1n\begin{align*} \sum_{n=1}^\infty \frac{4^n}{n}\bigg( -\frac{1}{2}+1 \bigg)^{2n} = \sum_{n=1}^\infty \frac{1}{n} \end{align*}

which is the divergent harmonic series. At the left endpoint x=−34x=-\frac{3}{4},

∑n=1∞4nn(−32+1)2n=∑n=1∞1n\begin{align*} \sum_{n=1}^\infty \frac{4^n}{n}\bigg( -\frac32+1 \bigg)^{2n} = \sum_{n=1}^\infty \frac{1}{n} \end{align*}

which is again the divergent harmonic series. Therefore the interval of convergence of the original series is −34<x<−14-\frac{3}{4}< x<-\frac{1}{4}, or (−34,−14)\big(-\frac{3}{4},-\frac{1}{4}\big).

Q13Stage 2Past exam · 2012A

Find, with explanation, the radius of convergence and the interval of convergence of the power series

∑n=0∞(−1)n(x−1)n2n(n+2)\begin{equation*} \sum_{n=0}^\infty (-1)^n\frac{(x-1)^n}{2^n(n+2)} \end{equation*}
Answer

The radius of convergence is 22. The interval of convergence is −1<x≤3-1< x\le3, or (−1,3]\big(-1,3\big].

Full solution

We determine the radius of convergence by applying (3.7) in the CLP-2 text. To do so, we set An=(−1)n2n(n+2)A_n= \frac{(-1)^n}{2^n(n+2)} and compute

A=lim⁡n→∞∣An+1An∣=lim⁡n→∞2n(n+2)2n+1(n+3)=lim⁡n→∞12 n+2n+3=12lim⁡n→∞1+2/n1+3/n=12\begin{align*} A &= \lim_{n\to\infty} \bigg| \frac{A_{n+1}}{A_n}\bigg| \\ &= \lim_{n\to\infty} \frac{2^n(n+2)}{2^{n+1}(n+3)} \\ &= \lim_{n\to\infty} \frac{1}{2}\ \frac{n+2}{n+3} \\ &= \frac{1}{2} \lim_{n\to\infty} \frac{1+2/n}{1+3/n} = \frac{1}{2} \end{align*}

So the radius of convergence is R=1A=2R=\frac{1}{A}=2 and the series converges if

∣x−1∣<2  ⟺  −2<(x−1)<2  ⟺  −1<x<3\begin{align*} |x-1|<2 \iff -2<(x-1)<2 \iff -1<x<3 \end{align*}

and diverges if ∣x−1∣>2|x-1|>2. Checking the left endpoint x=−1x=-1, we see that

∑n=0∞(−1)n(−1−1)n2n(n+2)=∑n=0∞1n+2\begin{align*} \sum_{n=0}^\infty (-1)^n\frac{(-1-1)^n}{2^n(n+2)} = \sum_{n=0}^\infty \frac{1}{n+2} \end{align*}

is the divergent harmonic series. At the right endpoint x=3x=3, so that

∑n=0∞(−1)n(3−1)n2n(n+2)=∑n=0∞(−1)nn+2\begin{align*} \sum_{n=0}^\infty (-1)^n\frac{(3-1)^n}{2^n(n+2)} = \sum_{n=0}^\infty \frac{(-1)^n}{n+2} \end{align*}

converges by the alternating series test. Therefore the interval of convergence of the original series is −1<x≤3-1< x\le 3, or (−1,3]\big(-1,3\big].

Q14Stage 2Past exam · 2014D

Find the interval of convergence for the series ∑n=1∞(−1)nn2(x−a)2n\displaystyle \sum_{n=1}^\infty (-1)^n n^2(x-a)^{2n} where aa is a constant.

Answer

The interval of convergence is a−1<x<a+1a-1< x<a+1, or (a−1,a+1)\big(a-1,a+1\big).

Full solution

We cannot determine the radius of convergence by applying (3.7), from the CLP-2 text, because the nthn^{\rm th} term is not in the form An(x−c)nA_n (x-c)^n required by (3.7). But we may still apply the ratio test with an=(−1)nn2(x−a)2na_n= (-1)^n n^2(x-a)^{2n}. Since

lim⁡n→∞∣an+1an∣=lim⁡n→∞∣(−1)n+1(n+1)2(x−a)2(n+1)(−1)nn2(x−a)2n∣=lim⁡n→∞∣x−a∣2(n+1)2n2=∣x−a∣2lim⁡n→∞(1+1 ⁣/n)2=∣x−a∣2⋅1.\begin{align*} \lim_{n\to\infty} \bigg| \frac{a_{n+1}}{a_n} \bigg| &= \lim_{n\to\infty} \bigg| \frac{(-1)^{n+1} (n+1)^2(x-a)^{2(n+1)}} {(-1)^n n^2(x-a)^{2n}} \bigg| \\ &= \lim_{n\to\infty} |x-a|^2 \frac{(n+1)^2}{n^2} \\ &= |x-a|^2 \lim_{n\to\infty} \big(1+\nicefrac{1}{n}\big)^2 = |x-a|^2\cdot1. \end{align*}

the series converges if

∣x−a∣2<1  ⟺  ∣x−a∣<1  ⟺  −1<x−a<1  ⟺  a−1<x<a+1\begin{equation*} |x-a|^2<1 \iff |x-a|<1 \iff-1<x-a<1 \iff a-1<x<a+1 \end{equation*}

and diverges if ∣x−a∣>1|x-a|>1. Checking both endpoints x−a=±1x-a=\pm 1, we see that

∑n=1∞(−1)nn2(x−a)2n∣x−a=±1= ∑n=1∞(−1)nn2\begin{align*} \sum_{n=1}^\infty (-1)^n n^2(x-a)^{2n}\bigg|_{x-a=\pm 1} =\ \sum_{n=1}^\infty (-1)^n n^2 \end{align*}

fails the divergence test — the nthn^{\rm th} term does not converge to zero as n→∞n\rightarrow\infty. Therefore the interval of convergence of the original series is a−1<x<a+1a-1< x<a+1, or (a−1,a+1)\big(a-1,a+1\big).

Q15Stage 2Past exam · 2015A

Find the interval of convergence of the following series:

  1. ∑k=1∞(x+1)kk29k{\displaystyle \sum_{k=1}^\infty \frac{(x+1)^k}{k^2 9^k}}.

  2. ∑k=1∞ak(x−1)k{\displaystyle \sum_{k=1}^\infty a_k(x-1)^k}, where ak>0a_k>0 for k=1,2,⋯k=1,2,\cdots and ∑k=1∞(akak+1−ak+1ak+2)=a1a2{\displaystyle \sum_{k=1}^\infty \Big(\frac{a_k}{a_{k+1}} -\frac{a_{k+1}}{a_{k+2}}\Big) =\frac{a_1}{a_2} }.

Hint

Start part (b) by computing the partial sums of ∑k=1∞(akak+1−ak+1ak+2){\displaystyle \sum_{k=1}^\infty \Big(\frac{a_k}{a_{k+1}} -\frac{a_{k+1}}{a_{k+2}}\Big)}

Answer

(a) ∣x+1∣≤9|x+1|\le 9 or −10≤x≤8-10\le x\le 8 or [−10,8][-10,8] (b) This series converges only for x=1x=1.

Full solution

(a) We could determine the radius of convergence by applying (3.7), from the CLP-2 text, with An=1n29nA_n = \frac{1}{n^2 9^n}. As an alternate method, we apply the ratio test for the series whose kthk^{\rm th} term is ak=(x+1)kk29ka_k=\frac{(x+1)^k}{k^2 9^k}. Then

lim⁡k→∞∣ak+1ak∣=lim⁡k→∞∣(x+1)k+1(k+1)29k+1 k29k(x+1)k∣=lim⁡k→∞∣x+1∣ 19 k2(k+1)2=lim⁡k→∞∣x+1∣ 19 1(1+1/k)2=∣x+1∣9\begin{align*} \lim_{k\rightarrow\infty}\bigg|\frac{a_{k+1}}{a_k}\bigg| &=\lim_{k\rightarrow\infty} \bigg|\frac{(x+1)^{k+1}}{(k+1)^2 9^{k+1}} \ \frac{k^2 9^k}{(x+1)^k}\bigg| \\ &=\lim_{k\rightarrow\infty} |x+1|\ \frac{1}{9}\ \frac{k^2}{(k+1)^2} \\ &=\lim_{k\rightarrow\infty} |x+1|\ \frac{1}{9}\ \frac{1}{(1+1/k)^2} \\ &=\frac{|x+1|}{9} \end{align*}

So the series must converge when ∣x+1∣<9|x+1|<9 and must diverge when ∣x+1∣>9|x+1|>9. When x+1=±9x+1=\pm 9, the series reduces to

∑k=1∞(±9)kk29k=∑k=1∞(±1)kk2\begin{align*} \sum_{k=1}^\infty \frac{(\pm 9)^k}{k^2 9^k} =\sum_{k=1}^\infty \frac{(\pm 1)^k}{k^2} \end{align*}

which converges (since, by the pp–test, ∑k=1∞1kp\sum_{k=1}^\infty\frac{1}{k^p} converges for any p>1p>1). So the interval of convergence is ∣x+1∣≤9|x+1|\le 9 or −10≤x≤8-10\le x\le 8 or [−10,8][-10,8].

(b) The partial sum

∑k=1N(akak+1−ak+1ak+2)=(a1a2−a2a3)+(a2a3−a3a4)+⋯+(aNaN+1−aN+1aN+2)=a1a2−aN+1aN+2\begin{align*} \sum_{k=1}^N\Big(\frac{a_k}{a_{k+1}} -\frac{a_{k+1}}{a_{k+2}}\Big) =\Big(\frac{a_1}{a_2}-\frac{a_2}{a_3}\Big) +\Big(\frac{a_2}{a_3}-\frac{a_3}{a_4}\Big) +\cdots+ \Big(\frac{a_N}{a_{N+1}}-\frac{a_{N+1}}{a_{N+2}}\Big) =\frac{a_1}{a_2}-\frac{a_{N+1}}{a_{N+2}} \end{align*}

We are told that ∑k=1∞(akak+1−ak+1ak+2)=a1a2{\displaystyle \sum_{k=1}^\infty \Big(\frac{a_k}{a_{k+1}} -\frac{a_{k+1}}{a_{k+2}}\Big) =\frac{a_1}{a_2} }. This means that the above partial sum converges to a1a2\frac{a_1}{a_2} as N→∞N\rightarrow\infty, or equivalently, that

lim⁡N→∞aN+1aN+2=0\begin{align*} \lim_{N\rightarrow\infty}\frac{a_{N+1}}{a_{N+2}}=0 \end{align*}

and hence that

lim⁡k→∞∣ak+1(x−1)k+1∣∣ak(x−1)k∣=∣x−1∣lim⁡k→∞∣ak+1∣∣ak∣\begin{align*} \lim_{k\rightarrow\infty}\frac{|a_{k+1}(x-1)^{k+1}|}{|a_k(x-1)^k|} =|x-1|\lim_{k\rightarrow\infty}\frac{|a_{k+1}|}{|a_k|} \end{align*}

is infinite for any x≠1x\ne 1. So, by the ratio test, this series converges only for x=1x=1.

Q16Stage 2Past exam · 2014A

Find a power series representation for x31−x\dfrac{x^3}{1-x}.

Hint

You should know a power series representation for 11−x\dfrac{1}{1-x}. Use it.

Answer

∑n=0∞xn+3=∑n=3∞xn\displaystyle\sum\limits_{n=0}^\infty x^{n+3} =\sum\limits_{n=3}^\infty x^{n}

Full solution

Using the geometric series ∑n=0∞xn=11−x\sum\limits_{n=0}^\infty x^n = \frac{1}{1-x},

x31−x=x3∑n=0∞xn=∑n=0∞xn+3=∑n=3∞xn\begin{align*} \frac{x^3}{1-x} =x^3 \sum_{n=0}^\infty x^n =\sum_{n=0}^\infty x^{n+3} =\sum_{n=3}^\infty x^{n} \end{align*}
Q17Stage 2

Suppose f′(x)=∑n=0∞(x−1)nn+2f'(x)=\displaystyle\sum_{n=0}^\infty \frac{(x-1)^{n}}{n+2}, and ∫5xf(t)dt=3x+∑n=1∞(x−1)n+1n(n+1)2\displaystyle \int_5^x f(t)\dee{t}=3x+\displaystyle\sum_{n=1}^\infty \frac{(x-1)^{n+1}}{n(n+1)^2}.

Give a power series representation of f(x)f(x).

Hint

You can safely ignore one of the given equations, but not the other.

Answer

f(x)=3+∑n=1∞(x−1)nn(n+1)f(x)=3+\displaystyle\sum_{n=1}^\infty \frac{(x-1)^{n}}{n(n+1)}

Full solution

We can find f(x)f(x) by differentiating its integral, or antidifferentiating its derivative. In the latter case, we'll have to solve for the arbitrary constant of integration; in the former case, we do not. (Remember that many different functions have the same derivative, but a single function has only one derivative.) To avoid the necessity of finding the arbitrary constant, we can ignore the given equation for f′(x)f'(x), which makes the problem much simpler. This is the method used in Solution 1.

  • Using the Fundamental Theorem of Calculus Part 1:

    ddx{∫5xf(t)dt}=f(x)So,f(x)=ddx{3x+∑n=0∞(x−1)n+1n(n+1)2}=3+∑n=1∞(n+1)(x−1)nn(n+1)2=3+∑n=1∞(x−1)nn(n+1)\begin{align*} \diff{}{x} \left\{\int_5^x f(t)\dee{t} \right\}&=f(x)\\ \text{So,}\qquad f(x)&=\diff{}{x} \left\{3x+\displaystyle\sum_{n=0}^\infty \frac{(x-1)^{n+1}}{n(n+1)^2} \right\}\\ &=3+\displaystyle\sum_{n=1}^\infty \frac{(n+1)(x-1)^{n}}{n(n+1)^2}\\ &=3+\displaystyle\sum_{n=1}^\infty \frac{(x-1)^{n}}{n(n+1)} \end{align*}

  • Suppose we had used f′(x)f'(x) instead. We would antidifferentiate to find:

    f(x)=∫(∑n=0∞(x−1)nn+2)dx=(∑n=0∞(x−1)n+1(n+1)(n+2))+C=(∑n=1∞(x−1)nn(n+1))+C\begin{align*} f(x)&=\int \left( \sum_{n=0}^\infty \frac{(x-1)^{n}}{n+2} \right)\dee{x}\\ &=\left(\sum_{n=0}^\infty \frac{(x-1)^{n+1}}{(n+1)(n+2)}\right)+C\\ &=\left(\sum_{n=1}^\infty \frac{(x-1)^{n}}{n(n+1)}\right)+C \end{align*}

    Notice f(1)=0+Cf(1)=0+C. So, to find CC, we must find f(1)f(1). We can't get that information from f′(x)f'(x), so our only option is to consider the given formula for ∫5xf(t)dt\int_5^x f(t)\dee{t}. Using the Fundamental Theorem of Calculus Part 1:

    f(1)=ddx{∫5xf(t)dt}∣x=1=ddx{3x+∑n=1∞(x−1)n+1n(n+1)2}∣x=1=[3+∑n=1∞(n+1)(x−1)nn(n+1)2]x=1=[3+∑n=1∞(x−1)nn(n+1)]x=1=3+∑n=1∞0nn(n+1)=3\begin{align*} f(1)&=\left.\diff{}{x}\left\{\int_5^x f(t)\dee{t} \right\}\right|_{x=1}\\ &=\left.\diff{}{x}\left\{ 3x+\displaystyle\sum_{n=1}^\infty \frac{(x-1)^{n+1}}{n(n+1)^2} \right\}\right|_{x=1}\\ &=\left[ 3+\displaystyle\sum_{n=1}^\infty \frac{(n+1)(x-1)^{n}}{n(n+1)^2} \right]_{x=1}\\ &=\left[ 3+\displaystyle\sum_{n=1}^\infty \frac{(x-1)^{n}}{n(n+1)} \right]_{x=1}\\ &= 3+\displaystyle\sum_{n=1}^\infty \frac{0^{n}}{n(n+1)}\\ &=3 \end{align*}

    So, f(x)=3+∑n=1∞(x−1)nn(n+1)f(x)=3+\displaystyle\sum_{n=1}^\infty \frac{(x-1)^{n}}{n(n+1)}.

    Note that in Solution 2, we did the same calculation as Solution 1, and more.

3Stage 3Application

Further than practice: several ideas at once, or an unfamiliar situation.

Q18Stage 3Past exam · M121 2002A

Determine the values of xx for which the series

∑n=2∞xn32nlog⁡n\begin{equation*} \sum_{n=2}^\infty\frac{x^n}{3^{2n}\log n} \end{equation*}

converges absolutely, converges conditionally, or diverges.

Hint

n≥log⁡nn\ge\log n for all n≥1n\ge 1.

Answer

The series converges absolutely for ∣x∣<9|x|<9, converges conditionally for x=−9x=-9 and diverges otherwise.

Full solution

We'll start by applying the ratio test twice,

  • once for the series whose nthn^{\rm th} term is an=xn32nlog⁡na_n=\frac{x^n}{3^{2n}\log n}, and

  • once for the series whose nthn^{\rm th} term is an=∣xn32nlog⁡n∣a_n=\big|\frac{x^n}{3^{2n}\log n}\big|.

For both series, i.e. for both ana_n's,

lim⁡n→∞∣an+1an∣=lim⁡n→∞∣xn+132(n+1)log⁡(n+1) 32nlog⁡nxn∣=lim⁡n→∞∣xlog⁡n32log⁡(n+1)∣=lim⁡n→∞∣xlog⁡n32[log⁡(n)+log⁡(1+1/n)]∣=lim⁡n→∞∣x32[1+log⁡(1+1/n)/log⁡(n)]∣=∣x∣9\begin{align*} \lim_{n\to\infty} \bigg| \frac{ a_{n+1} }{ a_n } \bigg| &= \lim_{n\to\infty} \bigg| \frac{x^{n+1}}{3^{2(n+1)}\log(n+1)} \ \frac{3^{2n}\log n}{x^n} \bigg| \\ &= \lim_{n\to\infty} \bigg| \frac{x\log n}{3^2\log(n+1)} \bigg| =\lim_{n\to\infty} \bigg| \frac{x\log n}{3^2[\log(n)+\log(1+1/n)]} \bigg| \\ &=\lim_{n\to\infty} \bigg| \frac{x}{3^2[1+\log(1+1/n)/\log(n)]} \bigg| \\ &=\frac{|x|}{9} \end{align*}

Therefore, by the ratio test, when ∣x∣<9|x|<9, the series ∑n=2∞∣xn32nlog⁡n∣\sum_{n=2}^\infty \big|\frac{x^n}{3^{2n}\log n}\big| converges and the series ∑n=2∞xn32nlog⁡n\sum_{n=2}^\infty \frac{x^n}{3^{2n}\log n} converges absolutely, and when ∣x∣>9|x|> 9 the series ∑n=2∞xn32nlog⁡n\sum_{n=2}^\infty \frac{x^n}{3^{2n}\log n} diverges. That just leaves x=±9x=\pm 9.

For x=−9x=-9, ∑n=2∞xn32nlog⁡n=∑n=2∞(−1)nlog⁡n\displaystyle\sum_{n=2}^\infty\frac{x^n}{3^{2n}\log n} =\sum_{n=2}^\infty\frac{(-1)^n}{\log n} which converges by the alternating series test.

For x=+9x=+9, ∑n=2∞xn32nlog⁡n=∑n=2∞1log⁡n\displaystyle\sum_{n=2}^\infty\frac{x^n}{3^{2n}\log n} =\sum_{n=2}^\infty\frac{1}{\log n} which is the same series as ∑n=2∞∣(−1)nlog⁡n∣\displaystyle\sum_{n=2}^\infty\Big|\frac{(-1)^n}{\log n}\Big|. We shall shortly show that n≥log⁡nn\ge\log n, and hence 1log⁡n≥1n\frac{1}{\log n}\ge \frac{1}{n} for all n≥1n\ge 1. This implies that the series ∑n=2∞1log⁡n\displaystyle\sum_{n=2}^\infty\frac{1}{\log n} diverges by comparison with the divergent series ∑n=2∞1np∣p=1\displaystyle\sum_{n=2}^\infty\frac{1}{n^p}\bigg|_{p=1}. This yelds both divergence for x=9x=9 and also the failure of absolute convergence for x=−9x=-9.

Finally, we show that n−log⁡n>0n-\log n > 0, for all n≥1n\ge 1. Set f(x)=x−log⁡xf(x)=x-\log x. Then f(1)=1>0f(1)=1>0 and

f′(x)=1−1x≥0for all x≥1\begin{equation*} f'(x) = 1 -\frac{1}{x} \ge 0\qquad\text{for all }x\ge 1 \end{equation*}

So f(x)f(x) is (strictly) positive when x=1x=1 and is increasing for all x≥1x\ge 1. So f(x)f(x) is (strictly) positive for all x≥1x\ge 1.

Q19Stage 3Past exam · 2013A

(a) Find the power–series representation for ∫11+x3 dx\displaystyle\int\frac{1}{1+x^3}\,\dee{x} centred at 00 (i.e. in powers of xx).

(b) The power series above is used to approximate ∫01/411+x3 dx\displaystyle\int_0^{1/4}\frac{1}{1+x^3}\,\dee{x}. How many terms are required to guarantee that the resulting approximation is within 10−510^{-5} of the exact value? Justify your answer.

Hint

See Example 3.5.21 in the

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

CLP-2 text.

Answer

(a) ∑n=0∞(−1)nx3n+13n+1+C\displaystyle\sum\limits_{n=0}^\infty(-1)^n \frac{x^{3n+1}}{3n+1} +C (b) We need to keep two terms (the n=0n=0 and n=1n=1 terms).

Full solution

(a) Applying 11+r=∑n=0∞(−1)nrn\displaystyle\frac{1}{1+r}=\sum\limits_{n=0}^\infty (-1)^nr^n with r=x3r=x^3 gives

∫11+x3 dx=∑n=0∞(−1)n∫x3n dx=∑n=0∞(−1)nx3n+13n+1+C\begin{align*} \int\frac{1}{1+x^3}\ \dee{x} =\sum_{n=0}^\infty(-1)^n \int x^{3n}\ \dee{x} =\sum_{n=0}^\infty(-1)^n \frac{x^{3n+1}}{3n+1} +C \end{align*}

(b) By part (a),

∫01/411+x3 dx=∑n=0∞(−1)nx3n+13n+1∣01/4=∑n=0∞(−1)n1(3n+1)43n+1\begin{align*} \int_0^{1/4}\frac{1}{1+x^3}\ \dee{x} =\sum_{n=0}^\infty(-1)^n \frac{x^{3n+1}}{3n+1} \bigg|_0^{1/4} =\sum_{n=0}^\infty(-1)^n \frac{1}{(3n+1)4^{3n+1}} \end{align*}

This is an alternating series with successively smaller terms that converge to zero as n→∞n\rightarrow\infty. So truncating it introduces an error no larger than the magnitude of the first dropped term. We want that first dropped term to obey

1(3n+1)43n+1<10−5=1105\begin{align*} \frac{1}{(3n+1)4^{3n+1}}<10^{-5}=\frac{1}{10^5} \end{align*}

So let's check the first few terms.

1(3n+1)43n+1∣n=0=14>11051(3n+1)43n+1∣n=1=145>11051(3n+1)43n+1∣n=2=17×47=17×214=17×16×1024=1112×1024<1105\begin{align*} \frac{1}{(3n+1)4^{3n+1}}\bigg|_{n=0}&=\frac{1}{4}>\frac{1}{10^5}\\ \frac{1}{(3n+1)4^{3n+1}}\bigg|_{n=1}&=\frac{1}{4^5}>\frac{1}{10^5}\\ \frac{1}{(3n+1)4^{3n+1}}\bigg|_{n=2}&=\frac{1}{7\times 4^7} =\frac{1}{7\times 2^{14}} =\frac{1}{7\times16\times 1024} =\frac{1}{112\times 1024}<\frac{1}{10^5} \end{align*}

So we need to keep two terms (the n=0n=0 and n=1n=1 terms).

Q20Stage 3Past exam · 2014A

(a) Show that ∑n=0∞nxn=x(1−x)2\displaystyle\sum_{n=0}^\infty nx^n =\frac{x}{(1-x)^2} for −1<x<1-1<x<1.

(b) Express ∑n=0∞n2xn\displaystyle\sum_{n=0}^\infty n^2x^n as a ratio of polynomials. For which xx does this series converge?

Hint

You know the geometric series expansion of 11−x\frac{1}{1-x}. What (calculus) operation(s) can you apply to that geometric series to convert it into the given series?

Answer

(a) See the solution.

(b) ∑n=0∞n2xn=x(1+x)(1−x)3\displaystyle\sum\limits_{n=0}^\infty n^2 x^n=\frac{x(1+x)}{(1-x)^3}. The series converges for −1<x<1-1<x<1.

Full solution

(a) Differentiating both sides of

∑n=0∞xn=11−x\begin{equation*} \sum_{n=0}^\infty x^n = \frac{1}{1-x} \end{equation*}

gives

∑n=0∞nxn−1=1(1−x)2\begin{equation*} \sum_{n=0}^\infty n x^{n-1}=\frac{1}{(1-x)^2} \end{equation*}

Now multiplying both sides by xx gives

∑n=0∞nxn=x(1−x)2\begin{equation*} \sum_{n=0}^\infty n x^n=\frac{x}{(1-x)^2} \end{equation*}

as desired.

(b) Differentiating both sides of the conclusion of part (a) gives

∑n=0∞n2xn−1=(1−x)2−2x(x−1)(1−x)4=(1−x)(1−x+2x)(1−x)4=1+x(1−x)3\begin{equation*} \sum_{n=0}^\infty n^2 x^{n-1}=\frac{(1-x)^2-2x(x-1)}{(1-x)^4} =\frac{(1-x)(1-x+2x)}{(1-x)^4} =\frac{1+x}{(1-x)^3} \end{equation*}

Now multiplying both sides by xx gives

∑n=0∞n2xn=x(1+x)(1−x)3\begin{equation*} \sum_{n=0}^\infty n^2 x^n=\frac{x(1+x)}{(1-x)^3} \end{equation*}

We know that differentiation preserves the radius of convergence of power series. So this series has radius of convergence 11 (the radius of convergence of the original geometric series). At x=±1x=\pm1 the series diverges by the divergence test. So the series converges for −1<x<1-1<x<1.

Q21Stage 3Past exam · 2015A

Suppose that you have a sequence {bn}\{b_n\} such that the series ∑n=0∞(1−bn)\sum_{n=0}^{\infty}(1-b_n) converges. Using the tests we've learned in class, prove that the radius of convergence of the power series ∑n=0∞bnxn\displaystyle\sum_{n=0}^{\infty}b_nx^n is equal to 11.

Hint

First show that the fact that the series ∑n=0∞(1−bn)\sum_{n=0}^{\infty}(1-b_n) converges guarantees that lim⁡n→∞bn=1\lim_{n\rightarrow\infty}b_n=1.

Answer

See the solution.

Full solution

By the divergence test, the fact that ∑n=0∞(1−bn)\sum\limits_{n=0}^{\infty}(1-b_n) converges guarantees that lim⁡n→∞(1−bn)=0\lim\limits_{n\rightarrow\infty}(1-b_n)=0, or equivalently, that lim⁡n→∞bn=1\lim\limits_{n\rightarrow\infty}b_n=1. So, by equation (3.7) in the

CLP-2 text, the radius of convergence is

R=[lim⁡n→∞∣bn+1bn∣]−1=[11]−1=1\begin{equation} R =\bigg[\lim_{n\rightarrow\infty}\Big|\frac{b_{n+1}}{b_n}\Big|\bigg]^{-1} =\bigg[\frac{1}{1}\bigg]^{-1} =1 \end{equation}
Q22Stage 3Past exam · M121 2014A

Assume {an}\big\{a_n \big\} is a sequence such that nanna_n decreases to CC as n→∞n \rightarrow\infty for some real number C>0C > 0

(a) Find the radius of convergence of ∑n=1∞anxn\displaystyle\sum\limits_{n=1}^\infty a_n x^n . Justify your answer carefully.

(b) Find the interval of convergence of the above power series, that is, find all xx for which the power series in (a) converges. Justify your answer carefully.

Hint

What does ana_n look like for large nn?

Answer

(a) 11. (b) The series converges for −1≤x<1-1\le x<1, i.e. for the interval [−1,1)[-1,1)

Full solution

(a) We know that the radius of convergence RR obeys

1R=lim⁡n→∞an+1an=lim⁡n→∞nn+1 (n+1)an+1nan=1CC=1\begin{align*} \frac{1}{R} = \lim_{n\rightarrow\infty}\frac{a_{n+1}}{a_n} = \lim_{n\rightarrow\infty}\frac{n}{n+1}\ \frac{(n+1)a_{n+1}}{na_n} = 1 \frac{C}{C} = 1 \end{align*}

because we are told that lim⁡n→∞nan=C\lim\limits_{n\rightarrow\infty} na_n=C. So R=1R=1.

(b) Just knowing that the radius of convergence is 11, we know that the series converges for ∣x∣<1|x|<1 and diverges for ∣x∣>1|x|>1. That leaves x±1x\pm 1.

When x=+1x=+1, the series reduces to ∑n=1∞an\sum\limits_{n=1}^\infty a_n. We are told that nanna_n decreases to C>0C>0. So an≥Cna_n\ge \frac{C}{n}. By the comparison test with the harmonic series ∑n=1∞1n\sum\limits_{n=1}^\infty\frac{1}{n}, which diverges by the pp–test with p=1p=1, our series diverges when x=1x=1.

When x=−1x=-1, the series reduces to ∑n=1∞(−1)nan\sum\limits_{n=1}^\infty (-1)^n a_n. We are told that nanna_n decreases to C>0C>0. So an>0a_n>0 and ana_n converges to 00 as n→∞n\rightarrow\infty. Consequently ∑n=1∞(−1)nan\sum\limits_{n=1}^\infty (-1)^n a_n converges by the alternating series test.

In conclusion ∑n=1∞anxn\sum\limits_{n=1}^\infty a_n x^n converges when −1≤x<1-1\le x< 1.

Q23Stage 3

An infinitely long, straight rod of negligible mass has the following weights:

  • At every whole number nn, a mass of weight 12n\dfrac{1}{2^n} at position nn, and

  • a mass of weight 13n\dfrac{1}{3^n} at position −n-n.

At what position is the centre of mass of the rod?

Figure from prob_s3.5, line 2

Figure from prob_s3.5, line 2

Hint

Equation 2.3.1 in the CLP-2 text tells us the centre of mass of a rod with weights {mn}\{m_n\} at positions {xn}\{x_n\} is xˉ=∑mnxn∑mn\displaystyle\bar x =\frac{\sum m_nx_n}{\sum m_n} .

Answer

56\dfrac{5}{6}

Full solution

Equation 2.3.1 in the CLP-2 text tells us the centre of mass of a rod with weights {mn}\{m_n\} at positions {xn}\{x_n\} is xˉ=∑mnxn∑mn\displaystyle\bar x =\frac{\sum m_nx_n}{\sum m_n} .

We find the combined mass of our weights using Equation 3.2.2 in the CLP-2 text with r=12r=\frac{1}{2} and r=13r=\frac{1}{3}, respectively.

∑n=1∞12n+∑n=1∞13n=∑n=0∞12⋅12n+∑n=0∞13⋅13n=12⋅11−12+13⋅11−13=1+12=32\begin{align*} \sum_{n=1}^{\infty}\frac{1}{2^n}+\sum_{n=1}^{\infty}\frac{1}{3^n}&= \sum_{n=0}^{\infty}\frac{1}{2}\cdot\frac{1}{2^n}+\sum_{n=0}^{\infty}\frac{1}{3}\cdot\frac{1}{3^n}\\ &=\frac{1}{2}\cdot\frac{1}{1-\frac12}+\frac{1}{3}\cdot\frac{1}{1-\frac13}\\ &=1+\frac12=\frac32 \end{align*}

Now, we want to calculate the sum of the products of the masses and their positions.

∑n=1∞12n⋅n+∑n=1∞13n⋅(−n)\sum_{n=1}^{\infty}\frac{1}{2^n}\cdot n +\sum_{n=1}^{\infty}\frac{1}{3^n}\cdot (-n)

We don't have such a nice formula for this, but we can make one by differentiating.

The following formula is true for any xx with ∣x∣<1|x|<1:

∑n=0∞xn=11−x\begin{align*}\sum_{n=0}^\infty x^n&=\frac{1}{1-x}\end{align*}

Differentiating both sides with respect to xx:

∑n=0∞nxn−1=1(1−x)2∑n=1∞nxn−1=1(1−x)2\begin{align*}\sum_{n=0}^\infty nx^{n-1}&=\frac{1}{(1-x)^2}\\ \sum_{n=1}^\infty nx^{n-1}&=\frac{1}{(1-x)^2}\end{align*}

Multiplying both sides by xx:

∑n=1∞nxn=x(1−x)2\begin{align*}\sum_{n=1}^\infty nx^{n}&=\frac{x}{(1-x)^2}\end{align*}

This allows us to evaluate our series.

∑n=1∞n2n−∑n=1∞n3n=12(1−12)2−13(1−13)2=2−34=54\begin{align*}\sum_{n=1}^{\infty}\frac{n}{2^n}-\sum_{n=1}^{\infty}\frac{n}{3^n} &=\frac{\frac12}{\left(1-\frac12\right)^2}-\frac{\frac13}{\left(1-\frac13\right)^2}\\ &=2-\frac{3}{4} = \frac{5}{4}\end{align*}

Therefore,

xˉ=5/43/2=56=0.833‾\begin{align*}\bar x &=\frac{5/4}{3/2}=\frac{5}{6} = 0.8\overline{33}\end{align*}

Remark: we can check that this makes some sense. Since the weights to the right of x=0x=0 are heavier than those to the left, but spaced the same, we would expect our rod to balance to the right of x=0x=0.

Q24Stage 3

Let f(x)=∑n=0∞An(x−c)nf(x)=\displaystyle\sum_{n=0}^{\infty}A_n(x-c)^n, for some constant cc and a sequence of constants {An}\{A_n\}. Further, let f(x)f(x) have a positive radius of convergence.

If A1=0A_1=0, show that y=f(x)y=f(x) has a critical point at x=cx=c. What is the relationship between the behaviour of the graph at that point and the value of A2A_2?

Hint

Use the second derivative test.

Answer

The point x=cx=c corresponds to a local maximum if A2<0A_2<0 and a local minimum if A2>0A_2>0.

Full solution

First, we differentiate.

f(x)=∑n=0∞An(x−c)nf′(x)=∑n=0∞nAn(x−c)n−1=∑n=1∞nAn(x−c)n−1f′(c)=∑n=1∞nAn⋅0n−1=A1⋅1+2A2⋅0+3A3⋅0+⋯=A1\begin{align*} f(x)&=\displaystyle\sum_{n=0}^{\infty}A_n(x-c)^n\\ f'(x)&=\sum_{n=0}^{\infty}nA_n(x-c)^{n-1}\\ &=\sum_{n=1}^{\infty}nA_n(x-c)^{n-1}\\ f'(c)&=\sum_{n=1}^\infty nA_n\cdot 0^{n-1}\\ &=A_1\cdot 1 + 2A_2\cdot0+3A_3\cdot0+\cdots\\ &=A_1 \end{align*}

So, if A1=0A_1=0, then f′(c)=0f'(c)=0. That is, f(x)f(x) has a critical point at x=cx=c.

To determine the behaviour of this critical point, we use the second derivative test.

f′(x)=∑n=1∞nAn(x−c)n−1f′′(x)=∑n=1∞n(n−1)An(x−c)n−2=∑n=2∞n(n−1)An(x−c)n−2f′′(c)=∑n=2∞n(n−1)An⋅0n−2=2(1)A2⋅00+3(2)A3⋅01+4(3)A4⋅02+⋯=2A2\begin{align*} f'(x)&=\sum_{n=1}^{\infty}nA_n(x-c)^{n-1}\\ f''(x)&=\sum_{n=1}^{\infty}n(n-1)A_n(x-c)^{n-2}\\ &=\sum_{n=2}^{\infty}n(n-1)A_n(x-c)^{n-2}\\ f''(c)&=\sum_{n=2}^{\infty}n(n-1)A_n\cdot 0^{n-2}\\ &=2(1)A_2\cdot 0^0+3(2)A_3\cdot 0^1+4(3)A_4\cdot 0^2+\cdots\\ &=2A_2 \end{align*}

Following the second derivative test, x=cx=c is the location of a local maximum if A2<0A_2<0, and it is the location of a local minimum if A2>0A_2>0. (If A2=0A_2=0, the critical point may or may not be a local extremum.)

Q25Stage 3

Evaluate ∑n=3∞n5n−1\displaystyle \sum_{n=3}^\infty \frac{n}{5^{n-1}}.

Hint

What function has ∑n=1∞nxn−1\displaystyle\sum_{n=1}^\infty nx^{n-1} as its power series representation?

Answer

1380\dfrac{13}{80}

Full solution

We recognize ∑n=3∞n5n−1\displaystyle \sum_{n=3}^\infty \frac{n}{5^{n-1}} as f(x)=∑n=3∞n⋅xn−1f(x)=\displaystyle \sum_{n=3}^\infty n\cdot x^{n-1}, evaluated at x=15x=\dfrac15. We should figure out what f(x)f(x) is in equation form (as opposed to power series form). Notice that this looks similar to the derivative of the geometric series ∑xn\displaystyle\sum x^n.

11−x=∑n=0∞xnwhen ∣x∣<1ddx{11−x}=ddx{∑n=0∞xn}1(1−x)2=∑n=1∞nxn−1=1x0+2x1+∑n=3∞nxn−1=1+2x+∑n=3∞nxn−1So,1(1−x)2−1−2x=∑n=3∞nxn−1Setting x=15:1(1−1/5)2−1−25=∑n=3∞n(15)n−1(54)2−1−25=∑n=3∞n5n−1\begin{align*} \frac{1}{1-x}&=\sum_{n=0}^\infty x^n \qquad \text{when }|x|<1\\ \diff{}{x}\left\{\frac{1}{1-x}\right\}&=\diff{}{x}\left\{\sum_{n=0}^\infty x^n\right\}\\ \frac{1}{(1-x)^2}&=\sum_{n=1}^\infty nx^{n-1}\\ &= 1x^0 + 2x^1+\sum_{n=3}^\infty nx^{n-1}\\ &= 1 + 2x+\sum_{n=3}^\infty nx^{n-1}\\ \text{So,}\qquad \frac{1}{(1-x)^2}-1-2x&=\sum_{n=3}^\infty nx^{n-1}\\ \text{Setting }x=\dfrac15\text{:}\qquad \frac{1}{(1-1/5)^2}-1-\frac25&=\sum_{n=3}^\infty n\left(\frac15\right)^{n-1}\\ \left(\frac{5}{4} \right)^2-1-\frac25&=\sum_{n=3}^\infty \frac{n}{5^{n-1}}\\ \end{align*}

So, our series evaluates to 2516−1−25=1380\displaystyle\frac{25}{16}-1-\frac{2}{5}=\frac{13}{80}.

Q26Stage 3

Find a polynomial that approximates f(x)=log⁡(1+x)f(x)=\log(1+ x) to within an error of 10−510^{-5} for all values of xx in (0,110)\left(0,\frac{1}{10}\right).

Then, use your polynomial to approximate log⁡(1.05)\log(1.05) as a rational number.

Hint

The power series representation in Example 3.5.20 is an alternating series when xx is positive.

Answer

x−x22+x33−x44x-\dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}

Full solution

As we saw in in Example 3.5.20 of the CLP-2 text,

log⁡(1+x)=∑n=0∞(−1)nxn+1n+1\log(1+ x) = \sum_{n=0}^\infty (-1)^n\frac{x^{n+1}}{n+1}

which is an alternating series when xx is positive. If we use its partial sum SNS_N to approximate log⁡(1+x)\log(1+ x), the absolute error involved is no more than

x(N+1)+1(N+1)+1=xN+2N+2\frac{x^{(N+1)+1}}{(N+1)+1} =\frac{x^{N+2}}{N+2}

We want this error to be at most 10−510^{-5} whenever 0<x<1100< x <\frac1{10}. For this range of xx values, xN+2N+2<1(N+2)10N+2\dfrac{x^{N+2}}{N+2}<\dfrac{1}{(N+2)10^{N+2}}, so we want NN that satisfies the inequality:

1(N+2)10N+2≤1105⇒(N+2)10N+2≥105\begin{align*} \frac{1}{(N+2)10^{N+2}}&\le \frac{1}{10^5}\\ \Rightarrow \qquad (N+2)10^{N+2}&\ge 10^5 \end{align*}

We see N=3N=3 suffices.

So, the partial sum

∑n=03(−1)nxn+1n+1=x−x22+x33−x44\sum_{n=0}^3 (-1)^n\frac{x^{n+1}}{n+1}=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}

approximates log⁡(1+x)\log(1+x) to within an error of x55\dfrac{x^5}{5}.

When xx is between 00 and 110\frac1{10}, that error is at most 15⋅105<10−5\dfrac{1}{5\cdot 10^5}<10^{-5}, as desired.

Now we can approximate log⁡(1.05)\log(1.05).

log⁡(1.05)=log⁡(1+120)≈(120)−(120)22+(120)33−(120)44=12×203−6×202+4×20−312×204=936771920000\begin{align*} \log(1.05)&=\log\left(1+\frac{1}{20}\right)\\ &\approx \left(\tfrac{1}{20}\right)-\dfrac{\left(\tfrac{1}{20}\right)^2}{2}+\dfrac{\left(\tfrac{1}{20}\right)^3}{3}-\dfrac{\left(\tfrac{1}{20}\right)^4}{4}\\ &=\frac{12\times 20^3-6\times 20^2+4\times 20-3}{12\times 20^4}=\frac{93677}{1920000} \end{align*}

We note that a computer approximates 936771920000≈0.04879010\frac{93677}{1920000}\approx0.04879010 and log⁡(1.05)≈0.04879016\log(1.05) \approx 0.04879016. So, our actual error is around 6×10−86\times 10^{-8}.

Q27Stage 3

Find a polynomial that approximates f(x)=arctan⁡xf(x)=\arctan x to within an error of 10−510^{-5} for all values of xx in (−14,14)\left(-\frac{1}{4},\frac{1}{4}\right).

Hint

The power series representation in Example 3.5.21 is an alternating series when xx is nonzero.

Answer

x−x33+x55x-\dfrac{x^3}{3}+\dfrac{x^5}{5}

Full solution

As we saw in in Example 3.5.21 of the CLP-2 text,

arctan⁡x=∑n=0∞(−1)nx2n+12n+1\arctan x = \sum_{n=0}^\infty (-1)^n\frac{x^{2n+1}}{2n+1}

which is an alternating series when xx is nonzero. If we use its partial sum SNS_N to approximate arctan⁡x\arctan x, the absolute error involved is no more than

∣x∣2(N+1)+12(N+1)+1=∣x∣2N+32N+3\frac{|x|^{2(N+1)+1}}{2(N+1)+1} =\frac{|x|^{2N+3}}{2N+3}

We want this error to be at most 10−510^{-5} whenever −14<x<14-\frac14< x <\frac14. For this range of xx values, ∣x∣2N+32N+3<1(2N+3)42N+3\dfrac{|x|^{2N+3}}{2N+3}<\dfrac{1}{(2N+3)4^{2N+3}}, so we want NN that satisfies the inequality:

1(2N+3)42N+3≤1105\frac{1}{(2N+3)4^{2N+3}}\le \frac{1}{10^5}

A quick check with a calculator shows that N=2N=2 suffices.

So, the partial sum

∑n=02(−1)nx2n+12n+1=x−x33+x55\sum_{n=0}^2 (-1)^n\frac{x^{2n+1}}{2n+1}=x-\frac{x^3}{3}+\frac{x^5}{5}

approximates arctan⁡x\arctan x to within an error of x77\dfrac{x^7}{7}.

When xx is between −14-\frac14 and 14\frac14, that error is at most 17⋅47=1114688<1100000=10−5\dfrac{1}{7\cdot 4^7}=\dfrac{1}{114688}<\dfrac{1}{100000}=10^{-5}, as desired. (When x=0x=0, our approximation is 0, the exact value of arctan⁡0\arctan 0.)

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From the UBC Math 101 open textbook project, © Joel Feldman, Andrew Rechnitzer, Elyse Yeager and others. Licensed CC BY-NC-SA 4.0; this HTML adaptation is released under the same licence. Full attribution.