Question

The Maclaurin series for ln(1+x) is given by

x - x^2/2 + x^3/3 - x^4/4 + ... + (-1)^(n+1) x^n/n + ...

On its interval of convergence, this series converges to ln(1+x). Let f be the function defined f(x) = x ln(1 + x/3).

(a) Write the first four nonzero terms and the general term of the Maclaurin series for f.

(b) Determine the interval of convergence of the Maclaurin series for f. Show the work that leads to your answer.

(c) Let P₄(x) be the fourth-degree Taylor polynomial for f about x = 0. Use the alternating series error bound to find an upper bound for |P₄(2) - f(2)|.

Answer 1

(a)

Replacing x with x/3 and then multiplying both sides by x

$\begin{aligned} \ln(1+x) &= x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots + (-1)^{n+1} \frac{x^n}{n} + \cdots \\ \\ \ln\left(1 + \tfrac{x}{3}\right) &= \frac{x}{3} - \frac{x^2}{3^2 \cdot 2} + \frac{x^3}{3^3 \cdot 3} - \frac{x^4}{3^4 \cdot 4} + \cdots + (-1)^{n+1} \frac{x^n}{3^n \cdot n} \\ \\ x\ln\left(1 + \tfrac{x}{3}\right) &= \frac{x^2}{3} - \frac{x^3}{3^2 \cdot 2} + \frac{x^4}{3^3 \cdot 3} - \frac{x^5}{3^4 \cdot 4} + \cdots + (-1)^{n+1} \frac{x^{n+1}}{3^n \cdot n} \end{aligned}$

The general term is

$\displaystyle(-1)^{n+1} \frac{x^{n+1}}{3^n \cdot n}$

____________

(b)

Due to the powers present in the general term, use the ratio test

$\displaystyle\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n}\right| = \lim_{n \to \infty} \left| \frac{ \frac{x^{n+2}}{3^{n+1}(n+1) } }{\frac{x^{n+1}}{3^n n} } \right| = \lim_{n\to\infty}\left|\frac{xn}{3n+3} \right| = \left|\frac{x}{n} \right| \ \textless \ 1$

The series converges on x ∈ (-3,3)
Determining the interval of convergence, test the endpoints

For x = -3, general term becomes

$\displaystyle \frac{(-1)^{n+1}(-3)^{n+1}}{3^n \cdot n} \\ \\ = \frac{(-1)^{n+1}(-1)^{n+1}(3)^{n+1}}{3^n \cdot n} \\ \\ = (-1)^{2n+2} \cdot \frac{3}{n}$

$(-1)^{2n+2} \implies 1 \text{ because 2n+2 is all even}$

$\sum_{n=1}^{\infty} \frac{3}{n} \text{ diverges via harmonic series}$

Diverges at x = -3.

For x = -3, general term becomes

$\displaystyle \frac{(-1)^{n+1}3^{n+1}}{3^n \cdot n} = (-1)^{n+1} \frac{3}{n} \\ \\ \implies 3 \sum_{n=1}^{\infty} \frac{3}{n}$

This is an alternate harmonic series, so it converges as the terms are decreasing in size (approaching zero as n approaches infinity)

The interval of convergence is

-3 < x ≤ 3

____________

(c)

We wrote all the terms above in part (a). The neglected term in the alternating series of f is

$\displaystyle - \frac{x^5}{3^4 \cdot 4}$

Thus

$\displaystyle \left| P_4(2) - f(2) \right| \le \left| - \frac{2^5}{3^4 \cdot 4} \right| = \frac{8}{81}$

Answer 2

Answer:

The stepwise solution of the above problem is given below.

Step-by-step explanation:

Given :

$\rm ln(1+x) = x - \dfrac {x^2}{2}+\dfrac {x^3}{3}-\dfrac {x^4}{4}+...... +(-1)^(^n^+^1^)\dfrac{x^n}{n}$     ------------ (1)

Calculation :

a) Repace x with x/3 in equation (1) and then multiply both the sides by x.

$\rm ln(1+\dfrac{x}{3}) = \dfrac{x}{3} - \dfrac {x^2}{3^2 \times2}+\dfrac {x^3}{3^3\times3}-\dfrac {x^4}{3^4 \times 4}+...... +(-1)^(^n^+^1^)\dfrac{x^n}{3^n \times n}$

$\rm xln(1+\dfrac{x}{3}) =\dfrac{ x^2}{3} - \dfrac {x^3}{3^2 \times2}+\dfrac {x^4}{3^3\times3}-\dfrac {x^5}{3^4 \times 4}+...... +(-1)^(^n^+^1^)\dfrac{x^n^+^1}{3^n \times n}$

General term,

$(-1)^n^+^1 \dfrac{x^n^+^1}{3^n\times n}$

b) Series converges at

$x\epsilon (-3,3)$

At x = -3, the general term becomes

$(-1)^n^+^1\dfrac{(-3)^n^+^1}{3^n\times n}$

$(-1)^2^n^+^2 \dfrac{3}{n}$

$(-1)^2^n^+^2$   is even

$\rm \sum_{n=1 }^{\infty} \dfrac {3}{n} (diverges\; through\; harmonic\; series)$

Diverges at x = -3

Now, at x = 3 general term becomes

$(-1)^n^+^1\dfrac{(3)^n^+^1}{3^n\times n}$

$(-1)^n^+^1 \dfrac{3}{n}$

$\rm 3\sum_{n=1 }^{\infty} \dfrac {3}{n}$

Therefore, interval of convergence is (-3,3)

c) In part a) we wrote all the terms. The neglecting term term in the alternating series of f is

$- \dfrac{x^5}{3^4 \times 4}$

hence,

$\rm |P_4(2)- f(2)| \leq \left | -\dfrac{2^5}{3^4\times4} \right | = \dfrac{8}{81}$

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