Question

8. Let f(x)- -132+1, n-1 (a) (10) Find the radius of convergence R of f. (b) (ao) Use the given power series to find an approximation of f(edt that has an error of less than 0.001. Don't simplify your answer.pproximationofhuamathathasanemor

Answer 1

Given the alternating power series

n 1 7l

Lets recall that for an alternating power series $S = \sum_{n=1}^{n=\infty}(-1)^n b_n\ , \quad b_n\ge 0$

to converge it must satisy two conditions

i.

limón-0

ii.

$\{b_n\}$ is a decreasing sequence.

From the problem

where $|\cdot|$ is the absolute value function

Applyin i. we have

lim bn-0

This is only possible iff

For being a decreasing sequence as in ii.

ntl <1 bn

Putting values we have

3+12n3 3"12n +1

Thus

V3 V3

Thus

$x \in (-1,1) \bigcap (\frac{-1}{\sqrt{3}},\frac{1}{\sqrt{3}}) \implies x\in (\frac{-1}{\sqrt{3}},\frac{1}{\sqrt{3}})$

Thus radius of convergence

$R = \frac{2}{\sqrt{3}}$

For integrating we integrate term by term, i.e.

$\int_{0}^{1/3}f(t)dt = \int_{0}^{1/3}\sum_{n=0}^{\infty}(-1)^n3^n x^{2n+1}dt = \sum_{n=0}^{\infty}(-1)^n3^n \int_{0}^{1/3}t^{2n+1} dt$

Thus,

1 (-1)'n 18 n+1) 181 2n+2 1/3 0

The above is another alternating seies satisfying i. & ii.

For the error term $E_N$ lets us write in the form as

$\int_{0}^{1/3}f(t)dt = \frac{1}{18} \sum_{n=0}^{\infty} \frac{(-1)^n}{n+1} = \frac{1}{18} \sum_{n=0}^{N-1}\frac{(-1)^n}{n+1} + E_N$

For

EN 0.001

we have

$|E_N| =\frac{1}{18} \left|\frac{1}{N+1} +\dots\right| = \frac{1}{18} \mathcal{O}\left(\frac{1}{N+1} \right) < 0.001 \implies N+1 \ge 56$

Thus

$\implies N \ge 55$

For the given error of 0.001 we have

$\int_{0}^{1/3}f(t)dt \sim \frac{1}{18} \sum_{n=0}^{55}\frac{(-1)^n}{n+1}$