Chapter 10, Problem 58RE

a.

To determine

To find the first four terms of the Taylor series and general term of Taylor series generated by $f(x)=\frac{1}{(1-2x)}$ at $x=0$ .

Answer to Problem 58RE

First four terms of series will be 1, $\frac{2x}{(1-2x)}$ , $\frac{4x^2}{{(1-2x)}^2}$ and $\frac{8x^3}{{(1-2x)}^3}$

General term will be $(2^{t-1}){(x)}^{t-1}$ for t^th term

Explanation of Solution

Given:

A function $f(x)=\frac{1}{(1-2x)}$ is given for evaluation.

Calculation:

A Taylor series is used to find the value of a function by conditional summation of its derivatives.

For any random function $f(x)$ the Taylor series is given by

  $f(x)={\displaystyle \sum _{n=0}^{\infty }\frac{f^n(a)}{n!}{(x-a)}^n}$

Here $f^n$ is the n^th derivative of the function.

  $f(x)$ is written as

  $f(x)=f_0+c_1{f}_1+c_2{f}_2+c_3{f}_3+c_4{f}_4+\mathrm{...}$

Here $c_1$ , $c_2$ , $c_3$ and $c_4$ represent coefficients of the terms.

These coefficients are equal to $\frac{f^n(a)}{n!}$ term in above formula.

To find these coefficients, first the function should be derived 3 to 4 times to develop a pattern so it can be generalized.

  $\begin{array}{}$ f(x)={(1-2x)}^{-1}$$ f^{\prime }(x)=(-1)(-2){(1-2x)}^{-2}$$ f^{″}(x)=(-1)(-2)(-2)(-2){(1-2x)}^{-3}$$ f^{‴}(x)=(-1)(-2)(-2)(-2)(-3)(-2){(1-2x)}^{-4}$\end{array}$

This when simplified can be written as

  $\begin{array}{}$ f(x)={(1-2x)}^{-1}$$ f^{\prime }(x)=(1)(2){(1-2x)}^{-2}$$ f^{″}(x)=(1)(2)(2)(2){(1-2x)}^{-3}$$ f^{‴}(x)=(1)(2)(2)(2)(3)(2){(1-2x)}^{-4}$\end{array}$

Which can be further simplified as

  $\begin{array}{}$ f(x)=(0!)(2^0){(1-2x)}^{-1}$$ f^{\prime }(x)=(1!)(2^1){(1-2x)}^{-2}$$ f^{″}(x)=(2!)(2^2){(1-2x)}^{-3}$$ f^{‴}(x)=(3!)(2^3){(1-2x)}^{-4}$\end{array}$

Now it can be generalized as

  $f^n(x)=(n!)(2^n){(1-2x)}^{-(n+1)}$

And for $x=0$ as given in the question, the equation can be further simplified to

  $f^n(0)=(n!)(2^n){(1)}^{-(n+1)}$

  $f^n(0)=(n!)(2^n)$

Here we get a part of coefficient.

The terms of Taylor series can be written as

  $f(x)={\displaystyle \sum _{n=0}^{\infty }\frac{f^n(a)}{n!}{(x-a)}^n}$

Here, it should be noted that 1^st term is undifferentiated function at $x=0$ .

Value of $f(0)$ is given as

  $\begin{array}{l}f(x)=\frac{1}{(1-2x)}\\ f(0)=\frac{1}{[1-2(0)]}\\ f(0)=\frac{1}{1-0}\\ f(0)=1\end{array}$

Now, the Taylor Series can be written as

  $f(x)=1+{\displaystyle \sum _{n=1}^{\infty }\frac{f^n(0)}{n!}{(x)}^n}$

For the given question value of $f^n(0)$ can be substituted as

  $\begin{array}{l}f(x)=1+{\displaystyle \sum _{n=1}^{\infty }\frac{(n!)(2^n)}{n!{(1-2x)}^n}{(x)}^n}\\ f(x)=1+{\displaystyle \sum _{n=1}^{\infty }\frac{(2^n){(x)}^n}{{(1-2x)}^n}}\\ f(x)={\displaystyle \sum _{n=0}^{\infty }\frac{(2^n){(x)}^n}{{(1-2x)}^n}}\end{array}$

  $f(x)=1+\frac{(2^1)(x^1)}{{(1-2x)}^1}+\frac{(2^2)(x^2)}{{(1-2x)}^2}+\frac{(2^3)(x^3)}{{(1-2x)}^3}+\mathrm{...}$

Hence, the first four terms of the Taylor series of the given function will be

1, $\frac{2x}{(1-2x)}$ , $\frac{4x^2}{{(1-2x)}^2}$ and $\frac{8x^3}{{(1-2x)}^3}$

And general term will be $\frac{(2^{t-1}){(x)}^{t-1}}{{(1-2x)}^{t-1}}$ for t^th term.

Conclusion:

Therefore, first n terms and general term of a Taylor series can be found out by generalizing the derivatives of its function.

b.

To determine

To find: the interval of convergence found in part (a).

Answer to Problem 58RE

When $\frac{2\left|x\right|}{\left|1-2x\right|}<1$ , the series converges

Explanation of Solution

Given:

The series found in part (a) is

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

Calculation:

To find the interval of convergence, one of the method that can be used is by using the ratio test.

It is done by finding the ratio of $\underset{x\to \infty }{\lim }\frac{f{(x)}_{n+1}}{f{(x)}_n}$

For the series in part (a) .

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

Thus on dividing two consecutive terms, it comes out as $\frac{2\left|x\right|}{\left|1-2x\right|}$ .

The series converges when $\frac{2\left|x\right|}{\left|1-2x\right|}<1$ by ratio test.

The answer shows that series converges when $\frac{2\left|x\right|}{\left|1-2x\right|}<1$ .

Conclusion:

The ratio test gives $\frac{2\left|x\right|}{\left|1-2x\right|}<1$ for convergence of series.

c.

To determine

To find $f(-\frac{1}{4})$ and number of terms adequate for approximating $f(-\frac{1}{4})$ with an error not exceeding one percent in magnitude.

Answer to Problem 58RE

  $\frac{2}{3}$ and 4 terms are required.

Explanation of Solution

Given:

The series found in part (a) is

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

Calculation:

Value of $f(-\frac{1}{4})$ can be found as,

  $f(-\frac{1}{4})=\frac{1}{1-2(-\frac{1}{4})}=\frac{1}{1+\frac{1}{2}}=\frac{1}{\frac{3}{2}}=\frac{2}{3}=\mathrm{0.666...}$

Now for finding adequate terms for error not exceeding for than 1 percent

The 1^st term of series of $f(-\frac{1}{4})$ will be 1.

The 2^nd term of series of $f(-\frac{1}{4})$ will be given as

  $f(-\frac{1}{4})=1+\frac{2(-\frac{1}{4})}{[1-2(-\frac{1}{4})]}=1+\frac{-\frac{1}{2}}{(\frac{3}{2})}=1-\frac{2}{2\times 3}=1-\frac{1}{3}=\frac{2}{3}=\mathrm{0.666..}$

The 3^rdterm of series of $f(-\frac{1}{4})$ will be given as

  $f(-\frac{1}{4})=\frac{2}{3}+\frac{4{(-\frac{1}{4})}^2}{{[1-2(-\frac{1}{4})]}^2}=\frac{2}{3}+\frac{\frac{1}{4}}{{(\frac{3}{2})}^2}=\frac{2}{3}+\frac{4}{4\times 9}=\frac{2}{3}+\frac{1}{9}=\frac{7}{9}=\mathrm{0.777...}$

The 4^th term of series of $f(-\frac{1}{4})$ will be given as

  $f(-\frac{1}{4})=\frac{7}{9}+\frac{8{(-\frac{1}{4})}^3}{{[1-2(-\frac{1}{4})]}^3}=\frac{7}{9}+\frac{\frac{1}{8}}{{(\frac{3}{2})}^3}=\frac{7}{9}+\frac{8}{8\times 27}=\frac{7}{9}+\frac{1}{27}=\frac{22}{27}=\mathrm{0.8148...}$

Thus, it requires minimum 4 terms to get an adequate result with error not more than 1 percent.

The reason is for 1^st, 2^nd and 3^rd terms the answer had recurring values after decimal point, thus the error can exceed more than 1 if 1^st three terms are considered.

However, if the 4^th term is also considered then the result is obtained where there is a non-recurring value is observed after decimal point.

Conclusion:

Therefore, value of $f(-\frac{1}{4})$ comes out to be $\frac{2}{3}$ and 4 terms are required for finding a value where the error doesn’t exceeds more than 1 percent.