Chapter 10.2, Problem 27E

(a)

To determine

To find:First four nonzero terms in the Taylor series generated by f ( x ) at $x=0$ .

Answer to Problem 27E

First four nonzero terms in the Taylor series are $1,\frac{x}{2},\frac{-x^2}{8},\text{and}\frac{x^3}{16}$ .

Explanation of Solution

Given information:

  $f\left(x\right)=\sqrt{1+x}$

At $x=0$

For nonzero terms:

We have

  $f\left(x\right)=\sqrt{1+x}={\left(1+x\right)}^{1/2}=1$

First derivative:

  $f\text{'}\left(x\right)=\frac{1}{2}{\left(1+x\right)}^{-1/2}=\frac{1}{2}$

Second derivative:

  $f\text{'}\text{'}\left(x\right)=-\frac{1}{4}{\left(1+x\right)}^{-3/2}=-\frac{1}{4}$

Third derivative:

  $f\text{'}\text{'}\text{'}\left(x\right)=\frac{3}{8}{\left(1+x\right)}^{-5/2}=\frac{3}{8}$

Thus,

  $P_4\left(x\right)=1+\frac{1}{2}x-\frac{1}{4}\frac{x^2}{2!}+\frac{3}{8}\frac{x^3}{3!}$

Therefore,

First four nonzero terms in the Taylor series are $1,\frac{x}{2},\frac{-x^2}{8},\text{and}\frac{x^3}{16}$ .

(b)

To determine

To find:First four nonzero terms in the Taylor series for g ( x ) at $x=0$ .

Answer to Problem 27E

For $g\left(x\right)=f\left(x^2\right)$ :

The first four non − zero terms in Taylor series expansion of are $1,\frac{x^2}{2},\frac{-x^4}{8},\text{and}\frac{x^6}{16}$ .

Explanation of Solution

Given information:

  $g\left(x\right)=\sqrt{1+x^2}$

At $x=0$

From Part (a) result,

We have

First four nonzero terms in the Taylor series are $1,\frac{x}{2},\frac{-x^2}{8},\text{and}\frac{x^3}{16}$ .

Now,

Replace x in Taylor series expansion of f(x) with $x^2$ , to get the Taylor series expansion of $g\left(x\right)=f\left(x^2\right)$ .

Thus,

  $P_4\left(x\right)=1+\frac{1}{2}{x}^2-\frac{1}{4}\frac{x^4}{2!}+\frac{3}{8}\frac{x^6}{3!}$

Therefore,

The first four non − zero terms in Taylor series expansion of

  $g\left(x\right)=f\left(x^2\right)$ are $1,\frac{x^2}{2},\frac{-x^4}{8},\text{and}\frac{x^6}{16}$ .

(c)

To determine

To find:First four nonzero terms in the Taylor series at $x=0$ for the function h .

Answer to Problem 27E

The first four non − zero terms in Taylor series expansion of h ( x ) are:

  $5x,\frac{x^3}{6},\frac{-x^5}{40},\text{and}\frac{x^7}{112}$

Explanation of Solution

Given information:

  $h\text{'}\left(x\right)=\sqrt{1+x^2}$

And

  $h\left(0\right)=5$

We can get the Taylor series expansion of h ( x ) by integrating the terms of the Taylor series expansion of $h\text{'}\left(x\right)$ .

The first four non − zero terms in Taylor series expansion of $h\text{'}\left(x\right)$ :

  $5,\frac{x^2}{2},\frac{-x^4}{8},\text{and}\frac{x^6}{16}$

Now,

Integrate above first four non − zero terms to get Taylor series expansion of h ( x ):

  ${\displaystyle \int 5},{\displaystyle \int \frac{x^2}{2}},{\displaystyle \int \frac{-x^4}{8}},\text{and}{\displaystyle \int \frac{x^6}{16}}$

Therefore,

The first four non − zero terms in Taylor series expansion of h ( x ) are:

  $5x,\frac{x^3}{6},\frac{-x^5}{40},\text{and}\frac{x^7}{112}$