Solutions to Problem Set 28

Math 211-03

11-14-2017

[Taylor series applications]

1. (a) Find the first 4 nonzero terms of the Taylor series at $c = 0$ for $e^{x^2}$ .

(b) Use the series in (a) to obtain the value of the limit $\displaystyle \lim_{x \to 0}
   \dfrac{e^{x^2} - 1}{x^2}$ .

(a) Use

$$e^u = 1 + u + \dfrac{u^2}{2!} + \dfrac{u^3}{3!} + \cdots + \dfrac{u^n}{n!} + \cdots, \quad -\infty < u < \infty.$$

Letting $u = x^2$ , I get

$$e^{x^2} = 1 + x^2 + \dfrac{x^4}{2!} + \dfrac{x^6}{3!} + \cdots.\quad\halmos$$

(b) Using the series in (a),

$$\dfrac{e^{x^2} - 1}{x^2} = \dfrac{1}{x^2} \left[\left(1 + x^2 + \dfrac{x^4}{2!} + \dfrac{x^6}{3!} + \cdots\right) - 1\right] =$$

$$\dfrac{1}{x^2} \left(x^2 + \dfrac{x^4}{2!} + \dfrac{x^6}{3!} + \cdots\right) = 1 + \dfrac{1}{2} x^2 + \dfrac{1}{6} x^4 + \cdots.$$

Hence,

$$\lim_{x \to 0} \dfrac{e^{x^2} - 1}{x^2} = \lim_{x \to 0} \left(1 + \dfrac{1}{2} x^2 + \dfrac{1}{6} x^4 + \cdots\right) = 1.\quad\halmos$$


2. (a) Find the first 4 nonzero terms of the Taylor series for $\sin (t^2)$ at $c = 0$ .

(b) Use the 4 terms of the series you found in (a) to find the first 4 nonzero terms of the series for $\displaystyle \int_0^x \sin (x^2)\,dx$ at $c = 0$ .

(a) Use

$$\sin u = u - \dfrac{u^3}{3!} + \dfrac{u^5}{5!} - \dfrac{u^7}{7!} + \cdots.$$

Let $u = t^2$ . This gives

$$\sin (t^2) = t^2 - \dfrac{1}{6} t^6 + \dfrac{1}{120} t^{10} - \dfrac{1}{5040} t^{14} + \cdots.\quad\halmos$$

(b) Using the first four terms, I get

$$\int_0^x \sin (t^2)\,dt = \int_0^x \left(t^2 - \dfrac{1}{6} t^6 + \dfrac{1}{120} t^{10} - \dfrac{1}{5040} t^{14} + \cdots\right)\,dx =$$

$$\left[\dfrac{1}{3} t^3 - \dfrac{1}{42} t^7 + \dfrac{1}{1320} t^{11} - \dfrac{1}{75600} t^{15} + \cdots\right]_0^x = \dfrac{1}{3} x^3 - \dfrac{1}{42} x^7 + \dfrac{1}{1320} x^{11} - \dfrac{1}{75600} x^{15} + \cdots.\quad\halmos$$


3. (a) Assume that the following function is defined and differentiable at $c = 0$ :

$$f(x) = \cases{ 1 & if $x = 0$ \cr \noalign{\vskip2pt} \dfrac{\sin x}{x} & if $x \ne 0$ \cr}.$$

Find the first 4 nonzero terms of the Taylor series for $f(x)$ at $c = 0$ .

(b) Use the 4 terms of the series you found in (a) to approximate $\displaystyle \int_{-2}^2
   \dfrac{\sin x}{x}\,dx$ .

(a)

$$\sin x = x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!} - \dfrac{x^7}{7!} + \cdots = x - \dfrac{1}{6} x^3 + \dfrac{1}{120} x^5 - \dfrac{1}{5040} x^7 + \cdots.$$

$$\dfrac{\sin x}{x} = 1 - \dfrac{1}{6} x^2 + \dfrac{1}{120} x^4 - \dfrac{1}{5040} x^6 + \cdots.\quad\halmos$$

(b) Using the first four terms, I get

$$\int_{-2}^2 \dfrac{\sin x}{x}\,dx \approx \int_{-2}^2 \left(1 - \dfrac{1}{6} x^2 + \dfrac{1}{120} x^4 - \dfrac{1}{5040} x^6\right)\,dx =$$

$$\left[x - \dfrac{1}{18} x^3 + \dfrac{1}{600} x^5 - \dfrac{1}{35280} x^7\right]_{-2}^2 = \dfrac{35396}{11025} = 3.21052 \ldots.\quad\halmos$$


[Differentiating and integrating series]

4. (a) Use a known series to construct the power series centered at $c = 0$ for $\dfrac{1}{1
   - x^4}$ . Write your answer in summation form, or give the first four nonzero terms of the series.

(b) By differentiating the series in (a), construct the power series centered at $c = 0$ for $\dfrac{x^3}{(1 - x^4)^2}$ . Write your answer in summation form, or give the first four nonzero terms of the series.

(a) I have

$$\dfrac{1}{1 - u} = \sum_{n=0}^\infty u^n = 1 + u + u^2 + u^3 + u^4 + \cdots.$$

Set $u = x^4$ :

$$\dfrac{1}{1 - x^4} = \sum_{n=0}^\infty x^{4 n} = 1 + x^4 + x^8 + x^{12} + x^{16} + \cdots.\quad\halmos$$

(b) Note that

$$\der {} x \left(\dfrac{1}{1 - x^4}\right) = \dfrac{4 x^3}{(1 - x^4)^2}.$$

Hence,

$$\dfrac{x^3}{(1 - x^4)^2} = \dfrac{1}{4} \der {} x \left(\dfrac{1}{1 - x^4}\right) = \dfrac{1}{4} \sum_{n=1}^\infty 4 n x^{4 n - 1} = \dfrac{1}{4} \left(4 x^3 + 8 x^7 + 12 x^{11} + 16 x^{15} + \cdots\right).\quad\halmos$$


5. (a) Use a known series to construct the power series centered at $c = 0$ for $\dfrac{1}{2
   - x}$ . Write your answer in summation form, or give the first four nonzero terms of the series.

(b) By differentiating the series in (a), construct the power series centered at $c = 0$ for $\dfrac{1}{(2 - x)^2}$ . Write your answer in summation form, or give the first four nonzero terms of the series.

(a) I have

$$\dfrac{1}{1 - u} = \sum_{n=0}^\infty u^n = 1 + u + u^2 + u^3 + u^4 + \cdots.$$

Set $u = \dfrac{x}{2}$ :

$$\dfrac{1}{2 - x} = \dfrac{1}{2} \left(\dfrac{1}{1 - \dfrac{x}{2}}\right) = \dfrac{1}{2} \sum_{n=0}^\infty \dfrac{x^n}{2^n} = \dfrac{1}{2} \left(1 + \dfrac{x}{2} + \dfrac{x^2}{4} + \dfrac{x^3}{8} + \dfrac{x^4}{16} + \cdots\right).\quad\halmos$$

(b) I have

$$\der {} x \left(\dfrac{1}{2 - x}\right) = \dfrac{1}{(2 - x)^2}.$$

So

$$\dfrac{1}{(2 - x)^2} = \dfrac{1}{2} \sum_{n=1}^\infty \dfrac{n x^{n - 1}}{2^n} = \dfrac{1}{2} \left(\dfrac{1}{2} + \dfrac{x}{2} + \dfrac{3 x^2}{8} + \dfrac{x^3}{4} + \cdots\right).\quad\halmos$$


6. (a) Write down the first 3 nonzero terms of the binomial series expansion for $(1 -
   t^2)^{-1/2}$ at $c = 0$ .

(b) By integrating the series in (a) from $t = 0$ to $t = x$ , find the first three nonzero terms of the power series centered at $c = 0$ for $sin^{-1} x$ .

(a) I have

$$(1 + u)^a = 1 + \sum_{n=1}^\infty \dfrac{a(a - 1) \cdots (a - n + 1)}{n!} u^n = 1 + a u + \dfrac{a(a - 1)}{2!} u^2 + \dfrac{a(a - 1)(a - 2)}{3!} u^3 + \cdots.$$

Set $a = -\dfrac{1}{2}$ and $u = -t^2$ :

$$(1 - t^2)^{-1/2} = 1 + \dfrac{1}{2} t^2 + \dfrac{3}{8} t^4 + \dfrac{5}{16} t^6 + \cdots.\quad\halmos$$

(b) I have

$$\sin^{-1} x = \int_0^x \dfrac{1}{\sqrt{1 - t^2}}\,dt = \int_0^x (1 - t^2)^{-1/2}\,dt = \int_0^x \left(1 + \dfrac{1}{2} t^2 + \dfrac{3}{8} t^4 + \dfrac{5}{16} t^6 + \cdots\right)\,dt =$$

$$\left[t + \dfrac{1}{6} t^3 + \dfrac{3}{40} t^5 + \dfrac{5}{112} t^7 + \cdots \right]_0^x = x + \dfrac{1}{6} x^3 + \dfrac{3}{40} x^5 + \dfrac{5}{112} x^7 + \cdots.\quad\halmos$$


7. (a) Use a known series to construct the power series centered at $c = 1$ for $\dfrac{1}{t}$ . Write your answer in summation form, or give the first four nonzero terms of the series.

(b) By integrating the series in (a) from $t = 1$ to $t = x$ , find the power series centered at $c = 1$ for $\ln x$ . Write your answer in summation form, or give the first four nonzero terms of the series.

(a) I have

$$\dfrac{1}{1 - u} = \sum_{n=0}^\infty u^n = 1 + u + u^2 + u^3 + u^4 + \cdots.$$

Set $u = -(t - 1)$ :

$$\dfrac{1}{t} = \dfrac{1 + (t - 1)} = \dfrac{1}{1 -[-(t - 1)]} = \sum_{n=0}^\infty (-1)^n (t - 1)^n = 1 - (t - 1) + (t - 1)^2 - (t - 1)^3 + (t - 1)^4 - \cdots.\quad\halmos$$

(b) I have

$$\ln x = \int_1^x \dfrac{1}{t}\,dt = \int_1^x \left(1 - (t - 1) + (t - 1)^2 - (t - 1)^3 + (t - 1)^4 - \cdots\right)\,dt =$$

$$\left[t - \dfrac{1}{2} (t - 1)^2 + \dfrac{1}{3} (t - 1)^3 - \dfrac{1}{4} (t - 1)^4 + \cdots\right]_1^x = (x - 1) - \dfrac{1}{2} (x - 1)^2 + \dfrac{1}{3} (x - 1)^3 + \dfrac{1}{4} (x - 1)^4 + \cdots.$$

In summation form,

$$\ln x = \sum_{n=1}^\infty (-1)^{n+1} \dfrac{(x - 1)^n}{n}.\quad\halmos$$


Courage consists of the power of self-recovery. - Ralph Waldo Emerson


Contact information

Bruce Ikenaga's Home Page

Copyright 2017 by Bruce Ikenaga