Power Series - Review

Example. Expand $f(x) = \dfrac{2}{4 - 5 x}$ in a power series at $c = 2$ and find the interval of convergence.

$$\dfrac{2}{4 - 5 x} = \dfrac{2}{-6 - 5(x - 2)} = -\dfrac{1}{3}\dfrac{1}{1 + \dfrac{5}{6}(x - 2)} = -\dfrac{1}{3}\dfrac{1}{1 - \left(-\dfrac{5}{6}(x - 2)\right)} =$$

$$-\dfrac{1}{3}\left(1 - \left(\dfrac{5}{6}(x - 2)\right) + \left(\dfrac{5}{6}(x - 2)\right)^2 - \left(\dfrac{5}{6}(x - 2)\right)^3 + \cdots \right).$$

For the interval of convergence,

$$-1 < \dfrac{5}{6}(x - 2) < 1 \quad\hbox{gives}\quad \dfrac{4}{5} < x < \dfrac{16}{5}.\quad\halmos$$


Example. Find the interval of convergence of $\displaystyle
   \sum_{n=0}^\infty \dfrac{3^n(x - 5)^n}{2^n}$ .

Apply the Root Test:

$$\lim_{n\to \infty} \left(\dfrac{3^n|x - 5|^n}{2^n}\right)^{1/n} = \lim_{n\to \infty} \dfrac{3}{2} |x - 5| = \dfrac{3}{2} |x - 5|.$$

The series converges for

$$\dfrac{3}{2}|x - 5| < 1, \quad\hbox{i.e. for}\quad \dfrac{13}{3} < x < \dfrac{17}{3}.$$

At $x = \dfrac{17}{3}$ , the series is

$$\sum_{n=0}^\infty \dfrac{3^n}{2^n}\left(\dfrac{17}{3} - 5\right)^n = \sum_{n=0}^\infty 1.$$

This series diverges by the Zero Limit Test.

At $x = \dfrac{13}{3}$ , the series is

$$\sum_{n=0}^\infty \dfrac{3^n}{2^n}\left(\dfrac{13}{3} - 5\right)^n = \sum_{n=0}^\infty (-1)^n.$$

This series also diverges by the Zero Limit Test.

The power series converges for $\dfrac{13}{3} < x < \dfrac{17}{3}$ and diverges elsewhere.


Example. Expand $f(x) = e^{-3 x}$ in a power series at $c = 4$ and find the interval of convergence.

$$e^{-3 x} = e^{-3(x-4)-12} = e^{-12}e^{-3(x - 4)}.$$

Set $u = -3(x - 4)$ in

$$e^u = 1 + u + \dfrac{u^2}{2!} + \dfrac{u^3}{3!} + \cdots.$$

This gives

$$e^{-3 x} = e^{-12}\left(1 - 3(x - 4) + \dfrac{3^2(x - 4)^2}{2!} - \dfrac{3^3(x - 4)^3}{3!} + \cdots\right).$$

The interval of convergence for the $e^u$ series is $-\infty < u < \infty$ . So for the $e^{-3 x}$ series,

$$-\infty < -3(x - 4) < +\infty, \quad -\infty < x < +\infty.\quad\halmos$$


Example. Expand $(\cos 5 x)^2$ in a Taylor series at $c = 0$ .

Using the double angle formula

$$(\cos 5 x)^2 = \dfrac{1}{2}(1 + \cos 10 x) = \dfrac{1}{2} + \dfrac{1}{2}\cos 10 x = \dfrac{1}{2} + \dfrac{1}{2}\left(1 - \dfrac{10^2 x^2}{2!} + \dfrac{10^4 x^4}{4!} - \dfrac{10^6 x^6}{6!} + \cdots\right) =$$

$$1 - \dfrac{1}{2}\dfrac{10^2 x^2}{2!} + \dfrac{1}{2}\dfrac{10^4 x^4}{4!} - \dfrac{1}{2}\dfrac{10^6 x^6}{6!} + \cdots.\quad\halmos$$


Example. (a) Use the first four nonzero terms of the Taylor series for $e^u$ at $c = 0$ to approximate $\displaystyle \int_0^1 e^{-x^3}\,dx$ .

(b) Use the Alternating Series Test to estimate the error in part (a).

(a)

$$e^{-x^3} = 1 - x^3 + \dfrac{x^6}{2!} - \dfrac{x^9}{3!} + \dfrac{x^{12}}{4!} - \cdots .$$

Hence,

$$\int_0^1 e^{-x^3}\,dx = \int_0^1 \left(1 - x^3 + \dfrac{x^6}{2!} - \dfrac{x^9}{3!} + \dfrac{x^{12}}{4!} - \cdots\right)\,dx = \left[x - \dfrac{x^4}{4} + \dfrac{x^7}{14} - \dfrac{x^{10}}{60} + \dfrac{x^{13}}{312} - \cdots\right]_0^1 =$$

$$1 - \dfrac{1}{4} + \dfrac{1}{14} - \dfrac{1}{60} + \dfrac{1}{312} - \cdots \approx 0.80476.$$

(I used the first four terms to get the approximation.)

(b) The error is no greater than the next term, which is $\dfrac{1}{312} = 0.00320
   \ldots$ .


Example. Use the Taylor series expansion of $\sin x$ at $c = 0$ to explain the fact that $\displaystyle
   \lim_{x\to 0} \dfrac{\sin x}{x} = 1$ .

The series for $\sin x$ at $c = 0$ is

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

Divide by x to obtain

$$\dfrac{\sin x}{x} = 1 - \dfrac{x^2}{3!} + \dfrac{x^4}{5!} - \cdots.$$

Then

$$\lim_{x\to 0} \dfrac{\sin x}{x} = \lim_{x\to 0} \left(1 - \dfrac{x^2}{3!} + \dfrac{x^4}{5!} - \cdots\right) = 1.\quad\halmos$$


Example. Find the first four nonzero terms of the Taylor expansion for $y
   = \sin x$ at $c = 3$ .

$$f'(x) = \cos x, \quad f''(x) = -\sin x, \quad f'''(x) = -\cos x.$$

The series is

$$\sin x = \sin 3 + (\cos 3)(x - 3) - \dfrac{\sin 3}{2!}(x - 3)^2 - \dfrac{\cos 3}{3!}(x - 3)^3 + \cdots.\quad\halmos$$


Example. Find the interval of convergence of $\displaystyle
   \sum_{n=1}^\infty \dfrac{(x - 5)^{2 n}}{n\cdot 9^n}$ .

$$\lim_{n\to \infty} \dfrac{\dfrac{|x - 5|^{2 n+2}}{(n + 1)\cdot 9^{n+1}}} {\dfrac{|x - 5|^{2 n}}{n\cdot 9^n}} = \lim_{n\to \infty} \dfrac{|x - 5|^{2 n+2}}{(n + 1)\cdot 9^{n+1}} \dfrac{n\cdot 9^n}{|x - 5|^{2 n}} =$$

$$\lim_{n\to \infty} \dfrac{9^n}{9^{n+1}}\cdot \dfrac{n}{n + 1}\cdot \dfrac{|x - 5|^{2 n+2}}{|x - 5|^{2 n}} = \lim_{n\to \infty} \dfrac{n}{n + 1}\cdot \dfrac{1}{9}|x - 5| = \dfrac{1}{9}|x - 5|^2.$$

The series converges for

$$\dfrac{1}{9}|x - 5|^2 < 1, \quad\hbox{i.e. for}\quad 2 < x < 8.$$

At $x = 8$ , the series is

$$\sum_{n=1}^\infty \dfrac{3^{2 n}}{n\cdot 9^n} = \sum_{n=1}^\infty \dfrac{1}{n}.$$

This is harmonic, so it diverges.

At $x = 2$ , the series is

$$\sum_{n=1}^\infty \dfrac{(-3)^{2 n}}{n\cdot 9^n} = \sum_{n=1}^\infty \dfrac{1}{n}.$$

This is harmonic, so it diverges.

The power series converges for $2 < x < 8$ and diverges elsewhere.


Example. $f(x)$ satisfies

$$f(2) = 5, \quad f'(2) = 2, \quad f''(2) = -3, \quad f'''(2) = 1.$$

Use the third degree Taylor polynomial for f at $c = 2$ to approximate $f(2.02)$ .

The third degree Taylor polynomial for f at $c = 2$ is

$$p_3(x) = 5 + 2(x - 2) - \dfrac{3}{2!}(x - 2)^2 + \dfrac{1}{3!}(x - 2)^3 = 5 + 2(x - 2) - \dfrac{3}{2}(x - 2)^2 + \dfrac{1}{6}(x - 2)^3.$$

So

$$f(2.02) \approx p_3(2.02) = 5 + (2)(0.02) - \left(\dfrac{3}{2}\right)(0.02)^2 + \left(\dfrac{1}{6}\right)(0.02)^3 = 5.039401 \ldots.\quad\halmos$$


Example. Suppose that $f^{(5)}(x) = \dfrac{2}{(1 -
   x)^7}$ . Use $R_4(x;0)$ to estimate the error in using the fourth degree Taylor polynomial at $c =
   0$ to approximate $f(x)$ for $0 \le
   x \le 0.1$ .

For some z between 0 and x,

$$R_4(x;0) = \dfrac{f^{(5)}(z)}{5!}x^5 = \dfrac{1}{120}\left(\dfrac{2}{(1 - z)^7}\right) x^5.$$

Since $0 \le x \le 0.1$ , $x^5 \le 0.1^5$ .

For the z-term, I have $0
   \le z \le x \le 0.1$ . Thus,

$$\eqalign{ 0 \le & z \le 0.1 \cr 0 \ge & -z \ge -0.1 \cr 1 \ge & 1 - z \ge 0.9 \cr 1 \ge & (1 - z)^7 \ge 0.9^7 \cr \noalign{\vskip2pt} 1 \le & \dfrac{1}{(1 - z)^7} \le \dfrac{1}{0.9^7} \cr \noalign{\vskip2pt} 2 \le & \dfrac{2}{(1 - z)^7} \le \dfrac{2}{0.9^7} \cr}$$

So $\dfrac{2}{(1 - z)^7}
   \le \dfrac{2}{0.9^7}$ . Therefore,

$$R_4(x;0) \le \left(\dfrac{1}{120}\right)\left(\dfrac{2}{0.9^7}\right)(0.1^5) \approx 3.4845 \cdot 10^{-7}.\quad\halmos$$


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