Solutions to Problem Set 26

Math 211-03

11-9-2017

[Constructing power series]

1. Construct the power series centered at $x = 0$ for $f(x) =
   \dfrac{1}{1 - 4 x^2}$ . Write your answer in summation form, or give the first 4 nonzero terms of the series. In addition, find the interval of convergence.

Let $u = 4 x^2$ in the series for $\dfrac{1}{1 - u}$ .

$$\dfrac{1}{1 - 4 x^2} = \sum_{n=0}^\infty 4^n x^{2 n} = 1 + 4 x^2 + 16 x^4 + 64 x^6 + \cdots.$$

For the interval of convergence, I have $-1 < 4 x^2 < 1$ . Since a square can't be negative, the relevant part of the inequality is $4 x^2 < 1$ . This gives $x^2 < \dfrac{1}{4}$ , or $-\dfrac{1}{2} < x < \dfrac{1}{2}$ .


2. Construct the power series centered at $x = 1$ for $f(x) =
   \dfrac{1}{6 - x}$ . Write your answer in summation form, or give the first 4 nonzero terms of the series. In addition, find the interval of convergence.

Letting $u = \dfrac{1}{5} (x -
   1)$ in the series for $\dfrac{1}{1 - u}$ , I get

$$\dfrac{1}{6 - x} = \dfrac{1}{5 - (x - 1)} = \dfrac{1}{5} \dfrac{1}{1 - \left(\dfrac{1}{5} (x - 1)\right)} = \dfrac{1}{5} \sum_{n=0}^\infty \dfrac{1}{5^n} (x - 1)^n =$$

$$\dfrac{1}{5} \left(1 + \dfrac{1}{5} (x - 1) + \dfrac{1}{25} (x - 1)^2 + \dfrac{1}{125} (x - 1)^3 + \cdots\right).$$

For the interval of convergence,

$$\eqalign{ -1 < & u < 1 \cr \noalign{\vskip2pt} -1 < & \dfrac{1}{5} (x - 1) < 1 \cr \noalign{\vskip2pt} -5 < & x - 1 < 5 \cr -4 < & x < 6 \quad\halmos \cr}$$


3. Construct the power series centered at $x = -1$ for $f(x) =
   \dfrac{1}{-2 - 3 x}$ . Write your answer in summation form, or give the first 4 nonzero terms of the series. In addition, find the interval of convergence.

Letting $u = 3(x + 1)$ in the series for $\dfrac{1}{1 - u}$ , I get

$$\dfrac{1}{-2 - 3 x} = \dfrac{1}{1 - 3(x + 1)} = \sum_{n=0}^\infty 3^n (x + 1)^n = 1 + 3(x + 1) + 9(x + 1)^2 + 27(x + 1)^3 + \cdots.$$

For the interval of convergence,

$$\eqalign{ -1 < & u < 1 \cr -1 < & 3(x + 1) < 1 \cr \noalign{\vskip2pt} -\dfrac{1}{3} < & x + 1 < \dfrac{1}{3} \cr \noalign{\vskip2pt} -\dfrac{4}{3} < & x < -\dfrac{2}{3} \quad\halmos \cr}$$


4. Construct the power series centered at $x = 2$ for $f(x) =
   \dfrac{x - 1}{3 - x}$ . Write your answer in summation form, or give the first 4 nonzero terms of the series. In addition, find the interval of convergence.

Letting $u = x - 2$ in the series for $\dfrac{1}{1 - u}$ , I get

$$\dfrac{x - 1}{3 - x} = \dfrac{(x - 2) + 1}{1 - (x - 2)} = [(x - 2) + 1] \dfrac{1}{1 - (x - 2)} = [(x - 2) + 1] \sum_{n=0}^\infty (x - 2)^n = \sum_{n=0}^\infty (x - 2)^n + \sum_{n=0}^\infty (x - 2)^{n+1} =$$

$$1 + 2 \sum_{n=1}^\infty (x - 2)^n = 1 + 2(x - 2) + 2(x - 2)^2 + 2(x - 2)^3 + \cdots.$$

For the interval of convergence,

$$\eqalign{ -1 < & u < 1 \cr -1 < & x - 2 < 1 \cr 1 < & x < 3 \quad\halmos \cr}$$


5. Construct the power series centered at $x = 2$ for $f(x) = e^{3
   x}$ . Write your answer in summation form, or give the first 4 nonzero terms of the series. In addition, find the interval of convergence.

Letting $u = 3(x - 2)$ in the series for $e^u$ , I get

$$e^{3 x} = e^{(6 + 3(x - 2)} = e^6 e^{3(x - 2)} = e^6 \sum_{n=0}^\infty \dfrac{3^n (x - 2)^n}{n!} =$$

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

For the interval of convergence,

$$\eqalign{ -\infty < & u < \infty \cr -\infty < & 3(x - 2) < \infty \cr -\infty < & x - 2 < \infty \cr -\infty < & x < \infty \quad\halmos \cr}$$


6. Construct the power series centered at $x = -2$ for $f(x) = \ln
   (3 + x)$ . Write your answer in summation form, or give the first 4 nonzero terms of the series. In addition, find the interval of convergence.

Letting $u = x + 2$ in the series for $\ln (1 + u)$ , I get

$$\ln (3 + x) = \ln [1 + (x + 2)] = \sum_{n=1}^\infty (-1)^{n+1} \dfrac{(x + 2)^n}{n} =$$

$$(x + 2) - \dfrac{1}{2} (x + 2)^2 + \dfrac{1}{3} (x + 2)^3 - \dfrac{1}{4} (x + 2)^4 + \cdots.$$

For the interval of convergence,

$$\eqalign{ -1 < & x + 2 \le 1 \cr -3 < & x \le -1 \quad\halmos \cr}$$


7. Construct the power series centered at $x = 0$ for $f(x) = \sin
   x^3$ . Write your answer in summation form, or give the first 4 nonzero terms of the series. In addition, find the interval of convergence.

The series is centered at $x =
   0$ , so I want powers of $x - 0 = x$ .

Letting $u = x^3$ in the series for $\sin u$ , I get

$$\sin x^3 = \sum_{n=0}^\infty (-1)^n \dfrac{x^{6 n + 3}}{(2 n + 1)!} = x^3 - \dfrac{x^9}{3!} + \dfrac{x^{15}}{5!} - \dfrac{x^{21}}{7!} + \cdots.$$

For the interval of convergence,

$$\eqalign{ -\infty < & x^3 < \infty \cr -\infty < & x < \infty \quad\halmos \cr}$$


8. Find the function to which the power series $\displaystyle \sum_{n=0}^\infty
   \dfrac{(x - 5)^n}{2^n}$ converges.

Letting $u = \dfrac{1}{2} (x -
   5)$ in the series for $\dfrac{1}{1 - u}$ , I find that

$$\sum_{n=0}^\infty \dfrac{(x - 5)^n}{2^n} = \dfrac{1}{1 - \left(\dfrac{1}{2} (x - 5)\right)} = \dfrac{2}{2 - (x - 5)} = \dfrac{2}{7 - x}.\quad\halmos$$


9. Find the function to which the power series $\displaystyle \sum_{n=0}^\infty
   \dfrac{4^n (x + 3)^n}{n!}$ converges.

Letting $u = 4(x + 3)$ in the series for $e^u$ , I find that

$$\sum_{n=0}^\infty \dfrac{4^n (x + 3)^n}{n!} = e^{4(x + 3)}.\quad\halmos$$


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