User:Egm4313.s12.team11.sheider/R5.1: Lecture 7 p.7-32 pb. 1-5

Problem Statement
Given: Find $$ R_{c} \!$$ for the following series:

1. $$ r(x) = \sum_{k=0}^\infty (k+1) k x^k \!$$

2. $$ r(x) = \sum_{k=0}^\infty \frac{(-1)^k}{\gamma^k} x^{2k} \!$$

Find $$ R_{c} \!$$ for the Taylor series of

3. $$ sin(x) \!$$ at  $$ x = 0 \!$$

4. $$ log(1+x) \!$$ at  $$ x=0 \!$$

5. $$ log(1+x)\!$$ at  $$ x=1 \!$$

Solution
The radius of convergence $$ R_{c} \!$$ is defined as

$$ R_{c} = [\lim_{k \to \infty} |\frac{d_{k+1}}{d_{k}}|]^{-1} \!$$

1. $$ R_{c} = [\lim_{k \to \infty} |\frac{(k+2)(k+1)}{(k+1)(k)}|]^{-1} \!$$

$$ R_{c} = [\lim_{k \to \infty} |\frac{(k+2)}{(k)}|]^{-1} \!$$

$$ R_{c} = [\lim_{k \to \infty} |\frac{k(1+\frac{2}{k})}{k}|]^{-1} \!$$

$$ R_{c} = [\lim_{k \to \infty} |1+\frac{2}{k}|]^{-1} \!$$

$$ R_{c} = [1+0]^{-1} \!$$

$$ R_{c} = 1 \!$$

2. $$ R_{c} = [\lim_{k \to \infty} |\frac{\frac{(-1)^{k+1}}{\gamma^{k+1}}}{\frac{(-1)^{k}}{\gamma^{k}}}|]^{-1} \!$$

$$ R_{c} = [\lim_{k \to \infty} |\frac{-1}{\gamma}|]^{-1} \!$$

$$ R_{c} = [\lim_{k \to \infty} \frac{1}{\gamma}]^{-1} \!$$

$$ R_{c} = [\frac{1}{\gamma}]^{-1} \!$$

$$ R_{c} = \gamma \!$$

However, in this problem, the series $$ x \!$$ term is $$ x^{2k} \!$$ not $$ x^k \!$$, as is the general form. Therefore, this implies:

$$ |x^2| = \gamma \!$$

$$ |x| = \sqrt{\gamma} \!$$

$$ R_{c} = \sqrt{\gamma} \!$$

3. The Taylor series for $$ sin(x) \!$$ is expressed as $$ sin(x) = \sum_{k=0}^\infty \frac{(-1)^{k}}{(2k+1)!}x^{2k+1} \!$$

$$\lim_{k \to \infty} \frac{\frac{(-1)^{k+1}}{(2(k+1)+1)!}x^{2(k+1)+1}}{\frac{(-1)^{k}}{(2k+1)!}x^{2k+1}} \!$$

$$\lim_{k \to \infty} \frac{\frac{1}{(2k+3)!}x^{2k+3}}{\frac{1}{(2k+1)!}x^{2k+1}} \!$$

$$\lim_{k \to \infty} \frac{\frac{1}{(2k+3)}x^{3}}{x} \!$$

$$\lim_{k \to \infty} \frac{1}{(2k+3)} x^{2} \!$$

Therefore: $$R_{c} = [\lim_{k \to \infty} \frac{1}{(2k+3)} ]^{-1}\!$$

$$R_{c} = \infty \!$$

4. The Taylor series for $$ log(1+x) \!$$ at $$ x = 0 \!$$ is expressed as $$ log(1+x) = \sum_{k=1}^\infty \frac{(-1)^{k+1}}{k}x^{k} \!$$

$$R_{c} = [\lim_{k \to \infty} |\frac{\frac{(-1)^{k+2}}{k+1}}{\frac{(-1)^{k+1}}{k}} |]^{-1} \!$$

$$R_{c} = [\lim_{k \to \infty} |\frac{\frac{(-1)}{k+1}}{\frac{(1)}{k}} |]^{-1} \!$$

$$R_{c} = [\lim_{k \to \infty} |\frac{\frac{(-1)}{k(1+\frac{1}{k})}}{\frac{1}{k}} |]^{-1} \!$$

$$R_{c} = [\lim_{k \to \infty} |\frac{(-1)}{(1+\frac{1}{k})}|]^{-1} \!$$

$$R_{c} = [1/1]^{-1} \!$$

$$R_{c} = 1 \!$$

5. The Taylor series for $$ log(1+x) \!$$ at $$ x = 1 \!$$ is expressed as $$ log(1+x) = log(2) + \sum_{k=1}^\infty \frac{(-1)^{k+1}}{2^{k} k}(x-1)^{k} \!$$

$$\lim_{k \to \infty} \frac{\frac{(-1)^{k+2}}{2^{k+1} (k+1)}(x-1)^{k+1}}{\frac{(-1)^{k+1}}{2^{k} k}(x-1)^{k}} \!$$

$$\lim_{k \to \infty} \frac{\frac{(-1)^{2}}{2(k+1)}(x-1)}{\frac{1}{k}} \!$$

$$\lim_{k \to \infty} \frac{\frac{1}{2k(1+\frac{1}{k})}(x-1)}{\frac{1}{k}} \!$$

$$\lim_{k \to \infty} \frac{\frac{1}{2(1+\frac{1}{k})}(x-1)}{1} \!$$

$$\frac{1}{2}(x-1) \!$$

For convergence: $$ \frac{1}{2}(x-1) < 1 \!$$

$$ x-1 < 2 \!$$

$$ x < 3 \!$$

Therefore, $$R_{c} = 3 \!$$