User:Egm4313.s12.team8.colocar/R5.1

Problem
$$ \displaystyle \text{Find } R_c \text{ for the following series:}$$ $$ \displaystyle \text{1. } r(x) = \sum_{k=0}^\infty(k+1)kx^k $$ $$ \displaystyle \text{2. } r(x) = \sum_{k=0}^\infty\frac{(-1)^k}{\gamma^k}x^{2k} $$ $$ \displaystyle \gamma = \text{constant} $$

$$ \displaystyle \text{Use (2)-(3) p.7-31 to find } R_c \text{ for the Taylor series of}$$

$$ \displaystyle \text{3. } sinx \text{ at } \hat{x}=0 $$ $$ \displaystyle \text{4. } log(1+x) \text{ at } \hat{x}=0 $$ $$ \displaystyle \text{5. } log(1+x) \text{ at } \hat{x}=1 $$

Solution
$$ \displaystyle \text{From 7-31, }$$ $$ \displaystyle r(x) = \sum_{k=0}^\infty d_k x^k $$ $$ \displaystyle \text{The radius of convergence of the series (1) is given by } $$ $$ \displaystyle R_c = \left[ \lim_{k \to \infty} \left| \frac{d_{k+1}}{d_k} \right| \right]^{-1}$$ $$ \displaystyle \text{or} $$ $$ \displaystyle R_c = \left[ \lim_{k \to \infty} \sqrt[k] {| d_k |} \right]^{-1}$$

$$ \displaystyle \text{1. } r(x) = \sum_{k=0}^\infty(k+1)kx^k $$ $$ \displaystyle d_k = (k+1)k $$ $$ \displaystyle d_{k+1} = (k+2)(k+1) $$ $$ \displaystyle R_c = \left[ \lim_{k \to \infty} \left| \frac{(k+2)(k+1)}{(k+1)k} \right| \right]^{-1}$$ $$ \displaystyle R_c = \left[ \lim_{k \to \infty} \left| 1+ \frac{2}{k} \right| \right]^{-1}$$ $$ \displaystyle R_c = \left[ 1+0 \right]^{-1}$$ $$ \displaystyle R_c = 1$$

$$ \displaystyle \text{2. } r(x) = \sum_{k=0}^\infty\frac{(-1)^k}{\gamma^k}x^{2k} $$ $$ \displaystyle d_k = \frac{(-1)^k}{\gamma^k} $$ $$ \displaystyle d_{k+1} = \frac{(-1)^{k+1}}{\gamma^{k+1}} $$ $$ \displaystyle R_c = \left[ \lim_{k \to \infty} \left| \frac{\frac{(-1)^{k+1}}{\gamma^{k+1}}}{\frac{(-1)^k}{\gamma^k}} \right| \right]^{-1}$$ $$ \displaystyle R_c = \left[ \lim_{k \to \infty} \left| \frac{(-1)^{k+1} \gamma^k}{(-1)^k \gamma^{k+1}} \right| \right]^{-1}$$ $$ \displaystyle R_c = \left[ \lim_{k \to \infty} \left| \frac{-1}{\gamma} \right| \right]^{-1}$$ $$ \displaystyle R_c = \gamma $$ $$ \displaystyle \text{Now since we have } x^{2k} \text{ instead of } x^k $$ $$ \displaystyle R_c = \sqrt{\gamma} $$

$$ \displaystyle \text{3. Taylor Series of } sinx \text{ at } \hat{x}=0 \text{ is}$$ $$ \displaystyle sin(x) = \sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)!} x^{1+2k} $$ $$ \displaystyle d_k = \frac{(-1)^k}{(2k+1)!} $$ $$ \displaystyle d_{k+1} = \frac{(-1)^{k+1}}{(2(k+1)+1)!} $$

$$ \displaystyle \text{Then,} $$ $$ \displaystyle R_c = \left[ \lim_{k \to \infty} \left| \frac{d_{k+1}}{d_k} \right| \right]^{-1}$$ $$ \displaystyle R_c = \left[ \lim_{k \to \infty} \left| \frac{\frac{(-1)^{k+1}}{(2(k+1)+1)!}}{\frac{(-1)^k}{(2k+1)!}} \right| \right]^{-1}$$ $$ \displaystyle R_c = \left[ \lim_{k \to \infty} \left| \frac{(-1)^{k+1}[(2k+1)!]}{(-1)^k[(2k+3)!]} \right| \right]^{-1}$$ $$ \displaystyle R_c = \left[ \lim_{k \to \infty} \left| \frac{(2k+1)!}{(2k+3)!} \right| \right]^{-1}$$ $$ \displaystyle R_c = \lim_{k \to \infty} \frac{(2k+3)!}{(2k+1)!}$$ $$ \displaystyle R_c = \infty$$

$$ \displaystyle \text{4. Taylor Series of } log(1+x) \text{ at } \hat{x}=0 \text{ is}$$ $$ \displaystyle log(1+x) = \sum_{k=1}^\infty \frac{(-1)^{k+1}}{k} x^{k} $$ $$ \displaystyle d_k = \frac{(-1)^{k+1}}{k} $$ $$ \displaystyle d_{k+1} = \frac{(-1)^{k+2}}{k+1} $$

$$ \displaystyle R_c = \left[ \lim_{k \to \infty} \left| \frac{\frac{(-1)^{k+2}}{k+1}}{\frac{(-1)^{k+1}}{k}} \right| \right]^{-1}$$ $$ \displaystyle R_c = \left[ \lim_{k \to \infty} \left| \frac{(-1)^{k+2}k}{(-1)^{k+1}(k+1)} \right| \right]^{-1}$$ $$ \displaystyle R_c = \left[ \lim_{k \to \infty} \frac{k}{k+1} \right]^{-1}$$ $$ \displaystyle R_c = \frac{\infty}{\infty} = 1$$

$$ \displaystyle \text{5. Taylor Series of } log(1+x) \text{ at } \hat{x}=1 \text{ is}$$ $$ \displaystyle log(1+x) = log(2) + \sum_{k=1}^\infty \frac{(-1)^{k+1}}{2^kk} (x-1)^{k} $$ $$ \displaystyle d_k = \frac{(-1)^{k+1}}{2^kk} (x-1)^{k} $$ $$ \displaystyle d_{k+1} = \frac{(-1)^{k+2}}{2^{k+1}k+1} (x-1)^{k+1} $$

$$ \displaystyle R_c = \left[ \lim_{k \to \infty} \left| \frac{(-1)^{k+2}(x-1)^{k+1}2^kk}{2^{k+1}k+1(-1)^{k+1}(x-1)^k} \right| \right]^{-1}$$ $$ \displaystyle R_c = \left[ \lim_{k \to \infty} \left| \frac{(-1)(x-1)}{2} \right| \right]^{-1}$$ $$ \displaystyle R_c = \frac{2}{(x-1)} \text{, which, to converge, is less than 1. So, }$$ $$ \displaystyle \frac{2}{(x-1)}<1 $$ $$ \displaystyle x>3 $$ $$ \displaystyle R_c = 3 $$