Laurent Series for $f(z)=frac{2}{(z+1)^2}-frac{5}{z-5}$
up vote
4
down vote
favorite
I am trying to find the Laurent series for the function $$f(z)=frac{2}{(z+1)^2}-frac{5}{z-5},$$ in powers of $(z-1)$ that converges when $z=4$.
The series will converge when $z=4$ is in the region $2<|z-1|<4$. Hence,
begin{align}
f(z)&=-2frac{d}{dz}left(frac{1}{z+1}right)+5left(frac{1}{5-z}right) \
&=-2frac{d}{dz}left(frac{1}{(z-1)}frac{1}{1+frac{2}{(z-1)}}right)+frac{5}{4}left(frac{1}{1-frac{(z-1)}{4}}right) \
&=-2frac{d}{dz}left(frac{1}{(z-1)}sum_{n=0}^{infty}frac{(-1)^n2^n}{(z-1)^n}right)+frac{5}{4}sum_{n=0}^{infty}frac{(z-1)^n}{4^n} \
&=sum_{n=0}^{infty}frac{(-1)^n2^{n+1}(n+1)(z-1)^n}{(z-1)^{2(n+1)}}+5sum_{n=0}^{infty}frac{(z-1)^n}{4^{n+1}}.
end{align}
Is this solution correct?
complex-analysis proof-verification laurent-series
asked Oct 29 at 22:28
Bell
167316
add a comment |
up vote
4
down vote
favorite
I am trying to find the Laurent series for the function $$f(z)=frac{2}{(z+1)^2}-frac{5}{z-5},$$ in powers of $(z-1)$ that converges when $z=4$.
The series will converge when $z=4$ is in the region $2<|z-1|<4$. Hence,
begin{align}
f(z)&=-2frac{d}{dz}left(frac{1}{z+1}right)+5left(frac{1}{5-z}right) \
&=-2frac{d}{dz}left(frac{1}{(z-1)}frac{1}{1+frac{2}{(z-1)}}right)+frac{5}{4}left(frac{1}{1-frac{(z-1)}{4}}right) \
&=-2frac{d}{dz}left(frac{1}{(z-1)}sum_{n=0}^{infty}frac{(-1)^n2^n}{(z-1)^n}right)+frac{5}{4}sum_{n=0}^{infty}frac{(z-1)^n}{4^n} \
&=sum_{n=0}^{infty}frac{(-1)^n2^{n+1}(n+1)(z-1)^n}{(z-1)^{2(n+1)}}+5sum_{n=0}^{infty}frac{(z-1)^n}{4^{n+1}}.
end{align}
Is this solution correct?
complex-analysis proof-verification laurent-series
asked Oct 29 at 22:28
Bell
167316
add a comment |
up vote
4
down vote
favorite
up vote
4
down vote
favorite
I am trying to find the Laurent series for the function $$f(z)=frac{2}{(z+1)^2}-frac{5}{z-5},$$ in powers of $(z-1)$ that converges when $z=4$.
The series will converge when $z=4$ is in the region $2<|z-1|<4$. Hence,
begin{align}
f(z)&=-2frac{d}{dz}left(frac{1}{z+1}right)+5left(frac{1}{5-z}right) \
&=-2frac{d}{dz}left(frac{1}{(z-1)}frac{1}{1+frac{2}{(z-1)}}right)+frac{5}{4}left(frac{1}{1-frac{(z-1)}{4}}right) \
&=-2frac{d}{dz}left(frac{1}{(z-1)}sum_{n=0}^{infty}frac{(-1)^n2^n}{(z-1)^n}right)+frac{5}{4}sum_{n=0}^{infty}frac{(z-1)^n}{4^n} \
&=sum_{n=0}^{infty}frac{(-1)^n2^{n+1}(n+1)(z-1)^n}{(z-1)^{2(n+1)}}+5sum_{n=0}^{infty}frac{(z-1)^n}{4^{n+1}}.
end{align}
Is this solution correct?
complex-analysis proof-verification laurent-series
asked Oct 29 at 22:28
Bell
167316
I am trying to find the Laurent series for the function $$f(z)=frac{2}{(z+1)^2}-frac{5}{z-5},$$ in powers of $(z-1)$ that converges when $z=4$.
The series will converge when $z=4$ is in the region $2<|z-1|<4$. Hence,
begin{align}
f(z)&=-2frac{d}{dz}left(frac{1}{z+1}right)+5left(frac{1}{5-z}right) \
&=-2frac{d}{dz}left(frac{1}{(z-1)}frac{1}{1+frac{2}{(z-1)}}right)+frac{5}{4}left(frac{1}{1-frac{(z-1)}{4}}right) \
&=-2frac{d}{dz}left(frac{1}{(z-1)}sum_{n=0}^{infty}frac{(-1)^n2^n}{(z-1)^n}right)+frac{5}{4}sum_{n=0}^{infty}frac{(z-1)^n}{4^n} \
&=sum_{n=0}^{infty}frac{(-1)^n2^{n+1}(n+1)(z-1)^n}{(z-1)^{2(n+1)}}+5sum_{n=0}^{infty}frac{(z-1)^n}{4^{n+1}}.
end{align}
Is this solution correct?
complex-analysis proof-verification laurent-series
complex-analysis proof-verification laurent-series
asked Oct 29 at 22:28
Bell
167316
asked Oct 29 at 22:28
Bell
167316
edited Nov 18 at 20:21
asked Oct 29 at 22:28
Bell
167316
asked Oct 29 at 22:28
Bell
167316
asked Oct 29 at 22:28
Bell
167316
167316
add a comment |
add a comment |
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2976785%2flaurent-series-for-fz-frac2z12-frac5z-5%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
