Laurent Series in powers of $z$ and $frac{1}{z}$
I'm working on a few problems from my textbook and have a bit of trouble figuring out a few things.
In a particular case, suppose $R$ is a rational function all of whose poles in the plane have order one and which has no pole at the origin.
After long division:
$R(z) = S(z) + frac{P(z)}{Q(z)}$
Where $S$, $P$ and $Q$ are polynomials, and the degree of $P$ is strictly less than the degree of $Q$, concentrating on expanding $f = frac{P}{Q}$
Then we can write:
$f(z) = sum_{-infty}^{infty} a_k z^k$
for $r <|z| < R$
Where
$a_k = sum_{|z_j<r|} z_j^{k-1} Res(f; z_j)$ for $k leq -1$
and $= -sum_{|z_j|>R} z_j^{-k-1} Res(f; z_j)$ for $k geq 0$
Now for the question:
Suppose we want to find the Laurent Series expansion for $R(z)$ in the region $|z|<1$
Let $R(z) = frac{z^3 - 3z^2 + 3} {(z-1)(z-3)}$
After long division we get:
$R(z) = (z + 1) + frac{z}{(z-1)(z-3)}$
In this case $r = 0$, $R = 1$, so the only terms that are used are from the negative part of $a_k$ so:
$frac {z} {(z-1)(z-3)} = sum_{k=0}^{infty} a_k z^k$, where $a_k = - [-frac{1}{2} + frac{3}{2} frac{1}{3^{k+1}}]$ for $k geq 0$
I'm confused on how the terms for $a_k$ were found. I know the definition was given, but that's what I'm having trouble with. First off, what does $z_j$ represent? How do we find the Residue before finding the Laurent Series?
complex-analysis laurent-series
asked Dec 10 '18 at 0:46
user2965071
1016
add a comment |
I'm working on a few problems from my textbook and have a bit of trouble figuring out a few things.
In a particular case, suppose $R$ is a rational function all of whose poles in the plane have order one and which has no pole at the origin.
After long division:
$R(z) = S(z) + frac{P(z)}{Q(z)}$
Where $S$, $P$ and $Q$ are polynomials, and the degree of $P$ is strictly less than the degree of $Q$, concentrating on expanding $f = frac{P}{Q}$
Then we can write:
$f(z) = sum_{-infty}^{infty} a_k z^k$
for $r <|z| < R$
Where
$a_k = sum_{|z_j<r|} z_j^{k-1} Res(f; z_j)$ for $k leq -1$
and $= -sum_{|z_j|>R} z_j^{-k-1} Res(f; z_j)$ for $k geq 0$
Now for the question:
Suppose we want to find the Laurent Series expansion for $R(z)$ in the region $|z|<1$
Let $R(z) = frac{z^3 - 3z^2 + 3} {(z-1)(z-3)}$
After long division we get:
$R(z) = (z + 1) + frac{z}{(z-1)(z-3)}$
In this case $r = 0$, $R = 1$, so the only terms that are used are from the negative part of $a_k$ so:
$frac {z} {(z-1)(z-3)} = sum_{k=0}^{infty} a_k z^k$, where $a_k = - [-frac{1}{2} + frac{3}{2} frac{1}{3^{k+1}}]$ for $k geq 0$
I'm confused on how the terms for $a_k$ were found. I know the definition was given, but that's what I'm having trouble with. First off, what does $z_j$ represent? How do we find the Residue before finding the Laurent Series?
complex-analysis laurent-series
asked Dec 10 '18 at 0:46
user2965071
1016
You are presumably looking for a series in powers of $z$. Since your function has no pole at $z=0$, there is no residue, your series is not merely a Laurent series but a regular power series. The specific numbers come from the Partial Fraction expansion of $z/((z-1)(z-3))$ as $-1/(2(z-1))+3/(2(z-3))$.
– Lubin
Dec 10 '18 at 1:57
After doing the partial fraction decomposition, what would be the next step? I'm confused as I don't fully understand how each of the coefficients are derived.
– user2965071
Dec 10 '18 at 3:25
add a comment |
I'm working on a few problems from my textbook and have a bit of trouble figuring out a few things.
In a particular case, suppose $R$ is a rational function all of whose poles in the plane have order one and which has no pole at the origin.
After long division:
$R(z) = S(z) + frac{P(z)}{Q(z)}$
Where $S$, $P$ and $Q$ are polynomials, and the degree of $P$ is strictly less than the degree of $Q$, concentrating on expanding $f = frac{P}{Q}$
Then we can write:
$f(z) = sum_{-infty}^{infty} a_k z^k$
for $r <|z| < R$
Where
$a_k = sum_{|z_j<r|} z_j^{k-1} Res(f; z_j)$ for $k leq -1$
and $= -sum_{|z_j|>R} z_j^{-k-1} Res(f; z_j)$ for $k geq 0$
Now for the question:
Suppose we want to find the Laurent Series expansion for $R(z)$ in the region $|z|<1$
Let $R(z) = frac{z^3 - 3z^2 + 3} {(z-1)(z-3)}$
After long division we get:
$R(z) = (z + 1) + frac{z}{(z-1)(z-3)}$
In this case $r = 0$, $R = 1$, so the only terms that are used are from the negative part of $a_k$ so:
$frac {z} {(z-1)(z-3)} = sum_{k=0}^{infty} a_k z^k$, where $a_k = - [-frac{1}{2} + frac{3}{2} frac{1}{3^{k+1}}]$ for $k geq 0$
I'm confused on how the terms for $a_k$ were found. I know the definition was given, but that's what I'm having trouble with. First off, what does $z_j$ represent? How do we find the Residue before finding the Laurent Series?
complex-analysis laurent-series
asked Dec 10 '18 at 0:46
user2965071
1016
I'm working on a few problems from my textbook and have a bit of trouble figuring out a few things.
In a particular case, suppose $R$ is a rational function all of whose poles in the plane have order one and which has no pole at the origin.
After long division:
$R(z) = S(z) + frac{P(z)}{Q(z)}$
Where $S$, $P$ and $Q$ are polynomials, and the degree of $P$ is strictly less than the degree of $Q$, concentrating on expanding $f = frac{P}{Q}$
Then we can write:
$f(z) = sum_{-infty}^{infty} a_k z^k$
for $r <|z| < R$
Where
$a_k = sum_{|z_j<r|} z_j^{k-1} Res(f; z_j)$ for $k leq -1$
and $= -sum_{|z_j|>R} z_j^{-k-1} Res(f; z_j)$ for $k geq 0$
Now for the question:
Suppose we want to find the Laurent Series expansion for $R(z)$ in the region $|z|<1$
Let $R(z) = frac{z^3 - 3z^2 + 3} {(z-1)(z-3)}$
After long division we get:
$R(z) = (z + 1) + frac{z}{(z-1)(z-3)}$
In this case $r = 0$, $R = 1$, so the only terms that are used are from the negative part of $a_k$ so:
$frac {z} {(z-1)(z-3)} = sum_{k=0}^{infty} a_k z^k$, where $a_k = - [-frac{1}{2} + frac{3}{2} frac{1}{3^{k+1}}]$ for $k geq 0$
I'm confused on how the terms for $a_k$ were found. I know the definition was given, but that's what I'm having trouble with. First off, what does $z_j$ represent? How do we find the Residue before finding the Laurent Series?
complex-analysis laurent-series
complex-analysis laurent-series
asked Dec 10 '18 at 0:46
user2965071
1016
asked Dec 10 '18 at 0:46
user2965071
1016
asked Dec 10 '18 at 0:46
user2965071
1016
asked Dec 10 '18 at 0:46
user2965071
1016
asked Dec 10 '18 at 0:46
user2965071
1016
1016
You are presumably looking for a series in powers of $z$. Since your function has no pole at $z=0$, there is no residue, your series is not merely a Laurent series but a regular power series. The specific numbers come from the Partial Fraction expansion of $z/((z-1)(z-3))$ as $-1/(2(z-1))+3/(2(z-3))$.
– Lubin
Dec 10 '18 at 1:57
After doing the partial fraction decomposition, what would be the next step? I'm confused as I don't fully understand how each of the coefficients are derived.
– user2965071
Dec 10 '18 at 3:25
add a comment |
You are presumably looking for a series in powers of $z$. Since your function has no pole at $z=0$, there is no residue, your series is not merely a Laurent series but a regular power series. The specific numbers come from the Partial Fraction expansion of $z/((z-1)(z-3))$ as $-1/(2(z-1))+3/(2(z-3))$.
– Lubin
Dec 10 '18 at 1:57
After doing the partial fraction decomposition, what would be the next step? I'm confused as I don't fully understand how each of the coefficients are derived.
– user2965071
Dec 10 '18 at 3:25
You are presumably looking for a series in powers of $z$. Since your function has no pole at $z=0$, there is no residue, your series is not merely a Laurent series but a regular power series. The specific numbers come from the Partial Fraction expansion of $z/((z-1)(z-3))$ as $-1/(2(z-1))+3/(2(z-3))$.
– Lubin
Dec 10 '18 at 1:57
You are presumably looking for a series in powers of $z$. Since your function has no pole at $z=0$, there is no residue, your series is not merely a Laurent series but a regular power series. The specific numbers come from the Partial Fraction expansion of $z/((z-1)(z-3))$ as $-1/(2(z-1))+3/(2(z-3))$.
– Lubin
Dec 10 '18 at 1:57
After doing the partial fraction decomposition, what would be the next step? I'm confused as I don't fully understand how each of the coefficients are derived.
– user2965071
Dec 10 '18 at 3:25
After doing the partial fraction decomposition, what would be the next step? I'm confused as I don't fully understand how each of the coefficients are derived.
– user2965071
Dec 10 '18 at 3:25
add a comment |
1 Answer
1
active
oldest
votes
You asked for a fuller explanation than I gave in my comment. We get the following:
begin{align}
frac z{(z-1)(z-3)}&=frac z{(1-z)(3-z)}\
&=frac{1/2}{1-z}-frac{3/2}{3-z}=frac{1/2}{1-z} - frac{1/2}{1-frac z3}\
&=frac12Bigl(1+z+z^2+z^3+cdotsBigr)\
&qquad-frac12Bigl(1+frac z3+frac{z^2}9+frac{z^3}{27}+cdotsBigr),,
end{align}
which should agree with your numbers.
answered Dec 10 '18 at 5:52
Lubin
43.7k44585
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
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%2f3033265%2flaurent-series-in-powers-of-z-and-frac1z%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
You asked for a fuller explanation than I gave in my comment. We get the following:
begin{align}
frac z{(z-1)(z-3)}&=frac z{(1-z)(3-z)}\
&=frac{1/2}{1-z}-frac{3/2}{3-z}=frac{1/2}{1-z} - frac{1/2}{1-frac z3}\
&=frac12Bigl(1+z+z^2+z^3+cdotsBigr)\
&qquad-frac12Bigl(1+frac z3+frac{z^2}9+frac{z^3}{27}+cdotsBigr),,
end{align}
which should agree with your numbers.
answered Dec 10 '18 at 5:52
Lubin
43.7k44585
add a comment |
You asked for a fuller explanation than I gave in my comment. We get the following:
begin{align}
frac z{(z-1)(z-3)}&=frac z{(1-z)(3-z)}\
&=frac{1/2}{1-z}-frac{3/2}{3-z}=frac{1/2}{1-z} - frac{1/2}{1-frac z3}\
&=frac12Bigl(1+z+z^2+z^3+cdotsBigr)\
&qquad-frac12Bigl(1+frac z3+frac{z^2}9+frac{z^3}{27}+cdotsBigr),,
end{align}
which should agree with your numbers.
answered Dec 10 '18 at 5:52
Lubin
43.7k44585
add a comment |
You asked for a fuller explanation than I gave in my comment. We get the following:
begin{align}
frac z{(z-1)(z-3)}&=frac z{(1-z)(3-z)}\
&=frac{1/2}{1-z}-frac{3/2}{3-z}=frac{1/2}{1-z} - frac{1/2}{1-frac z3}\
&=frac12Bigl(1+z+z^2+z^3+cdotsBigr)\
&qquad-frac12Bigl(1+frac z3+frac{z^2}9+frac{z^3}{27}+cdotsBigr),,
end{align}
which should agree with your numbers.
answered Dec 10 '18 at 5:52
Lubin
43.7k44585
You asked for a fuller explanation than I gave in my comment. We get the following:
begin{align}
frac z{(z-1)(z-3)}&=frac z{(1-z)(3-z)}\
&=frac{1/2}{1-z}-frac{3/2}{3-z}=frac{1/2}{1-z} - frac{1/2}{1-frac z3}\
&=frac12Bigl(1+z+z^2+z^3+cdotsBigr)\
&qquad-frac12Bigl(1+frac z3+frac{z^2}9+frac{z^3}{27}+cdotsBigr),,
end{align}
which should agree with your numbers.
answered Dec 10 '18 at 5:52
Lubin
43.7k44585
answered Dec 10 '18 at 5:52
Lubin
43.7k44585
answered Dec 10 '18 at 5:52
Lubin
43.7k44585
answered Dec 10 '18 at 5:52
Lubin
43.7k44585
43.7k44585
add a comment |
add a comment |
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%2f3033265%2flaurent-series-in-powers-of-z-and-frac1z%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
You are presumably looking for a series in powers of $z$. Since your function has no pole at $z=0$, there is no residue, your series is not merely a Laurent series but a regular power series. The specific numbers come from the Partial Fraction expansion of $z/((z-1)(z-3))$ as $-1/(2(z-1))+3/(2(z-3))$.
– Lubin
Dec 10 '18 at 1:57
After doing the partial fraction decomposition, what would be the next step? I'm confused as I don't fully understand how each of the coefficients are derived.
– user2965071
Dec 10 '18 at 3:25