Finding simple fraction decomposition with help of Taylor's Theorem and Residues theorem

Clash Royale CLAN TAG#URR8PPP

up vote
0
down vote

favorite

1

I have this problem:
$F(s)=fraca_1s+a_2s^2(s^2+12s+37)$

I thought in Taylor's expansion of a function f(s) in s=0:

$f(s)= f(0)+f'(0)cdot s+f''(0)cdot s^2+cdots$
and then I defined:

$f(s)=fraca_1s+a_2(s^2+12s+37)$

Then:
$F(s)=fracf(s)s^2= fracf(0)s^2+fracf'(0)s+f''(0)cdot s^2+cdots$

But by simple fractions the function F(s) should look like this:

$F(s)=fracAs^2+fracBs+fracCs+D(s^2+12s+37)$

So $A=f(0)$ and $B=f'(0)$

So, for example with $a_1=25$ and $a_2=37$, by the "Taylor Method" I get $A= 1$ and $B=13/37$.

Is there a way using the Residues theorem or the Taylor expantion or another trick to obtain C and D getting rid of making distribution of the factors? With the distribution I got $C=frac-1337$ and $D=frac-1437$.

If this hint of the exercise helps: I'm trying to write that in this way to calculate de inverse of the Laplace transform.
Thanks.

taylor-expansion laplace-transform residue-calculus

share|cite|improve this question

edited Jul 31 at 22:28

Bernard

110k635102

asked Jul 31 at 21:59

Verónica

235

add a comment | 

up vote
0
down vote

favorite

1

I have this problem:
$F(s)=fraca_1s+a_2s^2(s^2+12s+37)$

I thought in Taylor's expansion of a function f(s) in s=0:

$f(s)= f(0)+f'(0)cdot s+f''(0)cdot s^2+cdots$
and then I defined:

$f(s)=fraca_1s+a_2(s^2+12s+37)$

Then:
$F(s)=fracf(s)s^2= fracf(0)s^2+fracf'(0)s+f''(0)cdot s^2+cdots$

But by simple fractions the function F(s) should look like this:

$F(s)=fracAs^2+fracBs+fracCs+D(s^2+12s+37)$

So $A=f(0)$ and $B=f'(0)$

So, for example with $a_1=25$ and $a_2=37$, by the "Taylor Method" I get $A= 1$ and $B=13/37$.

Is there a way using the Residues theorem or the Taylor expantion or another trick to obtain C and D getting rid of making distribution of the factors? With the distribution I got $C=frac-1337$ and $D=frac-1437$.

If this hint of the exercise helps: I'm trying to write that in this way to calculate de inverse of the Laplace transform.
Thanks.

taylor-expansion laplace-transform residue-calculus

share|cite|improve this question

edited Jul 31 at 22:28

Bernard

110k635102

asked Jul 31 at 21:59

Verónica

235

add a comment | 

up vote
0
down vote

favorite

1

up vote
0
down vote

favorite

1

1

I have this problem:
$F(s)=fraca_1s+a_2s^2(s^2+12s+37)$

I thought in Taylor's expansion of a function f(s) in s=0:

$f(s)= f(0)+f'(0)cdot s+f''(0)cdot s^2+cdots$
and then I defined:

$f(s)=fraca_1s+a_2(s^2+12s+37)$

Then:
$F(s)=fracf(s)s^2= fracf(0)s^2+fracf'(0)s+f''(0)cdot s^2+cdots$

But by simple fractions the function F(s) should look like this:

$F(s)=fracAs^2+fracBs+fracCs+D(s^2+12s+37)$

So $A=f(0)$ and $B=f'(0)$

So, for example with $a_1=25$ and $a_2=37$, by the "Taylor Method" I get $A= 1$ and $B=13/37$.

Is there a way using the Residues theorem or the Taylor expantion or another trick to obtain C and D getting rid of making distribution of the factors? With the distribution I got $C=frac-1337$ and $D=frac-1437$.

If this hint of the exercise helps: I'm trying to write that in this way to calculate de inverse of the Laplace transform.
Thanks.

taylor-expansion laplace-transform residue-calculus

share|cite|improve this question

edited Jul 31 at 22:28

Bernard

110k635102

asked Jul 31 at 21:59

Verónica

235

I have this problem:
$F(s)=fraca_1s+a_2s^2(s^2+12s+37)$

I thought in Taylor's expansion of a function f(s) in s=0:

$f(s)= f(0)+f'(0)cdot s+f''(0)cdot s^2+cdots$
and then I defined:

$f(s)=fraca_1s+a_2(s^2+12s+37)$

Then:
$F(s)=fracf(s)s^2= fracf(0)s^2+fracf'(0)s+f''(0)cdot s^2+cdots$

But by simple fractions the function F(s) should look like this:

$F(s)=fracAs^2+fracBs+fracCs+D(s^2+12s+37)$

So $A=f(0)$ and $B=f'(0)$

So, for example with $a_1=25$ and $a_2=37$, by the "Taylor Method" I get $A= 1$ and $B=13/37$.

Is there a way using the Residues theorem or the Taylor expantion or another trick to obtain C and D getting rid of making distribution of the factors? With the distribution I got $C=frac-1337$ and $D=frac-1437$.

If this hint of the exercise helps: I'm trying to write that in this way to calculate de inverse of the Laplace transform.
Thanks.

taylor-expansion laplace-transform residue-calculus

share|cite|improve this question

edited Jul 31 at 22:28

Bernard

110k635102

asked Jul 31 at 21:59

Verónica

235

share|cite|improve this question

share|cite|improve this question

edited Jul 31 at 22:28

Bernard

110k635102

edited Jul 31 at 22:28

Bernard

110k635102

edited Jul 31 at 22:28

Bernard

110k635102

110k635102

asked Jul 31 at 21:59

Verónica

235

asked Jul 31 at 21:59

Verónica

235

asked Jul 31 at 21:59

Verónica

235

235

add a comment | 

add a comment | 

1 Answer
1

active

oldest

votes

up vote
0
down vote

If you want to use residues, you could say $frac Cs + Ds^2 + 12s + 37 = frac Qs+6+i + frac Rs+6-i$

In which case $Q$ is the residue at $-6-i$ and $R$ is the residue at $-6+i$ and $D = i(Q-R)$ and $C = 6(Q+R)$

But ultimately I think it is easier to say

$Bs^3 + Cs^3 = 0\
As^2 + 12Bs^2 + Ds^2 = 0$

And for the inverse Laplace transform, you are going to want to say $s^2 + 12s + 37 = (s+6)^2 + 1$

share|cite|improve this answer

answered Jul 31 at 22:22

Doug M

39k31749

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',
convertImagesToLinks: true,
noModals: false,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

 

draft saved

draft discarded

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

Name Email

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2868539%2ffinding-simple-fraction-decomposition-with-help-of-taylors-theorem-and-residues%23new-answer', 'question_page');

);

Post as a guest

Name Email

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
0
down vote

If you want to use residues, you could say $frac Cs + Ds^2 + 12s + 37 = frac Qs+6+i + frac Rs+6-i$

In which case $Q$ is the residue at $-6-i$ and $R$ is the residue at $-6+i$ and $D = i(Q-R)$ and $C = 6(Q+R)$

But ultimately I think it is easier to say

$Bs^3 + Cs^3 = 0\
As^2 + 12Bs^2 + Ds^2 = 0$

And for the inverse Laplace transform, you are going to want to say $s^2 + 12s + 37 = (s+6)^2 + 1$

share|cite|improve this answer

answered Jul 31 at 22:22

Doug M

39k31749

add a comment | 

up vote
0
down vote

If you want to use residues, you could say $frac Cs + Ds^2 + 12s + 37 = frac Qs+6+i + frac Rs+6-i$

In which case $Q$ is the residue at $-6-i$ and $R$ is the residue at $-6+i$ and $D = i(Q-R)$ and $C = 6(Q+R)$

But ultimately I think it is easier to say

$Bs^3 + Cs^3 = 0\
As^2 + 12Bs^2 + Ds^2 = 0$

And for the inverse Laplace transform, you are going to want to say $s^2 + 12s + 37 = (s+6)^2 + 1$

share|cite|improve this answer

answered Jul 31 at 22:22

Doug M

39k31749

add a comment | 

up vote
0
down vote

up vote
0
down vote

If you want to use residues, you could say $frac Cs + Ds^2 + 12s + 37 = frac Qs+6+i + frac Rs+6-i$

In which case $Q$ is the residue at $-6-i$ and $R$ is the residue at $-6+i$ and $D = i(Q-R)$ and $C = 6(Q+R)$

But ultimately I think it is easier to say

$Bs^3 + Cs^3 = 0\
As^2 + 12Bs^2 + Ds^2 = 0$

And for the inverse Laplace transform, you are going to want to say $s^2 + 12s + 37 = (s+6)^2 + 1$

share|cite|improve this answer

answered Jul 31 at 22:22

Doug M

39k31749

If you want to use residues, you could say $frac Cs + Ds^2 + 12s + 37 = frac Qs+6+i + frac Rs+6-i$

In which case $Q$ is the residue at $-6-i$ and $R$ is the residue at $-6+i$ and $D = i(Q-R)$ and $C = 6(Q+R)$

But ultimately I think it is easier to say

$Bs^3 + Cs^3 = 0\
As^2 + 12Bs^2 + Ds^2 = 0$

And for the inverse Laplace transform, you are going to want to say $s^2 + 12s + 37 = (s+6)^2 + 1$

share|cite|improve this answer

answered Jul 31 at 22:22

Doug M

39k31749

share|cite|improve this answer

share|cite|improve this answer

answered Jul 31 at 22:22

Doug M

39k31749

answered Jul 31 at 22:22

Doug M

39k31749

answered Jul 31 at 22:22

Doug M

39k31749

39k31749

add a comment | 

add a comment | 

 

draft saved

draft discarded

 

draft saved

draft discarded

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

Name Email

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2868539%2ffinding-simple-fraction-decomposition-with-help-of-taylors-theorem-and-residues%23new-answer', 'question_page');

);

Post as a guest

Name Email

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

Name Email

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

Name Email

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

Name Email

Name Email

Name

Email

Name Email

Name

Email