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## Partial Fraction Expansion

• NerdLifter

Posts: 1510
UPDATE 9/13/2011: I found the solution; I should have it posted as a thread response below shortly. It is the correct answer and was confirmed via Matlab.

Original Message:
Need to take the partial fraction expansion of the following:

We have to do it by hand, step by step. But I'm kinda stuck since the denominator is a nasty third order polynomial with only complex roots.

Any help would be much appreciated.

You can copy and past the following Matlab code I made which illustrates my problem:

%%
%Problem 4
clear all
nomy = [1 3];
denomy = [1 11 12 18];
[residues, poles, DirectTerms] = residue(nomy, denomy)
%Note: We get no direct terms
%The output we recieved is:
%-0.0765/(x+9.9782)+(0.0383 - 0.1109i)/(x-(-0.5109 + 1.2421i))+ (0.0383 +
%0.1109i)/(x-(-0.5109 - 1.2421i))

Output:

residues =

-0.0765
0.0383 - 0.1109i
0.0383 + 0.1109i

poles =

-9.9782
-0.5109 + 1.2421i
-0.5109 - 1.2421i

DirectTerms =

[]

• Posted by a hidden member.
• Posted by a hidden member.
awww this makes me nostalgic for the days when I knew how to do this and more importantly I understood the ins and out of calculus. I've forgotten how to start now.
• NerdLifter

Posts: 1510
SkinnyBitch saidawww this makes me nostalgic for the days when I knew how to do this and more importantly I understood the ins and out of calculus. I've forgotten how to start now.

Aww, if it makes you feel any better, this isn't calculus. But I still send you hugs.
• Posted by a hidden member.
but we did this shit in Calculus. What is it then?

I can't even remember how to factor a quadratic equation anymore.

Thanks for the hugs
• Posted by a hidden member.
I like pie.
• NerdLifter

Posts: 1510
SkinnyBitch saidbut we did this shit in Calculus. What is it then?

I can't even remember how to factor a quadratic equation anymore.

Thanks for the hugs

It's true that they teach Partial Fraction Expansion in most calculus classes, because you need it in order to break up integrands and for Laplace transforms (and other really nasty things).

But partial fraction expansion in itself is just an algebraic technique, nothing more. You were probably taught it in high school algebra and/or pre-Calculus, then revisited it in Calculus. I was exposed to it originally in 8th grade but then forgot about it until I went to college.
• NerdLifter

Posts: 1510
Scruffypup saidI like pie.

Me too!
• Posted by a hidden member.
Studinprogress said
SkinnyBitch saidbut we did this shit in Calculus. What is it then?

I can't even remember how to factor a quadratic equation anymore.

Thanks for the hugs

It's true that they teach Partial Fraction Expansion in most calculus classes, because you need it in order to break up integrands and for Laplace transforms (and other really nasty things).

But partial fraction expansion in itself is just an algebraic technique, nothing more. You were probably taught it in high school algebra and/or pre-Calculus, then revisited it in Calculus. I was exposed to it originally in 8th grade but then forgot about it until I went to college.

Makes sense, all of my memory of math came from high school classes. University was like a quick and passionless fuck where high school was an in-depth, meaningful lovemaking to the numbers and xs and weird symbols. The romance was gone, everyone was told to memorize formulas for the exam, then forget about them. Understanding was a luxury we weren't afforded.
• NerdLifter

Posts: 1510
SkinnyBitch said
Studinprogress said
SkinnyBitch saidbut we did this shit in Calculus. What is it then?

I can't even remember how to factor a quadratic equation anymore.

Thanks for the hugs

It's true that they teach Partial Fraction Expansion in most calculus classes, because you need it in order to break up integrands and for Laplace transforms (and other really nasty things).

But partial fraction expansion in itself is just an algebraic technique, nothing more. You were probably taught it in high school algebra and/or pre-Calculus, then revisited it in Calculus. I was exposed to it originally in 8th grade but then forgot about it until I went to college.

Makes sense, all of my memory of math came from high school classes. University was like a quick and passionless fuck where high school was an in-depth, meaningful lovemaking to the numbers and xs and weird symbols. The romance was gone, everyone was told to memorize formulas for the exam, then forget about them. Understanding was a luxury we weren't afforded.

The University system right now is really f*d up, and has been getting worse with each passing year. We throw money at them to get a piece of paper that 'potentially' helps us in the job market. Understanding for academic's sake is completely gone from most Universities.
• Posted by a hidden member.
Wait, are you supposed to factor the denominator by hand as well, or is it acceptable to use the output from Matlab that you've posted? If it's OK for Matlab to factor the cubic polynomial, then you've got three distinct roots and the partial fraction expansion has the simple form:

T(s) = A/(s-a) + B/(s-b) + C/(s-c)

Once you've gotten to this point, it's a matter of solving for A,B,C. You can do this by multiplying out the RHS to obtain the form Q(s)/R(s), where Q will be a quadratic in s and R will be the original (cubic) denominator, and then equating powers of s, yielding three linear equations in the three unknowns A, B, C (this is the familiar method taught in first year calculus). Much faster is the residue method. There's also the cover-up method to obtaining the coefficients.

This is a complex analysis course or a control theory course, right? If so, I doubt you're being asked to derive the roots of the cubic equation by hand. In the unfortunate event that you do need to solve the cubic by hand (sucks to be you!) you can use Cardano's method; the results are not pretty. But seriously, the solution of cubic equations by radicals is a curiosity that's rarely taught anymore.
• Posted by a hidden member.
Studinprogress, just write exactly what yourname2000 posted and you should get an A!
• Posted by a hidden member.
zotamorf62 said In the unfortunate event that you do need to solve the cubic by hand (sucks to be you!) you can use Cardano's method; the results are not pretty. But seriously, the solution of cubic equations by radicals is a curiosity that's rarely taught anymore.

http://en.wikipedia.org/wiki/Quartic_functionLodovico Ferrari is attributed with the discovery of the solution to the quartic in 1540, but since this solution, like all algebraic solutions of the quartic, requires the solution of a cubic to be found, it couldn't be published immediately.[1] The solution of the quartic was published together with that of the cubic by Ferrari's mentor Gerolamo Cardano in the book Ars Magna (1545).

It is reported that even earlier, in 1486, Spanish mathematician Paolo Valmes was burned at the stake for claiming to have solved the quartic equation. Inquisitor General Tomás de Torquemada allegedly told him that it was the will of God that such a solution be inaccessible to human understanding.[2] However, attempts to find corroborating evidence for this story, or for the existence of Paolo Valmes, have not succeeded.[3]

Thank god for Galois theory (it made so much sense once...).
• Posted by a hidden member.
LA LA LA LA LA
• Posted by a hidden member.
Studinprogress saidNeed to take the partial fraction expansion of the following:

We have to do it by hand, step by step. But I'm kinda stuck since the denominator is a nasty third order polynomial with only complex roots.

Any help would be much appreciated.

You can copy and past the Matlab code I made which illustrates my problem:

%%
%Problem 4
clear all
nomy = [1 3];
denomy = [1 11 12 18];
[residues, poles, DirectTerms] = residue(nomy, denomy)
%Note: We get no direct terms
%The output we recieved is:
%-0.0765/(x+9.9782)+(0.0383 - 0.1109i)/(x-(-0.5109 + 1.2421i))+ (0.0383 +
%0.1109i)/(x-(-0.5109 - 1.2421i))

Output:

residues =

-0.0765
0.0383 - 0.1109i
0.0383 + 0.1109i

poles =

-9.9782
-0.5109 + 1.2421i
-0.5109 - 1.2421i

DirectTerms =

[]

does not compute
• NerdLifter

Posts: 1510
zotamorf62 saidWait, are you supposed to factor the denominator by hand as well, or is it acceptable to use the output from Matlab that you've posted?

Yes, we must factor the denominator by hand, and then confirm that answer with Matlab.
• Posted by a hidden member.
i need a drink now
• NerdLifter

Posts: 1510
No one?
• Posted by a hidden member.
2!
• UVaRob9

Posts: 282
Ugh, that's messy. You've got one real irrational root and (naturally) two complex conjugate roots. Maple gives me an output, but I have no idea how you'd do the work for this. The other possibility wouldbbe to factor an s+3 out of the polynomial and then cancel the one in the numerator. You end up with something that's unpleasant, but those coefficients sucked anyway.

It does beg the question: why does you professor hate you so much? These coefficients are not at all conducive to showing your work. This is what computer algebra systems are for.

• Posted by a hidden member.
42
• Posted by a hidden member.
Studinprogress said
zotamorf62 saidWait, are you supposed to factor the denominator by hand as well, or is it acceptable to use the output from Matlab that you've posted?

Yes, we must factor the denominator by hand, and then confirm that answer with Matlab.

Matlab? A mere abacus. Mention it not.
• Posted by a hidden member.