Partial Fraction Expansion

  • NerdLifter

    Posts: 1509

    Sep 12, 2011 4:00 AM GMT
    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:

    PFE.png

    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.
    Log in to view his profile

    Sep 12, 2011 4:04 AM GMT
    SIMON-CONFUSED-GIF.gif
  • Posted by a hidden member.
    Log in to view his profile

    Sep 12, 2011 4:05 AM GMT
    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: 1509

    Sep 12, 2011 4:08 AM GMT
    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.
    Log in to view his profile

    Sep 12, 2011 4:15 AM GMT
    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.
    Log in to view his profile

    Sep 12, 2011 4:18 AM GMT
    I like pie.
  • NerdLifter

    Posts: 1509

    Sep 12, 2011 4:21 AM GMT
    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: 1509

    Sep 12, 2011 4:26 AM GMT
    Scruffypup saidI like pie.

    Me too! icon_biggrin.gif
    Pi-Pie.jpg
  • Posted by a hidden member.
    Log in to view his profile

    Sep 12, 2011 4:36 AM GMT
    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: 1509

    Sep 12, 2011 4:43 AM GMT
    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.
    Log in to view his profile

    Sep 12, 2011 7:56 AM GMT
    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.
    Log in to view his profile

    Sep 12, 2011 10:17 AM GMT
    Studinprogress, just write exactly what yourname2000 posted and you should get an A!
  • Posted by a hidden member.
    Log in to view his profile

    Sep 12, 2011 11:52 AM GMT
    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]


    icon_eek.gif

    Thank god for Galois theory (it made so much sense once...).
  • Posted by a hidden member.
    Log in to view his profile

    Sep 12, 2011 11:54 AM GMT
    LA LA LA LA LA
  • Posted by a hidden member.
    Log in to view his profile

    Sep 12, 2011 12:00 PM GMT
    Studinprogress saidNeed to take the partial fraction expansion of the following:

    PFE.png

    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: 1509

    Sep 12, 2011 4:41 PM GMT
    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.
  • smudgedude

    Posts: 260

    Sep 12, 2011 4:57 PM GMT
    i need a drink now
  • NerdLifter

    Posts: 1509

    Sep 13, 2011 2:46 AM GMT
    No one? icon_neutral.gif
  • Posted by a hidden member.
    Log in to view his profile

    Sep 13, 2011 2:51 AM GMT
    2! icon_lol.gif
  • UVaRob9

    Posts: 282

    Sep 13, 2011 4:13 AM GMT
    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.
    Log in to view his profile

    Sep 13, 2011 4:16 AM GMT
    42
  • Posted by a hidden member.
    Log in to view his profile

    Sep 13, 2011 4:19 AM GMT
    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.
    Log in to view his profile

    Sep 13, 2011 6:42 PM GMT
    OK, try this: Every cubic with real coefficients has at least one real root (plot a graph to see that yours has exactly one), so for sure there exist real coefficients a, b, c such that

    s^3 + 11s^2 + 12s + 18 = (s + a)(s^2 + bs + c).

    Expand the RHS, match powers of s, and obtain a system of three nonlinear equations in three unknowns a, b, c. There's NO F*CKING WAY that you won't be allowed to use Matlab or Wolfram Alpha to solve this system numerically. Solving yields one real root, s = -a. The other two roots will be complex conjugates of each other and can be derived from b and c via the trusty quadratic formula. Now you can proceed to partial fractions, expressing the decomp with either three linear denominators and complex coefficients, or with real coefficients on one linear denominator and one quadratic denominator.
  • Posted by a hidden member.
    Log in to view his profile

    Sep 13, 2011 6:46 PM GMT
    I clicked on this thread thinking I would have a clue...
    icon_eek.gif LOL no.
  • Posted by a hidden member.
    Log in to view his profile

    Sep 13, 2011 6:55 PM GMT