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MathGroup Archive 2001

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Re: Partial fraction expansion

  • To: mathgroup at smc.vnet.net
  • Subject: [mg30288] Re: [mg30271] Partial fraction expansion
  • From: Andrzej Kozlowski <andrzej at tuins.ac.jp>
  • Date: Sun, 5 Aug 2001 16:18:33 -0400 (EDT)
  • Sender: owner-wri-mathgroup at wolfram.com

I assume that your question concerns Mathematica, doesn' t it?
Actually, this issue of "partial fractions" is a bit confusing because 
different people mean different things by "partial fractions".  The 
ambiguity concerns the coefficient field. Basically, Mathematica's Apart 
function will expand rational functions into partial fractions using 
only rational coefficients. So in the case f= 1/(1+x^4) you can get the 
following three results:

In[1]:=
Apart[1/(1 + x^4)]

Out[1]=
1/(1 + x^4)

In[2]:=
Apart[1/Factor[1 + x^4, GaussianIntegers -> True]]

Out[2]=
-(I/(2*(-I + x^2))) + I/(2*(I + x^2))

In[3]:=
Apart[1/Factor[1 + x^4, Extension -> Sqrt[2]]]

Out[3]=
-(1/(2*(-1 + Sqrt[2]*x - x^2)*(1 + x^2))) +   1/(2*(1 + x^2)*(1 + 
Sqrt[2]*x + x^2))

The first is the "partial fractions" expansion over the rationals, the 
second over the complex numbers and the third over the reals. The 
following function will perform partial fraction expansions over the 
complex numbers, which is what I assume you wanted:

In[4]:=
partialFractionsC[f_, x_] := Module[{h = Simplify[f], n, d}, n = 
Numerator[h]; d = Denominator[h];
     Apart[n/Factor[d, Extension -> (x /. Solve[d == 0, x])]]]

The easiest way to write a function which will expand rational 
expressions over the reals is to make use of the built in ability to do 
this in Integrate:

partialFractionsR[f_, x_] := D[Integrate[Simplify[f], x], x]

Now let's look at a couple of cases:

In[6]:=
f = (1 - x)/(1 + x^3);

In[7]:=
Apart[f]

Out[7]=
2/(3*(1 + x)) + (1 - 2*x)/(3*(1 - x + x^2))

In[8]:=
partialFractionsC[f, x]

Out[8]=
-(2/((-2 + (-1)^(1/3))*(1 + (-1)^(1/3))*(1 + x))) +
   (1 - (-1)^(1/3))/((1 + (-1)^(1/3))*(-1 + 2*(-1)^(1/3))*(-(-1)^(1/3) + 
x)) +
   (-1)^(1/3)/((-2 + (-1)^(1/3))*(-1 + 2*(-1)^(1/3))*(-1 + (-1)^(1/3) + 
x))

In[9]:=
partialFractionsR[f, x]

Out[9]=
2/(3*(1 + x)) - (-1 + 2*x)/(3*(1 - x + x^2))

In this case there is no difference between Apart and partialFractionsR 
but in the next one there is:

In[10]:=
g = (1 - x)/(1 + x^4);

In[11]:=
Apart[g, x]

Out[11]=
(1 - x)/(1 + x^4)

In[12]:=
partialFractionsC[g, x]

Out[12]=
-(I/(2*(1 + (-1)^(1/4))*((-1)^(1/4) + (-1)^(3/4))*(-(-1)^(1/4) + x))) -
   ((1/4 - I/4)*(-1)^(1/4))/((-1 + (-1)^(1/4))*((-1)^(1/4) + x)) -
   ((1/4 - I/4)*(-1)^(3/4)*(1 + I + (-1)^(1/4)))/((1 + (-1)^(1/4))*(-
(-1)^(3/4) + x)) -
   ((1/4 - I/4)*(-1)^(3/4)*(1 + I - (-1)^(1/4)))/((-1 + (-1)^(1/4))*
((-1)^(3/4) + x))

In[13]:=
partialFractionsR[g, x]

Out[13]=
-((Sqrt[2] - 2*x)/(4*Sqrt[2]*(-1 + Sqrt[2]*x - x^2))) +
   (Sqrt[2] + 2*x)/(4*Sqrt[2]*(1 + Sqrt[2]*x + x^2)) +
   (-2 + Sqrt[2])/(2*Sqrt[2]*(1 + (1/2)*(-Sqrt[2] + 2*x)^2)) +
   (2 + Sqrt[2])/(2*Sqrt[2]*(1 + (1/2)*(Sqrt[2] + 2*x)^2))


Andrzej Kozlowski
Toyama International University
JAPAN
http://platon.c.u-tokyo.ac.jp/andrzej/

On Sunday, August 5, 2001, at 09:02  AM, Lee Lin Yap wrote:

> Hi,
>
> I'm currently doing a software programming for my final year project. I 
> need
> to solve a partial fraction expansion in my program. I have a numerator 
> and
> I got the roots for it. But I can't solve the problem to get the answer 
> that
> I need.
>
> _________________________________________________________________
> Get your FREE download of MSN Explorer at 
> http://explorer.msn.com/intl.asp
>
>
>


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