Laurent Series & Partial Fraction Decomposition.

  • Thread starter binbagsss
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  • #1
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Okay so the partial fraction decomposition theorem is that if f(z) is a rational function, f(z)=sum of the principal parts of a laurent expansion of f(z) about each root.

I'm working through an example in my book, I am fine to follow it. (method 1 below)

But instinctively , I would have thought you would have started with a laurent expansion rather than a taylor expansion - i.e- a expansion of z^-1 terms, as we are after the principal part.(method 2 below)

f(z)=[itex]\frac{1}{z^{2}+1}[/itex]=[itex]\frac{1}{z+i}[/itex][itex]\frac{1}{z-i}[/itex]

I am attaining the pricipal part corresponding to the root z=i :

Method 1

[itex]\frac{1}{z+i}[/itex][itex]\frac{1}{z-i}[/itex]=[itex]\frac{1}{z-i}[/itex][itex]\frac{1}{z-i+2i}[/itex]=[itex]\frac{1}{z-i}[/itex][itex]\frac{1}{2i}[/itex][itex]\frac{1}{\frac{z-i}{2i}+1}[/itex]

=[itex]\frac{1}{z-i}[/itex][itex]\frac{1}{2i}[/itex][itex]^{\infty}_{0}[/itex][itex]\sum[/itex]((-1)([itex]\frac{z-i}{2i}[/itex]))[itex]^{n}[/itex] [1]

From which I can observe the principal part is given by n=0: [itex]\frac{1}{2i(z-i)}[/itex]

Method 2

f(z)=[itex]\frac{1}{z-i}[/itex][itex]\frac{1}{z-i+2i}[/itex]=[itex]\frac{1}{z-i^{2}}[/itex][itex]\frac{1}{\frac{2i}{z-i}+1}[/itex]= [itex]\frac{1}{z-i^{2}}[/itex][itex]^{\infty}_{0}[/itex][itex]\sum[/itex]((-1)[itex]\frac{2i}{z-i}[/itex])[itex]^{n}[/itex] [2]

From which I can see that to attain a (z-i)[itex]^{-1}[/itex] we require n=-3, which by equality [2], is not valid as n runs from n=0. (it would also yield the answer to method 1 multiplied by -(2i)[itex]^{-3}[/itex], so they don't agree anyway.

I am not entirely sure, but I think which summation includes the zero n power may be causing some confusion. But I can't see how this would be a strict rule - are my equivalent GP summation expressions [1] and [2] okay?

(I can see that conventionally the ≥ 0 is for positive powers and ≥ 1 for negative powers. But doing this I still get an incorrect answer with method 2)


So I don't understand why you would not attain the same answer expanding in terms of negative z powers.


Many thanks for any assistance, greatly appreciated !
 

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  • #3
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anyone?
 
  • #4
vela
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A function has different Laurent series for different regions of the complex plane. For example, the series
$$\frac{1}{1-z} = 1+z+z^2+\cdots$$ only converges for |z|<1. For points in the plane where ##|z|>1##, you have the series
$$\frac{1}{1-z} = -\frac{1}{z} \times \frac{1}{1-1/z} = -\frac{1}{z} - \frac{1}{z^2} - \frac{1}{z^3} - \cdots.$$ One function, but two Laurent series.

Given a function, you have only one partial fractions expansion, so apparently, only one of its Laurent series is applicable with regard to the theorem. You need to find out which series the theorem is referring to.
 
  • #5
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Thanks. The question is just to find the partial fraction decomposition of 1/(z^2+1). There is no mention of any region.

I have the theorem stated as this:

I am struggling to interpret it, to see whether it is referring to the 1/z expansion or the z expansion.
 

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  • #6
vela
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How is Princ(f(x), bj) defined?
 
  • #7
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How is Princ(f(x), bj) defined?
Ahh okay thanks I think I see: Princ (f(x)) is define as the a_1 coeffient where the series takes the form of [itex]^{∞}_{0}[/itex][itex]\sum[/itex] a[itex]_{n}z^{n}[/itex], so the z^n rather than the (1/z)^n expansion is the Laurent Series required.
 
  • #8
vela
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That's not correct.
 
  • #9
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It's not the z^n expansion rather than the (1/z)^n expansion?

Sorry , my summation should be from -∞ to ∞
 
  • #10
vela
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The principal part consists of all of the terms with a negative exponent. It's not just one coefficient.

The problem here is that you need to find out which Laurent series applies in the definition of Princ(f(x),bj).
 

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