MathGroup Archive 2008

[Date Index] [Thread Index] [Author Index]

Search the Archive

Re: Coefficient[ ] quirks

  • To: mathgroup at
  • Subject: [mg93326] Re: [mg93310] Coefficient[ ] quirks
  • From: Andrzej Kozlowski <akoz at>
  • Date: Tue, 4 Nov 2008 06:16:05 -0500 (EST)
  • References: <>

On 3 Nov 2008, at 19:28, AES wrote:

> If you have a function  f = a + b (z-c)^2 and want to extract out the
> powers of the "form" (z-c), you can use either Coefficient[f, (z-c),  
> n]
> or Coefficient[f, (z-c)^2, n].  The first one gives you {a, 0, b} for
> the first three coefficients, while the second one gives you {a, b,  
> 0},
> both of these of course being entirely correct.
> Suppose however that you add a denominator to the function, e.g.
> f = (a + b (z-c)^2 )/d.  Then the first approach (looking for powers  
> of
> (z-c)) no longer gives you the information you're seeking, but instead
> stuffs the entire f into the first position, i.e. {f, 0, 0} -- which  
> at
> least doesn't throw away any information about f.
> The second approach (looking for powers of (z-c)^2) still works fine
> even with a denominator added, however, and gives you {a/d, b/d, 0}.
> If out of restless curiosity, you then expand out the square of (z-c)
> into either of the forms  f = (a + b (z^2 - 2 c z + c^2) or
> f = (a + b z^2 - 2 b c z + + b c^2), with or without the denominator,
> but still seek powers of (z-c) (which are no longer explicitly  
> present),
> the first approach continues to give you {f, 0, 0}, which it can be
> argued is entirely correct, since there is no longer any explicit  
> power
> of (z-c) in f.
> The second approach, however, now gives you either {a, 0, 0} or {a/ 
> d, 0,
> 0}, which in a certain sense is arguably correct for the first or n=0
> term -- any powers of z or c have apparently been chopped off, even
> though you asked Coefficient to look only for powers of (z-c).
> The only problem is that they seem to have been thrown out totally,
> beyond recovery.  You could write
>   f = Sum[Coefficient[f, (z-c), n] * (z-c)^n, {n,0,Infinity}]
> for f with or without denominator, and never get all your function  
> back.
> It all seems rather quirky . . .

Because it is not the way to do this at all...

There are a number possible approaches but only one that is completely  
reliable (and very standard in this sort of situation, not just in  
Mathematica) is :

f = (a + b (z^2 - 2 c z + c^2))

SeriesCoefficient[f + O[z, c]^10, 2]

This will work no matter how you express your f.

Andrzej Kozlowski

  • Prev by Date: Trouble solving system of first order, two variable PDEs in
  • Next by Date: Re: 3D Graphics in a Palette or Dialog
  • Previous by thread: Coefficient[ ] quirks
  • Next by thread: Re: Coefficient[ ] quirks