Services & Resources / Wolfram Forums / MathGroup Archive
-----

MathGroup Archive 2008

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

Search the Archive

Coefficient[ ] quirks

  • To: mathgroup at smc.vnet.net
  • Subject: [mg93310] Coefficient[ ] quirks
  • From: AES <siegman at stanford.edu>
  • Date: Mon, 3 Nov 2008 05:28:38 -0500 (EST)
  • Organization: Stanford University

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 . . .


  • Prev by Date: Re: Re: ODE solution problem
  • Next by Date: Re: Eliminating common factors?
  • Previous by thread: Re: Re: Re: ODE solution problem
  • Next by thread: Re: Coefficient[ ] quirks