       Re: Coefficient[ ] quirks

• To: mathgroup at smc.vnet.net
• Subject: [mg93326] Re: [mg93310] Coefficient[ ] quirks
• From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
• Date: Tue, 4 Nov 2008 06:16:05 -0500 (EST)
• References: <200811031028.FAA05229@smc.vnet.net>

```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]
b

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