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Re: Coefficient[ ] quirks
*To*: mathgroup at smc.vnet.net
*Subject*: [mg93325] Re: [mg93310] Coefficient[ ] quirks
*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>
*Date*: Tue, 4 Nov 2008 06:15:54 -0500 (EST)
*References*: <200811031028.FAA05229@smc.vnet.net> <6BD8F5AE-B50C-4C87-B939-35B7C2163844@mimuw.edu.pl>
On 3 Nov 2008, at 20:34, Andrzej Kozlowski wrote:
>
> 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
>
Actually, there is a slightly better way, based on the same idea.
Above I expanded f as a Taylor series about c wit remainder of order
10, which I chose essentially arbitrarily , making sure that it is
large enough. But this is not needed, Mathematica is cleverer than
that. All you need is
SeriesCoefficient[f, {z, c, 2}]
b
In other words, one does not need to expand f to some fixed order, it
is enough to tell Mathematica that we want the Coefficient of the term
of degree 2 and Mathematica will do the expanding itself. This is much
more convenient and, as far as I know, the best way to deal with this
problem.
Andrzej Kozlowski
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