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Re: Coercion into series
*To*: mathgroup at smc.vnet.net
*Subject*: [mg71458] Re: [mg71449] Coercion into series
*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>
*Date*: Mon, 20 Nov 2006 02:43:35 -0500 (EST)
*References*: <200611190610.BAA23542@smc.vnet.net>
On 19 Nov 2006, at 15:10, dimitris wrote:
> Any time that an object like O[x] appears in a sum of terms,
> Mathematica will convert the whole sum into a power series.
>
> E.g.
>
> a x + Exp[x] + O[x]^3
> SeriesData[x, 0, {1, 1 + a, 1/2}, 0, 3, 1]
>
> Cos[z] + O[z, Pi/2]^4
> SeriesData[z, Pi/2, {-1, 0, 1/6}, 1, 4, 1]
>
> and so on...
>
> Consider the following equation
>
> eq=(y + b)*(y + c) == a^4 + y^2 + 3;
>
> Obviously
>
> Solve[eq, {b, c}]
> Solve::svars: Equations may not give solutions for all "solve"
> variables.
> {{b -> (3 + a^4 - c*y)/(c + y)}}
>
> However how the appearance of O[y]^3 below "makes" Solve give solution
> for {b,c}
>
> Solve[(y + b)*(y + c) == (a^4 + y^2 + 3) + O[y]^3, {b, c}]
> {{b -> -Sqrt[-3 - a^4], c -> Sqrt[-3 - a^4]}, {b -> Sqrt[-3 - a^4], c
> -> -Sqrt[-3 - a^4]}}
>
> Thanks a lot.
>
> Dimitris
>
This is really just the same as
LogicalExpand[(y + b)*(y + c) == (a^4 + y^2 + 3) + O[y]^3]
-b - c == 0 && a^4 - b*c + 3 == 0
or
SolveAlways[(y + b)*(y + c) == a^4 + y^2 + 3, y]
{{b -> -Sqrt[-a^4 - 3], c -> Sqrt[-a^4 - 3]}, {b -> Sqrt[-a^4 - 3], c
-> -Sqrt[-a^4 - 3]}}
and simply amounts to equating the coefficients of two power series
up to a specified order (or of polynomials in the second example).
Andrzej Kozlowski
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