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MathGroup Archive 2006

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Re: Coercion into series

  • To: mathgroup at smc.vnet.net
  • Subject: [mg71469] Re: Coercion into series
  • From: bghiggins at ucdavis.edu
  • Date: Mon, 20 Nov 2006 02:43:50 -0500 (EST)
  • References: <ejosoj$n41$1@smc.vnet.net>

Dimitris,

You can write your equation as

-3 - a^4 + b*c + (b + c)*y=0

Now a series expansion of the above equation about y=0 gives the above
equation  for O[x]^1

Thus it is clear that Mathematica has found the solution for the
special case

b+c=0

Why Solve does this is not clear to me. Though it is consistent as

Solve[b*c + (b + c)*y^4 - (a^4 + 3) == 0 + O[y]^5, {b, c}]

gives the same solution. Whereas,


Solve[b*c + (b + c)*y^4 - (a^4 + 3) == 0 + O[y]^4, {b, c}]

 gives

{{b -> (3 + a^4)/c}}

Note that the series expansion of the solution


Normal[Series[b /. Solve[3 + a^4 + b*c - (b + c)*y^4 ==
      0, b], {y, 0, 5}]]

 is

{-(3/c) - a^4/c + (1 - 3/c^2 - a^4/c^2)*y^4}

which of course agrees with Solve[b*c + (b + c)*y^4 - (a^4 + 3) == 0 +
O[y]^5, {b, c}],  if b+c=0

Cheers,

Brian


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


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