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