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

**References**:**Coercion into series***From:*"dimitris" <dimmechan@yahoo.com>