Re: problem with Expand

*To*: mathgroup at smc.vnet.net*Subject*: [mg26259] Re: problem with Expand*From*: Jens-Peer Kuska <kuska at informatik.uni-leipzig.de>*Date*: Sun, 10 Dec 2000 00:19:34 -0500 (EST)*Organization*: Universitaet Leipzig*References*: <90kqsc$r3g@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

Hi, a bit of basic algebra -a - b == -(a+b) and Expand[pn1 - pn - k/2*fn*fn] is Expand[pn1 - (pn + k/2*fn*fn)] Expand[pn1]- Expand[pn+k/2*fn*fn] Don't use a computer algebra before you have finished the primary school. Regards Jens Peter Pollner wrote: > > hi, > > I got a strange answer from Expand in the following situation: > fn1, pn1, fn and pn are polynomials upto order to in x. I want to check > the identity: > > pn1 - pn - k/2*fn*fn = 0 + O(x^3) > > the answer depends on the usage of Expand: > Expand[pn1 - pn - k/2*fn*fn] > and > Expand[pn1] - Expand[pn - k/2*fn*fn] > are not the same (the first expansion contains O(x^2) terms)! > > Here is the full input: > > pn1 = p0*x - k/12*n(n + 1)(2n + 1)x^2p0^2 - k/2*n(n + 1)p0*f0*x^2 - > k/2*(n + 1)x^2f0^2 > > fn1 = f0*x + (n + 1)p0*x - k/24*n(n + 1)(n + 1)(n + 2)x^2 p0^2 - > k/6*n(n + 1)(n + 2)x^2p0*f0 - k/4*(n + 1)(n + 2)x^2f0^2 > > pn = p0*x - k/12*(n - 1)n(2(n - 1) + 1)x^2p0^2 - k/2*n(n - 1)p0*f0*x^2 - > k/2*n x^2f0^2 > > fn = f0*x + n p0*x - k/24*(n - 1)n^2(n + 1)x^2 p0^2 - > k/6*(n - 1)n(n + 1)x^2p0*f0 - k/4*n(n + 1)x^2f0^2 > > Expand[pn1 - pn - k/2*fn*fn] > > Expand[pn1] - Expand[pn - k/2*fn*fn] > > I am using Mathematica 4.0.1.0, Platform : X > > Is it a bug? > > Please send me answers to the email address: > Peter.Pollner at Physik.Uni-Marburg.De > > Thanks > Peter Pollner