Re: Re: Bug in interpretation of mma Series[] command?
- To: mathgroup at smc.vnet.net
- Subject: [mg2326] Re: [mg2268] Re: Bug in interpretation of mma Series[] command?
- From: Paul Abbott <paul at earwax.pd.uwa.edu.au>
- Date: Tue, 24 Oct 1995 02:14:50 -0400
- Organization: Dept of Physics, University of WA
Richard Mercer <richard at seuss.math.wright.edu> wrote: >From the viewpoint of most users, I think the ideal situation would be >for > >Exp[-a y^2] * Series[1/(1 + y^2), {y,0,5}] > >to act like > > >Exp[-a y^2] * Normal[Series[1/(1 + y^2), {y,0,5}]] > >Reasons: >(1) Most average (nonexpert) users do not have a mental model >corresponding to a SeriesData object; they think of the result of a >Series command as being a polynomial and expect it to behave that way >in calculations. When you compute a series, say, f[x] + O[x]^3 2 f''[0] x 3 f[0] + f'[0] x + --------- + O[x] 2 then you expect to be able to do operations on this series, e.g. 1/% // Simplify 2 1 f'[0] x f'[0] f''[0] 2 3 ---- - ------- + (------ - -------) x + O[x] f[0] 2 3 2 f[0] f[0] 2 f[0] If you think of the result of a Series command as being a polynomial you lose this functionality. Also, the fact that the syntax f[x] + O[x]^3 coerces Taylor series expansion of f[x] is elegant and useful. It also means that if you write: Exp[-a y^2] 1/(1 + y^2) + O[y]^6 you get 2 2 a 4 6 1 + (-1 - a) y + (1 + a + --) y + O[y] 2 >(2) The structure Series[Exp[-a y^2] * 1/(1 + y^2), {y,0,5}] is >available and much more natural if you want the exponential converted >to a series. So is the syntax Exp[-a y^2] Normal[1/(1 + y^2) + O[y]^6] >The best solution would seem to be a user-settable "switch" that would >apply Normal to the output of all Series commands. This would >presumably satisfy all those who expect Series objects to act like >polynomials in this and other situations. You do have a user-settable "switch". It is Normal and you can use this to turn a Series into a polynomial whenever you need this behaviour. Here is an example that demonstrates what a nice design feature this syntax is. Suppose that you want to factor off some (exponential) asymptotic behaviour. You can use the syntax: Exp[a r] Normal[Exp[-a r] f[r] + O[r]^2] a r E (f[0] + r (-(a f[0]) + f'[0])) Cheers, Paul