Re: Generate polynomial of specified degree

*To*: mathgroup at smc.vnet.net*Subject*: [mg60757] Re: Generate polynomial of specified degree*From*: Maxim <ab_def at prontomail.com>*Date*: Tue, 27 Sep 2005 03:45:28 -0400 (EDT)*References*: <dgtj9q$28g$1@smc.vnet.net> <dgtuja$6ni$1@smc.vnet.net> <dh0f27$qaf$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

On Fri, 23 Sep 2005 08:40:39 +0000 (UTC), Paul Abbott <paul at physics.uwa.edu.au> wrote: > In article <dgtuja$6ni$1 at smc.vnet.net>, dh <dh at metrohm.ch> wrote: > >> it is not too hard: >> >> f[x_Symbol,n_Integer]:= Array[c,{n+1},0].x^Range[0,n] > > Personally, I think it is better to separate parameters from variables, > and also to localise the coefficient name: > > f[n_Integer,c_:c][x_Symbol]:= With[{p = Range[0, n]}, (c /@ p) . x^p] > f[0, c_:c][x_Symbol]:= c[0] > > Now try > > f[0][x] > f[3][x] > f[3,a][x] > > Sometimes using pure functions is advantageous: > > Clear[f] > f[n_Integer,c_:c] := Function[x, (c /@ Range[0, n]) . x^Range[0, n]] > f[0, c_:c]:= Function[x, c[0]] > > As a particular advantage of this syntax, try > > f[2]'[x] > > Cheers, > Paul > > > Differentiation of pure functions in Mathematica is more like a lottery: In[1]:= Function[x, {1, 1}.x^{1, 2}]'[x] Out[1]= {3, 3*x} In[2]:= Function[x, Total[x^{1, 2}]]'[x] Out[2]= {{0, 0}, {0, 0}} Note that the functions are scalar-valued (and equivalent) for scalar x. It probably would be better if Mathematica always evaluated f' as Module[{var}, Function @@ {var, D[f[var], var]}] That would get the two above examples right. Mathematica seems to do roughly that if the function is defined as f[x_] : = ..., except that it uses Slot instead of creating unique formal parameters. This means that there may be a conflict if the definition of f already contains a Function without named formal parameters: In[3]:= f[x_] := x + x^2&[1]; f'[x] Out[3]= 0 This time it's the other way around: the result would be correct if f were defined in a pure function form as Function[x, x + x^2&[1]]. Also see http://forums.wolfram.com/mathgroup/archive/2005/Jun/msg00059.html . Maxim Rytin m.r at inbox.ru