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How to ask mathematica to compute higher order derivatives evaluated at 0

开发者 https://www.devze.com 2023-02-11 07:21 出处:网络
I have a function, let\'s say for example, D[x^2*Exp[x^2], {x, 6}] /. x -> 0 And I want to replace 6 by a general integer n,

I have a function, let's say for example,

D[x^2*Exp[x^2], {x, 6}] /. x -> 0

And I want to replace 6 by a general integer n,

Or cases like the following:

 Limit[Limit[D[D[x /((-1 + x) (1 - y) (-1 + x + x y)), {x, 3}], {y, 5}], 开发者_C百科{x -> 0}], {y -> 0}]

And I want to replace 3 and 5 by a general integer m and n respectively.

How to solve these two kinds of problems in general in mma?

Many thanks.


Can use SeriesCoefficient, sometimes.

InputForm[n! * SeriesCoefficient[x^2*Exp[x^2], {x,0,n}]]

Out[21]//InputForm= n!*Piecewise[{{Gamma[n/2]^(-1), Mod[n, 2] == 0 && n >= 2}}, 0]

InputForm[mncoeff = m!*n! *
  SeriesCoefficient[x/((-1+x)*(1-y)*(-1+x+x*y)), {x,0,m}, {y,0,n}]]

Out[22]//InputForm= m!*n!*Piecewise[{{-1 + Binomial[m, 1 + n]*Hypergeometric2F1[1, -1 - n, m - n, -1], m >= 1 && n > -1}}, 0]

Good luck extracting limits for m, n integer, in this second case.

Daniel Lichtblau Wolfram Research


No sure if this is what you want, but you may try:

D[x^2*Exp[x^2], {x, n}] /. n -> 4 /. x -> 0  

Another way:

f[x0_, n_] := n! SeriesCoefficient[x^2*Exp[x^2], {x, x0, n}]  
f[0,4]  
24   

And of course, in the same line, for your other question:

f[m_, n_] := 
 Limit[Limit[
   D[D[x/((-1 + x) (1 - y) (-1 + x + x y)), {x, m}], {y, n}], {x -> 
     0}], {y -> 0}]  

These answers don't give you an explicit form for the derivatives, though.

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