当Mathematica中的r为符号时，如何找到函数的rth导数? [英] How to find a function's rth derivative when r is symbolic in Mathematica?

问题描述

我有一个功能f(t)=2/(2-t).在不使用Mathematica的情况下，很难在t = 0处获得rth导数(即2^(-r)*r!).在Mathematica计算的情况下，当r = 4时，我可以得到第r个导数，如下所示:D[2/(2-t), {t, 4}].但是，当r为ANY整数时，如何在Mathematica中的t = 0处获得rth导数?我尝试使用此表达式，但未按预期工作:

I have a function f(t)=2/(2-t). It is not so hard to get the rth derivative at t=0 (i.e. 2^(-r)*r!) without using Mathematica. In the case of Mathematica calculation, I can get the r-th derivative when r=4 like this: D[2/(2-t), {t, 4}]. But how can I get the rth derivative at t=0 in Mathematica when r is ANY integer? I tried to use this expression, but it didn't work as expected:

Simplify[D[2/(2 - t), {t, r}], Assumptions -> Element[r, Integers]]  /. {t->0}

是否有可能像人类一样在Mathematica中象征性地进行上述数学运算?

Is it possible to do the above math symbolically in Mathematica just as we humans do?

推荐答案

f = FindSequenceFunction[Table[D[2/(2 - t), {t, n}], {n, 1, 5}], r]

(*
-> -((2 (2 - t)^-r Pochhammer[1, r])/(-2 + t))
*)
g[r_, t_] := f
FullSimplify@FindSequenceFunction[Table[g[r, t], {r, 1, 5}] /. t -> 0]

 (*
 -> 2^-#1 Pochhammer[1, #1] &
 *)

修改

或者只是

FindSequenceFunction[Table[D[2/(2 - t), {t, n}], {n, 1, 5}], r] /. t -> 0
(*
-> 2^-r Pochhammer[1, r]
*)

* 编辑*

注意:尽管FindSequenceFunction[]在这种简单情况下可以工作，但在更一般的情况下请勿押注.

Note: While FindSequenceFunction[] works in this simple situation, don't bet on it in more general cases.

修改

要获得以阶乘函数表示的结果，只需执行以下操作:

To get the result expressed in terms of the factorial function, just do:

FunctionExpand@FindSequenceFunction[Table[D[2/(2-t),{t, n}],{n,1,5}], r] /.t->0
(*
-> 2^-r Gamma[1 + r]
*)

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