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}
Is it possible to do the above math symbolically in Mathematica just as we humans do?
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] &
*)
Edit
Or just
FindSequenceFunction[Table[D[2/(2 - t), {t, n}], {n, 1, 5}], r] /. t -> 0
(*
-> 2^-r Pochhammer[1, r]
*)
*Edit *
Note: While FindSequenceFunction[]
works in this simple situation, don't bet on it in more general cases.
Edit
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]
*)