$f(x)= \tan x; f^n(0)-{n\choose 2}f^{n-2}(0)+ {n\choose 4}f^{n-4}(0)+...=\sin\frac{n\pi}{2}$

Question

How should I go about proving following relation $f(x)=\tan(x)$, then $$f^n(0)-{n\choose 2}f^{n-2}(0)+ {n\choose 4}f^{n-4}(0)+...=\sin\frac{n\pi}{2}$$

I tried Maclaurin expansion but not able to get any recursive relation for the expression of $f^n(x)$. Thanks in advance.

score 5 · Accepted Answer · edited Jul 31 '19 at 01:54

5

Hint:

$$\sin x=f(x)\cos x$$

Use General Leibniz rule and Is the $n^{th}$ derivative of $\sin(x)$ just a translation of $\sin(x)$?

edited Jul 31 '19 at 01:54

Pang

399
5
8

answered Jun 22 '18 at 13:12

lab bhattacharjee

274,582

$f(x)= \tan x; f^n(0)-{n\choose 2}f^{n-2}(0)+ {n\choose 4}f^{n-4}(0)+...=\sin\frac{n\pi}{2}$

1 Answers1