Proof with derivatives (most likely induction)

Question

Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be given by an equation $$f(x)=(\sin(x^3))^3$$ With use of the fact that function f is odd, show that all derivatives in a form $f^{(2n)}(0)$ for $n=0,1,2, \ldots$ are equal. Maybe inductional proof will be the most suitable, but i do not know i would appreciate detailed explaination.

Answer 1

Hint:

$$f''(x)=\displaystyle{\lim_{h \to 0}}\frac{f(x+h)-2f(x)+f(x-h)}{h^2}$$

Once you see what happens, induction should follow easily.

Answer 2

You can show using induction that for any function $f : \Bbb R \to \Bbb R$ given in the form of $f(x) = p[\sin(x^3), x] $ where $ p(s, t) $ is a bivariate polynomial of $s$ and $t$ $f^{(n)} (x)$ is also a polynomial of $\sin(x^3)$ and $x$.

To be a it more formal you can write down the general $f$ in the form, $$ f(x) = p[\sin(x^3), x] = \sum_m \sum_n (\sin(x^3))^m \cdot x^n \;\;\;\; m, n \in \Bbb N$$

And it is easy to see that any bivariate polynomial whose terms are multiples of powers of $x$ and $\sin (x^3)$ takes the value $0$ at the origin.

Answer 3

Show that any odd differentiable function $f\colon (a,b)\to \mathbb R$ has even derivative $f'\colon (a,b)\to \mathbb R$ and that any even differentiable function $g\colon (a,b)\to\mathbb R$ has odd derivative $g'\colon (a,b)\to\mathbb R$.

Then $f$ is odd, $f'$ is even, $f''$ is odd, $f'''$ is even...