Proving that: $\forall n \in \mathbb{N} :|f^{(n)}(x)|\leq \frac{1}{n+1}$

Question

Suppose $x>0$ and we have function $f(x)=\frac{\sin x}{x}$ how can we show $$\forall n \in \mathbb{N} :|f^{(n)}(x)|\leq \frac{1}{n+1}$$ I need a hint to show this property .Thanks in advance . I tried for $n=1 ,2$ by finding maximum of $|f'| ,|f''|$ but I get stuck to show for $n$

Answer 1

Since $$ \frac{\sin(x)}{x} = \int_{0}^{1} \cos(tx) dt$$

with proper justifications (differentiation under the integral sign) you may derive

$$\left(\frac{\sin(x)}{x}\right)^{(n)} = \int_{0}^{1} t^n \cos(tx+n\pi/2) dt$$ which immediately yields the wanted estimate.

Answer 2

See that, $\displaystyle\frac{\sin x}{x} =f(x) = \frac{1}{2}\int_{-1}^{1} e^{-itx} dt$ Then $$|f^{(n)}(x)| =\left|\frac{1}{2}\int_{-1}^{1} (-it)^ne^{-itx} dt\right| \le\frac{1}{2}\int_{-1}^{1} |t|^n dt=\int_0^1t^n\,dt=\frac1{n+1}.$$

Answer 3

Hint : Combine the Dirichlet kernel to the following inequality :

$$\frac{1}{2n+1}\geq\frac{1}{(2n+1)^2}|\frac{sin((n+0.5)x)}{sin(0.5x)}|\geq \frac{1}{(2n+1)^2}|\frac{sin((n+0.5)x)}{0.5x}|$$

And use the same inductive reasoning as Emil Artin for this proof related to the Gamma function .