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

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 $n$

• I was wondering if you could bring this into a complex integral of some sort, and bring Cauchy's estimates into it. That's all I wondered for now, but I'll get back if I find anything more. – Teresa Lisbon Oct 30 '17 at 6:07
• @астонвіллаолофмэллбэрг Oh i was wondering whether we can use the general leibniz rule: en.wikipedia.org/wiki/General_Leibniz_rule and use some elementary inequalities to get an estimate :D – crskhr Oct 30 '17 at 6:20
• Yes, this is also very interesting. I will look through it, and see if we can come up with some induction argument. – Teresa Lisbon Oct 30 '17 at 6:20
• Hello maybe it could help math.stackexchange.com/questions/130192/… .Have a good day. – max8128 Oct 30 '17 at 6:54

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.

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

• This is basically Gabriel's answer. – user223391 Oct 30 '17 at 23:18
• – Guy Fsone Oct 30 '17 at 23:20

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 .