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 $n$

  • $\begingroup$ 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. $\endgroup$ – Teresa Lisbon Oct 30 '17 at 6:07
  • $\begingroup$ @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 $\endgroup$ – crskhr Oct 30 '17 at 6:20
  • $\begingroup$ Yes, this is also very interesting. I will look through it, and see if we can come up with some induction argument. $\endgroup$ – Teresa Lisbon Oct 30 '17 at 6:20
  • $\begingroup$ Hello maybe it could help math.stackexchange.com/questions/130192/… .Have a good day. $\endgroup$ – 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 $$|f^{(n)}(x)| =\left|\frac{1}{2}\int_{-1}^{1} (-it)^ne^{-itx} dt\right| \le\frac{1}{2}\int_{-1}^{1} |t|^n dt=\int_0^1t^n\,dt=\frac1{n+1}.$$


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 .


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.