Let $f(x) =\dfrac{\sin x}{x}$ prove that : $|f^{(n)}(x)|<\frac{1}{n+1}$

Question

let $x>0$ And $f(x) =\dfrac{\sin x}{x}$ prove that for every $n$ :

$$|f^{(n)}(x)|<\frac{1}{n+1}$$

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for $f^{(1)}$ we have :

$$\frac{x\cos x-\sin x}{x^2}<\frac{1}{2}$$

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for $f^{(2)}$ we have :

$$\frac{x^2(-x\sin x)-2x(x\cos x-\sin x)}{x^2}<\frac{1}{3}$$

Now what ?

Answer 1

$$f(x)=\int_0^1\cos(tx)\,dt.$$ Write $c(y)=\cos y$ $$f^{(n)}(x)=\int_0^1 t^n c^{(n)}(tx)\,dt.$$ As $c^{(n)}$ is $\pm\sin$ or $\pm\cos$ then $|c^{(n)}(y)|\le 1$ so $$|f^{(n)}(x)|\le\int_0^1t^n\,dt=\frac1{n+1}.$$ I'll leave it to you to prove equality cannot occur.

Answer 2

See that,
$$\frac{\sin x}{x} =f(x) = \frac{1}{2}\int_{-1}^{1} e^{-itx} dt$$

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