# $f(x)=\frac{\sin x}{x}$, prove that $|f^{(n)}(x)|\le \frac{1}{n+1}$ [duplicate]

This is a homework question.

Let $$f: (0, \infty) \to \mathbb{R}$$ with $$f(x)=\dfrac{\sin x}{x}$$. I have to prove that

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

where $$f^{(n)}$$ is the nth derivative of $$f$$.

I started to do a few derivatives:

$$f^{(1)}(x)=\dfrac{1}{x^2}(x \cos x-\sin x)$$

$$f^{(2)}(x)=\dfrac{1}{x^3}((2-x^2) \sin x-2x\cos x)$$

$$f^{(3)}(x)=\dfrac{1}{x^4}(3(x^2-2) \sin x-x(x^2-6)\cos x)$$

$$f^{(4)}(x)=\dfrac{1}{x^5}(4x(x^2-6)\cos x + (x^4-12x^2+24)\cos x)$$

I noticed that in denominator, there is always $$x^{n+1}$$ (because there is $$x$$ in denominator of $$f$$), but I couldn't spot a pattern for the numerator that would help me prove the inequality. Can I get a hint or a clue, please?

I will assume (since it's not specified) that $$n$$ is a non-negative integer.

Claim: For $$f:(0,\infty)\to \mathbb{R}$$, defined by $$f(x)=\dfrac{\sin x}{x}$$ and any non-negative integer $$n$$, we have:

$$f^{(n)}(x)=\frac{1}{x^{n+1}}\int_0^xu^n\cos\left(u+\frac{n\pi}{2}\right)\,du$$

Proof: I will prove this by induction. The case $$n=0$$ is obvious:

$$f^{(0)}(dx)=f(x)=\frac{\sin x}{x}=\frac{1}{x}\int_0^x\cos u\,du$$

Now assume it is true for some positive integer $$n \geq 1$$. Then, integrating by parts:

\begin{aligned} f^{(n+1)}(x)&=(f^{(n)})'(x)\\ &=\frac{1}{x^{n+2}}\left(-(n+1)\int_0^xu^n\cos\left(u+\frac{n\pi}{2}\right)+x\cdot x^n\cos\left(x+\frac{n\pi}{2}\right)\right)\\ &=\frac{1}{x^{n+2}}\left[-u^{n+1}\cos(u+\frac{n\pi}{2})\bigg|_0^x+\int_0^xu^{n+1}(\cos\left(u+\frac{n\pi}{2}\right))'\,du+x^{n+1}\cos(x+\frac{n\pi}{2})\right] \\ &=\frac{1}{x^{n+2}}\int_0^xu^{n+1}\cos\left(u+\frac{(n+1)\pi}{2}\right)\,du \end{aligned}

Claim proved. Now, the inequality follows immediately:

$$|f^{(n)}(x)|=\left|\frac{1}{x^{n+1}}\int_0^xu^n\cos\left(u+\frac{n\pi}{2}\right)\,du\right|\leq \frac{1}{x^{n+1}}\int_0^xu^n\,du=\frac{1}{n+1}$$

• Note that with the substitution $u=xy$, $\frac{1}{x^{n+1}}\int_0^x u^n \cos\left(u+\frac{n\pi}{2}\right)\,du$ becomes $\int_0^1 y^n\cos\left(xy+\frac{n\pi}{2}\right)\,dy$, which might simplify the induction step. Compare math.stackexchange.com/a/2496305/42969. Mar 2, 2020 at 7:53
• @MartinR, Yes, I've seen your initial link to AOPS. I overcomplicated some the form of $f^{(n)}(x)$, as I was looking mainly at the numerator (probably influenced by the OP's work) and I didn't see it can be written in a more simple form.
– LHF
Mar 2, 2020 at 8:01

Here is a suggestion:

Try proving the Liebniz rule for differentiating the product of two functions on $$\mathbb R$$:

$$(fg)^{(n)}(x) = \sum_{k=0}^n{n\choose k}f^{(k)}(x)g^{(n-k)}(x),$$ and apply this to $$f(x) = 1/x$$ and $$g(x) = \sin(x)$$ to show that $$\left|\left(\frac{\sin x}{x}\right)^{(n)}\right| \le n!|x|^{-n-1}\sum_{k=0}^n\frac{|x|^k}{k!}.$$ Now try to use this to show the inequality you want.

Edit: This likely isn't enough to show the inequality on its own, as the bound we get still blows up at least like $$1/|x|$$ on the right-hand side as $$x \to 0^+$$. We aren't taking advantage of the cancellations in the sum over $$k$$, which evidently matter here.

• This doesn't seem to be helpful. The RHS goes to $\infty$ as $x\to 0$. Mar 1, 2020 at 19:04
• @WETutorialSchool: Good point, we need to take more advantage of the cancellations in the sum to prove the inequality. I'll leave the answer as I think it's instructive to see that merely using the triangle inequality isn't enough here. Mar 1, 2020 at 19:07