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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?

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2 Answers 2

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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:

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

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  • $\begingroup$ 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. $\endgroup$
    – Martin R
    Mar 2, 2020 at 7:53
  • $\begingroup$ @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. $\endgroup$
    – LHF
    Mar 2, 2020 at 8:01
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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.

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  • $\begingroup$ This doesn't seem to be helpful. The RHS goes to $\infty$ as $x\to 0$. $\endgroup$ Mar 1, 2020 at 19:04
  • $\begingroup$ @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. $\endgroup$
    – Alex Ortiz
    Mar 1, 2020 at 19:07