How can we prove the derivative inequality for f(x)=sin(x)/x?

[Fallen Angel]

Hi,

My first challenge was not very popular so I bring you another one.

Let us define $$f(x)=\dfrac{sin(x)}{x}$$ for $$x>0$$.

Prove that for every $$n\in \mathbb{N}$$, $$|f^{(n)}(x)|<\dfrac{1}{n+1}$$ where $$f^{n}(x)$$ denotes the n-th derivative of $$f$$

 

[Opalg]

Hi,

My first challenge was not very popular so I bring you another one.

Let us define $$f(x)=\dfrac{\sin(x)}{x}$$ for $$x>0$$.

Prove that for every $$n\in \mathbb{N}$$, $$\bigl|\,f^{(n)}(x)\bigr| < \dfrac{1}{n+1}$$ where $$f^{n}(x)$$ denotes the n-th derivative of $$f$$

[sp]The $n$th derivative of $\dfrac{\sin x}x = x^{-1}\sin x$ is given by Leibniz's formula as $$\sum_{k=0}^n {n\choose k} \frac{d^k}{dx^k}(\sin x)\frac{d^{n-k}}{dx^{n-k}}(x^{-1}).$$

The derivatives of $\sin x$ oscillate between $\pm\sin x$ and $\pm\cos x$. I found it most conveniant to regard $\sin x$ as the imaginary part of $e^{ix}$, so that its $k$th derivative can be written as $\mathrm{Im}\bigl(i^ke^{ix}\bigr)$. The $k$th derivative of $x^{-1}$ is $(-1)^kk!x^{-k-1}.$ So the Leibniz formula becomes $$\frac{d^n}{dx^n}\Bigl(\frac{\sin x}x\Bigr) = \sum_{k=0}^n {n\choose k} \mathrm{Im}\bigl(i^ke^{ix}\bigr) (-1)^{n-k}(n-k)!x^{-n+k-1}.$$ We want to show that the absolute value of this is less than $\dfrac1{n+1}$. After multiplying both sides by $x^{n+1}$ that becomes $$\left| \sum_{k=0}^n {n\choose k} \mathrm{Im}\bigl(i^ke^{ix}\bigr) (-1)^{n-k}(n-k)!x^k \right| <\frac{x^{n+1}}{n+1}.\qquad(*)$$ So we want to show that (*) holds for all $x>0$. Notice that both sides of (*) are zero when $x=0$. The idea will be to differentiate the expression on the left of (*) and show that this is less than the derivative of the right side. That will show that the left side of (*) is always less than the right side. By the product rule, $$\frac d{dx} \sum_{k=0}^n {n\choose k} \mathrm{Im}\bigl(i^ke^{ix}\bigr) (-1)^{n-k}(n-k)!x^k = \sum_{k=0}^n \frac{n!}{k!}(-1)^{n-k}\Bigl( \mathrm{Im}\bigl(i^{k+1}e^{ix}\bigr)x^k + \mathrm{Im}\bigl(i^ke^{ix}\bigr)kx^{k-1} \Bigr).\qquad(**)$$ (When $k=0$ the second term in the big parentheses becomes $0$.) In fact, the right side of (**) is a telescoping sum. For each power of $x$ apart from $x^0$ and $x^n$, the coefficient in the second term in the big parentheses is the negative of the coefficient of the same power of $x$ occurring in the first term in the parentheses. So the only term that survives is the term in $x^n$, namely $\mathrm{Im}\bigl(i^{n+1}e^{ix}\bigr)x^n$. But that is less in absolute value than $ \dfrac d{dx}\Bigl(\dfrac{x^{n+1}}{n+1}\Bigr) = x^n$, just as we wanted.

(Strictly speaking, the absolute value of $\mathrm{Im}\bigl(i^{n+1}e^{ix}\bigr)x^n$ is actually equal to $x^n$ when $x$ is equal to some multiples of $\pi/2$. But that does not affect the argument.)[/sp]

 

[Fallen Angel]

Good work Opalg!

My solution is slightly different.

Spoiler

Let us denote $f(x)=\dfrac{sin(x)}{x}=\displaystyle\int_{0}^{1}cos(xt)dt$.
Then $f^{(n)}(x)=\displaystyle\int_{0}^{1}\dfrac{\partial ^{n}}{\partial x^{n}}cos(xt)dt=\displaystyle\int_{0}^{1}t^{n}g_{n}(xt)dt $

Where $g_{n}(xt)=\pm cos(xt), \pm sin(xt)$, but $|g_{n}(xt)|\leq 1$ and the equality holds only for finitely many points. Hence

$|f^{n}(x)|\leq \displaystyle\int_{0}^{1}t^{n}|g_{n}(xt)|dt<\displaystyle\int_{0}^{1}t^{n}dt=\dfrac{1}{n+1}$

 

[Opalg]

Good work Opalg!

My solution is slightly different.

Spoiler

Let us denote $f(x)=\dfrac{sin(x)}{x}=\displaystyle\int_{0}^{1}cos(xt)dt$.
Then $f^{(n)}(x)=\displaystyle\int_{0}^{1}\dfrac{\partial ^{n}}{\partial x^{n}}cos(xt)dt=\displaystyle\int_{0}^{1}t^{n}g_{n}(xt)dt $

Where $g_{n}(xt)=\pm cos(xt), \pm sin(xt)$, but $|g_{n}(xt)|\leq 1$ and the equality holds only for finitely many points. Hence

$|f^{n}(x)|\leq \displaystyle\int_{0}^{1}t^{n}|g_{n}(xt)|dt<\displaystyle\int_{0}^{1}t^{n}dt=\dfrac{1}{n+1}$

Very neat! (Rock)