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

AI Thread Summary
The discussion focuses on proving the inequality |f^(n)(x)| < 1/(n+1) for the function f(x) = sin(x)/x, defined for x > 0. The approach involves using Leibniz's formula to express the n-th derivative and analyzing the oscillatory nature of the derivatives of sin(x). The participants explore differentiating the left side of the inequality and demonstrating that it remains less than the derivative of the right side. The argument hinges on the behavior of the terms involved, particularly at specific values of x, ensuring the inequality holds universally for all x > 0. The conversation highlights different methods of arriving at the solution, showcasing collaborative problem-solving.
Fallen Angel
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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 $$f$$
 
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Fallen Angel said:
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]
 
Good work Opalg!

My solution is slightly different.

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

$|f^{n}(x)|\leq \displaystyle\int_{0}^{1}t^{n}|g_{n}(xt)|dt<\displaystyle\int_{0}^{1}t^{n}dt=\dfrac{1}{n+1}$
 
Fallen Angel said:
Good work Opalg!

My solution is slightly different.

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)
 
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