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#1 h Mar 21, 2025, 4:55 AM

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Let $f:[0,1]->R$ continously function with $f'$ bounded
Find $\lim_{n \rightarrow \infty}n(n\int_{0}^{1}x^{n}f(x)dx-f(1))$

This post has been edited 1 time. Last edited by teomihai, Mar 21, 2025, 4:57 AM

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#4 h Mar 21, 2025, 8:24 AM • 2 Y

Y by YuLuo, Safal

The answer is $-f(1)-f'(1).$

Let
$g(x):=f(x)-f(1)-(x-1)f'(1).$ Then $g(1)=g'(1)=0$ and we only need to show that $\lim_{n\to\infty}n^2\int_0^1x^ng(x) \ dx=0.$ It's clear that
$\lim_{n\to\infty}n^2\int_0^{\delta}x^ng(x) \ dx=0, \ \ \ \ \ \ \ \ \ (1)$ for all $\delta \in (0,1),$ because $g$ is bounded. Now, let $\epsilon > 0$ and $h(x):=\frac{g(x)}{1-x}.$ We define $h(1)=0$ so that $h$ becomes continuous on $[0,1].$ Since $\lim_{x\to{1^-}}h(x)=-g'(1)=0,$ there exists $0 < \delta < 1$ such that $|h(x)| < \epsilon$ for all $x \in [\delta,1].$ Finally,
$n^2\left|\int_{\delta}^1x^ng(x) \ dx\right| \le n^2\int_{\delta}^1x^n|g(x)| \ dx \le n^2\int_{\delta}^1x^n(1-x)|h(x)| \ dx \le \frac{n^2}{(n+1)(n+2)}\epsilon < \epsilon, \ \ \ \ \ \ \ \ (2)$ and the result follows from $(1),(2).$

This post has been edited 1 time. Last edited by ysharifi, Mar 21, 2025, 1:26 PM
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#5 h Mar 21, 2025, 8:37 AM

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ysharifi wrote:

The answer is $-f(1)-f'(1).$

Let
$g(x):=f(x)-f(1)-(x-1)f'(1).$ Then $g(1)=g'(1)=0$ and we only need to show that $\lim_{n\to\infty}n^2\int_0^1x^ng(x) \ dx=0.$ It's clear that
$\lim_{n\to\infty}n^2\int_0^{\delta}x^ng(x) \ dx=0, \ \ \ \ \ \ \ \ \ (1)$ for all $\delta \in (0,1),$ because $g$ is bounded. Now, let $\epsilon > 0$ and $h(x):=\frac{g(x)}{1-x}.$ We define $h(1)=0$ so that $h$ becomes continuous on $[0,1].$ Since $\lim_{x\to{1^-}}h(x)=-g'(1)=0,$ there exists $0 < \delta < 1$ such that $|h(x)| < \epsilon$ for all $x \in [\delta,1].$ Finally,
$n^2\left|\int_{\delta}^1x^ng(x) \ dx\right| \le n^2\int_{\delta}^1x^n|g(x)| \ dx \le n^2\int_{\delta}^1x^n(1-x)|h(x)| \ dx \le \frac{n^2}{(n+1)(n+2)}\epsilon < \epsilon, \ \ \ \ \ \ \ \ (2)$ and the result follows from $(1),(2).$

Thanks very much Sir, :10:

:10:

Existe more problems with this?
Ex:
$\lim_{n \rightarrow \infty}n(n(n\int_{0}^{1}x^{n}f(x)dx-f(1))+f(1)+f'(1))$ ?
i mean existe one generalization?

This post has been edited 3 times. Last edited by teomihai, Mar 22, 2025, 4:21 AM

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#6 h Mar 21, 2025, 9:48 AM • 1 Y

Y by teomihai

I think we need $f'$ is integrable on $[0,1]$ and $f'$ is continuous at $1$ :
Lemma: If $f$ is integrable on $[0,1]$ and is continuous at $1$ , then we have
$\lim_{n \to\infty}n\int_{0}^{1}x^{n}f(x)dx=f(1).$ We will use the lemma to prove your question: Integral by part, we can get
$\int_{0}^{1}x^{n}f(x)dx=\frac1{n+1}\int_{0}^{1}f(x)d(x^{n+1})=\frac{f(1)}{n+1}-\frac1{n+1}\int_{0}^{1}x^{n+1}f'(x)dx =\frac{f(1)}n-\frac{f(1)}{n(n+1)}-\frac1{n+1}\int_{0}^{1}x^{n+1}f'(x)dx.$ So
$n\left(n\int_{0}^{1}x^{n}f(x)dx-f(1)\right)=-\frac{n^2}{n(n+1)}f(1)-\frac{n^2}{n+1}\int_{0}^{1}x^{n+1}f'(x)dx.\qquad (*)$ By the lemma for the function $f'$ , we can get
$\lim_{n\to\infty}\frac{n^2}{n+1}\int_{0}^{1}x^{n+1}f'(x)dx=\lim_{n\to\infty}\frac{n^2}{(n+1)^2}\cdot(n+1)\int_{0}^{1}x^{n+1}f'(x)dx =f'(1).$ Let $n\to\infty$ for equation $(*)$ , we get
$\lim_{n\to\infty}n\left(n\int_{0}^{1}x^{n}f(x)dx-f(1)\right)=-f(1)-f'(1).$
Actually, if we suppose $f\in C^{(2)}[0,1]$ , we can get
$\int_{0}^{1}x^{n}f(x)dx=\frac{f(1)}n-\frac{f(1)+f'(1)}{n^2}+o(1/n^2).$
Added: For your extra question:
$\lim_{n \rightarrow \infty}n(n(n\int_{0}^{1}x^{n}f(x)dx-f(1))+f(1)+f'(1)),$ you can do IBP more times(also need your $f$ has high order derivatives).

This post has been edited 1 time. Last edited by Alphaamss, Mar 21, 2025, 10:04 AM
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#7 h Mar 21, 2025, 10:53 AM

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Alphaamss wrote:

I think we need $f'$ is integrable on $[0,1]$ and $f'$ is continuous at $1$ :
Lemma: If $f$ is integrable on $[0,1]$ and is continuous at $1$ , then we have
$\lim_{n \to\infty}n\int_{0}^{1}x^{n}f(x)dx=f(1).$ We will use the lemma to prove your question: Integral by part, we can get
$\int_{0}^{1}x^{n}f(x)dx=\frac1{n+1}\int_{0}^{1}f(x)d(x^{n+1})=\frac{f(1)}{n+1}-\frac1{n+1}\int_{0}^{1}x^{n+1}f'(x)dx =\frac{f(1)}n-\frac{f(1)}{n(n+1)}-\frac1{n+1}\int_{0}^{1}x^{n+1}f'(x)dx.$ So
$n\left(n\int_{0}^{1}x^{n}f(x)dx-f(1)\right)=-\frac{n^2}{n(n+1)}f(1)-\frac{n^2}{n+1}\int_{0}^{1}x^{n+1}f'(x)dx.\qquad (*)$ By the lemma for the function $f'$ , we can get
$\lim_{n\to\infty}\frac{n^2}{n+1}\int_{0}^{1}x^{n+1}f'(x)dx=\lim_{n\to\infty}\frac{n^2}{(n+1)^2}\cdot(n+1)\int_{0}^{1}x^{n+1}f'(x)dx =f'(1).$ Let $n\to\infty$ for equation $(*)$ , we get
$\lim_{n\to\infty}n\left(n\int_{0}^{1}x^{n}f(x)dx-f(1)\right)=-f(1)-f'(1).$
Actually, if we suppose $f\in C^{(2)}[0,1]$ , we can get
$\int_{0}^{1}x^{n}f(x)dx=\frac{f(1)}n-\frac{f(1)+f'(1)}{n^2}+o(1/n^2).$
Added: For your extra question:
$\lim_{n \rightarrow \infty}n(n(n\int_{0}^{1}x^{n}f(x)dx-f(1))+f(1)+f'(1)),$ you can do IBP more times(also need your $f$ has high order derivatives).

splendid thanks! :10:

:10:

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#9 h Mar 21, 2025, 1:44 PM • 1 Y

Y by teomihai

ysharifi wrote:

$g(x):=f(x)-f(1)-(x-1)f'(x).$

That $f'(x)$ on the RHS was a typo; I wrote $f'(x)$ instead of $f'(1).$

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#10 h Mar 21, 2025, 2:11 PM

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Thanks Sir.

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