MHB How is the proof of the hard limit problem related to continuity of a function?

[Cephal]

Hello everybody,

I have proved that:

$$\displaystyle \lim_{n\to+\infty} n\int_{0}^1 x^ng(x)\mathrm{d}x=g(1)$$
with $$g \in\mathcal{C}^0(\left[0,1\right],\mathbb{R})$$.But I don't know how to prove this:

$$\displaystyle \int_0^1 x^n f(x)\mathrm{d}x=\dfrac{f(1)}{n}-\dfrac{f(1)+f'(1)}{n^2}+o_{+\infty}(\dfrac{1}{n^2})$$
with $$f\in\mathcal{C}^1(\left[0,1\right],\mathbb{R})$$.
(Tongueout)
Thank you for your answers.

 

[Opalg]

Hello everybody,

I have prove that:

$$\displaystyle \lim_{n\to+\infty} n\int_{0}^1 x^ng(x)\mathrm{d}x=g(1)$$
with $$g \in\mathcal{C}^0(\left[0,1\right],\mathbb{R})$$.But I don't know how to prove this:

$$\displaystyle \int_0^1 x^n f(x)\mathrm{d}x=\dfrac{f(1)}{n}-\dfrac{f(1)+f'(1)}{n^2}+o_{+\infty}(\dfrac{1}{n^2})$$
with $$f\in\mathcal{C}^1(\left[0,1\right],\mathbb{R})$$.
(Tongueout)
Thank you for your answers.

Hi Cephal, and welcome to MHB!

What you know about $g$ is (1) $g$ is continuous and therefore bounded on $[0,1]$, say $|g(x)| \leqslant M$ for some $M$; (2) $g(x)$ is continuous at $x=1$ and therefore given $\varepsilon>0$ there exists $\delta>0$ such that $|g(x) - g(1)| < \varepsilon$ whenever $1- \delta \leqslant x \leqslant 1$.

Now write $$n\int_0^1 x^n g(x)\,dx = n\int_0^{1-\delta} x^n g(x)\,dx + n\int_{1-\delta}^1 x^n g(x)\,dx.$$ Show that if $n$ is large then the first of those integrals is small and the second one is close to $g(1)$.

 

[Cephal]

Thank you Opalg for your answer.

In fact I have already proved that:

$$\displaystyle \lim_{n\to+\infty} n\int_{0}^1 x^ng(x)\mathrm{d}x=g(1)$$
with $$g \in\mathcal{C}^0(\left[0,1\right],\mathbb{R})$$.My problem is to prove this:

$$\displaystyle \int_0^1 x^n f(x)\mathrm{d}x=\dfrac{f(1)}{n}-\dfrac{f(1)+f'(1)}{n^2}+o_{+\infty}(\dfrac{1}{n^2})$$
with $$f\in\mathcal{C}^1(\left[0,1\right],\mathbb{R})$$.

Thank you for your help.

 

[chisigma]

Hello everybody,

I have proved that:

$$\displaystyle \lim_{n\to+\infty} n\int_{0}^1 x^ng(x)\mathrm{d}x=g(1)$$
with $$g \in\mathcal{C}^0(\left[0,1\right],\mathbb{R})$$.But I don't know how to prove this:

$$\displaystyle \int_0^1 x^n f(x)\mathrm{d}x=\dfrac{f(1)}{n}-\dfrac{f(1)+f'(1)}{n^2}+o_{+\infty}(\dfrac{1}{n^2})$$
with $$f\in\mathcal{C}^1(\left[0,1\right],\mathbb{R})$$.
(Tongueout)
Thank you for your answers.

You can proceed with integration by parts obtaining... $\displaystyle \int_{0}^{1} x^{n}\ f(x)\ dx = \frac{f(1)}{n+1} - \frac{1}{n+1}\ \int_{0}^{1} x^{n + 1}\ f^{\ '} (x)\ dx =$

$\displaystyle = \frac{f(1)}{n+1} - \frac{f^{\ '}(1)}{(n+1) (n+2)} + \frac{1}{(n+1) (n+2)}\ \int_{0}^{1} x^{n+2}\ f^{\ ''} (x)\ d x\ (1)$

Now if You consider that...

$\displaystyle \frac{f(1)}{n + 1} = \frac{f(1)}{n} - \frac{f(1)}{n^{2}} + \mathcal{o}\ (\frac{1}{n^{2}})\ (2)$

... and...

$\displaystyle \frac{f^{\ '}(1)}{(n+1) (n+2)} = \frac{f^{\ '}(1)}{n^{2}} + \mathcal{o}\ (\frac{1}{n^{2}})\ (3)$

... You arrive to the result...Kind regards $\chi$ $\sigma$

 

[Cephal]

You can proceed with integration by parts obtaining... $\displaystyle \int_{0}^{1} x^{n}\ f(x)\ dx = \frac{f(1)}{n+1} - \frac{1}{n+1}\ \int_{0}^{1} x^{n + 1}\ f^{\ '} (x)\ dx =$

$\displaystyle = \frac{f(1)}{n+1} - \frac{f^{\ '}(1)}{(n+1) (n+2)} + \frac{1}{(n+1) (n+2)}\ \int_{0}^{1} x^{n+2}\ f^{\ ''} (x)\ d x\ (1)$

Now if You consider that...

$\displaystyle \frac{f(1)}{n + 1} = \frac{f(1)}{n} - \frac{f(1)}{n^{2}} + \mathcal{o}\ (\frac{1}{n^{2}})\ (2)$

... and...

$\displaystyle \frac{f^{\ '}(1)}{(n+1) (n+2)} = \frac{f^{\ '}(1)}{n^{2}} + \mathcal{o}\ (\frac{1}{n^{2}})\ (3)$

... You arrive to the result...Kind regards $\chi$ $\sigma$

Thank you very very much $$\chi\sigma$$.(Clapping)

 

[Opalg]

Apologies for not reading the original question carefully enough. Chisigma's approach is certainly the right way to go. My only comment on it is that you are only told that $$f\in\mathcal{C}^1(\left[0,1\right],\mathbb{R})$$. So you should not assume that $f$ is twice differentiable. Instead, use chisigma's method to integrate by parts just once, getting $$\int_{0}^{1} x^{n}f(x)\, dx = \frac{f(1)}{n+1} - \frac{1}{n+1} \int_{0}^{1} x^{n + 1} f^{\, '} (x)\, dx. $$ Then write that as $$\int_{0}^{1} x^{n}f(x)\, dx = \frac{f(1)}{n+1} - \frac{1}{(n+1)^2}\left[(n+1) \int_{0}^{1} x^{n + 1} f^{\, '} (x)\, dx\right], $$ and use the previous result about $g(x)$ to conclude that the part in the large brackets is $f^{\, '} (1) + o_{n\to\infty}(1).$

 

[chisigma]

Now is my turn to apologies because I have had a little hurry in my last post, so that some essential step has been omitted. The most important step I have omitted is that continuity is not neccesary condition for integrability of a function but is necessary condition for univocal definition of a function in a closed interval. Now if we write again...

$\displaystyle \int_{0}^{1} x^{n}\ f(x)\ d x = \frac{f(1)}{n+1} - \frac{f^{\ '}(1)}{(n + 1) (n + 2)} + \frac{1}{(n+1) (n+2)}\ \int_{0}^{1} x^{n+2}
f^{\ ''}(x)\ dx\ (1)$

... taking into account the relations...

$\displaystyle \frac{1}{n + 1} = \frac{1}{n}\ (1 - \frac{1}{n} + \frac{1}{n^{2}} - ...)\ (2)$

$\displaystyle \frac{1}{n+2} = \frac{1}{n}\ (1 - \frac{2}{n} + \frac{4}{n^{2}} - ...)\ (3)$

all that we can say is...

a) if f(*) is continuous in [0,1] then f(*) is 'well defined' in x=1 and is...

$\displaystyle \int_{0}^{1} x^{n}\ f(x)\ dx = \frac{f(1)}{n} + \mathcal {o}\ (\frac{1}{n})\ (4)$

b) if f(*) and its derivative are continuous in [0,1] then f(*) and its derivative are 'well defined' in x=1 and is...

$\displaystyle \int_{0}^{1} x^{n}\ f(x)\ dx = \frac{f(1)}{n} - \frac{f(1) + f^{\ '}(1)}{n^{2}} + \mathcal {o}\ (\frac{1}{n^{2}})\ (5)$

Of course if all the derivatives of f(x) in x=1 are 'well defined', then we can extend the expansion without limits...

Kind regards

$\chi$ $\sigma$