Sum of limit

1. Oct 24, 2016

Hamal_Arietis

1. The problem statement, all variables and given/known data
Given the function in x
$$f_n(x)=sin^nx (n=1,2,3,...)$$
For this $f_n(x)$, consider the definite intergral
$$I_n=\int_{0}^{\pi/2}f_n(x)sin2xdx$$
a) Find $I_n$
b) Hence the obtain
$$lim_{n→∞}(I_{n-1}+I_n+I_{n+1}+...+I_{2n-2})=\int_0^W\frac{X}{Y+x}dx$$
Find X,Y,Z.

2. Relevant equations
b) I think we may use Riemann sum, and the answer in a, help we in b, but I can't do it.

3. The attempt at a solution
a) $I_n=\int_{0}^{\pi/2}f_n(x)sin2xdx=2\int_{0}^{\pi/2}sin^{n+1}cosxdx=2\int_{0}^{\pi/2}sin^{n+1}d(sinx)$
$$I_n=\frac{2}{n+2}$$
b) Dont have any attempt.

I need some books about the calculus I that have some problems like this
Thank you very much

Last edited: Oct 24, 2016
2. Oct 24, 2016

haruspex

Have you tried swapping the order of the steps, i.e. performing the sum before integrating?

3. Oct 25, 2016

Hamal_Arietis

It means:$$lim_{n→∞}\Sigma=lim_{n→∞}2(\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+...+\frac{1}{2n})(1)$$
The Riemann sum must have form: $$\int = lim_{n→∞}\Sigma _{i=0}^n f(i)\Delta x$$
I can't make (1) to appear element $\Delta x$.

4. Oct 25, 2016

haruspex

Can you use the harmonic sum bounds given by the log function and Euler's constant?
Trouble is, it does not give you an integral, but the actual answer.

5. Oct 25, 2016

Staff: Mentor

But your previous answer has been a good idea (it's been what I was looking for), to integrate the sum of sines which has a closed form. With the substitution of the OP ($u = \sin x$) one gets an integral of one quotient of polynomials in $u$.

6. Oct 25, 2016

haruspex

Yes, but I could not see how to massage it into the required form. Maybe I missed something.

7. Oct 25, 2016

Staff: Mentor

I haven't seen it either, but as long as nobody requires $X$ and $Y$ to be constants ...

8. Oct 25, 2016

haruspex

I would assume they have to be.

9. Oct 25, 2016

Hamal_Arietis

The answer W=1; X=2; Y=1 and the integral equals ln4. I think it may not use Euler's constant because It is the question for high school students. May be difficult but they dont learn about harmonic sum and Euler's constant.

10. Oct 25, 2016

haruspex

Well, you don't need Euler's constant as such. You just need the bounded form of the Riemann sum: with suitable facts about f, $\Sigma_m^{n-1}f(r)<=\int_m^nf <= \Sigma_{m+1}^n f(r)$, or something like that.

11. Oct 25, 2016

Staff: Mentor

One could use the Euler-Mascheroni constant and write the sum $I_{n-1}+\ldots +I_{2n-2} = I_{-1}+\ldots +I_{2n-2} - I_{-1} - \ldots - I_{n-2}$.

12. Oct 25, 2016

haruspex

Yes, but Hamal wants to avoid appealing to that as a standard result. As I posted, you can effectively achieve the same using Riemann sums, which Hamal wanted to do in the first place.

13. Oct 25, 2016

Hamal_Arietis

If I use Euler-Mascheroni constant, how to make $lim_{n→∞}∑=∫$?
If that can make answer, can you show me? The Riemann sum is just my idea. Maybe have another method.

14. Oct 25, 2016

Hamal_Arietis

In part A of problem, they give some question, And that is anwer I solved, I think maybe help you.
Given the funtion of x $$f_n(x)=sin^nx(n=1,2,3...)$$
Consider the cases in which the equality
$$lim_{x→0}\frac{a-x^2-(b-x^2)^2}{f_n(x)}=c$$
holds for three real numbers a,b,c.
a) We have $a=b^{O}$. Find O
b) When n=2, if c=6. Find b
c) When n=4 find b and c.
My solution:
a) Because c is real number so
$$lim_{x→0}\frac{a-x^2-(b-x^2)^2}{f_n(x)}=c⇔lim_{x→0}(a-x^2-(b-x^2)^2)=c.lim_{x→0}f_n(x)=0⇒a=b^2$$
b) $$lim_{x→0}\frac{a-x^2-(b-x^2)^2}{f_n(x)}=lim_{x→0}\frac{-x^2(x^2+2b-1)}{sin^2x}$$
$$lim_{x→0}\frac{-x^2(x^2+2b-1)}{sin^2x}=6⇔b=\frac{7}{2}$$
c) b=2 ;c=-1

15. Oct 25, 2016

haruspex

I urge you to try what I proposed in post #10. Draw a diagram to illustrate the inequalities.

16. Oct 25, 2016

Hamal_Arietis

I solved this problem. Firstly, I see
$I_{n-1}=\frac{2}{n+1};I_n=\frac{2}{n+2};...;I_{2n-2}=\frac{2}{n+n}$
So let $$P_i=\frac{2}{n+i}=\frac{1}{n}\frac{2}{1+\dfrac{i}{n}}=\Delta x .f(x_i)$$
If $i=1;n→∞⇒x→0$ and $i=n;n→∞;x=1$
So we have the Riemann sum:
$$lim_{n→∞}(I_{n-1}+I_n+I_{n+2}+...)=lim_{n→∞}\Sigma_{i=1}^{n}\frac{1}{n}\frac{2}{1+\dfrac{i}{n}}=∫_{0}^{1}\frac{2dx}{x+1}$$
Thanks for all helping

17. Oct 25, 2016

haruspex

Neat. Well done.