/sci/ - Science & Math

[Anonymous No.4076511]

<span class="math">\lim_{n \to \infty }\int_{0}^{1}\int_{0}^{1}\cdots \int_{0}^{1}cos^2\left [ \frac{\pi }{2n}(x_{1}+x_{2}+\cdots x_{n}) \right ]dx_{1}dx_{n}\cdots dx_{n}[/spoiler]

[Anonymous No.4076528]

1

[Anonymous No.4076540]

Assuming you meant
<span class="math">\lim_{n \to \infty }\int_{0}^{1}\int_{0}^{1}\cdots \int_{0}^{1}cos^2\left [ \frac{\pi }{2n}(x_{1}+x_{2}+\cdots +x_{n}) \right ]dx_{1}dx_{2}\cdots dx_{n}[/spoiler]
the answer is 1/2. You're integrating over the same interval so you can take <span class="math">x_i=x_j[/spoiler] for all i,j. It falls apart pretty easily after that.

[Anonymous No.4076544]

>>4076540

I love you /sci/

[Anonymous No.4076578]

>>4076540
I don't get it. Is there some theorem?

[Anonymous No.4076617]

>>4076578
Seems simple enough with a proof by induction.

[Anonymous No.4076639]

>>4076540
Let's see... If I wanted to attack this, I think I'd start with
http://en.wikipedia.org/wiki/Central_limit_theorem
So, I know we're effectively integrating over a single variable x, and the inside of the cos function contains a normal(x). Seems pretty straightforward from there.