Skip to content

testing inline latex

2 February, 2008

J_n = \int_0^{\pi/2} \cos^{2n}\theta\,d\theta

Expand \cos^{2n}\theta as a trigonometric polynomial in e^{i\theta} and e^{-i\theta}, using the binomial expansion. The constant term is

2^{-2n} {2n\choose n}

and the other terms are multiples of \cos (2k\theta), for k=1,2, \ldots n
Therefore

J_n = \int_0^{\pi/2} 2^{-2n} {2n\choose n}\,d\theta = \frac{\pi}{2} 2^{-2n} {2n\choose n}

We are done if we can prove that \sqrt{n} J_n \to \frac{1}{2}\sqrt{\pi} as n\to\infty

No comments yet

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: