4 6 2q 2 13 5 2q 1 2 d 2 d cos 2q

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: = 4∫∫ e −r 2 2m −1 rcos 2n −1 sin r drd = 00 2∞ 2 4 ∫ ∫ e−r cos 2m−1 r 2m2n −2 sin 2n −1 r drd = 0 0 ∞ 2∫ e −r 2 2 r 2m 2n −2 0 2n −1 2r dr ∫ sin cos 2m −1 d 0 : ‫ يصبح‬r 2 = Z ‫و بفرض‬ ∞ 2 −Z =∫e 0 Z m n − 1 2n − 2 dZ ∫ sin 2m −2 cos 2sin cos d 0 : ‫و بفرض‬ 2 sin = X −1 , cos = X 2 m n = m n B m , n ⇒ B m , n = 2 2 m n m n Wallis ‫دستورا واليس‬ 2 • m n ‫يمكن استخدام العلقة‬ ∫ sin 2n−1 cos 2m−1 d = mn 0 : ‫في استنتاج القانون عندما‬ 2n −1 =0 , 2m −1 = p 2n −1 = q , 2m −1 = 0 : ‫لنجد أن‬ p 1 2 2n −1 2m −1...
View Full Document

This document was uploaded on 01/18/2014.

Ask a homework question - tutors are online