tag:blogger.com,1999:blog-9004452969563620551.post2817277178068845773..comments2017-06-21T11:55:27.964+05:30Comments on Paramanand's Math Notes: π(PI) and the AGM: Evaluating Elliptic Integrals contd.testernoreply@blogger.comBlogger3125tag:blogger.com,1999:blog-9004452969563620551.post-83418981655408208562015-02-08T20:41:39.440+05:302015-02-08T20:41:39.440+05:30Hi, Paramanand,
I found the limits in this post i...Hi, Paramanand,<br /> I found the limits in this post interesting.<br /><br /> From the Landen transformation section:<br /><br /> $ \theta_{n+1} = \arctan(\frac{b_n}{a_n} tan(\theta_n)) + \theta_n $<br /><br /> Near $ \pi / 2 $, the $ \tan() $ approaches $ \infty $. Therefore, it doesn't matter much what $ b_n $ and $ a_n $ are, the $ \arctan() $ is still $ \pi / 2 $, and thus $ \theta_{n+1} = 2 \theta_n $.<br /><br /> Later on, it took me a while to see the double sign reversal with $ \psi(x) $ . <br /><br />As noted above, corresponding to $ t $ going up from $ b \to a $ , $ x $ goes down $ a_1 \to b_1 $ and then up again from $ b_1 \to a_1 $ . This is reflected in the 2 versions of $ \psi(x) $ .<br /> Since the overall integral is stated as $ \int_{b_1}^{a_1} $ , the result of the first part of the integral range above must be sign reversed. This is the 2nd $ \psi(x) $ in the expansion above. This corresponds to the part of the integral where the bottom $ \sqrt{..} \space $ is negative, resulting in a positive $ \psi(x) $ component. The other $ \psi(x) $ has no sign reversals.<br /><br />Thanks for the http://math.stackexchange.com/questions/ask hint for previewing. It was very helpful for checking the formatting. I hope it worked ok.<br /><br />Thanks, a very informative section. --gjk<br /> Glenn Kellerhttp://www.blogger.com/profile/03353030102510662096noreply@blogger.comtag:blogger.com,1999:blog-9004452969563620551.post-42093971706834279162013-03-20T14:17:37.051+05:302013-03-20T14:17:37.051+05:30Hi Bob,
I read the alternate proof in some paper o...Hi Bob,<br />I read the alternate proof in some paper online. It’s an old entry so I don’t remember the source now. Sorry.paramanandhttp://www.blogger.com/profile/03855838138519730072noreply@blogger.comtag:blogger.com,1999:blog-9004452969563620551.post-24259505084977756652013-03-20T14:16:55.770+05:302013-03-20T14:16:55.770+05:30Hi there,
I really like this blog, I’ve found it ...Hi there,<br /><br />I really like this blog, I’ve found it fascinating. I was wondering if you could tell me where you found the alternative proof above?<br /><br />Thanks.bobnoreply@blogger.com