tag:blogger.com,1999:blog-9004452969563620551.post6562152181655744572..comments2017-06-21T11:55:27.964+05:30Comments on Paramanand's Math Notes: Arithmetic-Geometric Mean of Gausstesternoreply@blogger.comBlogger8125tag:blogger.com,1999:blog-9004452969563620551.post-52854443136434074242016-07-16T11:48:09.520+05:302016-07-16T11:48:09.520+05:30@Anonymous,
If you read the comments you will find...@Anonymous,<br />If you read the comments you will find how one can compute the coefficients for the series of $1/M(1 - x, 1 + x)$ by hand calculation. But to calculate the general term $p_{n}$ you need to read the paper "Gauss, recurrence relations, and the agM" by Stacy G. Langton. Unfortunately this paper does not appear to available online for free now. It was available when I wrote this post, but I am unable to find that copy right now.<br /><br />Regards,<br />ParamanandParamanand Singhhttp://www.blogger.com/profile/03855838138519730072noreply@blogger.comtag:blogger.com,1999:blog-9004452969563620551.post-25630494970385087772016-07-16T10:08:48.061+05:302016-07-16T10:08:48.061+05:30It'll awesome if you could show how Gauss work...It'll awesome if you could show how Gauss worked out the power series. I've read quite a few papers on this area, and most authors skip that part. Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-9004452969563620551.post-35102440572501861322015-01-19T14:59:29.072+05:302015-01-19T14:59:29.072+05:30@Glenn Keller,
I am glad that you took some effort...@Glenn Keller,<br />I am glad that you took some effort to understand the calculations done in this post, especially going till $(1 + x)^{13}$. Given the effort you put, I think you will not have a serious difficulty in dealing with the elliptic integrals of second kind (especially the second proof is much easier to handle compared to the first proof for formula of elliptic integrals of second kind).<br /><br />Regards,<br />ParamanandParamanand Singhhttp://www.blogger.com/profile/03855838138519730072noreply@blogger.comtag:blogger.com,1999:blog-9004452969563620551.post-57051627740617369562015-01-19T08:28:32.457+05:302015-01-19T08:28:32.457+05:30Hello, Paramanand,
I was able to get through eve...Hello, Paramanand,<br /> I was able to get through everything on this page now, thanks. With the binominal multiplying it was not too bad once I got a pattern going, although I had to be very careful about sign mistakes when getting up around (1+x)^13.<br /><br /> It bothered me about the y=sqrt(2) not in the series convergence area. It took a long time to think of another way, but I think if you choose x=cos(theta) and y=1/sqrt(2) instead, you can get to the same place using:<br /> M(1,sqrt(2)) = M(sqrt(2),1) = M(1,1/sqrt(2)) * sqrt(2).<br /><br /> I was confused by the b*tan(theta) proof final conclusion, but when I read something in Borwein's "PI & the AGM" book the penny dropped. I didn't realize that the invariance under the AGM transformation proved this. However, it is obvious now, and probably was to you all along: <br /> M(a,b)= M( M(a,b),M(a,b) ) == q. <br /> Substituting q for a & b into the eqn with the "a^2*cos^2" gives the result desired. Such a wonderfully simple proof, or so it seems now.<br /><br /> Looks like it is going to take a bit to get through the 2nd kind ellipticals. The end result of the pi calc works pretty beautifully, it will take a while to understand the path to that reasonably.<br /><br />Best Regards, --Glenn Keller<br /><br /><br />Glenn Kellerhttp://www.blogger.com/profile/03353030102510662096noreply@blogger.comtag:blogger.com,1999:blog-9004452969563620551.post-89210317104843339092015-01-06T21:37:45.164+05:302015-01-06T21:37:45.164+05:30Hello, Paramanand,
Thanks for your help. I now...Hello, Paramanand,<br /> Thanks for your help. I now understand the t=b*tan(theta) substitution completely, and will follow that further later. Next I need to go back and understand the binomial theorem again. It has been quite a while, but I can do that piece on my own.<br /><br />Thanks again, --Glenn KellerGlenn Kellerhttp://www.blogger.com/profile/03353030102510662096noreply@blogger.comtag:blogger.com,1999:blog-9004452969563620551.post-86757545240152763922015-01-06T10:18:51.351+05:302015-01-06T10:18:51.351+05:30@Glenn Keller
Thanks for reading this post. The ca...@Glenn Keller<br />Thanks for reading this post. The calculation of $p_{1}, p_{2}, \ldots$ is difficult and Gauss did considerable algebraic manipulation to calculate $p_{n}$ in general. But finding first few coefficients is easy. For example the coefficient of $x$ on LHS is $0$. On RHS we have the coefficient of $x$ as $-1 + 4p_{1}$. This leads to $4p_{1} - 1 = 0$ so that $p_{1} = 1/4$.<br /><br />Again for $p_{2}$ we note that coefficient of $x^{2}$ on LHS is $p_{1}$. On RHS the coefficient of $x^{2}$ is $1 - 12p_{1} + 16p_{2}$ so we get $16p_{2} - 2 = 1/4$ or $p_{2} = 9/64$. Note that the calculation of coefficient of powers of $x$ on RHS requires the use of binomial theorem to handle $(1 + x)^{2n + 1}$ in the denominator.<br /><br />Regarding your second point about the substitution $t = b\tan \theta$, we can see that $dt = b\sec^{2}\theta\, d\theta$. And we have $$\begin{aligned}\frac{d\theta}{\sqrt{a^{2}\cos^{2}\theta + b^{2}\sin^{2}\theta}} &= \frac{\sec\theta\,\, d\theta}{\sqrt{a^{2} + b^{2}\tan^{2}\theta}}\\<br />&= \frac{b\sec^{2}\theta\,\, d\theta}{\sqrt{(a^{2} + b^{2}\tan^{2}\theta)(b^{2}\sec^{2}\theta)}}\\<br />&= \frac{b\sec^{2}\theta\,\, d\theta}{\sqrt{(a^{2} + b^{2}\tan^{2}\theta)(b^{2} + b^{2}\tan^{2}\theta)}}\\<br />&= \frac{dt}{\sqrt{(a^{2} + t^{2})(b^{2} + t^{2})}}\end{aligned}$$ In the above substitution the interval of integration changes to $[0, \infty)$ for $t$ corresponding to $[0, \pi/2]$ for $\theta$. Since the function of $t$ is even we can change the interval to $(-\infty, \infty)$ and add a factor of $1/2$.<br /><br />Regards,<br />ParamanandParamanand Singhhttp://www.blogger.com/profile/03855838138519730072noreply@blogger.comtag:blogger.com,1999:blog-9004452969563620551.post-60922152728291493792015-01-05T00:33:54.204+05:302015-01-05T00:33:54.204+05:30Paramanand,
Thank you for your clear comments o...Paramanand,<br /> Thank you for your clear comments on this subject, the best I've found after quite a lot of searching on the web. I am working to get a better understanding of the AGM<>elliptic integral connection, and am finding it hard going. I've gotten stuck at least 10 times on various dead ends, but with your blog I did make some progress. At the moment, however,, I'm stuck in 2 places. I don't see how you calculate the polynominal coefs in the 2nd part of gauss's 1st proof, and I can't figure out how the "t=b*sin(theta)" results in the transformed integral below it. I also tried gauss' original substitution and got completely lost on that one.<br /> I've ordered a couple of the books you recommended, maybe that will help. However, maybe I'm just missing some obvious (to you) techniques to get past the current stuck places.<br /><br />Thanks for your blog, --Glenn KellerGlenn Kellerhttp://www.blogger.com/profile/03353030102510662096noreply@blogger.comtag:blogger.com,1999:blog-9004452969563620551.post-29831580814352445172013-03-20T14:23:32.078+05:302013-03-20T14:23:32.078+05:30In the last stage of the first proof we establishe...In the last stage of the first proof we established that<br /><br />$\displaystyle K(x) = \frac{\pi}{2M(1 + x, 1 - x)}$<br /><br />Also comparing this with the identity established in stage 1 of first proof<br /><br />$\displaystyle \dfrac{1}{M(1 + x, 1 - x)} = \left(\dfrac{1}{1 + x}\right)\left(\dfrac{1}{M\left(1 + \dfrac{2\sqrt{x}}{1 + x}, 1 - \dfrac{2\sqrt{x}}{1 + x}\right)}\right)$<br /><br />we get<br /><br />$\displaystyle K(x) = \left(\dfrac{1}{1 + x}\right)K\left(\dfrac{2\sqrt{x}}{1 + x}\right)$<br /><br />This identity is called Landenâ€™s transformation and can be proved directly (with some heavy algebraical manipulation) using the substitution<br /><br />$\displaystyle x\sin\theta = \sin(2\phi - \theta)$<br /><br />in the integral defining $K(x)$<br /><br />$\displaystyle K(x) = \int_{0}^{\pi / 2}\frac{dx}{\sqrt{1 - x^{2}\sin^{2}\theta}}$paramanandhttp://www.blogger.com/profile/03855838138519730072noreply@blogger.com