tag:blogger.com,1999:blog-9004452969563620551.post523454933720654905..comments2019-05-15T12:04:42.529+05:30Comments on Paramanand's Math Notes: π(PI) and the AGM: Evaluating Elliptic IntegralsUnknownnoreply@blogger.comBlogger13125tag:blogger.com,1999:blog-9004452969563620551.post-87432590634792884262015-01-26T15:52:06.900+05:302015-01-26T15:52:06.900+05:30@Glenn Keller, Your reasoning and Latex looks good...@Glenn Keller,<br />Your reasoning and Latex looks good. You should try to ask/answer questions at http://math.stackexchange.com It uses the same Latex and Mathjax syntax as my blog and hence you will have a lot of practice there. Plus it has the live preview of whatever latex you type. I also have a profile at stack exchange http://math.stackexchange.com/users/72031/paramanand-singh<br /><br />Regards,<br />ParamanandParamanand Singhhttps://www.blogger.com/profile/03855838138519730072noreply@blogger.comtag:blogger.com,1999:blog-9004452969563620551.post-76564605663050434632015-01-26T15:18:18.696+05:302015-01-26T15:18:18.696+05:30Hi, Thanks for fixing my mistakes with the Late...Hi,<br /> Thanks for fixing my mistakes with the LateX. I&#39;ll follow your suggestions. The web site you gave doesn&#39;t accept the same syntax for multiline eqns. I&#39;ll look for some others later.<br /> I&#39;ve been thinking about the $\theta$ to $\phi$ correspondence a bit more:<br /> $\phi = \frac{\arcsin(k\sin(\theta))+\theta)}{2}$<br /> At k near 0, $\phi$ slope is about $\theta / 2$ , slightly more for $\theta = 0 : \pi /2$, less for $\pi / 2 : \pi$.<br /> At k near 1, $\phi$ slope is about $\theta$ for $0 : \pi / 2$ and drops abruptly to near 0 for $\pi/2 : \pi$.<br /> The closer to $k = 1$, the closer the initial slope to 1 and the faster the drop to 0 above $\pi / 2$. Initially I couldn&#39;t understand the behavior near $k = 1$, but now it makes sense.<br /><br /> Sorry if there are any LateX errors, I haven&#39;t been able to check my syntax as the site has stopped working this morning.<br /><br />Best Regards, --gjk<br />Glenn Kellerhttps://www.blogger.com/profile/03353030102510662096noreply@blogger.comtag:blogger.com,1999:blog-9004452969563620551.post-31577775813533315692015-01-26T09:48:49.571+05:302015-01-26T09:48:49.571+05:30@Glenn Keller, I am glad that you fixed the issues...@Glenn Keller,<br />I am glad that you fixed the issues of relation between $\theta$ and $\phi$ yourself. Also the relation between $a_{n}, b_{n}$ and $a_{n + 1}, b_{n + 1}$ is also correct. You are on right track.<br /><br />Regards,<br />ParamanandParamanand Singhhttps://www.blogger.com/profile/03855838138519730072noreply@blogger.comtag:blogger.com,1999:blog-9004452969563620551.post-55829571588903259572015-01-26T09:46:54.085+05:302015-01-26T09:46:54.085+05:30This comment was posted by Glenn Keller and has be...This comment was posted by Glenn Keller and has been reposted after fixing latex errors.<br /><br />I think I figured it out. $\phi$ is NOT symmetrical:<br />The case for $\theta = 3\pi / 2$ is as follows: \begin{aligned}<br />k\sin(3\pi / 2) = \sin(2\phi - 3\pi / 2) \\<br />\arcsin(k / \sqrt(2)) = 2\phi - 3\pi / 2 \\<br />2\phi = \arcsin(k / \sqrt(2)) + 3\pi / 2 \end{aligned}<br />As $\theta$ approaches $\pi$ the $\sin, 1 / \sqrt(2)$ above, gets smaller and approaches 0. So does the $\arcsin()$. Meantime, the $3 \pi / 2$ approaches $pi$ , and :<br />$2 \phi = \pi, \phi = \pi / 2$<br />just like you said. Sometimes I have to work through this with very small steps.<br />Best Regards, --gjk<br /><br />Paramanand Singhhttps://www.blogger.com/profile/03855838138519730072noreply@blogger.comtag:blogger.com,1999:blog-9004452969563620551.post-63861630918828362602015-01-26T09:44:12.473+05:302015-01-26T09:44:12.473+05:30The following comment was posted posted by Glenn K...The following comment was posted posted by Glenn Keller and had some latex errors hence corrected and reposted.<br /><br />Hello again,<br />Thanks ! I got through all the algebra on that page but am still confused about the limits.<br />The substitution at $\pi$ worked ok,<br /><br />$k \sin(\pi) = \sin(2\phi - \pi), 0 = \sin(2\phi - \pi) = 2\phi - \pi, \phi = \pi / 2$<br /><br />but I got lost looking at $\pi/2$ &amp; the ratios &amp; ranges.<br /><br />I must be missing something. It probably will be obvious to you. See below:<br /><br />I look at $k = 1$ : \begin{aligned}<br />\sin(\theta) = \sin(2\phi - \theta) \\<br />\theta = 2\phi - \theta, 2\theta = 2\phi<br />\end{aligned}<br />This doesn&#39;t match your $\theta / 2 \leq \phi \leq \theta$ for $0 \leq \theta \leq \pi / 2$ . I&#39;m confused.<br /><br />I think the function has a mirror symmetry about $\theta = \pi / 2$. At this point:\begin{aligned}<br />k\sin(\pi / 2) = \sin(2\phi - \pi / 2) \\<br />\arcsin(k) = 2\phi - \pi / 2 \\<br />2\phi = \arcsin(k) + \pi / 2 \\<br />k = 0 , 1, \arcsin(k) = 0 , \pi / 2, \phi = \pi / 2 , \pi / 4 \end{aligned}<br />Due to the mirroring, I would expect $\phi$ to always be within this range. Therefore, depending on $k$, there could be some range of $0 , \pi / 2$ not covered by $\phi$ . Maybe I&#39;m thinking about some aspect of the mirroring incorrectly.<br /><br />Please let me know what I&#39;m missing.<br /><br />Thanks, --gjk<br />Paramanand Singhhttps://www.blogger.com/profile/03855838138519730072noreply@blogger.comtag:blogger.com,1999:blog-9004452969563620551.post-61154714143845261522015-01-26T09:32:59.752+05:302015-01-26T09:32:59.752+05:30@Glenn Keller, Good to know that you are learning ...@Glenn Keller,<br />Good to know that you are learning Latex. Your first comment looks OK (as you can see). However the syntax for handling multiline equations in MathJax latex is bit complicated. You should use $\$\\\begin\{aligned\} multiline equation with lines separated by double backslash \\end\{aligned\}\$\$$for the same. This will put your multiline equation in center.<br /><br />Since blogger does not allow editing of comments, I will have to delete your last two comments (they are difficult to read because of the errors, but don&#39;t worry you will learn all the tricks of latex soon) and fix the errors there and publish on your behalf.<br /><br />Regards,<br />ParamanandParamanand Singhhttps://www.blogger.com/profile/03855838138519730072noreply@blogger.comtag:blogger.com,1999:blog-9004452969563620551.post-28794303161773400662015-01-25T18:12:01.878+05:302015-01-25T18:12:01.878+05:30also, thanks, the latex preview link was very help...also, thanks, the latex preview link was very helpful.Glenn Kellerhttps://www.blogger.com/profile/03353030102510662096noreply@blogger.comtag:blogger.com,1999:blog-9004452969563620551.post-81362226987572365982015-01-25T18:09:48.631+05:302015-01-25T18:09:48.631+05:30Hi, Paramanand, Thanks again for your help gett...Hi, Paramanand,<br /> Thanks again for your help getting unstuck. It seems so obvious now. I had to pick the eqn with only \theta on the left and only \phi on the right. <br /><br /> I found it helped my understanding on the first set of equations that the goal was to calculate the series backwards :<br /> a_n, b_n, c_n = f(a_{n+1},b_{n+1},c_{n+1}) <br /><br /> with a reminder from previous posts:<br /> (a_n, b_n) = (a_{n+1} + \sqrt{a_{n+1}^{2} - b_{n+1}^{2}},a_{n+1} - \sqrt{a_{n+1}^{2} - b_{n+1}^{2}}) <br /><br /> Please check if I got this correct. I am very new with LateX.<br /><br />Best Regards, --Glenn Keller<br />Glenn Kellerhttps://www.blogger.com/profile/03353030102510662096noreply@blogger.comtag:blogger.com,1999:blog-9004452969563620551.post-21279140035522471002015-01-25T09:47:14.318+05:302015-01-25T09:47:14.318+05:30@Glenn Keller, First of all your latex stuff works...@Glenn Keller,<br />First of all your latex stuff works (although you have a minor syntax error in first latex code so it does not show up). Also it will not be shown in preview. There is an online latex editor which gives preview as you type. please have a look at http://www.codecogs.com/latex/eqneditor.php<br /><br />Now we come to the derivative \dfrac{d\theta}{d\phi}. We have the relation$$\tan \theta = \frac{\sin 2\phi}{k + \cos 2\phi}$$and we need to differentiate this with respect to \phi. Clearly this will require the use of chain rule (and quotient rule too) and we get$$\sec^{2}\theta\frac{d\theta}{d\phi} = \dfrac{(k + \cos 2\phi)\dfrac{d}{d\phi}(\sin 2\phi) - \sin 2\phi\dfrac{d}{d\phi}(k + \cos 2\phi)}{(k + \cos 2\phi)^{2}}$$After some simplification you can get the value of$d\theta/d\phi$.<br /><br />Regards,<br />ParamanandParamanand Singhhttps://www.blogger.com/profile/03855838138519730072noreply@blogger.comtag:blogger.com,1999:blog-9004452969563620551.post-16138308188036904222015-01-25T09:23:16.563+05:302015-01-25T09:23:16.563+05:30Hi, I&#39;m confused about how to get$ \frac{d...Hi,<br /> I&#39;m confused about how to get $\frac{d\theta}{d\phi} &amp;=$ from the equations at hand. Actually, I don&#39;t even know quite how to approach it with having function with $\theta$ and $\phi$ on one side of the equation and a different function with just $\phi$ on the other side. You don&#39;t have to work it out in detail, just if you have time give me a hint of where to go to learn this kind of thing. I&#39;m probably missing something fairly fundamental and hopefully pretty simple.<br /> We&#39;ll see if the LateX works. The preview doesn&#39;t show it, and I don&#39;t know if I&#39;m doing the syntax wrong or if it just doesn&#39;t work in the preview.<br />Thanks, --gjk<br />Thanks, --Glenn KellerGlenn Kellerhttps://www.blogger.com/profile/03353030102510662096noreply@blogger.comtag:blogger.com,1999:blog-9004452969563620551.post-2829674202068572792013-03-20T14:21:21.448+05:302013-03-20T14:21:21.448+05:30@paramanand i got it, thanks very much@paramanand<br /><br />i got it, thanks very muchalphanoreply@blogger.comtag:blogger.com,1999:blog-9004452969563620551.post-78237146355973900312013-03-20T14:20:29.848+05:302013-03-20T14:20:29.848+05:30Hi alpha, The relation between $\phi$ and $\theta$...Hi alpha,<br />The relation between $\phi$ and $\theta$ is given by $\sin(2\phi - \theta) = k\sin\theta$ and on further analysis of this relation it follows that $\phi$ increases with $\theta$ (you can check that $d\theta / d\phi &gt; 0$) and also $\theta / 2 \leq \phi \leq \theta$ for $0 \leq \theta \leq \pi / 2$ and $\theta / 2 \leq \phi \leq \pi / 2$ for $\pi / 2 \leq \theta \leq \pi$. In that case if we put $\theta = \pi$ the suitable value of $\phi$ turns out to be $\pi / 2$.paramanandhttps://www.blogger.com/profile/03855838138519730072noreply@blogger.comtag:blogger.com,1999:blog-9004452969563620551.post-60216277610439988422013-03-20T14:18:52.827+05:302013-03-20T14:18:52.827+05:30hi : it’s very good of your blog , i have a questi...hi :<br />it’s very good of your blog , i have a question about your prove of the landen’s transformation at the equation after “Now with all the substitution formulas in place we have” .<br />i want to know why the upper bound of the integral by Φ is π/2 (half of the upper bound of integral by θ)alphanoreply@blogger.com