tag:blogger.com,1999:blog-9004452969563620551.post1453607918978394189..comments2019-05-15T12:04:42.529+05:30Comments on Paramanand's Math Notes: π(PI) and the AGM: Legendre's IdentityUnknownnoreply@blogger.comBlogger6125tag:blogger.com,1999:blog-9004452969563620551.post-55822890345191635312015-01-25T08:57:53.688+05:302015-01-25T08:57:53.688+05:30Ok, thanks for the hints. I&#39;ll try it on next...Ok, thanks for the hints. I&#39;ll try it on next question on following page.Glenn Kellerhttps://www.blogger.com/profile/03353030102510662096noreply@blogger.comtag:blogger.com,1999:blog-9004452969563620551.post-11940465925411300442015-01-24T10:49:52.231+05:302015-01-24T10:49:52.231+05:30@Glenn Keller, You can learn some latex by the equ...@Glenn Keller,<br />You can learn some latex by the equations already given in my post. Just do a right click any of the equations here and you will get a MathJax menu. Then in this menu use the &quot;Show Math As -&gt; Tex Commands&quot; option and it will give you the Latex code for that equation (without the dollar symbols, you need to add those at beginning and end yourself because those are part of the syntax to start Latex processing by MathJax). You can get more information by visiting MathJax web page (www.mathjax.org)<br /><br />Regards,<br />ParamanandParamanand Singhhttps://www.blogger.com/profile/03855838138519730072noreply@blogger.comtag:blogger.com,1999:blog-9004452969563620551.post-22035225179006312172015-01-24T10:44:35.616+05:302015-01-24T10:44:35.616+05:30@Glenn Keller, Glad that you found the problem and...@Glenn Keller,<br />Glad that you found the problem and fixed it yourself. BTW in case you want the mathematical equations to appear properly in the comments you need to learn a little bit of Latex. There are many online tutorials for the same. Also this blog uses MathJax software to handle Latex. For that you need to write all the mathematical latex code between two dollar symbols. For example if you write $\$$\dfrac{1}{2}\$$ then it will be seen as$\dfrac{1}{2}$and if you write$\$$\int_{0}^{1}f(x)\,dx\$$ then it will be seen as $\int_{0}^{1}f(x)\,dx$<br /><br />Regards,<br />ParamanandParamanand Singhhttps://www.blogger.com/profile/03855838138519730072noreply@blogger.comtag:blogger.com,1999:blog-9004452969563620551.post-67574696234775714462015-01-23T23:30:22.841+05:302015-01-23T23:30:22.841+05:30Sorry, Paramanand, I didn&#39;t explain my problem... Sorry, Paramanand, I didn&#39;t explain my problem correctly. I apologize for wasting your time. I had trouble deriving the initial derivative, the one you label with &quot;we have the equation&quot;. After making the same mistake 5 times, I eventually I found my problem by working backwards from the end.<br /><br /> After that, the rest of the page went pretty smoothly.<br /><br /> On another topic, do you have the syntax of whatever it is I have to type to make the equations come out nicely instead of plain text ?<br /><br /> Thanks for your time, --gjk Glenn Kellerhttps://www.blogger.com/profile/03353030102510662096noreply@blogger.comtag:blogger.com,1999:blog-9004452969563620551.post-42980707007389276032015-01-21T09:53:39.181+05:302015-01-21T09:53:39.181+05:30@Glenn Keller We have the equation $$\frac{d}{d\t...@Glenn Keller<br /><br />We have the equation$$\frac{d}{d\theta}\left(\frac{\sin\theta\cos\theta}{\sqrt{1 - m\sin^{2}\theta}}\right) = \frac{1}{m}\sqrt{1 - m\sin^{2}\theta} - \frac{m^{\prime}}{m}\frac{1}{(1 - m\sin^{2}\theta)^{3 / 2}}$$and integrating this we get$$\frac{\sin\theta\cos\theta}{\sqrt{1 - <br />m\sin^{2}\theta}} = \frac{1}{m}\int\sqrt{1 - m\sin^{2}\theta}\,d\theta - <br />\frac{m^{\prime}}{m}\int\frac{d\theta}{(1 - m\sin^{2}\theta)^{3 / 2}}$$Using limits 0, \pi/2 in the above integration we get the LHS as 0 (because of \sin\theta\cos\theta) and thus we get$$\frac{m^{\prime}}{m}\int_{0}^{\pi/2}\frac{d\theta}{(1 - m\sin^{2}\theta)^{3 / 2}} = \frac{1}{m}\int_{0}^{\pi/2}\sqrt{1 - m\sin^{2}\theta}\,d\theta = \frac{E}{m} This integral is then used in calculation of $\dfrac{d}{dm}(E - K)$. I hope this helps.<br /><br />Regards,<br />ParamanandParamanand Singhhttps://www.blogger.com/profile/03855838138519730072noreply@blogger.comtag:blogger.com,1999:blog-9004452969563620551.post-90525799896164768292015-01-20T21:27:20.978+05:302015-01-20T21:27:20.978+05:30Hi, I&#39;ve having trouble transforming the der...Hi,<br /> I&#39;ve having trouble transforming the derivative with sin * cos on top to the form you show. Can you help ? Thanks, --gjkGlenn Kellerhttps://www.blogger.com/profile/03353030102510662096noreply@blogger.com