tag:blogger.com,1999:blog-9004452969563620551.post2440215970708422854..comments2017-06-21T11:55:27.964+05:30Comments on Paramanand's Math Notes: A Continued Fraction for Error Function by Ramanujantesternoreply@blogger.comBlogger4125tag:blogger.com,1999:blog-9004452969563620551.post-47599699531067802292016-06-15T15:44:54.942+05:302016-06-15T15:44:54.942+05:30@excentrique,
From equation $(21)$ you will get a...@excentrique,<br /><br />From equation $(21)$ you will get an equation similar to $(12)$ namely $$Q_{n + 1}(x)P_{n}(x) - P_{n + 1}(x)Q_{n}(x) = (-2)^{n}n!$$ and there will some minor tweaks in the equations which follow accordingly. But end result will be as mentioned after equation $(21)$. You should try out these calculations and let me know if you face any problem.<br /><br />Regards,<br />ParamanandParamanand Singhhttp://www.blogger.com/profile/03855838138519730072noreply@blogger.comtag:blogger.com,1999:blog-9004452969563620551.post-49302894039762014202016-06-14T19:52:39.406+05:302016-06-14T19:52:39.406+05:30Hi Paramanand,
thanks for the explanation.
I work...Hi Paramanand,<br /><br />thanks for the explanation.<br />I worked my way through the proof and I'm almost there at the end where I cannot quite reproduce the proof for psi(x), the recurrence relations of (21) don't lead to the conclusion (12), so how can I proceed from these relations to a similar proof of phi(x). Is there a shorter way to explain oneself that the continued fraction in (3) is equivalent to the new one at the end?excentriquehttp://www.blogger.com/profile/15772048701347307294noreply@blogger.comtag:blogger.com,1999:blog-9004452969563620551.post-59416843617193176842016-06-14T00:01:28.378+05:302016-06-14T00:01:28.378+05:30@excentrique,
You are right and thanks for catchin...@excentrique,<br />You are right and thanks for catching the typo. I will fix it. Equation $(16)$ comes from $(15)$ and we are just using the fact that $P_{n}(x)$ is greater than one particular term in the series representation $(15)$ of $P_{n}(x)$ and this we do based on parity of $n$. This is possible because each term in series for $P_{2n}(x)$ and for $P_{2n + 1}(x)/x$ is positive.<br /><br />Regards,<br />ParamanandParamanand Singhhttp://www.blogger.com/profile/03855838138519730072noreply@blogger.comtag:blogger.com,1999:blog-9004452969563620551.post-10888235258521309162016-06-13T19:44:12.047+05:302016-06-13T19:44:12.047+05:30The substitution by u^2 = 2t^2 is wrong, it has to...The substitution by u^2 = 2t^2 is wrong, it has to be u^2 = 2t.<br /><br />My main question is though: I don't understand the conclusions in (16), what do you mean by these?excentriquehttp://www.blogger.com/profile/15772048701347307294noreply@blogger.com