I am reading the following paper:
http://www-biba.inrialpes.fr/Jaynes/cc07s.pdf
and cannot seem to figure out how Jaynes is deriving (P2) and below (specifically the log arithmetic log[f(x)/f(0)]....)
Can someone help me out?
$\begingroup$
$\endgroup$
1
-
$\begingroup$ Can you put some details in your question please, firstly so readers of your question are not required to go read a paper to understand the question, and secondly, to protect your question from the possibility that the link will disappear (at which point it becomes not so much a useful resource for others as junk cluttering up the site) $\endgroup$– Glen_bCommented Sep 1, 2014 at 23:01
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
1
Equation $$ f(x)f(y)=g(\sqrt{x^2+y^2}) \qquad (*) $$ holds for every $x,y$. If $y=0$, then $(*)$ gives $g(|x|)=f(x)f(0)$ for every $x$. Using this in $(*)$ to "eliminate" $g$, we have $f(x)f(y)=f(\sqrt{x^2+y^2})f(0)$. Dividing by $(f(0))^2$ and taking the log on both sides we find $$ \frac{f(x)f(y)}{f(0)f(0)}=\frac{f\left(\sqrt{x^2+y^2}\right)}{f(0)} \, , $$ $$ \log\left(\frac{f(x)}{f(0)}\right)+\log\left(\frac{f(y)}{f(0)}\right)= \log\left( \frac{f\left(\sqrt{x^2+y^2}\right)}{f(0)}\right) \, . $$
-
$\begingroup$ In this video, it's explained step-by-step in a very detalied manner: youtube.com/watch?v=cTyPuZ9-JZ0 $\endgroup$ Commented May 5, 2019 at 13:01