@robjohn I like this form much more
$$\lim_{n\to\infty} \frac{1}{n^2} \int_0^{\pi/2} \left(\sum_{k=-n}^{n} e^{\large 2k x i}\right)^3 \psi^{(-2)} (x+1) \ dx=\frac{3\pi}{4} \log(2\pi)$$
$$\lim_{n\to\infty} \frac{1}{n^2} \int_0^{\pi/2} \left(\sum_{k=-n}^{n} e^{\large 2k x i}\right)^3 \psi^{(-2)} (x+1) \ dx=\frac{3\pi}{4} \log(2\pi)$$
In polar coordinates $(r, \theta)$, the equation $$r = \sin\left(a \theta\right)$$ gives a rose with $a$ petals if $a$ is odd, or $2a$ petals if $a$ is even.
Thus, the number of petals generated for some values of $a$ are
a | petals
=======+========
1 | 1
2 | 4
3 | 3
...
In every field there's always that person who's just years ahead of their time. For instance, Paul Morphy (born 1837) is said to have retired from chess because he found no one to match his technique that very much resembles modern chess theory.
So, who were the Paul Morphys of mathematics?
I am having a problem with the derivation of the prolongation formula from PJ Olver's text :"Applications of Lie groups to differential equations" Page 105,106.
Considering a differential equation with independent variable(x) and one dependent variable(u).
(x,u) $\subset$ $X \times U$
The fir...
