Connection between Laplace Transforms and Pythagorean triples I was studying for a recent university exam when I realized that there appears to be a connection between the Laplace transforms of certain functions and Pythagorean triples.
Mainly the Laplace of $t\sin(wt) = \frac{2sw}{(s^2 + w^2)^2}$ and the Laplace of $t\cos(wt) = \frac{s^2 - w^2}{(s^2+w^2)^2}$.
These numbers $(2sw, s^2-w^2, s^2+w^2)$ match the definition of a Pythagorean triple and I was wondering if anyone could explain why. I've spent a little bit of time working on it and couldn't figure it out.
Just a question for curiosity, not that deep. Thank you in advance.
 A: Consider $f(t)=\cos(wt)+i\sin(wt)=e^{iwt}$. Then the Laplace transform of $f(t)$ is $$F(s)=\frac 1{s-iw},$$ defined for $s$ with $R(s)>0$. Using the properties of Laplace transform , the Laplace transform for $tf(t)$ will be given by $$-F’(s)=\frac 1{(s-iw)^2}=\frac{(s+iw)^2}{(s^2+w^2)^2}=\frac{(s^2-w^2)+i(2sw)}{(s^2+w^2)^2},$$ so the phenomenon you observed follows from the multiplication of complex numbers.
A: When we go from a real variable to a complex variable, we are dealing with two dimensions instead of just the real-number line. It is natural that locations, vectors, and whatever be represented as $\space x,y\space$ coordinates with respect to the origin.
Think of the odd side
$\space A \space$ of a right triangle as being horizontal, the even side $\space B \space$ being vertical, and the hypotenuse
$\space C \space$ as running diagonally from the origin to some $\space x,y\space$ coordinate. Your version of Euclid's formula has the odd/even sides as the opposite of this image.
If we show Euclid's formula here as
$$A=m^2-k^2\qquad B=2mk\qquad C=m^2+k^2$$
We can see that
$\space\sin =\dfrac{2mk}{m^2+n^2}
\space\text{ and }\space
\cos=\dfrac{m^2-k^2}{m^2+k^2}.$
I do not know why the denominator sum
$\space m^2+k^2\space$ is squared in your question. I do not know that much about Laplace transforms.
