Question

@dan815 factoring a number "fourier series coefficient" style

Answer 1

I thought we can only factor term by term with integration for fourier series?!

Answer 2

I was thinking what if we had a number like this \[\Large \log n = \sum _{i=1}^\infty a_i \log p_i\] Is there some kind of thing we can do to solve for the individual coefficients like \[\Large \int\limits \log p_i \log p_j =0 \text{ for i} \ne \text{j}\]

Answer 3

this sounds similar to the proof I had to do last semester.

Answer 4

very interesting

Answer 5

Yeah in a way this is kind of like inverting a matrix and multiplying both sides of the equation by it so that we get the identity matrix

Answer 6

first does this integral really hold true?

Answer 7

for i=/=j

Answer 8

Dan no, it doesn't even have bounds. I'm saying let's find out what it has to be. We might have to include some other things, or possibly even make it a summation instead of an integral.

Answer 9

\[i \neq j \]

Answer 10

Like for instance: \[\Large f(x) = \sum_{n=1}^\infty a_n \sin(nx)\]\[\Large \int\limits_0^{2\pi} \sin(nx) \sin(mx) \frac{dx}{\pi} = \delta_{nm}\] Do you see the resemblance to matrix multiplication here by its inverse?

Answer 11

im thinking maybe, lets try to determine a relationship with logs such that we have an orthonormal basis to work with

Answer 12

it doesn't have to be orthogonal or normal, we can always orthogonalize it with gram-schmidt and normalize it as well later. It just has to form a basis that spans the set fortunately. =D

Answer 13

i dont understand gram-schmidt process for functions

Answer 14

does it come from the descrete method of solving for the orthgonal vectors

Answer 15

Well just treat the integral as the dot product and just do the same process you would normally for gram schmidt, just like projections. I actually just basically made a guide for this the other day haha

Answer 16

okay i think i can work that out there, its just like doing an integral in place of the dot products right and doing multiple integrals

Answer 17

ill take a look at it after this