Question

I've been thinking about this the past few days but haven't been able to focus on it. I'm trying to find the similarity here...

Answer 1

\[\Large \Psi(x,y) = f(x,y)+g(x)+h(y)+k \\ \Large \Psi_x = f_x+g_x \\ \Large \Psi_y = f_y+h_y \\ \Large \Psi_{xy} = f_{xy}\] So you could imagine being given the exact equation \[\Large (f_x+g_x)dx + (f_y+h_y)dy = 0\] and solving it from there. So what I'm trying to compare this with is \[\Large t=a+bn+cm+dnm \\ \Large t \equiv a+cm \mod n \\ \Large t \equiv a+bn \mod m \\ \Large t \equiv (a \mod n) \mod m \equiv (a \mod m) \mod n\] Now we can imagine we're given the two statements of t with mod n and m and that they satisfy the final equation I've given to make it "exact" it seems we can solve it the same kind of way. But I just want to figure this out better.

Answer 2

http://openstudy.com/study#/updates/55045ce4e4b088d8e035d86b

Answer 3

Psi is basically just a predefined general function I made up. So like psi has a sum of a bunch of terms and all the terms that depend on only x or y alone are the g and h terms, so for example if I have \[\Large \Psi (x,y) = 1+y^7+x^3 \sin x +ye^{x^2}-y+\pi^9\] then we have these terms \[\Large f(x,y) = ye^{x^2} \\ \Large g(x) = x^3 \sin x \\ \Large h(x) = y^7 - y \\ \Large k = 1+\pi^9\] So you can see how every function can be broken up into these types of terms, which is kind of similar to how modular arithmetic breaks up numbers by divisibility which is kinda cool! =D

Answer 4

By similarity you're comparing
\[t\sim \Psi(x,y)\]
\[a\sim f(x,y)\]
\[bn\sim g(x)\]
\[cm\sim h(y)\]
\[dmn\sim k\]

?

Answer 5

that congruence in the end looks interesting, let me think a bit more..

Answer 6

Yes, exactly just like that. =) In my mind I was trying to extend this sort of thing to using an integrating factor to solve modular arithmetic problems and try to sort of mix and match different strategies from solving differential equations and modular arithmetic problems to help build a stronger understanding using ideas we already have.

I think what I'd like is some kind of modular arithmetic equivalent of \[\Large \bar \nabla \times ( \bar \nabla f) = \bar 0\] Which is essentially the more advanced version of exact equations.

Answer 7

\[ \Large t \equiv (\color{red}{a} \mod n) \mod m \equiv (\color{red}{a} \mod m) \mod n\]

it should be \(t\) right ?

Answer 8

\[ \Large t \equiv (\color{red}{t} \mod n) \mod m \equiv (\color{red}{t} \mod m) \mod n\]

dont you mean something like that ?

Answer 9

At this point I think these are all the same statements

Answer 10

Maybe this is what I mean? I'm not sure, I'm not very good with manipulating mods to be honest lol \[ \Large \color{red}{a} \equiv (\color{red}{t} \mod n) \mod m \equiv (\color{red}{t} \mod m) \mod n\]

Answer 11

that looks better

Answer 12

doing \(t \mod n\) gives us \(a+cm\)
after that taking \(\mod m\) we will be left with \(a\)


Okay i kinda see what you're doing now xD

Answer 13

As long as you _see_ the idea I'm thinking about that's what's important since I don't really want to get caught up in too many technical details cause I don't really know so much lol.

Answer 14

Although this is unrelated, to give an idea of something I've been considering as well is: \[\Large m^ \epsilon \equiv c \mod n\] this formula is represented outside of modular arithmetic as \[\Large e^{i \frac{2\pi}{n} m^ \epsilon}=e^{i \frac{2\pi}{n} c}\] So now I can sort of look at the idea from two separate angles. So I kind of want a similar thing here as well if that makes sense... in a weird kind of way... lol