i needz da math help...

abs(x)^n+abs(y)^n=k^n
solve for n in terms of k, x, and y.

i think i'm going to cry. i really want a solution for that but i'm too stupid. anyone willing to help me out with it?

You mihgt just wanna use newton/raphson or whittaker iteration if n is dynamic.

n=log(x^n+y^n)/log(k) is as far as I could figure to go, sorry. I'm no math whiz though.

As far as I know, you're not gonna solve that. You can approximate it though.

Like they said - I'm pretty sure that one isn't solvable - check on matlab or mathematica or maple or something, but I don't think it is solvable. You will have to do numeric approximations.

I tried as well. Anyone else thinks is looks a lot like Fermat's Last Theorem. I got Shock Value's formula and also this one:

n = - log ( k^-n + abs(y)^n * abs(x)^-n * k^-n ) / log (x)

Solving for x, y and k is easy though, but I guess you don't need that. I think you need to factor abs(x)^n + abs(y)^n, but I don't know how to do it. I did find a (crappy) method for splitting k^n - abs(y)^n in different parts, but that doesn't help at all. Maybe anyone here finds it interesting.

k^n - abs(y)^n = ( k^(n-1) - abs(y)^(n-1) )*( k + abs(y) ) - ( k^(n-2) - abs(y)^(n-2) ) * k * abs(y)

I would be very interested in hearing the solution. Maybe you could post it on a math forum.

fermat's theorem is only for integers isn't it................

/me is stupid and ignorant and stupid

and yes this is for a preset. i tried it on my ti-89 but it didn't work - but then again there's a lot of things that the 89 is bad at, like working right...anyone here have mathematica?

Yes, but it has a very similar form. x^n + y^n = z^n

Where did you find this problem?

Maybe filling us in on what you plan to do with this equation might help us supply ways of solving the equation more suitably?

Punch it in a cartesian formula simulator thingamajig, you'll find out.

*hinthint* superellipse

Nic01 is right. Take a look at this article.

The formula can be rewritten to:

(abs(x)/k)^n + (abs(y)/k)^n -1 = 0

Maybe this helps.

Well that's assuming z equals one. Like I said, if he tells us what he plans on doing with it then maybe we can help a little better.

No,

abs(x)^n + abs(y)^n = k^n
abs(x)^n/k^n + abs(y)^n/k^n = k^n/k^n
(abs(x)/k)^n + (abs(y)/k)^n = 1
(abs(x)/k)^n + (abs(y)/k)^n - 1 = 0

It works except for k=0.

OK, I asked my math teacher and he didn't know a method to solve it. He also said that he didn't know if it was impossible. He did help make the formula easier.

abs(x)^n + abs(y)^n = k^n
abs(x)^n/k^n + abs(y)^n/k^n = k^n/k^n
(abs(x)/k)^n + (abs(y)/k)^n = 1

this can be written as the seemingly easier calculation:
a^n + b^n = 1
where a = abs(x)/k and b = abs(y)/k

I've attached a plot of the above equation on my graphing calc:
the x-axis is a from 0 to 1, with a scale of 0.1
the y-axis is n from 0 to 5, with a scale of 0.5
the bottom line: b=0.25
the middle line: b=0.5
the top line: b=0.75

After another long talk with my math teacher, we have concluded this it's probably unsolvable. This theory is supported by this Ask Dr. Math article. I haven't found a real proof that it is impossible to solve, but I think it is.

*cough* approximation *cough*

Depending on what he wants to do he could use newton/raphson or whittaker or just good old Euler's method.

This is a very hard problem... the only way i can think of solving it is numerically by using either n-r/whittaker. You can also solve it by using log expansions but this is also an iterative process and is difficult and more expensive in terms of divisions.

well if he's just doing some 2D stuff he could use Euler's method as well to inch along.

BTW what i can tell you is that this is 100% insoluble in a finite fashion due to the use of real powers. The definition of a generic non integer power can be given only in terms of an infinite series... whether its some clever sum of sines and cosines or something based on e or calculus, also the instant you use log you are playing with an iterative solutions whether you realise it or not.

eeps. sorry guys, i've been busy the past few days and i've forgotten to check this thread...

anyway, basically what i'm looking for is, given a point (x,y), find the superellipse of power n:
|x|^n+|y|^n=r
which would intersect that point.

and i'm thinking the more that i look at it that i shouldn't have had k^n at all in the first place, so it should probably be rewritten as:
|x|^n+|y|^n=1
i doubt that'd make it possible to solve, but i guess it's worth a try....

thanks guys 🙂

I think atomic calculator can solve it.
My math is very dangerous.

An atomic calculator in avs, right.....

this thread was let go for a reason...

atomic calculators.. now this rings a bell.. wait..

ahh.. thats right, that was the thread in which a certain user got banned.

don't i have a good memory.

Hahahahaha.

laughs out very very very loud.... 😁

you said banned?

haha pwnt

pity tho 🙁

Originally posted by skupers

I think you need to factor abs(x)^n + abs(y)^n, but I don't know how to do it.

You can find a factor with the Striling-formula:

n! = sqrt(2*pi*n)*(n^n)*(e^(-n))

Don't know if this helps to find a solution but I thought to post it.

The original thread was posted 2 months ago! If you look at the dates on posts this thread has been revived twice... Just let this die already.