dirkdeftly

29th February 2004 07:17 UTC

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?

UIUC85

29th February 2004 08:50 UTC

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

Shock Value

29th February 2004 17:05 UTC

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

UIUC85

29th February 2004 18:56 UTC

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

anubis2003

29th February 2004 21:10 UTC

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.

skupers

1st March 2004 08:53 UTC

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)

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)

dirkdeftly

2nd March 2004 04:47 UTC

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?

skupers

2nd March 2004 07:03 UTC

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

Where did you find this problem?

UIUC85

2nd March 2004 07:35 UTC

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

Nic01

2nd March 2004 14:37 UTC

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

*hinthint* superellipse

skupers

2nd March 2004 17:01 UTC

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.

UIUC85

2nd March 2004 17:17 UTC

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.

http://forums.winamp.com/showthread....hreadid=164315
http://forums.winamp.com/showthread....hreadid=164753

skupers

3rd March 2004 06:28 UTC

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.

skupers

4th March 2004 08:17 UTC

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

skupers

5th March 2004 18:35 UTC

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.

UIUC85

5th March 2004 19:47 UTC

*cough* approximation *cough*

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

jheriko

9th March 2004 00:40 UTC

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.

UIUC85

9th March 2004 00:48 UTC

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

jheriko

10th March 2004 02:04 UTC

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.

dirkdeftly

11th March 2004 23:43 UTC

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 :)

D12

10th April 2004 07:17 UTC

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

Synth-C

10th April 2004 12:58 UTC

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

UIUC85

10th April 2004 20:09 UTC

this thread was let go for a reason...

sidd

10th April 2004 21:08 UTC

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.

Raz

11th April 2004 03:48 UTC

Hahahahaha.

Warrior of the Light

11th April 2004 15:25 UTC

laughs out very very very loud.... :D

you said banned?

sidd

11th April 2004 16:33 UTC

haha pwnt

pity tho :(

hboy

28th April 2004 10:39 UTC

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.

UIUC85

28th April 2004 16:51 UTC

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.