x^n + y^n = 1

random_fruit · 2008-12-25 10:51:33

So the formula above with n=2 generates a circle on the x,y plane, and can be demonstrated by use of Pythagoras' theorm. With n=1 it makes a straight line. But what happens when n=3 or n=4?

I used Excel to draw a graph of them. (I re-arranged the formula to be y = (1-x^n)^(1/n), then set x to values from 0.00 to 1.00 by 0.01 in column A, and computed y in column B, then asked the graph wizard to draw the B column.)

It's quite clear as n increases the line produced moves closer to the box formed by the lines x=1 and y=1.

Does anyone know of the way to differentiate and integrate the functions? Are there functions like sine and cosine which exist for the case n=2?

Where n=2 we get square roots which cause the arc to reflect about the x and y axes to make a full circle. Where n=3 does this make sense, or do we get lots of possible answers?

When I had (as a boy) a toy railway set (i.e., Hornby now in the UK) I was told that the sudden entry to a bend from straight to a fixed radius was not used on real, full-size railways (and probably is uncommon on UK highways now I think about it) so do any of the curves I generate have uses in road building?

When Christmas is over I will have less time, but for now the mind is rested and can spare some effort for this sort of amusement. javascript:insert_text('', '');

Your educative comments welcome. And please be gentle, this is my first real post here.

Kurre · 2008-12-26 09:03:24

We could maybe define a new type of sine function, s(A), as the y-component of a point (x,y) satisfying x^n+y^n=1, and same with the new Cosine function, c(A), as the x-component (A being the orinary angle in radians, altough this doesnt have the same "geometric intrepetation" in our "new unit circle"). With n=2 we get the ordinary Sine and Cosine. Now consider a point p= (x,y)=(c(A),s(A)) satisfying x^n+y^n=1. we have that x=CotA*y, which yields:
(y*CotA)^n+y^n=1
s^n(A)=y=1/(Cot^nA+1)
and similary we get:
c^n(A)=1/(Tan^nA+1)
altough im not sure how to define the "unit circle", since if n is odd, we will not have any symetry and particulary no solutions for x,y<0. Maybe we should consider the equation |x|^n+|y|^n=1 instead, maybe makes more sense.
Edit: after plotting in mathematica, I see that we obviously must consider |x|^n+|y|^n=1, otherwise we get some ridiculous function. This would give:

Last edited by Kurre (2008-12-26 09:13:58)

random_fruit · 2008-12-29 08:52:47

By the use of chain rule dy/dx = dy/dh . dh/dx and also by writing
y = sqrt (1 - x^2)
I got
dy/dx = (-x) / sqrt(1-x^2)
which satisfied the trig rules, ie I had found an equivalent of -cot(alpha).

By the chain rule on x^4+y^4=1 I had
y=sqrt(sqrt(1 - x^4))
which differentiated to
dy/dx = (-4 x^3) / sqrt(sqrt((1 - x^4) ^3))
and seems to give a gradient on the tanget on my lines when I plot on Excel.

Looking at wikipedia (see http://en.wikipedia.org/wiki/Trig_functions) I see what remember,
sin x = x - x^3/3! + x^5/5! - ...
and I have wondered if there can be a similar infinite series for the X^4 form I've considered.

I looked at the Co-tangent observations made by Kurre (above) but couldn't follow his (or, her) step from
(y*CotA)^n+y^n=1
to
s^n(A)=y=1/(Cot^nA+1)
So thanks for the reply but sorry I didn't "get" it.

bossk171 · 2008-12-29 14:41:56

I've spent at least an hour working on this and I've accomplished: nothing.

I wanted to isolate n so that some function N could be written in terms of x and why and then set N(x,y) = z and graph N(x,y) in three dimensions. If it's possible to isolate n, then it's well beyond my abilities (which isn't saying much). But I'm very much interested in where this might lead.

random_fruit, look into a freeware program called graphcalc (http://www.graphcalc.com/) you can graph functions pretty easily with it.

Might this be better suited in Euler's Avenue or Cafe Infinity?

Last edited by bossk171 (2008-12-29 14:44:30)

Kurre · 2008-12-30 02:45:43

random_fruit wrote:

I looked at the Co-tangent observations made by Kurre (above) but couldn't follow his (or, her) step from
(y*CotA)^n+y^n=1
to
s^n(A)=y=1/(Cot^nA+1)
So thanks for the reply but sorry I didn't "get" it.

First, we must define the new Sine and Cosine. The definition for the ordinary sin(x) and cos(x) are defined (the geometric definitions, some books start by defining them as their infinite series) as:
Cos(a) is the x coordinate of a point (x,y) on the unit circle, where a is the angle from the x axis in the positive direction. See following link for a better explanation:
http://en.wikipedia.org/wiki/Sine#Unit-circle_definitions

So I thought we could define the new "Cosine" and "Sine" in a similar way. Let c(a) be the new Cosine function, and s(a) be the new Sine function, which will be the x respectively y coordinate for a point (x,y)=(c(a),s(a)) lying on our "new unit circle", defined by |x|^n+|y|^n=1, and a is the same angle as in the definition above. If n=2, we get the functions Cos(a) and Sin(a), as expected. Now we want a good expression for this, and my first thought was to try to describe it with the ordinary trigonometric functions. If we draw a right angle triangle, where the hypotenuse is from the origin to a point on the "unit circle". The attached picture shows an example for n=3. Now x and y are related as |x|^n+|y|^n=1, and y/x=tan(a). Solving this system of equations for x=c(a) in the following way yields:
y=tan(a)*x.
substituting into |x|^n+|y|^n=1:
|x|^n+|tan(a)*x|^n=1
|x|^n+|tan(a)|^n*|x|^n=1
|x|^n*(1+|tan(a)|^n)=1
x=c(a)=±1/(1+|tan(a)|^n)^(1/n)
where the ± sign is determined from the quadrant the point is in. First and fourth quadrant yields +, second and third yields -. Similar for y=s(a).

Hopefully this made my post clearer

Last edited by Kurre (2008-12-30 02:58:44)

random_fruit · 2008-12-30 08:36:20

My thanks to Bossk171, for pointing out the program graphcalc. This is new to me, but I love it and have started using it as well as MS Excel to draw graphs of the functions.

I agree, having seen some more of the site now, that my post might have been better in Euler's Avenue or Cafe Infinity. However, I'm still a novice (joined Christmas day 2008) so I trust I'm excused. If a moderator wishes to move all this to another part of the site, fine by me.

My thanks too to Kurre for his formula c(a) = cos(a) / ((abs(cos(a))^n + abs(sin(a))^n) ^ (1/n)). I found the explanation easier this time. It's cool, that s(a) uses the same formula but has sin(a) above the line, which I guessed for myself. I could flatter myself and call it maths intuition. Cool also that c(a) for n=4 can be expressed in terms of sin and cos re-using n=2.

Between the two of you I now have the formulae above in graphcalc drawing the lines by parametric and also by Euclidian versions - and it pleases me that they look just the same (at least above the x-axis)

random_fruit · 2008-12-30 09:10:37

Back to my original post: "Does anyone know of the way to differentiate and integrate the functions?" I can begin to differentiate the function given by Kurre using product, quotient and chain rules, but I will have to nest them and I will need a BIG piece of paper to deal with all the 'recursion' or 'nesting' of pieces in each other.

I will start with quotient rule: [u/v]' = (u'v - uv') / v^2
where u = cos(a) and v = (cos^n(a) + sin^n(a))^(1/4)
(I'm aware of ditching the unsigned or absolute, but I'm doing that for simplicity, and will limit my results to the a from 0 to pi/2)

If anyone gets there with a differential then I'd be interested to see it. I suspect that a few trig simplifications which I don't know may result in a much tidier formula.

Go on, you know you want to...

random_fruit · 2009-01-01 09:50:13

I get a real-headache result, which I may try to LaTeX for you all.

I was pleased when I used Excel to differentiate the c(a) taking Δx as 0.01 radians, then plotting the graph. The graph of d/dx(c(a)) from excel was excitingly the same shape as I got when I put my headache into graphcalc.

bossk171 · 2009-01-01 10:21:00

Are you trying to differentiate:

With respect to x with n as a constant? If so it's something like:

Which can be confirmed with this site: http://calc101.com/webMathematica/derivatives.jsp

(I did it long hand before cheating though)

The integral is way over my head and it involves a function I've never heard of called the hypergeometric function: http://integrals.wolfram.com/index.jsp? … ndom=false

If that's not what you were looking for, sorry. I really hope that someone much smarter than I (which is most people here) might get involved and work some calculus magic.

Also it would be great if a mod moved this to Euler Avenue.

random_fruit · 2009-01-01 10:49:11

OK, here's the "headache" formula I wrote about in the last post...

where, as an abbreviation

The GraphCalc program shows me that

which I don't think I could have found for myself, and makes the expression above the line a lot easier!

Math Is Fun Forum

#1 2008-12-25 10:51:33

x^n + y^n = 1

#2 2008-12-26 09:03:24

Re: x^n + y^n = 1

#3 2008-12-29 08:52:47

Re: x^n + y^n = 1

#4 2008-12-29 14:41:56

Re: x^n + y^n = 1

#5 2008-12-30 02:45:43

Re: x^n + y^n = 1

#6 2008-12-30 08:36:20

Re: x^n + y^n = 1

#7 2008-12-30 09:10:37

Re: x^n + y^n = 1

#8 2009-01-01 09:50:13

Re: x^n + y^n = 1

#9 2009-01-01 10:21:00

Re: x^n + y^n = 1

#10 2009-01-01 10:49:11

Re: x^n + y^n = 1

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