Definition of a special sequence of curves?

Question

As n->infinity, the plot of the curve x^n+y^n=1 approaches a square. Does anyone know of a sequence of curves (preferably with a simple definition) which approaches an equilateral triangle?

I'm looking for a sequence of smooth (differentiable) closed curves.

The square approach occurs as even values of n approach infinity. To put it another way, the plot of x^(2*n)+y^(2*n)=1 approaches a square as the positive integer n approaches infinity.

The orientation and size of the approached triangle is not important, since I can rotate it by suitable substitutions for x and y.

I found one candidate, except it is not smooth at (0,0). I can live with this, but can any of you improve on it?

2*(2*x)^(2*n)=(1-sqrt(2)*y)*(1+sqrt(2)*y)^(2*n)

Oops, got truncated. Here it is again:

2*(2*x)^(2*n)
=
(1-sqrt(2)*y)
*(1+sqrt(2)*y)^(2*n)

Edit: not smooth at (1/2,0).

Answer 1

Try playing with families of curves of this type:

x^n + y^n + (x+y+1)^n = 1

This will only get you a rignt triangle, but I think it can be fine tuned into making an equilateral triangle.

The idea is to begin with trilinear coordinates.

Try plotting it in 3D, you'll get a fun result.

Addendum: The following equation will approach a perfect equilateral triangle of sides = 1, base centered at (0,0), as n -> infinity. Smooth, clean corners, no gaps, no spurious lines, for any value n.

(1-√(3/2) y)^n + (-(3/√2) x-√(3/2) y)^n + ((3/√2) x-√(3/2) y)^n = 1

Answer 2

I am thinking but its kind of hard first you need to define which way the triangle is facing (do you want an angle to be at the top or do you want a side to be at the top)

in other words do you want it pointing down or pointing up?

I must first understand why x^n+y^n=1 approaches a square

I may edit my answer

Answer 3

I really don't understand all of it but I looked it up on google.
I found a website: http://www.jstor.org/view/00361429/di976165/97p0031e/0
for you to look at.

I hope I helped some. Good Luck!