I have found that the below implicite function can be approximated to an ellipse within a high level of accuracy. However I can only prove it numerically that the enclosed area is within 1.5% error for 1.8 < n < 6, when compared to an ellipse.
( (x+d/2)^2 + y^2 ) ^ (n/2) + (x-d/2)^2 + y^2 ) ^ (n/2) = d^n
(1) Is there any analytical method to prove that the above function can be approximated to an ellipse for this range of (n)?
And / Or (2) Is there any close form method to obtain the area enclosed by this curve?
Akram,
The general equation of an ellipse is given as: (x/a)^2 + (y/b)^2 = 1 where a and b are the semi-major and semi-minor axis.
We also define a "hyper-ellipse" of the form (x/a)^n + (y/b)^n = 1 where (n>2).
Obviously when n is large the closed curve approaches the rectangle (+a,-a,+b,-b).
The approximation of a hyper-ellipse by an ellipse is "rough". Why do you need to do so?
If you want the area enclosed by the hyper-ellipse, it is given on the wikipedia page of the hyperellipse in terms of Euler Gamma function
http://en.wikipedia.org/wiki/Hyperellipse
Regards
Hi Akram,
The equation you have given is not always an eliipse or a hyperellipse!
For example, for the case, n = 2,
Scaling: X = x/d and Y = y/d
We obtain, X^2 + Y^2 = 1/4 which repersents a unit circle of radius 1/2
Thus f(x,y) is actually a circle of center (0,0) and radius (d/2). thus the area is pi*(d^2)/4.
My suggestion, work the problem for INTEGER n. You can always find the area of the curve using Euler Gamma function.
This question somehow resembles you previuos questions on solving the Algebraic equation which Hermann suggested to solve it numerically and use an interpolation.
Here, to find the area inside the curve, you have to express y = g(x) and integrate, something you cannot always do unless n is integer or "for some rational".
Again you are stuck with numerical methods, sove y in terms of x numerically, the integrate the data set (x_n,y_n) numerically by spline method, cubic splines is the best one!!!
Hope that targets your inquiry!!
Thanks Hady Joumaa for your valuable answer. You are correct about n=2 the curve becomes a circle (which is a special case of an ellipse).
The approximating ellipse have a semi horizontal axis of a= d/2 and a semi vertical axis of b=k d/2, where k = sqrt ( 2^(2-2/n) - 1). Where the equation of an ellipse is (x/a)^2 +(y/b)^2 =1.
As mentioned before this approximation is very close to the original equation, i can only proove it (graphically). But how to prove this analytically?
Here is the plot showing the great similarity with the ellipse, any analytical way to prove this similarity?
In the attached maple sheet, in rescaled coordinates X=x/d, Y=y/d, I consider introducing new variables (ranges in parentheses)
Z=sqrt(X^2+Y^2+1/4), ( 1/2
Thanks very much Herbert for your comprehensive answer, I think it is a quite good method for the analytical proof.
I am wondering if the area enclosed by the exact curve of eq1:=( (x+d/2)^2 + y^2 ) ^ (n/2) + ((x-d/2)^2 + y^2 )^(n/2) - d^n; could be analytically calculated as well?
The whole reason for the approximation I performed is that the area enclosed by a perfect ellipse is quite easy to obtain as (Pi * a * b) = ( Pi * d/2 * k *d/2).
Any analytical method to obtain the exact area enclosed by "eq1"?
- Badges
- Science topic
- Similar topics
- Mathematics
- Applied Mathematics