Number crunching algorithm

> I've run into a programming challenge that involves

> number crunching and I'm stumped on an equation.

> Here's what I have that works so far:

> > double y = (Math.sqrt(x) - 1/x) / 2;

>

> It basically takes x as a given and computes y,

> producing a 2D curve. Where I'm stumped is I want to

> take y as a given and compute x to produce the

> identical curve. Is this possible? I've managed to

> reformulate the equation several different ways on

> paper, but haven't arrived at anything useful. (I'm

> kind of rusty on algebra.) Any help on how to go about

> this would be appreciated.

let a = sqr(x), and rewrite your function:

y = (a- 1/a^2)/2

simple above, you get

2y = a - (1/a^2)

then

a = 2y + 1/a^2

then

a = ((2a^2)*y + 1)/a^2

then

a^3 = (2a^2)*y + 1

then

(2a^2)*y = a^3 - 1

therefore

y = (a^3 - 1)/2a^2

then substitute sqr(x) back, you get

y = (sqr(x)-1)/2

hmm...hope I have done everything correctly....

yue42 at 2007-7-7 0:58:03 >

Difficult problem! Cheating, here's what matlab's symbolic toolbox gives me:

>> f=sym('y = (sqrt(x) - 1/x) / 2');

>> finv = solve(f, x)

finv =

[(1/6*(108+64*y^3+12*(81+96*y^3)^(1/2))^(1/3)+8/3*y^2/(108+64*y^3+12*(81+96*y^3)^(1/2))^(1/3)+2/3*y)^2]

[ (-1/12*(108+64*y^3+12*(81+96*y^3)^(1/2))^(1/3)-4/3*y^2/(108+64*y^3+12*(81+96*y^3)^(1/2))^(1/3)+2/3*y+1/2*i*3^(1/2)*(1/6*(108+64*y^3+12*(81+96*y^3)^(1/2))^(1/3)-8/3*y^2/(108+64*y^3+12*(81+96*y^3)^(1/2))^(1/3)))^2]

[ (-1/12*(108+64*y^3+12*(81+96*y^3)^(1/2))^(1/3)-4/3*y^2/(108+64*y^3+12*(81+96*y^3)^(1/2))^(1/3)+2/3*y-1/2*i*3^(1/2)*(1/6*(108+64*y^3+12*(81+96*y^3)^(1/2))^(1/3)-8/3*y^2/(108+64*y^3+12*(81+96*y^3)^(1/2))^(1/3)))^2]

That's 3 separate equations, btw...

Good luck.

RadcliffePike at 2007-7-7 0:58:03 >

Thanks. I experimented with java.awt.geom.CubicCurve2D to try to solve the equation for x using the solveCubic() method and it gives unexpected results--maybe I'm not ordering the coefficients right, I'll have to keep fiddling with it. It crunches them fairly fast I think; about a million per second on my four year old wintel laptop, so that's good at least.

drpeterjones at 2007-7-7 0:58:03 >

It's some consolation to discover that the CubicCurve2D implementation fails my unit test, so I didn't entirely waste my time.

As to ordering of coefficients: the documentation is far more confusing than it should be. To solve ax^3 + bx^2 + cx + d you call CubicCurve2D.solveCubic(new double[]{d, c, b, a});

YATArchivist at 2007-7-7 0:58:03 >