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Solving a quartic equation

asked 2015-06-01 04:51:26 +0200

nicholasbishop gravatar image

I'm attempting to rearrange an equation from an answer on the Mathematics StackExchange.

The answer given is this equation:

$$L^2 = (-ab(t)+p)^2-\left(\frac{(-ab(t)+p).(cd(t)-ab(t))}{(cd(t)-ab(t))^2}(cd(t)-ab(t))\right)^2$$

Where $a$, $b$, $c$, $d$, and $p$ are known 2D points, $L$ is a known length, and $t$ is an unknown scalar. $ab(t)$ indicates interpolation between $a$ and $b$.

I am interested in rearranging this to solve for $t$. Here's what I've tried in Sage:

def sqr(var): return var.dot_product(var)
var('ax bx cx dx px ay by cy dy py t L')
a = vector([ax, ay])
b = vector([bx, by])
c = vector([cx, cy])
d = vector([dx, dy])
p = vector([px, py])
g = a - at + bt
h = c - ct + dt
u = p - g
v = h - g
eq = L^2 == sqr(u) - sqr((u.dot_product(v)/sqr(v)) * v)

At the solve step I have observed it to sit for quite a while without producing a result. Two questions:

  1. Am I inputting the problem correctly?

  2. Is there any way to know if this is likely to terminate in a reasonable time? I have no idea what the solver looks like under the hood, and wouldn't want to wait for some O(n!) calculation to terminate :)

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The problem per se seems quite involved. The r.h.s when expanded displays 3 million characters:

sage: len(str(eq.rhs().expand()))
rws gravatar imagerws ( 2015-06-01 08:04:35 +0200 )edit

1 Answer

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answered 2015-06-02 15:02:15 +0200

rws gravatar image

updated 2015-06-02 15:32:42 +0200

There is a simpification possible. Instead of expanding the expression (which makes for a nice Pynac benchmark) we transform the equation into a fraction and get the numerator:

sage: g = a - a*t + b*t
sage: h = c - c*t + d*t
sage: u = p - g
sage: v = h - g
sage: eq = L^2 == sqr(u) - sqr((u.dot_product(v)/sqr(v)) * v)
sage: rhs = eq.rhs()
sage: num = (rhs.rational_simplify()-L^2).numerator()
sage: solve(num, t)

This is much smaller but unfortunately still takes forever to solve. Using collect and coefficient you can let Sage show you the coefficient of t^0, t, t^2, t^3, and t^4 to get an idea of what's involved and which equations determine the solution process of the quartic. For example this is the constant coefficient:

sage: num.coefficient(t,0)
-L^2*ax^2 - L^2*ay^2 + 2*L^2*ax*cx - L^2*cx^2 + ay^2*cx^2 + 2*L^2*ay*cy - 2*ax*ay*cx*cy - L^2*cy^2 + ax^2*cy^2 - 2*ay^2*cx*px + 2*ax*ay*cy*px + 2*ay*cx*cy*px - 2*ax*cy^2*px + ay^2*px^2 - 2*ay*cy*px^2 + cy^2*px^2 + 2*ax*ay*cx*py - 2*ay*cx^2*py - 2*ax^2*cy*py + 2*ax*cx*cy*py - 2*ax*ay*px*py + 2*ay*cx*px*py + 2*ax*cy*px*py - 2*cx*cy*px*py + ax^2*py^2 - 2*ax*cx*py^2 + cx^2*py^2

You could also get from Sage the general solution to the quartic via:

sage: (a,b,c,d,e)  = var('a,b,c,d,e')
sage: sol = solve(a*x^4+b*x^3+c*x^2+d*x+e,x)

This returns a vector of 4 solutions. You could then simply substitute your coefficients for a,b,c,d,e in one of the solutions:

sage: sol1 = sol[0]
sage: sol1 = sol1.subs(e==num.coefficient(t,0)).subs(d==num.coefficient(t,1)).subs(c==num.coefficient(t,2)).subs(b==num.coefficient(t,3)).subs(a==num.coefficient(t,4))
sage: len(str(sol1))

This solution however is still too big to handle. It needs 500 pages to print! Simplifying is out of the question.

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Asked: 2015-06-01 04:51:26 +0200

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Last updated: Jun 02 '15