I'll need to think of a problem more like these, I guess. Nobody likes my integrals
(And if you're done - can you make one for p(x,y,z), or even one with an arbitrary number of integer inputs? What's the lowest degree polynomial you can think of, for the 2- and 3-variable cases? Is it possible to make a generalized set of polynomials, one for each number of inputs, so that if we encode some finite ordered list of integers, then *without knowing which polynomial we used* the final number will tell us how long the list was as well as letting us figure out the list's contents?)I like this one
X1 = {1, 0}; X2 = {-1, 0}; X3 = {0, Sqrt[3]};
tri = ListPlot[{X1, X2, X3, X1}, AspectRatio -> 1, Axes -> False,
PlotRange -> {{-2, 2}, {-1, 3}}, Joined -> True];
sol = NDSolve[{
x1'[t] == (x2[t] - x1[t])/
Sqrt[(x2[t] - x1[t])^2 + (y2[t] - y1[t])^2],
y1'[t] == (y2[t] - y1[t])/
Sqrt[(x2[t] - x1[t])^2 + (y2[t] - y1[t])^2],
x2'[t] == (x3[t] - x2[t])/
Sqrt[(x3[t] - x2[t])^2 + (y3[t] - y2[t])^2],
y2'[t] == (y3[t] - y2[t])/
Sqrt[(x3[t] - x2[t])^2 + (y3[t] - y2[t])^2],
x3'[t] == (x1[t] - x3[t])/
Sqrt[(x1[t] - x3[t])^2 + (y1[t] - y3[t])^2],
y3'[t] == (y1[t] - y3[t])/
Sqrt[(x1[t] - x3[t])^2 + (y1[t] - y3[t])^2],
x1[0] == X1[[1]], x2[0] == X2[[1]], x3[0] == X3[[1]],
y1[0] == X1[[2]], y2[0] == X2[[2]], y3[0] == X3[[2]]
}, {x1, x2, x3, y1, y2, y3}, {t, 0, 4/3}][[1]];
Show[
tri,
ParametricPlot[
{{x1[t], y1[t]}, {x2[t], y2[t]}, {x3[t], y3[t]}} /. sol,
{t, 0, 4/3}]
]
NIntegrate[Sqrt[
((x2[t] - x1[t])/Sqrt[(x2[t] - x1[t])^2 + (y2[t] - y1[t])^2])^2 +
((x2[t] - x1[t])/Sqrt[(x2[t] - x1[t])^2 + (y2[t] - y1[t])^2])^2
] /. sol, {t, 0, 4/3}]/2
Chestey wins!Jimmy:
\( \int_a^b\frac{\sqrt{x-a}\sqrt{b-x}}{x}\,dx=\left(\frac{a+b}{2}-\sqrt{ab}\right)\pi \)
Hope it's right because I cheated
NIntegrate[
Sqrt[((x2[t] - x1[t])/
Sqrt[(x2[t] - x1[t])^2 + (y2[t] - y1[t])^2])^2 + ((y2[t] -
y1[t])/Sqrt[(x2[t] - x1[t])^2 + (y2[t] - y1[t])^2])^2] /.
sol,
{t, 0, 4/3}]/2How do you know that the ratio is a constant?
How do you know that?
