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After all this time I came up with a very nice tensor calculus proof of the Hairy Ball Theorem. It only depends on Stokes theorem and standard laws of tensor calculus like the Ricci identity and symmetries of curvature tensors. All the topology is done by Stokes theorem. The remainder of the proof is equational, local and geometrical. It is coordinate/basis ...


3

You're looking for TagSetDelayed I believe: Unprotect@Integrate; Integrate /: Plus[Integrate[ft_, {t_, a_, b_}], Integrate[ft_, {t_, b_, c_}]] := Integrate[ft, {t, a, c}]; Integrate /: Plus[Integrate[ft_, {t_, b_, a_}], Integrate[ft_, {t_, b_, c_}]] := Integrate[ft, {t, a, c}]; Protect@Integrate; But be careful when you unprotect system ...


2

I would make use of Mathematica patterns in the assumptions: Assuming[{Dm[a_, a_][x_] == Dm[a_, a_][-x_]}, Simplify[Transpose[dmat[q]] == dmat[-q]]] (* True *) This tells Simplifythat a matrix element with equal indices (i.e. a diagonal element) is an even function. The problem with your code is that with Apply, you are replacing the equal sign with an ...


2

Mathematica might need some help with this one. One of the problems is surely that you are asking for y - try first to get a solution for u(x)=y''(x), this makes things a bit easier. You can always integrate twice at the end. In this case, it might be an idea to split up the 'source term' (i.e. the right hand side). As the differential equation is linear, ...


2

If Mathematica can't decide if argument to HeavisideTheta is positive or negative, this remains unevaluated. Type HeavisideTheta[x - L/2] on your computer and see the result. So, the result will contain this even if it can solve it. I think it just can't find a particular solution because of this. One way to help M, is to break it to 2 ODE to avoid the ...


2

Try this: Flatten@Position[Sqrt[2 Range[512] + 1], _Integer, 1] resulting in {4, 12, 24, 40, 60, 84, 112, 144, 180, 220, 264, 312, 364, 420, 480} You'll notice that the quasi-periodic oscillations are centered at the numbers in the list above. This is due to the fact that $\sqrt{2i+1}\in\mathbb{Z}$ for those values of $i$, and presumably Integrate's ...


1

One may also use Inactivate/Activateconstruct. For example, try this expr = Inactivate[ Integrate[f[x], {x, 0, 1}] + Integrate[f[x], {x, 1, 2}], Integrate] yielding this: Then make the replacement: expr /. Inactivate[ Integrate[g_, {x, a_, b_}] + Integrate[g_, {x, b_, c_}], Integrate] -> Inactivate[Integrate[g, {x, a, c}], Integrate] ...



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