Wikipedia:Reference desk/Archives/Mathematics/2016 December 6

= December 6 =

Vector Problem
If dU/dt=WxU and dV/dt=WxV, prove that d(UxV) /dt=Wx(UxV).

This is not a homework problem. 47.29.88.170 (talk) 08:52, 6 December 2016 (UTC)


 * You need the identity


 * $$\mathbf{V}\times (\mathbf{W}\times \mathbf{U}) + \mathbf{U}\times (\mathbf{V}\times \mathbf{W}) + \mathbf{W}\times (\mathbf{U}\times \mathbf{V}) = 0$$


 * I can't immediately spot this on either the Vector algebra relations or Vector calculus identities page, but you can easily prove it by expanding each of the terms according to


 * $$\mathbf{U}\times (\mathbf{V}\times \mathbf{W}) = (\mathbf{U}\cdot \mathbf{W}) \mathbf{V} - (\mathbf{U}\cdot \mathbf{V}) \mathbf{W}$$


 * and observing that they then cancel. --catslash (talk) 10:26, 6 December 2016 (UTC)
 * See Triple product (Jacobi identity) --catslash (talk) 10:33, 6 December 2016 (UTC)


 * This is not a homework problem. Then what is it? Tigraan Click here to contact me 12:29, 6 December 2016 (UTC)

Angle quintisection
What formula is there for the sine of quintuple angle as a function of sine of simple angle? Can it be inverted to find out the sine of a fifth of an angle?--82.137.11.135 (talk) 14:09, 6 December 2016 (UTC)
 * [EDITED 17:28, 6 December 2016 (UTC) per RDBury] Thanks to List_of_trigonometric_identities, one can compute $$\sin(5\theta)=\sin(\theta)^5+5\sin(\theta)\cos(\theta)^4 - 10\sin(\theta)^3\cos(\theta)^2 $$ (I asked WolframAlpha, actually, as any lazy person would do). That formula cannot be reformulated as a "normal" function purely of the sine (you can convert cosine in sine only if you know the sign), and certainly not be inverted; by that I mean there is no function such that $$\sin(\theta)=f(\sin(\theta)))$$ (because $$\sin(0)=\sin(\pi)$$ so the right hand side results must be equal for these values, but $$\sin(0/5)\neq \sin(\pi/5)$$).
 * This is no guarantee of the possibility of "angle quintisection" (in the sense that you could construct the fifth of a given angle with straight lines and circles). See Angle_trisection (I think a similar argument would apply to quintisection as well, but hunting down the polynomial's roots could prove harder). Tigraan Click here to contact me 15:02, 6 December 2016 (UTC)
 * Slight copy error, it should be
 * $$\sin(5\theta)=\sin(\theta)^5+5\sin(\theta)\cos^4(\theta) - 10\sin^3(\theta)\cos^2(\theta). $$
 * You then get a polynomial in sin using cos2θ=1-sin2θ. The proof of the impossibility of angle trisection actually works for any odd number. So angle n-section is only possible (for general angles) when n is a power of 2. In fact, one only has to prove that the regular n2-gon is impossible to construct. --RDBury (talk) 15:55, 6 December 2016 (UTC)
 * In other words, you have $$\sin(5\theta) = 16 \sin ^5(\theta )-20 \sin ^3(\theta )+5 \sin (\theta )$$.
 * However the OP didn't ask about geometric construction, only about extracting from this a formula for $$\sin(\theta)$$ given $$\sin(5\theta)$$ (equivalent to finding $$\sin(\tfrac{\theta}5)$$ given $$\sin\theta$$).
 * The only real problem with this is that it requires solving a quintic polynomial, which is impossible by usual means - this is the Abel–Ruffini theorem. -- Meni Rosenfeld (talk) 16:17, 6 December 2016 (UTC)
 * True, I only mentioned the geometry thingie because of the thread title. (Note that although a third-degree polynomial can be factorized algebraically, it involves cubic roots which are non-constructible generally speaking.)
 * As for the Abel–Ruffini theorem, it only means there is no general factorization method for quintic polynomials (with "usual means"). Maybe this particular quintic polynomial, $$16X^5-20X^3+5X-a$$, can be factorized. I doubt it, because it would mean finding a factorization that works for every a. The proof section of the ABT says it remains valid for a given set of algebraically independent coefficients, but the set $$\{16,0,-20,0,5,a\}$$ is not that (the first four coefficients are rational). Tigraan Click here to contact me 17:28, 6 December 2016 (UTC)


 * Thanks for your answers. Is the situation similar for cosine case?--82.137.9.227 (talk) 19:56, 6 December 2016 (UTC)
 * What is the situation when applying a numerical root-finding algorithm for the mentioned quintic equation for every a?--82.137.9.227 (talk) 20:02, 6 December 2016 (UTC)
 * I made a mistake in the polynomial above, I've corrected it now. It doesn't materially change the results. I've also taken the liberty of updating Tigraan's response accordingly.
 * It's true that it doesn't immediately follow from the theorem that this particular polynomial is unsolvable, but it's almost certain.
 * For cosine it's quite similar - $$\cos(5\theta)=16 \cos ^5(\theta )-20 \cos ^3(\theta )+5 \cos (\theta )$$.
 * There's no real problem with solving it numerically, but you can only do that once a is given (if you do it with a symbolic a, you quickly get unwieldy expressions), and it might be easier to calculate the sine directly with Taylor series or whatnot. -- Meni Rosenfeld (talk) 11:16, 7 December 2016 (UTC)
 * (I'm an amateur here, so feel free to correct me.) I could easily have messed up typing all these numbers into Mathematica, but I just checked the Cayley resolvent, and it appears to have a rational root at $$z=125/16$$ for all $$a$$. This would suggest that this is a family of rare quintics which are in fact solvable. 129.234.195.173 (talk) 16:20, 7 December 2016 (UTC)
 * A little more digging suggests that it's actually an example of de Moivre's quintic. 19:36, 7 December 2016 (UTC) — Preceding unsigned comment added by 129.234.195.173 (talk)
 * Interesting.
 * I double-checked the resolvent and it checks out. I only wonder why Mathematica won't return the roots in radical form.
 * I'm not an expert on this myself but I'd say Bo (in a comment below) is on to something, the solvability of this is probably connected to the fact that in complex numbers you can find it by simply taking a fifth root. -- Meni Rosenfeld (talk) 01:09, 8 December 2016 (UTC)
 * See Chebyshev polynomials. — 86.125.199.30 (talk) 00:45, 7 December 2016 (UTC)

Let $$a=\cos(\theta )+i \sin(\theta )$$. The problem is to solve the quintic equation $$z^5=a$$. The solution can be expressed by roots ($$z=\sqrt[5]a$$), but cannot be constructed by compas and straightedge. Bo Jacoby (talk) 22:39, 7 December 2016 (UTC). Refer to the Durand-Kerner method for solving the equation $$z^5=a$$. Bo Jacoby (talk) 20:58, 9 December 2016 (UTC).
 * Hats off! It is so obvious in hindsight... Tigraan Click here to contact me 16:41, 9 December 2016 (UTC)
 * But my interpretation of the OP's question is that he wants an expression in real radicals for the sine of the one-fifth angle. In general none exists. Loraof (talk) 16:56, 9 December 2016 (UTC)