What is the determinant of the Wronskian of the functions $\{\cos\ x, \sin\ x, \cos\ 2x, \sin\ 2x,\ldots, \cos\ nx, \sin\ nx\}$? This determinant seems to be an integer, and the sequence starts with 1, 18, 86400, 548674560000... It is not in the Encyclopedia.

Question What is this sequence? I guess it is enough to prove that it consists of integers (constants), because then to compute it, one can simply put $x=0$.

Update 1. The fact that the determinant of the Wronskian matrix is a constant is obvious. Take the derivative of the determinant. It is a sum of determinants of matrices each of which has two proportional columns.

Update 2. The determinant is equal to the square of the Vandermonde determinant of $1, 2^2,\ldots, n^2$ times $n!$ (alternatively see Felipe Voloch's answer below). It is interesting that for $n=1$ we get just the equality $\cos^2 x+\sin^2 x=1$ (the equation of the circle). So the equality for $n > 1$ can be considered as the generalization of this equation. What is the geometry behind this identity? Of course the parametrization $(\cos x,\sin x,\ldots, \cos nx, \sin nx)$ defines some curve in $\mathbb{R}^{2n}$. What is known about that curve?

Update 3. Here is an easier formula for the determinant. It is equal to $$(1! 3!\ldots (2n-1)!)^2/n!$$

Update 4 I found a related paper: Larsen, Mogens Esrom, Wronskian harmony. Math. Mag. 63 (1990), no. 1, 33–37. He considers the Wronskian of $\sin x, \ldots, \sin nx$.

Update 5. The sequence is in the Encyclopedia now.

  • $\begingroup$ Should be easy to compute by changing basis to exponentials. $\endgroup$ – Felipe Voloch Oct 11 '11 at 0:04
  • $\begingroup$ @Felipe: you are most probably right. But that must have been done 150 years ago! $\endgroup$ – Mark Sapir Oct 11 '11 at 0:13
  • $\begingroup$ You are probably aware that Mathematica's Wronskian[] function computes these number pretty easily, so it must follow from basic trig identities. The next element, for $n=5$, is 14450101093977292800000; then for $n=6$, 3837346984091658732083085312000000000... $\endgroup$ – Joseph O'Rourke Oct 11 '11 at 0:30
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    $\begingroup$ It seems worth writing the sequence in factored form. I get 1, $2\cdot3^2$, $2^6\cdot3^2\cdot5$, $2^{13}\cdot3^7\cdot5^4\cdot7^2$, $2^{27}\cdot3^{15}\cdot5^5\cdot7^4$, $2^{42}\cdot3^{22}\cdot5^9\cdot11^2$. $\endgroup$ – Barry Cipra Oct 11 '11 at 0:57
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    $\begingroup$ Great that you got it in the OEIS! $\endgroup$ – Joseph O'Rourke Oct 11 '11 at 20:48

If you only need that it doesn't depend on $x$ consider the following argument. For any $\pi \in S_{2n}$ written as $\pi(1)\pi(2)\cdots \pi(n)$, let $[\pi]$ be the set of all permutations $\sigma \in S_{2n}$ which satisfy $\lbrace\sigma(2k-1),\sigma(2k)\rbrace =\lbrace\pi(2k-1),\pi(2k)\rbrace$ as sets for all $k$. Then $S_{2n}$ can be written as a disjoint union of such classes $[\pi]$ and so the terms in the determinant can be grouped accordingly. One can see that for each $[\pi]$ the correpsonding terms in the determinant add up to zero if $\pi(2j-1)+\pi(2j)$ is even for some $j$, and to $$ (-1)^{\text{something}}\prod_{j=1}^{n}\left( j^{\pi(2j-1)+\pi(2j)}(\sin^2(jx)+\cos^2(jx))\right)$$ otherwise, which makes it clear that the determinant doesn't depend on $x$.


These functions are solutions of the homogeneous differential equation with characteristic equation $(r^2+1)(r^2+4)(r^2+9)\dots(r^2+n^2)$. By Abel's identity, the Wronskian of a fundamental set of solutions is its value at zero times the integral $\int_0^x -p_{2n-1}(t) dt$, where $p_{2n-1}(t)$ is the coefficient at $y^{(2n-1)}$. But in our case $p_{2n-1}(t)=0$, so we have that the Wronskian is constant.

  • $\begingroup$ Is it written somewhere? It is hard to imagine that was not considered before. $\endgroup$ – Mark Sapir Oct 11 '11 at 0:37
  • $\begingroup$ I do not know about this determinant. Of course, Abel's identity is well known. $\endgroup$ – nekrashevych Oct 11 '11 at 0:46
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    $\begingroup$ @Volodya: Are you sure you wrote Abel's identity correctly? Did you forget $\exp$? $\endgroup$ – Mark Sapir Oct 11 '11 at 20:44

Is something similar to $(i/2)^n$ times the Vandermonde of $i,2i,\ldots,ni,-i,-2i,\ldots,-ni$.

  • $\begingroup$ I got a square of the Vandermonde determinant of $1, 2^2, 3^2,...,n^2$ times $n!$ (but I could have made an error). That is why I wrote that it is probably enough to show that the Wronskian is a constant. $\endgroup$ – Mark Sapir Oct 11 '11 at 0:52
  • $\begingroup$ The constancy of the wronskian is trivial once you write it as exponentials. $\endgroup$ – Felipe Voloch Oct 11 '11 at 0:55
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    $\begingroup$ I just realized that it is trivial without using exponents. Just take the derivative of the determinant. The derivative is obviously 0 (it is a sum of determinants of matrices each of which has two proportional columns). $\endgroup$ – Mark Sapir Oct 11 '11 at 2:52

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