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The following Wronskian identity can be proved by expanding both sides and checking that two sides are the same. But how to prove it more elegantly?

Let $u_1(x), u_2(x), u_3(x), u_4(x)$ be four functions. Define q-shift Wronskian as follows: $$W(u_1, u_2, u_3)(x)=\det \begin{bmatrix} u_1(x) & u_1(xq^{-2}) & u_1(xq^{-4}) \\ u_2(x) & u_2(xq^{-2}) & u_2(xq^{-4}) \\ u_3(x) & u_3(xq^{-2}) & u_3(xq^{-4}) \end{bmatrix}.$$

Similarly for $W(u_1, u_2)(x)$ and $W(u_1, u_2, u_3, u_4)(x)$. Then we have a Wronskian identity:

$$ W(W(u_1, u_3, u_4)(x), W(u_2, u_3, u_4)(x))(x) = W(u_1, u_2, u_3, u_4)(x) \cdot W(u_3, u_4)(xq^{-2}). $$ Thank you very much.

Edit: The general version of the identity is the following. Let $W_s(i)=W(u_1, \ldots, \hat{u_i}, \ldots, u_{s+1})$, where $\hat{u_i}$ means without $u_i$. Given functions $u_1, \ldots, u_{s+1}$.

$$ W_{k+1}(W_s(s-1)(x), W_s(s-2)(x), \ldots, W_{s}(s-k-1)(x))(x) = \\ \left(\prod_{j=1}^{k} W_{s+1}(u_1, \ldots, u_{s+1})(xq^{-2(j-1)})\right) \cdot W_{s-k}(u_{k+2}, \ldots, u_{s+1})(xq^{-2k}). $$

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Missing a $\det$. Is there a more general version of the formula that you know about? –  anon Jan 29 '12 at 22:36
    
@anon, thank you. det has been added. Yes, there is a more general version of the formula. –  LJR Jan 30 '12 at 1:26
    
user, it might be easier to see a route to proving this if you posted the general form. Also, I'd like to see it, out of curiosity.. –  anon Jan 30 '12 at 1:28
    
Cool, thanks.$\text{ }$ –  anon Jan 30 '12 at 18:40
    
Fascinating identities. Curious, where do they come from? –  Bill Cook Feb 1 '12 at 2:29
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1 Answer

Disclaimer: This is not an answer...just an idea which doesn't fit in a comment.

I can't come up with a nice trick to get a slick proof, but here's a suggestion for organizing the data you're working with.

Let $u_{ij} = u_i(xq^{-2(j-1)})$. Then $u_{21} = u_2(x)$ and $u_{23} = u_2(xq^{-4})$ etc. Then

$$W(u_1,u_2,\dots,u_n)(x) = \mathrm{det} \begin{bmatrix} u_1(x) & u_1(xq^{-2}) & \cdots & u_1(xq^{-2n-2}) \\ u_2(x) & u_2(xq^{-2}) & \cdots & u_2(xq^{-2n-2}) \\ \vdots & \vdots & \vdots & \vdots \\ u_n(x) & u_n(xq^{-2}) & \cdots & u_n(xq^{-2n-2}) \end{bmatrix} $$ $$ = \mathrm{det} \begin{bmatrix} u_{11} & u_{12} & \cdots & u_{1n} \\ u_{21} & u_{22} & \cdots & u_{2n} \\ \vdots & \vdots & \vdots & \vdots \\ u_{n1} & u_{n2} & \cdots & u_{nn} \end{bmatrix} = W(u_{11},u_{21},u_{31},u_{41})$$

Notice that, for example, $u_{23}(x) = u_{21}(xq^{-4}) = u_{22}(xq^{-2})$. So $W(u_3,u_4)(xq^{-2})=$ $W(u_{31},u_{41})(xq^{-2})=$ $W(u_{32},u_{42})(x)$

Again, I know this doesn't solve the problem, but maybe reorganizing the data this way will yield a reasonable proof.

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