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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?2012-01-29
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    @anon, thank you. det has been added. Yes, there is a more general version of the formula.2012-01-30
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    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..2012-01-30
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    Cool, thanks.$\text{ }$2012-01-30
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    Fascinating identities. Curious, where do they come from?2012-02-01
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    @Bill, they are from a paper. The version of the identity in that paper is the usual Wronskian.2012-02-01
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    I think we can use induction to prove the identity. But I don't know whether there is a better way to prove the identity.2012-02-01
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    Excuse me, my mathematics is a bit rusty. Why is W called a Wronskian, when no differentiation is involved?2012-11-15

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