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Does the Wronskian have anything to do with the product rule in calculus. I ask this because i noticed the form looking similar to the product rule. $$W=g(x)f'(x)-g'(x)f(x)$$ where as the product rule is $$(f(x)g(x))' = g(x)f'(x)+g'(x)f(x)$$ they only differ by a plus operation.

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If you assume that $g\neq 0$, then it's $g(x)^2$ times the derivative of the quotient, and when it's $0$ it means that $f$ and $g$ are proportional. – Davide Giraudo Nov 2 '11 at 19:28
If $W = \left( \begin{array}{ccccc} f_1(x) & f_2(x) & f_3(x) & \dots & f_n(x) \\ f_1'(x) & f_2'(x) & f_3'(x)& \dots & f_n'(x) \\ 1 & 1 & 1 & \dots & 1 \\ \vdots & & &\ddots & \\ 1 & 1 & 1 &\dots & 1 \end{array} \right)$, the product rule is determined by the permanent of the matrix $W$. – JavaMan Nov 2 '11 at 19:33
@DavideGiraudo: Careful. A classic example is $f(x)=x^2$ and $g(x)=x|x|$. Then $g'(x) = 2x$ if $x\geq 0$, $g'(x)=-2x$ if $x\lt 0$, $W(f,g)=0$, but $f$ and $g$ are not proportional (they are equal on the nonnegative numbers, and $f=-g$ on the nonpositive ones). – Arturo Magidin Nov 2 '11 at 19:34
@ArturoMagidin yes you're right. So we have to assume that $g$ never vanishes. – Davide Giraudo Nov 2 '11 at 19:40
@DavideGiraudo: There are any number of conditions you can put to ensure the implication"$W(f,g)=0$ implies $f$ and $g$ proportional"; for example, if they are both analytic (due to Peano). See this paper, and Wikipedia for other situations. – Arturo Magidin Nov 2 '11 at 19:45
up vote 1 down vote accepted

Not really. The Wronskian is a determinant of functions and their derivatives. In your case you have that, for two functions $f$ and $g$,

$$W(f,g) = \det\begin{pmatrix}f & g \\ f' & g'\end{pmatrix} = fg'-f'g$$

In general, the Wronskian of $n$ different functions is the determininant of the square $n\times n$ matrix where each column is composed of the $f$ and its derivatives up to the $(n-1)$ th derivative (such that every derivative exists in an interval $I$), ordered from lower to higher order from top to bottom. In general, $$W(f,g) \neq W(g,f)$$ as opposed to $(f·g)'$ and the same goes for higher order Wronskians.

Here you have an example of the Wronskian of two functions, and here one of three.

Wikipedia can give you some info on the Wronskian, and you can visualize what the layout of a Wronskian is. (I don't know how to code matrices here. I tried LaTeX but it didn't work.)

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I had to look up what a Wronskian is, so I didn't feel that I could evaluate either the question or any answers without putting in some work. I only noticed your comment regarding TeX at the end. This, I think, is why some (good) answers are under-appreciated, particularly if the writer is not well known. It's hard to think of a good solution for that problem. – Dylan Moreland Feb 5 '12 at 6:45

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