Talk:Wronskian

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This article incorporates material from Wronskian determinant on PlanetMath, which is licensed under the GFDL.

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[edit] Does the Wronskian need to vanish at every point?

f1(x) = x, f2(x) = x^2, are linearly independent.

W(f_1, f_2) = \begin{vmatrix} x & x^2 \\ 1 & 2x \end{vmatrix} = 2x^2 - x^2 = x^2

W(f1, f2) is 0 at x=0, but it doesn't matter, because we only need one point where W is non-zero. x=1, W=1 is non-zero, so the functions are linearly independent.

The converse is not necessarily true: W=0 everywhere does not imply that the functions are dependent. Example: f1 = x^3, f2 = abs(x^3). They are linearly independent but the Wronskian is uniformly zero.

Dzhim 02:10, Feb 4, 2005 (UTC)

You're absolutely right. I thought about it for some time before I reverted you, but now I don't know what I was thinking. Sorry about that. -- Jitse Niesen 11:56, 4 Feb 2005 (UTC)

The converse is true if the two functions are real, differentiable and have only simple zero in the interval; or if the two functions are analytic

Davide

[edit] Linear dependence at a point

I removed the sentence about linear dependence at a point. As it stoodthat didn't say anything (for n > 1 you will always have dependence at a given point).

Charles Matthews 06:49, 23 Aug 2003 (UTC)

[edit] Homogeneous comment

I removed the condition that "the functions are solutions of some homogenous linear differential equation" from the statement "If the Wronskian is non-zero at some point in an interval, then the functions are linearly independent in that interval." If the Wronskian is nonzero at some point x, then the vectors

[f_k, f'_k, \ldots, f^{(n-1)}_k], \quad k=1,\ldots,n

are linearly independent, so every linear combination of the vectors will be a nonzero vector, so every linear combination of f_1, ..., f_n will have a nonzero derivative at x, hence it is not zero in a neighbourhood of x. -- Jitse Niesen (talk) 13:01, 12 October 2005 (UTC)

[edit] On calculators

I moved the following text from the article.

"Wronskian on the HP 49 series
You can simply edit the equations into the hp49, just use the matrix editor to enter the equations into the 49 and then do the det command to work out the Wronskian of a set of equations, having said this there must be an easyier way and I am looking into making a small program to do this, if you are interested please send me a message! user:Mick8882003"

The first bit (how to evaluate the Wronskian) does not give much useful information; it is too specific and it is rather straightforward, in my opinion. The second bit clearly does not belong in an encyclopaedia. -- Jitse Niesen (talk) 13:45, 18 December 2005 (UTC)

[edit] Proof

The motto of the supplied proof on 'Wronskian and linear independence' is mechanical but true. I think Wikepedia should seriously consider allowing special PROOF subpages for mathematical articles.

Ben Spinozoan 08:08, 29 December 2005 (UTC)

This has been discussed before. My view: proofs here have to justify their place, like any other content. Charles Matthews 09:57, 29 December 2005 (UTC)
But anyway this proof is over the top, no? A linear dependence gives a vanishing determinant for more obvious reasons. The column rank is less than n. Charles Matthews 11:43, 29 December 2005 (UTC)
Touché.....I never said it was elegant! Be my guest and supplant at will. Ben Spinozoan 14:47, 29 December 2005 (UTC)


Actually... the proof is incorrect. Just because differentiation is a linear operation, it doesn't mean that linearly independent functions will map to linearly independent columns, since differentiation is a nontrivial kernel (the set of all constant functions); so, the proof needs some elaboration. —Preceding unsigned comment added by 18.243.2.126 (talk) 20:23, 6 October 2007 (UTC)

Would this be a proof? http://tutorial.math.lamar.edu/Classes/DE/FundamentalSetsofSolutions.aspx It makes sense to me. —Preceding unsigned comment added by 201.62.109.74 (talk) 20:58, 26 November 2007 (UTC)