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When proving this, my textbook first writes down the determinant, and then states:

This will be tedious to compute directly, and we only need to know that the Wronskian is nonzero at a single point, so we evaluate it at $x = 0$

Why is this?

I recall that there's a similar theorem that states if these functions are solutions to a differential equation

then they are linearly independent $\iff$ Wronskian = 0 at one point in the interval $\iff$ Wronskian = 0 at all points in the interval.

However, the problem in the header does not state that they are solutions to any differential equation.

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2 Answers

$\rm Theorem$. If the Wronskian of a set of $n$ functions defined on the interval $[a,b]$ is nonzero for at least one point in this interval, then the set of functions is linearly independent on the interval. If the Wronskian is identically zero on this interval and if each of the functions is a solution to the same linear differential equation, then the set of functions is linearly dependent.

Note that the assumption that the these functions are the solution to a linear differential equation is not necessary for the part of the theorem you wish to invoke.

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(One minor detail: you've written the theorem kind of backwards. Either it should be "linearly dependent", or your zero Wronskians should be nonzero.)

There are two different ways you could think about this.

First, any set of sufficiently differentiable functions whose Wronskian is nonzero at a point will be linearly independent; it's not necessary that those functions satisfy a nonsingular differential equation. (The hypothesis that they satisfy a DE is required for the "only if" half of the quoted theorem. For example, the set of functions $\{x^2, x^3\}$ has a Wronskian which vanishes at $0$, but is linearly independent.) Hopefully you should be able to look at the proof of quoted theorem in your textbook and convince yourself of this, or maybe it's even proved somewhere else in the book.

Second, you could also just apply the theorem as you've stated it; all you need to do is concoct a differential equation which has your functions as solutions! For example, if $D=\frac{d}{dx}$ is the differentiation operator, $1$ is a solution to $Dy=0$, and $\{\sin x, \cos x\}$ are solutions to $(D^2+1)y=0$, so $\{1, \sin x, \cos x\}$ must all be solutions to $D(D^2+1)y=0$. You can extend this idea to get an equation which works for $\sin 2x$ and $\cos 2x$ as well...

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Thanks. Can you prove the theorem or point me to a proof? –  Covi Nov 16 '12 at 19:03
    
The part of the theorem you need is basically just linear algebra. If $\{f_1,\dots,f_n\}$ are linearly dependent, there's some set of constants $a_1,\dots,a_n$ with $a_1f_1+\dots+a_nf_n=0$ identically zero. But then you can check that, if $v_i$ is the $i^{\rm th}$ column of the Wronskian matrix, it's also true that $a_1v_1+\dots+a_n=0$. That is, the columns of the matrix are linearly dependent, so the matrix's determinant (the Wronskian itself) vanishes everywhere. –  Micah Nov 16 '12 at 19:12
    
Thanks. What do you mean by the "only if half"? –  Covi Nov 16 '12 at 22:07
    
I mean [the functions are linearly independent] $\implies$ [the functions' Wronskian is nonzero (everywhere)]. This is false unless you throw in the additional hypothesis that the functions solve a nonsingular ODE, and $\{x^2, x^3\}$ is a counterexample. –  Micah Nov 16 '12 at 23:02
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