- #1
dogma
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Hello out there.
I'm working on a proof by induction of the Wronskian and need a little boost to get going.
So, here goes:
If [tex]y_1,...,y_n \in C^n[a,b][/tex], then their Wronskian is:
[tex]Wr(y_1,...,y_n)=det\left(\begin{array}{ccc}y_1&\cdots&y_n\\y_1\prime&\cdots&y_n\prime\\\vdots&\vdots&\vdots\\y_1^{(n-1)}&\cdots&y_n^{(n-1)}\end{array}\right)[/tex]
In general, a set of functions will be linearly independent IFF the Wronskian is not identically zero.
Prove this for n = 2, that is [tex]f(x), g(x)[/tex] are independent IFF
[tex]\left\vert\begin{array}{cc}f(x)&g(x)\\f^\prime(x)&g^\prime(x)\end{array}\right\vert[/tex] is not identically zero.
Okay, I think I understand how to prove this in one direction. That is, assuming the [tex]Wr(f,g) \neq 0[/tex] and showing that [tex]f(x), g(x)[/tex] are independent.
But I'm a little stuck in the assumptions for proving the other direction.
This is what I have for the proof in one direction:
Assume [tex]Wr(f(x),g(x)) \neq 0[/tex].
Then there exists some [tex]x_o[/tex] such that [tex]Wr(f(x_o),g(x_o)) \neq 0[/tex].
Assume:
[tex]c_1 f(x) + c_2 g(x) \equiv 0[/tex]
[tex]c_1 f^\prime(x) + c_2 g^\prime(x) \equiv 0[/tex]
then
[tex]c_1 f(x_o) + c_2 g(x_o) = 0[/tex]
[tex]c_1 f^\prime(x_o) + c_2 g^\prime(x_o) = 0[/tex]
We have two equations in the unknowns [tex]c_1[/tex] and [tex] c_2[/tex].
[tex]\left \vert \begin{array}{cc}f(x_o) & g(x_o) \\ f^ \prime (x_o) & g ^\prime (x_o) \end{array} \right \vert \neq 0 [/tex]
Therefore there are unique solutions.
One such solution is: [tex]c_1 = c_2 = 0[/tex]
This turns out to be the only solution.
Therefore [tex]f(x)[/tex] and [tex]g(x)[/tex] are independent.
I'm sure my wording is a little off. But I'm still confused about how I would prove it the other way.
Any help would be greatly appreciated.
dogma
I'm working on a proof by induction of the Wronskian and need a little boost to get going.
So, here goes:
If [tex]y_1,...,y_n \in C^n[a,b][/tex], then their Wronskian is:
[tex]Wr(y_1,...,y_n)=det\left(\begin{array}{ccc}y_1&\cdots&y_n\\y_1\prime&\cdots&y_n\prime\\\vdots&\vdots&\vdots\\y_1^{(n-1)}&\cdots&y_n^{(n-1)}\end{array}\right)[/tex]
In general, a set of functions will be linearly independent IFF the Wronskian is not identically zero.
Prove this for n = 2, that is [tex]f(x), g(x)[/tex] are independent IFF
[tex]\left\vert\begin{array}{cc}f(x)&g(x)\\f^\prime(x)&g^\prime(x)\end{array}\right\vert[/tex] is not identically zero.
Okay, I think I understand how to prove this in one direction. That is, assuming the [tex]Wr(f,g) \neq 0[/tex] and showing that [tex]f(x), g(x)[/tex] are independent.
But I'm a little stuck in the assumptions for proving the other direction.
This is what I have for the proof in one direction:
Assume [tex]Wr(f(x),g(x)) \neq 0[/tex].
Then there exists some [tex]x_o[/tex] such that [tex]Wr(f(x_o),g(x_o)) \neq 0[/tex].
Assume:
[tex]c_1 f(x) + c_2 g(x) \equiv 0[/tex]
[tex]c_1 f^\prime(x) + c_2 g^\prime(x) \equiv 0[/tex]
then
[tex]c_1 f(x_o) + c_2 g(x_o) = 0[/tex]
[tex]c_1 f^\prime(x_o) + c_2 g^\prime(x_o) = 0[/tex]
We have two equations in the unknowns [tex]c_1[/tex] and [tex] c_2[/tex].
[tex]\left \vert \begin{array}{cc}f(x_o) & g(x_o) \\ f^ \prime (x_o) & g ^\prime (x_o) \end{array} \right \vert \neq 0 [/tex]
Therefore there are unique solutions.
One such solution is: [tex]c_1 = c_2 = 0[/tex]
This turns out to be the only solution.
Therefore [tex]f(x)[/tex] and [tex]g(x)[/tex] are independent.
I'm sure my wording is a little off. But I'm still confused about how I would prove it the other way.
Any help would be greatly appreciated.
dogma