Proving Independent Functions: Wronskian for n=2

In summary: R.In summary, the Wronskian is a determinant that can be used to determine the linear independence of a set of functions. It can be calculated using a specific formula and is not identically zero if and only if the set of functions is linearly independent. This can be proven for a set of two functions by showing that the determinant of their derivative is not identically zero, and vice versa.
  • #1
dogma
35
0
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
 
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  • #2
Looks good to me.

And the other way is easier!

Suppose f and g are NOT independent. Then there exist a, b such that af(x)+ bg(x)= 0 for all x. If b is not 0, then g(x)= (a/b)f(x) and the Wronskian if f(x)g'(x)- f'(x)g(x)= f(x)(b/a)f'(x)- f'(x)(b/a)f(x)= (b/a)(f(x)f'(x)- f'(x)f(x))= 0 for all x.

(If b is 0, then f(x) must be identically 0 so the Wronskian is 0(g'(x))- 0(g(x))= 0.)
 
  • #3
Thanks, Mr. Ivy!

That makes things a lot clearer! I can see the light.


dogma
 
  • #4
Ok, so Ivy proved the way you wanted : (Linearly dependent)->Wr(x)=0 forall x

The problem is that : this is equivalent to "exists t Wr(t)!=0->Lin independent" which is what dogma proved.

Since you said it's IFF, then remains to prove :

"Lin indep->exists t Wr(t)!=0" which is equivalent to "W(x)=0 \forall x->lin dep"

Since W(x)=0=f(x)g'(x)-f'(x)g(x).

If f(x)!=0 and g(x)!=0 for a certain domain of x...then clearly f(x)=Cg(x) from the diff. equ...(f'/f)=(g'/g)...but only on this domain

If exist a domain with f(y)=0, f'(y)=0 then g(x) can be anything

Hence the theorem is only in one direction since you can find f,g lin indep such that Wr(f,g)=0 forall x

For counterexample :

[tex] f(x)=\left\{\begin{array}{cc} 0 & |x|\leq 1\\ (x+1)^2(x-1)^2 & else\end{array}\right.[/tex]
[tex] g(x)=(x-1)^2(x+1)^2 [/tex]

Then cleary f,g are in C^1(R) :

[tex] f'(x)=\left\{\begin{array}{cc} 0 & |x|\leq 1 \\4x(x-1)(x+1) & |x|\geq 1\end{array}\right.[/tex]

You see that f'(\pm 1+)=0=f'(\pm 1-) so f'(x) is continuous on R.

If |x|<1 then W=0 clearly since f=0
if |x|>1 then W=0 clearly since f,g are dep on |x|>1

However f and g are lin indep on R, since if f-g=0 on |x|>1 then f-g!=0 in [-1;1]
 

Related to Proving Independent Functions: Wronskian for n=2

1. What is the Wronskian for n=2?

The Wronskian for n=2 is a mathematical tool used to determine if two functions are independent, meaning that one function cannot be written as a multiple of the other. It is a determinant of a 2x2 matrix that is formed by taking the derivatives of the two functions.

2. How do you calculate the Wronskian for n=2?

To calculate the Wronskian for n=2, you first need to take the derivatives of the two functions and write them in a 2x2 matrix. Then, calculate the determinant of this matrix by multiplying the elements on the main diagonal and subtracting the product of the elements on the other diagonal. The resulting value is the Wronskian.

3. What does a Wronskian value of zero mean?

If the Wronskian value for n=2 is zero, it means that the two functions are not independent. This could be due to them being linearly dependent, meaning that one function can be written as a multiple of the other, or that one function is a constant multiple of the other.

4. Can the Wronskian be used to determine if more than two functions are independent?

Yes, the Wronskian can be used for any number of functions, not just two. For n functions, the Wronskian is a determinant of an nxn matrix formed by taking the derivatives of the n functions.

5. What is the significance of the Wronskian in mathematics?

The Wronskian is an important tool in mathematics as it can be used to determine the linear independence of functions. It is also used in differential equations to check for the existence of a general solution by ensuring that the Wronskian is non-zero.

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