Linear independence with differentiable functions

[kbrono]

I don't this this is an overly complicated proof but it is one I have never seen or done before.

Let f be a polynomial with atleast two non-zero terms having different degrees. Prove that the set {f(x),xf'(x)} is linearly independent in P

Proof:

With out loss of generality we can take f₁(x) = 1 and f₂(x) then xf₁'(x) = 0 and xf₂'(x)= x. Thus we have the matrix A=[[1,x],[0,x]] and rref(A)=[[1,0],[0,1]] and is therefore linearly independent.

 

[micromass]

Uh, I don't really understand your proof... what are f₁ and f₂? And why is there no loss in generality in showing this??
I really don't see what this proof has to do with the theorem...

 

[fzero]

You might want to expand a general polynomial f(x) and x f'(x) in a suitable basis.

 

[kbrono]

Sorry i use a weird notation, f_1(x) denotes the the first term in f(x) f_2(x) denotes second term in f(x).

im breaking down f(x) = 1+x and f'(x)=0 + 1 into a term by term format.

What I'm doing is setting up a matrix where the first like is f(x) and the second line is x(f'(x)) then using rref to show they are linearly independent. I say without loss of generality because we only need atleast 2 different terms with different degrees so it doesn't matter which two I pick.

Hope the clarify's what I am saying...

Looks like i forgot to add that {f(x),xf'(x)} is linearly independent

I could be way off mark but since we only need two different, doesn't it suffice to prove for just two different degrees?

Thank you

 

[fzero]

You've shown that 1+x and x (1+x)' are linearly independent, not that this is true of all polynomials.

 

[kbrono]

ah, ok so i should expand for each term out to the nth?

 

[fzero]

If you mean "write down a general nth order polynomial," then yes.