hint anyone?

[Sarahisme]

would anyone be able to give me a hint as to what to do with this quesiton?

 

what i have said so far is

[math]

f(x) = ax^n, \ n \geq 1

[/math]

[math]

g(x) = bx^m, \ m \geq 1

[/math]

and

[math]

h(y,z) = cy^p + dz^q, \ p,q \ \geq 1

[/math]

[Sarahisme]

oops, here is the question

[Sarahisme]

ok, so now i am thinkning that it has something to do with matrices and linear transformations..... is that the right line of thought?

[Sarahisme]

hello?

[Severian]

Write

 

[math]f(x)=\sum_n a_n x^n[/math]

 

[math]g(x)=\sum_m b_m x^m[/math]

 

[math]h(y,z)=\sum_{k,l} c_{kl} y^k z^l[/math]

 

Now stick your definitions for f(x) and g(x) into h(y,z) to give h(f(x),g(x)) and find the condition that it is zero for all x (ie. that all the coefficients vanish). This will give you the form of h(y,z) which you need.

[Sarahisme]

So to do this problem do i need to know stuff about difference equations ?? (its a chapter in my linear algebra book)

 

??

[gnpatterson]

proof would, I assume, be by iteration, and that is where your knowledge of difference equations would come in.

 

if you make f(x) and g(x) be of degree n then make the proof work if for some f' and g' of degree n-1 ( f' related to f in some way and g' related to g in some way) if h' exists

 

hope this helps

[CPL.Luke]

couldn't you just make h(x,y) equal to 0 for all values of x and y?

 

I have a feeling that there is some big reason why you couldn't do this, it just seems logical

 

Edit: woops misread the question, the polynomial had to be non-zero.

 

Thats what I get for posting at 3:00 in the morning

[gnpatterson]

Sorry I looked at the problem and I cant prove it, the method i suggested is rubbish. please post if you do get the proof

[Sarahisme]

yeah its cool, i got it in the end, you do it by trying to get lineraly dependent set sort of thing

 

thanks for trying though