Finding f(f(2)) from the functional equation 2+f(x)f(y)=f(x)+f(y)+f(xy)

[utkarshakash]

Exactly. So, what is the general form of g(x), and thus, the general form of f(x)?

f(x)=g(x)+1

 

[haruspex]

f(x)=g(x)+1

Well, yes, that's the relationship between f(x) and g(x), but what is the general form of g(x)? We have shown that all its roots are 0, right? So what does that polynomial look like?

 

[utkarshakash]

Well, yes, that's the relationship between f(x) and g(x), but what is the general form of g(x)? We have shown that all its roots are 0, right? So what does that polynomial look like?

You have already stated that in an earlier post

 

[haruspex]

You have already stated that in an earlier post

Indeed I did, but from your post #29 it seemed like you'd not been reading all those exchanges, perhaps because you wanted to figure it for yourself with a few hints.
So, do you understand why g(x) = ax^k for some a and k? Do you understand how to determine a from the equation for g(), and then the value of k from the given datapoints?

 

[utkarshakash]

Indeed I did, but from your post #29 it seemed like you'd not been reading all those exchanges, perhaps because you wanted to figure it for yourself with a few hints.
So, do you understand why g(x) = ax^k for some a and k? Do you understand how to determine a from the equation for g(), and then the value of k from the given datapoints?

I do not know how to derive a but assuming a=1, I can find k.

 

[Dick]

I do not know how to derive a but assuming a=1, I can find k.

You would able to derive it if you used some other properties of f(x) you figured out before.

 

[haruspex]

I do not know how to derive a but assuming a=1, I can find k.

In your post 27 you wrote, correctly,

g(x)g(y)=g(xy)

Substitute g(x) = ax^k in there.

 

[utkarshakash]

In your post 27 you wrote, correctly,

Substitute g(x) = ax^k in there.

Thanks!

PS-This was the longest thread I have ever started.