Find f(2) of a Polynomial Function | R and f(2)=5

[utkarshakash]

Homework Statement

If f(x) is a polynomial function satisfying 2+f(x)f(y)=f(x)+f(y)+f(xy), x,y belongs to R and if f(2)=5, then find the value of f(f(2))

Homework Equations

The Attempt at a Solution

The question clearly seeks the value of f(5). I put x=0 and y=2. Then
2+f(0)f(2)=f(0)+f(2)+f(0)
2+5f(0)=2f(0)+5
f(0)=1

Now I put x=0 and y=5
2+f(0)f(5)=f(0)+f(5)+f(0)
f(5)=f(5)

 

[ehild]

f(x) is a polynomial function. What does it mean?
ehild

 

[haruspex]

Consider a substitution f(x) = g(x) + c. Can you find a value of c that simplifies the equation?

 

[utkarshakash]

f(x) is a polynomial function. What does it mean?



ehild

Nothing special I can think of.

 

[Office_Shredder]

Consider a substitution f(x) = g(x) + c. Can you find a value of c that simplifies the equation?

I'm going to second this suggestion

 

[ehild]

Nothing special I can think of.

f(x) is a polynomial function, of form f(x)=a₀+a₁x+a₂x²+a₃x³+...

What relations do you get for the coefficients from the given equation and data?


ehild

 

[utkarshakash]

f(x) is a polynomial function, of form f(x)=a₀+a₁x+a₂x²+a₃x³+...

What relations do you get for the coefficients from the given equation and data? ehild

I still don't know the degree of polynomial

 

[utkarshakash]

Consider a substitution f(x) = g(x) + c. Can you find a value of c that simplifies the equation?

Ok following your method I arrive at this

2+g(x)g(y)+(c-1){g(x)+g(y)}=3c-c²+g(xy)

 

[haruspex]

Ok following your method I arrive at this

2+g(x)g(y)+(c-1){g(x)+g(y)}=3c-c²+g(xy)

Right, so what value of c will simplify that greatly?

 

[ehild]

I still don't know the degree of polynomial

That is you need to figure out. From the condition f(2)=5 you get a relation between f(y) and f(2y), and that can be fulfilled with polynomials of a certain degree.

ehild

 

[utkarshakash]

Right, so what value of c will simplify that greatly?

The only number I can think of is 1

 

[haruspex]

Right, so what equation do you get for g()? When you have that, suppose α is a root of g(x). What other root(s) can you then deduce?

 

[ehild]

I show my way as I think it is quite simple and straightforward.

2+f(x)f(y)=f(x)+f(y)+f(xy)

From f(2)=5 follows: 2+5f(x)=5+f(x)+f(2x) ----> -3+4f(x)=f(2x)*

f(x) is a polynomial f(x)=a₀+a₁x+a₂x²+...+a_kx^k+...

Plug into * and compare the coefficients of powers of x on both sides

-3+4(a₀+a₁x+a₂x²+...+a_kx^k+...)=a₀+2a₁x+4a₂x²+...+2^ka_kx^k+...-3+4a₀=a₀
4a₁=2a₁
4a₂=4a₂
.
.
.
4a_k=2^ka_k

What is the degree of the polynomial?

ehild

 

[haruspex]

I show my way as I think it is quite simple and straightforward.

Ah, but mine is so elegant

 

[Dick]

Ah, but mine is so elegant

Elegant and cute, but less obvious.

 

[ehild]

I must be dumb but still do not know your solution.

ehild

 

[Dick]

I must be dumb but still do not know your solution.

ehild

I saw it. g(x)g(y)=g(xy) means if x is root of g then xy must be a root of g for ANY y. Severely limits the choice of roots. I think this is little too subtle.

 

[ehild]

I saw it. g(x)g(y)=g(xy) means if x is root of g then xy must be a root of g for ANY y. Severely limits the choice of roots. I think this is little too subtle.

I reached here, but what after? f(x)=1+xh(x). But it is obvious as f(x) is polynomial, and f(0)=1 (obtained by the OP already). Find the possible root of h?

ehild

 

[Office_Shredder]

I reached here, but what after? f(x)=1+xh(x). Find the possible root of h?

ehild

I'm not sure what h is supposed to be here. But once you get to the post you quoted you say "ah, g(x) must be x^k" and use f(2) = 5 to figure out what k is

 

[ehild]

I'm not sure what h is supposed to be here. But once you get to the post you quoted you say "ah, g(x) must be x^k"

I do not see that "ah" Perhaps I stick to my version too much which gives the degree at once.

ehild

 

[haruspex]

I do not see that "ah" Perhaps I stick to my version too much which gives the degree at once.

ehild

If α is a root of g(x) then so is every multiple of α. So α = 0, and g(x) = ax^k, some a, k. From the equation involving g, a = 1.

 

[ehild]

If α is a root of g(x) then so is every multiple of α. So α = 0, and g(x) = ax^k, some a, k. From the equation involving g, a = 1.

Why can not have g(x) other roots than zero?

ehild

 

[Office_Shredder]

Why can not have g(x) other roots than zero?

ehild

If g(b) = 0, then
g(b)g(y) = g(by)
0 = g(by)

If b is not equal to zero, I can pick y to make b*y any arbitrary number, so g = 0 always. Therefore the only possible value b can be is zero

 

[ehild]

I awoke at last (it is 8 am here). So from g(b)=0 follows that b=0; then all b-s have to be zero if g(b)=0. Thanks :)

But my solution is also nice

ehild

 

[haruspex]

I awoke at last (it is 8 am here). So from g(b)=0 follows that b=0; then all b-s have to be zero if g(b)=0. Thanks :)

But my solution is also nice

ehild

Sure. One reason I like my solution is that it solves the functional equation in general. You only have to plug in the datapoints given at the end.

 

[ehild]

I have to admit that your solution is really elegant and cute

ehild

 

[utkarshakash]

Right, so what equation do you get for g()? When you have that, suppose α is a root of g(x). What other root(s) can you then deduce?

g(x)g(y)=g(xy). Ok I assume that g(α)=0. So g(α)g(y)=g(αy) => g(αy)=0. This means any multiple of α is a root of g(x). But how is this result useful to me?

 

[haruspex]

g(x)g(y)=g(xy). Ok I assume that g(α)=0. So g(α)g(y)=g(αy) => g(αy)=0. This means any multiple of α is a root of g(x). But how is this result useful to me?

A polynomial has only finitely many roots. So you can deduce the value of α.

 

[utkarshakash]

A polynomial has only finitely many roots. So you can deduce the value of α.

Ok I think the value of α is 0.

 

[haruspex]

Ok I think the value of α is 0.

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

 

[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.