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)?