Functional Equation

Find all functions $$f: \mathbb R \to \mathbb R$$ such that for arbitrary real numbers $$x$$ and $$y$$,

$f(x+y) + f(x) f(y) = f(xy) + f(x) + f(y) .$

Note by Shrihari B
2 years, 4 months ago

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Put $$x=y=0$$ $f(0)+f(0)^{2}=3f(0)$ This implies $$f(0)=0$$ or $$f(0)=2$$.

CaseI : $$f(0)=2$$ Now put $$x=0$$ $3f(y)=4+f(y)$ $\boxed{f(y)=2}$

Case II: $$f(0)=0$$ Put $$x=y=2$$ $f(4)+f(2)^{2}=f(4)+2f(2)$ $$f(2)=2$$ or $$f(2)=0$$.

Since the main functional equation all terms are only monic functions so another solution is $$f(y)=0$$.

Other solution is $$f(y)=y$$ which can be proved by induction.

- 2 years, 4 months ago

As mentioned by Ivan, I don't understand what you are doing here. I agree up the Case 2, but fail to follow form "Since the main functional equation ..."

Staff - 2 years, 4 months ago

I mean that both sides have terms which have 1 as coefficient so each term can be zero.

- 2 years, 4 months ago

Yea Ivan has a point. I don't understand how u can use induction when the domain is reals ?

- 2 years, 4 months ago

The induction part is wrong, but it's a minor mistake; verifying that $$f(x) = x$$ works is simply plugging it into the equation. The problem is that other potential solutions haven't been ruled out.

- 2 years, 4 months ago

You haven't shown that there is no other solution. In fact, as my solution suggests, if we restrict $$f$$ only to the integers, there exists another solution, and you haven't ruled it out.

- 2 years, 4 months ago

Let $$P(x,y)$$ be the statement $$f(x+y) + f(x)f(y) = f(xy) + f(x) + f(y)$$.

$$P(x,0) \implies f(x) + f(x)f(0) = f(0) + f(x) + f(0) \implies f(x)f(0) = 2f(0)$$. If $$f(0) \neq 0$$, we can cancel it to give $$f(x) = 2$$ for all $$x$$, which is a solution. Otherwise, $$f(0) = 0$$.

$$P(2,2) \implies f(4) + f(2)f(2) = f(4) + f(2) + f(2) \implies f(2)^2 = 2f(2)$$. This gives $$f(2) = 0$$ or $$f(2) = 2$$.

$$P(1,1) \implies f(2) + f(1)^2 = 3f(1)$$. If $$f(2) = 0$$, this leads to $$f(1) = 0$$ or $$f(1) = 3$$. If $$f(2) = 2$$, this leads to $$f(1) = 1$$ or $$f(1) = 2$$.

$$P(-1,1) \implies f(-1)f(1) = 2f(-1) + f(1)$$. $$f(1) = 2$$ leads to a contradiction, so we are left with three cases. To recap, our current cases are $$(f(1), f(2)) = (0,0), (1,2), (3,0)$$.

$$P(x,1) \implies f(x+1) + f(x)f(1) = 2f(x) + f(1)$$. We can now apply induction (in both ways, going on the positive integers and the negative integers) to determine the value of $$f$$ on the integers:

Case 1: $$f(1) = 0,$$. Then $$P(x,1) \implies f(x+1) = 2f(x)$$. Inducting gives $$f(x) = 0$$ for all integer $$x$$.

Case 2: $$f(1) = 1$$. Then $$P(x,1) \implies f(x+1) = f(x) + 1$$. Inducting gives $$f(x) = x$$ for all integer $$x$$.

Case 3: $$f(1) = 3$$. Then $$P(x,1) \implies f(x+1) = 3 - f(x)$$. Inducting gives $$f(x) = 3$$ if $$x$$ is odd and $$f(x) = 0$$ if $$x$$ is even.

Let $$x = \frac{p+1}{p}, y = p+1$$ for nonzero integer $$p$$. Note that $$x+y = xy$$. Thus $$P(x,y) \implies f(x)f(y) = f(x) + f(y)$$. Since we know the value of $$f(y)$$, we can determine the value of $$f(x)$$. That is, the value of $$f \left( 1 + \frac{p}{1} \right)$$.

Using $$P(x,1)$$, we can induct again to determine the values of $$f \left( n + \frac{1}{p} \right)$$ for all integer $$n$$, including $$n = 0$$.

$$P \left( \frac{1}{p}, n \right) \implies f \left( \frac{n}{p} \right) + f \left( \frac{1}{p} \right) f(n) = f \left( n + \frac{1}{p} \right) + f \left( \frac{1}{p} \right) + f(n)$$. Since we know the values of $$f \left( \frac{1}{p} \right)$$, $$f \left( n + \frac{1}{p} \right)$$, and $$f(n)$$, we know the value of $$f \left( \frac{n}{p} \right)$$. That is, we know $$f$$ to the rationals. I haven't actually worked it out to figure out whether Case 3 still persists.

I'm not sure how to extend this to the reals. I'm also pretty sure this is too convoluted (that is, there should be a simpler solution). You might want to first try to solve the following: find all $$f$$ on the reals such that $$f(x+y) = f(x) + f(y)$$ and $$f(xy) = f(x)f(y)$$ for all reals $$x,y$$.

- 2 years, 4 months ago

I believe that you're missing the constant solution $$f(x) = 2$$.

Staff - 2 years, 4 months ago

I already wrote that one near the top.

- 2 years, 4 months ago

Can u too please post some functional equations to solve ? I am quite weak at that

- 2 years, 4 months ago

Try these:

$$1$$. If $$f$$ is a polynomial function satisying $$2+f(x)f(y)=f(x)+f(y)+f(xy)$$ for all real $$x,y$$ and if $$f(2)=5$$, find $$f(f(2))$$.

$$2$$. A polynomial function $$f(x)$$ satisfies the condition $$f(x)f(\dfrac{1}{x})=f(x)+f(\dfrac{1}{x})$$ if $$f(12)=1729$$, then find $$f(10)$$.

$$3$$.Find all functions $$f$$ from $$R{0,1}$$ to $$R$$ satisfying the functional relation $$f(x)+f(\dfrac{1}{1-x})=\dfrac{2(1-2x)}{x(1-x)}$$. Source:INMO

Buy the book Functional Equations by B.J. Venkatachala. It has very nice functional equations problem .

- 2 years, 4 months ago

Hey should we make a thread for "Brilliant Functional Equation Contest" ? This is just an excuse for me to improve our functional equation as one problem in INMO is definitely on functional equations.

- 2 years, 4 months ago

A functional equation problem shouldn't have the statement "if $$f$$ is a polynomial..."; that statement makes it a polynomial equation problem (which has quite a different method to solve them).

- 2 years, 4 months ago