Functional equations problem 3

Question

Find the functions that satisfy the relation

$$f(x)f\left(\frac{1}{x}\right)=f(x)+f\left(\frac{1}{x}\right).$$

Did not get any idea how to do this.

Answer 1

Define $g(x)=\frac{x}{x-1}$ for $x\neq 1$. Then, $$g(g(x))=\frac{\frac x{x-1}}{\frac x{x-1}-1}=x$$ that is, this function is the inverse of itself.

There are two numbers $x$ such that $g(x)=x$, namely $0$ and $2$. Thus, we can define a partition in $\Bbb R-\{0,1,2\}$ that consists on all (not ordered) pairs $\{a,b\}$ such that $g(a)=b$, or equivalently, $g(b)=a$. Then, for each $x$ such that $0<|x|<1$, pick any of these pairs $\{a,b\}$ and define $f(x)=a$ and $f(1/x)=b$. You can see that $$f(x)+f\left(\frac 1x\right)=a+\frac{a}{a-1}=\frac{a^2}{a-1}=f(x)f\left(\frac 1x\right)$$

Now, assign to $f(-1)$ any of the numbers $\{0,2\}$ and assign to $f(1)$ the other, or the same, as you wish.

Any function $f$ so defined will satisfy the condition.

Answer 2

One of the function satisfying this condition will be $f(x) = 1+x$

Then $f(1/x) = 1 + 1/x$;

therefore, $f(x)f(1/x) = (1+x)(1+1/x) = 2 + x + 1/x$, and $f(x)+f(1/x) = 1 + x + 1 + 1/x = 2 + x + 1/x$

It is a hit and trial approach and to get more precise function some additional conditions must surely be satisfied by the function. For the above example $f(0) = 1$ is satisfied.

If f(x) is some polynomial function

the polynomial can be written as:

$f(x)$=a0+a1${\times {x^1}}$+a2${\times x^2}$+a3${\times x^3}$+...

$f(1/x)$=a0+${\frac{a1}{{x^1}}}$+${\frac{a2}{{x^2}}}$+${\frac{a3}{{x^3}}}$+... we are given:

$f(x).f(1/x)=f(x)+f(1/x)$

therefore:

$f(1/x)={f(x)/(f(x)−1)}$ and we can write:

$f(1/x)$=a0+${\frac{a1}{{x^1}}}$+${\frac{a2}{{x^2}}}$+${\frac{a3}{{x^3}}}$+...+${\frac{an}{{x^n}}}$

=${\frac{1}{{x^n}}}$(a0${\times {x^n}}$+a1${\times {x^{n-1}}}$+a2${\times {x^{n-2}}}$+a3${\times {x^{n-3}}}$+...+${{an^n}}$)

=$f(x)/(f(x)−1)$from above

∴$(f(x)−1)$(a0${\times {x^n}}$+a1${\times {x^{n-1}}}$+a2${\times {x^{n-2}}}$+a3${\times {x^{n-3}}}$+...+${{an^n}}$)=${x^n}$f(x)

∴(a0−1+a1${\times {x^1}}$+...+an${\times {x^n}}$)(a0${\times {x^n}}$+a1${\times {x^{n-1}}}$+...+an)

=${{x^n}}$(a0+a1${\times {x^1}}$+...+an${\times {x^n}}$)

=a0${\times {x^n}}$+a1${\times {x^{n+1}}}$+...+an${\times {x^{2n}}}$

from inspection, we can infer:

an(a0−1)=0 => an=0 or a0=1

1. if we pursue the case of an=0 then we will successively get a(n−1)=0, a(n−2)=0, ..., a1=0 and will end up with a0=2. this gives us $f(x)=f(1/x)=2$ and satisfies the condition $f(x).f(1/x)=f(x)+f(1/x)$ this can be one of the case.

2. if we pursue the case a0=1 we will successively be able to prove that a1=0, a2=0, ...an−1=0, and will end up with: an$\times{x^n}$(${x^n}$+an)−${x^n}$+an${\times{x^{2n}}}$ from which we can prove an=1. this gives us $f(x)$=1+${x^n}$ and satisfies the condition $f(x).f(1/x)=f(x)+f(1/x)$.