A Nice Functional Equation is Investigated Here!

Algebra Level 5

What can we say about the cardinality of the set of solutions of the functional equation f ( 1 x − 1 ) f ( x ) = f ( 1 x − 1 ) + 1 ? f\left(\frac{1}{x-1}\right)f(x)=f\left(\frac{1}{x-1}\right)+1?

1 0 Finite greater than 1 Countable Uncountable

Arturo Presa by Arturo Presa , 62, USA

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1 solution

Arturo Presa
Nov 30, 2019

Let us assume that a possible solution of the given functional equation has the form f ( x ) = x + a b x + c . f(x)=\frac{x+a}{bx+c}. By substituting into the de equation, we can see that we might choose a = b = 1 − c a=b=1-c for the equation to hold. Therefore, any function f c ( x ) = x + 1 − c ( 1 − c ) x + c f_c (x) =\frac{x+1-c}{(1-c)x+c} is a solution of the functional equation for any real value of c c . Besides that, it easy to see that if c 1 ≠ c 2 , c_1 \neq c_2, then f c 1 ≠ f c 2 . f_{c_1} \neq f_{c_2}. Then the cardinality of the set of all possible solutions must be uncountable. \boxed{\text{uncountable.}}

Can you show your own solution below?

3 Helpful 0 Interesting 0 Brilliant 1 Confused

How is the assumption of the possible solution made? By observation or is there a standard theorem/method ? Thank you.

Jacob Sony - 1 year, 5 months ago

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Jacob, in my opinion, this is mostly trials and errors in this case. You might assume, for example, that the solution is a polynomial and then by substitution into the equation you can find the coefficients. When I was solving this problem, I assumed that the solution is a rational function that is also a Mobius transformation, so a function of the form y = a x + b c x + d . y=\frac{ax+b}{cx+d}. Once I found the solutions, I realized that the procedure that I used works even better (easier to explain) if you assume that the function has the form y = x + a b x + c . y=\frac{x+a}{bx+c}. Of course, this does not prove that all possible solutions can be represented as an f c , f_c, but at least I found a family of solutions big enough to make the conclusion that the set of all possible solutions must be uncountable.

Arturo Presa - 1 year, 5 months ago

An ersatz is when you “try out” a solution known in advance to be the answer. The ersatz here is a special case of the form a x + b c x + d \frac{ax+b}{cx+d} , known as a Möbius transformation.

Jake Lai - 1 year, 5 months ago

f(x)=x is solution

Praveen Kumar - 1 year, 3 months ago

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