A continuous function equation.

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[AlexandreRibeiroXRV]

Uhh... something doesn't seem to add up. If f(x) = 1/(1 - x), for all x such that 0 <= x < 1 then f(x) >= 1 and therefore it falls out of the proposed range. Have I gotten something wrong?

[arkitson]

How to solve function equations:
Step 1. Guess the correct answer.
Step 2. Show that the answer that you guessed is correct.

[blackpenredpen]

Beautiful solution! And I am also going to compliment your well-designed thumbnail!

[BrollyyLSSJ]

My first thought was basically the same approach, but without "guessing" the solution form.

Using the functional equation, we can show by induction that for all x in (-1, 1) range and natural n we have:
f(x^(2^n)) = ((1-x)f(x))/(1-x^(2^n)),
with the base case given by
f(x^2) = f(x)/(1+x) = ((1-x)f(x))/(1-x^2).

Using continuity of f, with n approaching infinity, we arrive in the limit at equation:
f(0) = (1-x)f(x)
f(x) = 1/(1-x)

[cormacfarrell]

I think I'm missing something here but how do we have uniqueness of the solution?

[noahtaul]

Well if the codomain is really (-1, 1) then there aren’t any functions because f blows up at 1. Was that a typo?

[Happy_Abe]

How do we know this is the only function that can answer the question?

[anshumanagrawal]

If the values in (-1, 1) map to (-1, 1), then how can 0(which belongs to (-1, 1)) map to 1(which doesn't belong to (-1, 1)) ?

[nandoaires]

I didn't quite get how you prove this is the only solution. As far as I understood, you proved this is a solution, not all of those.

[martinepstein]

How can f(0) = 1 if the range doesn't include 1?

[Lembo]

Thanks for your tireless work with the video output. These problems are far more stimulating than crosswords or sudokus with my morning coffee. Feels like I'm back in undergrad.

[skylardeslypere]

How would you check that there are no other solutions? I.e. no solutions where f is NOT of the form f*(x)/(1+x)?

[Sam-fvyw]

I guess we can also solve this question by simply assuming that the function f(x) = summation from i=0 to infinity CiX^i and doing the same with f(X^2), and then we know F(X) = F(X^2)(1+x). Then we just need to equate the coefficient. the coefficient came out to be equal i.e. C0 = C1 = C2 then we can assume Ci = c and solve for it giving us the beautiful series of F(X) = C(1 + X + X^2 + which is F(X) = C/(1 - X) and upon seeing that the domain is also in the range (-1, 1) we know the geometric progression is applicable. the F(0) = 1 gives us C = 1 thus we get solution without thinking a lot.

[briemann]

Just finished one video and get to see another! :)

[cycklist]

In a maths contest will you get marks for making a guess based on the phrasing of the question like that? Or would you have to come up with another approach for full marks?

[TheNiTeMaR]

Here is my different solution in the reply (Youtube won't show it as a comment for some reason?). Great video!

[michaelz]

Rather than guessing most of the answer: You have f(x) = (1 + x)f(x^2) = (1 + x)(1 + x^2) f(x^4) = (1 + x)(1 + x^2)(1 + x^4)f(x^8), etc. so taking limits one gets f(x) = f(0) times the product from n = 0 to infinity of (1 + x^(2^n)). Since the product from n = 0 to n = m of (1 + x^(2^n)) is the sum from n = 0 to n = 2^(m+1) - 1 of x^n, the infinite product is 1/(1 - x) and you're done.

[mriel]

Because of the (1+x) at the bottom, I assumed F(x) was probably a fraction of two polynomials, let's say P(x)/Q(x). If we assume the "asymptotic degree" of F(x) to be:

degree( F(x) ) = degree( P ) - degree( Q ) = N, then
degree( F(x²) ) = 2N, and
degree( F(x)/(1+x) ) = N - 1

leading to 2N = N - 1 => N = -1

The 3 simplest functions F that satisfy this are 1/x, 1/(1+x) and 1/(1-x). I chose the latter considering the difference of two squares, and that turned out to be the answer.

That, of course, does not account for the uniqueness of the solution.

[toriknorth]

Since f(x) only differs from f^(x) by a polynomial factor the number of possible functions for f(x) is identical to the number of possible functions for f^(x). The limit argument showed that there's only a single possible function for f^(x) that satisfies the modified functional equation, thus there's only a single possible function for f(x) that satisfies the original functional equation.

[gibbogle]

I did it crudely: guessed f(x) = (ax + b)/(cx + d) then found b = d from f(0) = 1, then from f(x^2) = f(x)/(x+1) by equating coefficients that a = 0 and c = -b,
giving f(x) = b/(-bx + b) = 1/(1-x).