A Functional Equation from Crux Mathematicorum

[@roberttelarket4934]

Two like terms. They like each other. I love it!

[@roberttelarket4934]

Two like terms. They like each other. I just don't only like it I love it!

[@SyberMath]

😍

[@darkphoenixzyzz]

I did it in my head strait when I saw it but I only found the first solution y = x. Nice !

[@scottleung9587]

Nice!

[@maxhagenauer24]

As soon as you see it, you can very easily see how the right side can be written as how it looks on the left side if f(x) were x so they must be equal.

[@WahranRai]

Just rewrite : f(x) + 1/f(x) = (x^2 + 1)/x = x + 1/x ----> f(x) = x

[@darkphoenixzyzz]

missing 2nd solution here tho

[@mystychief]

Here the solution f(x)=1/x is also easy to see: if you fill in 1/x for x then on both sides of the "=" the two terms swap and the equality remains basically the same.

[@VanamaliParthasarathi]

​@@darkphoenixzyzzsince addition is commutative, swap right side as 1/x+x, then you get other solution

[@kikilolo6771]

Nice guess of a solution. Now you can search for the trivial proof

[@pipilu3055]

u observed a solution how does that show u found all the solutions

[@MichaelJamesActually]

symmetry for the win

[@joseluishablutzelaceijas928]

Aren't there in fact 16 solutions? One can choose, for each of the intervals (-infty, -1), (-1, 0), (0, 1) and (1, infty) the function x or the function 1/x respectively, with f(1) = 1 and f(-1) = -1. One has two choices for each interval and there are four intervals, giving a total of 2^4 = 16 solutions.

[@naharmath]

Yes, if f is supposed to be continuous. But here there are an infinite number of solutions

[@ehtibardzhafarov5108]

@@naharmaththis sybernath guy really doesn’t know much, does he? He could have added differentiability at all points except for zero to justify the “two solutions”, but he clearly unaware of the problem.

[@naharmath]

@@ehtibardzhafarov5108 even then we will have 4 solutions because of the non existence of f at 0

[@mircoceccarelli6689]

👍👍👍😁🤪👋

[@MochooCheung-pu8js]

A systematic way: let y=f(x) to save notations. y^2-(x+1/x)y+1=0. Solving this quadratic equation for y in terms of x yields y=x or y=1/x. Done.

[@CriticSimon]

Did you watch the video?

[@duccline]

f(x) = x or 1/x

[@gelbkehlchen]

Solution: f(x)+1/f(x) = (x²+1)/x |*f(x) ⟹ [f(x)]²+1 = (x²+1)/x*f(x) |-(x²+1)/x*f(x) ⟹ [f(x)]²-(x²+1)/x*f(x)+1 = 0 |p-q-formula ⟹ f1/2(x) = (x²+1)/(2x)±√[(x²+1)²/(4x²)-1] = (x²+1)/(2x)±√[(x^4+2x²+1-4x²)/(4x²)] = (x²+1)/(2x)±√[(x^4-2x²+1)/(4x²)] = (x²+1)/(2x)±√[(x²-1)²/(4x²)] = (x²+1)/(2x)±(x²-1)/(2x) ⟹ f1(x) = (x²+1)/(2x)+(x²-1)/(2x) = (x²+1+x²-1)/(2x) = 2x²/(2x) = x and f2(x) = (x²+1)/(2x)-(x²-1)/(2x) = (x²+1-x²+1)/(2x) = 2/(2x) = 1/x Checking the result by setting the result into the equation: 1st result f1(x) = x left side: f(x)+1/f(x) = x+1/x = (x²+1)/x right side: (x²+1)/x 2nd result f2(x) = 1/x left side: f(x)+1/f(x) = 1/x+x = (x²+1)/x right side: (x²+1)/x everything o.k.

[@kikilolo6771]

Actually the two proofs you present are false. You MUST check for multigraphs functions

[@tixanthrope]

KUDOZ!

[@CriticSimon]

What are you talking about?

[@kikilolo6771]

@@CriticSimon if you have (f(x))^2=1 it DOESN'T mean that there are two solutions. You also have f such that f(x)=1 if x>0 and f(x)=-1 if x

[@PooshanHalder]

3rd method: guess it like i did

[@giuseppemalaguti435]

x ,1/x

[@naharmath]

there are more than two solutions.

[@armacham]

name three solutions

[@benjaminvatovez8823]

Do you mean it could be multigraph functions?

[@darkphoenixzyzz]

Not real for sure. Maybe complex.

[@naharmath]

@@armacham without any condition of continuity a we can have f(x)=x for some values of x and f(x)=1/x for the others.

[@armacham]

@@naharmath f(x) = x is one solution. f(x) = 1/x is a second solution That's two. Name a third solution. We're all waiting.