Solving a Nice Functional Equation

Показать описание

If you need to post a picture of your solution or idea:
#ChallengingMathProblems #FunctionalEquations

PLAYLISTS 🎵 :

Рекомендации по теме

Комментарии

f(x·y) = f(x)·f(y)/[f(x) + f(y)] assumes f(x) + f(y) = 0 is false for all real x, y, since the equation is assumed to be true for all real x, y. Suppose there exists some real t such that f(t) = 0. Hence f(t·y) = f(t)·f(y)/[f(t) + f(y)] = 0·f(y)/[0 + f(y)] = 0 for all real y, but this is impossible, since this contradicts f(x) + f(y) = 0 being false for real x, y. Therefore, 0 is not in the range of f. Also, if f(t) > 0 for some real t, then there cannot exist some real s such that f(s) = –f(t), because otherwise, f(s) + f(t) = 0 occurs, and again, this contradicts f(x) + f(y) = 0 being false for all real x, y. If f is continuous, then this implies the range of f is (0, ♾) or (–♾, 0), but not both. With this knowledge, the functional equation implies f(x^2) = f(x)/2 for all real x, which can be rewritten as f(x) = 2·f(x^2) = f[(–x)^2] = f(–x). Here is an issue: the equation implicitly assumes the domain is the field of real numbers, since no other domain is explicitly stated in the postulation of the exercise. However, the above result means f(0) = 2·f(0^2) = 2·f(0), which implies f(0) = 0, but we know this is impossible, because 0 is not in the range of f, even if it is discontinuous. So technically, the equation has no solutions. However, if we repostulate the exercise, then we can consider f to be an unknown function R\{0} —> R instead. We still have f(x) = f(–x) = 2·f(x^2), so f is an even function, but by the same token as before, we have f(1) = 0 = f(–1), which again, is impossible: 0 is not in the range of f. So in fact, if we want any solutions to the functional equation, then we must require f : R\{–1, 0, 1} —> R.

As in the video, f(x·y) = f(x)·f(y)/[f(x) + f(y)] for all real x, y not in {–1, 0, 1} implies 1/f(x·y) = [f(x) + f(y)]/[f(x)·f(y)] = 1/f(y) + 1/f(x), but the reason we can say this is because the possibility of 0 being in the image of f is discarded. This allows g(x) = 1/f(x) for all x in the domain of f, so that g(x·y) = g(x) + g(y) for all real x, y not in {–1, 0, 1}. Also, this means g(x) = g(–x) for all real x not in {–1, 0, 1}, so g is an even function.

Remember, g is a function R\{–1, 0, 1} —> R. The equation g(x·y) = g(x) + g(y) for all real x, y not in {–1, 0, 1}, is not an equation that this channel has analyzed in the cases where the domain is R\{–1, 0, 1}. All the previous videos have assumed that 1 is in the domain, for example. This complicates it quite tremendously. Why? For example, it may seem harmless to consider both 2 and 1/2 to be in the domain. However, if so, the functional equation implies g(1) = g(2) + g(1/2), but this contradicts the fact that 1 is not in the domain. Similarly, things such as g(2) + g(–1/2) = g(–1) contradict -1 not being in the domain, and then g(–2) + g(–1/2) = g(1) and g(–2) + g(1/2) = g(–1) cause similar problems. So this makes it clear that it is necessary that the domain be restricted such that |x| < 1 is satisied, or |x| > 1 is satisfied. So there are only two families of functions g that can satisfy the functional equation: g : (–♾, –1)&(1, ♾) —> R, or g : (–1, 1)\{0} —> R. With this restriction, there are no more issues, and now, the analysis that applies to g(x·y) = g(x) + g(y) for all nonzero real x, y applies to these restricted g as well. So for continuous g, g(x) = k·ln(|k|), where k is nonzero real, since k = 0 implies g(x) = 0 for all x in the domain, which is false. Therefore, f(x) = 1/[k·ln(|x|)] = λ/ln(|x|). The continuous solution families for the functional equation are f : (–1, 1)\{0} —> R such that f(x) = λ/ln(|x|), and f : (–♾, 1)&(1, ♾) —> R such that f(x) = λ/ln(|x|), as well as any restrictions thereof.

Interestingly, we must be careful to note that f : R\{–1, 0, 1} —> R such that f(x) = λ/ln(|x|) does not actually solve the functional equation, despite combining both restrictions, because it violates the condition that f(x) + f(y) = 0 must always be false, and because otherwise, you could produce f(1) from the functional equation, which again, is not possible. Also, f : (–1, 1)\{0} —> R such that f(x) = λ/ln(|x|) does have a removable singularity at 0, so if the functional equation was rewritten using limits, 0 could definitely be made part of the domain.

angelmendez-rivera

Its interesting the other way, put y=1
We have after cross multiplication: f(x) = 0.
I thought this is gonna be the solution.. But f(x) = 0 doesn't satisfy the given equation as value goes undefined.

abhinavbhutadab

6:00 *No! Honesty, you're a Mathematician* 😉💖

jimmykitty

Hey, can you suggest some books with quality function problems? That would be really helpful!!

minutemotivation

The full answer should be
f(x) = C/ln(x) for all x∊R+\{1}, C∊R\{0}

For example:
When you put x=y=1, you will see f(1) = 0 and you can become confused.
But, if you looked back to the original functional equation, put back f(1) = 0 to right hand side
The right hand side becomes undefined (0/0).
Similarly, C cannot be zero too.

mokouf

my solution.
f(xy)= 1/(1/fx + 1/fy)
ie 1/f(xy) = 1/f(x) +1/f(y)
if g(z) =1/f(z)
then g(xy) = g(x) +g(y)
it is well known that the function with the property of g() is log() to an arbitrary base.
hence f(x) = 1/log(x)
try log base 2 x=2, y=4
lhs =1/log(8) =1/3
rhs=(1*1/2)/(1+1/2) =(1/2)/(3/2) =1/3
so f(x) =1/log(x) to an arbitrary base
or f(x) =c/ln(x) with c being an arbitrary constant

davidseed

We can also further writr it as f(x)=c. ln e base x

RandomGuy-pezs

From the equation it is evident, when x=y, that f(x^2)=f(x)/2 => f(-x)=f(x)=f(|x|)
Either, you take g(x)=k*Ln(|x|) as solution, or if you restricted g to x>0, set f(x)=1/g(|x|) as end solution.
Curve of f has the shape _/U\_ with two vertical asymptotes at -1 and +1, respectively.

antonyqueen

Hey!!, Nice Problem, but I have a doubt?
If we put x=y, then equation becomes f(x^2)=f(x)/2
Of which if we substitute x=1, then f(1)=0.
Then substituting, x=x, y=1 in the original equation we get f(x)=0?
Why?Can you please sort it?

venkatsriramr

You should mention the domain and range before solving the problem. Otherwise f=0 would be the only solution here. (First insert 0 for both x, y and obtain f(0)=0 in the same way you get f(1)=0. Now let x=0 and y arbitrary, then you would get f(y)=0 for any y)

yt-