Find a function satisfying these conditions: f(x)f(f(x)) = 1 and f(2020) = 2019

[LCSphysicist]

Homework Statement

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Relevant Equations

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Let a continuous function f: R -> R such that f(x)f(f(x)) = 1 and f(2020) = 2019. What is the value of f(2018)?

I am having problem with this question.
I already tried, through various attempts, to find the function explicitly which satisfy this condition and f(0) different of 0. But it leads me to nothing. I am really thinking there is no way to find a explicit function, so i would appreciate any tip to this problem

 

[fresh_42]

Try to start at the other end. We know $f(2020)=2019$, so $1=f(2020)f(f(2020))=2019\cdot f(2019)$ and we have another value. Since we only want to know $f(2018)$ you just have to add another step.

 

[Office_Shredder]

Fresh, I don't see how you work backwards from there. I think you need to use this fact:

Let y be a point in the range of f. What is f(y)?

Also, they didn't tell us this function is continuous just for fun.

 

[LCSphysicist]

It is funny because if we call f(x) = 2018, we got f(2018) = 1/2018. It maybe has an interesting graph

 

[William Crawford]

Not sure if this is going to be of any help. But doesn't the condition $f(2020)=2019$ seem to be contradictory? I mean, let $y=f(x)$ for some arbitrary chosen $x\in\mathbb{R}$. Then
$$yf(y) = 1,$$
according to the functional equation. Or equivalently that
$$ f(x) = \frac{1}{x}.$$
This clearly doesn't fulfill the aforementioned condition. Though it does satisfy the conditions derived from it (fresh_42 in #2), that is $f(2019) = \frac{1}{2019}$ and $f(1/2019) = 2019$.

It might of course be that there are more functions (besides $x\mapsto 1/x$) that satisfy the functional equation.

EDIT: The function $f(x) = \frac{1}{x}$ isn't continuous at $x=0$ neither.

 

[William Crawford]

It is funny because if we call f(x) = 2018, we got f(2018) = 1/2018. It maybe has an interesting graph

Sure! However call $f(x) = 2020$. Then, by an identical argument, we instead get that $f(2020) = 1/2020$, which clearly contradicts the condition $f(2020) = 2019$.

I agree. If the function exist then its graph is intersting. It clearly isn't monotone.

 

[fresh_42]

Fresh, I don't see how you work backwards from there.

I didn't do the homework, I just wanted to give a hint to a possible way out of the dead end street. I hoped a quadratic equation would show up.

 

[Office_Shredder]

It is funny because if we call f(x) = 2018, we got f(2018) = 1/2018. It maybe has an interesting graph

This is exactly the point. If y is in the range, f(y)=1/y.

So is 2018 in the range of f? This is not obvious! For example, we know 2020 is not in the range of f, since f(2020) is not 1/2020.

I agree the graph of f is interesting. Once you have described why 2018 is in the range of f I think we can discuss what possible graphs of f look like and it will not be complicated to construct them.

 

[Vanadium 50]

First, the problem does not ask us to find f(x). It asks us to find f(2018). Answer the problem asked.

Second, it's really hard to figure out if this problem is subtle or just a mistake.

 

[Office_Shredder]

First, the problem does not ask us to find f(x). It asks us to find f(2018). Answer the problem asked.

Second, it's really hard to figure out if this problem is subtle or just a mistake.

I think this problem is intentionally subtle. If there was supposed to be a recursive solution like fresh proposed thinking about, they wouldn't have needed f to be continuous.

 

[LCSphysicist]

Not sure if this is going to be of any help. But doesn't the condition $f(2020)=2019$ seem to be contradictory? I mean, let $y=f(x)$ for some arbitrary chosen $x\in\mathbb{R}$. Then
$$yf(y) = 1,$$
according to the functional equation. Or equivalently that
$$ f(x) = \frac{1}{x}.$$
This clearly doesn't fulfill the aforementioned condition. Though it does satisfy the conditions derived from it (fresh_42 in #2), that is $f(2019) = \frac{1}{2019}$ and $f(1/2019) = 2019$.

It might of course be that there are more functions (besides $x\mapsto 1/x$) that satisfy the functional equation.

EDIT: The function $f(x) = \frac{1}{x}$ isn't continuous at $x=0$ neither.

I don't think the function 1/x is not continuous at x=0. Since it is not definite at this point, it does not need to be, i think. But yes, i think there are more functions, but i am not able to find them.

First, the problem does not ask us to find f(x). It asks us to find f(2018). Answer the problem asked.

Second, it's really hard to figure out if this problem is subtle or just a mistake.

The function is continuous and, certain values, we can see that, vary between 2019 (which is f(2020)) and 1/2019, as Fresh showed. So, to be more specified, the intermediate theorem allow us to search for values between 2019 and 1/2019. 2018 need to be image of some x, f(x). I am just seeing what it implies. I think this is what @Office_Shredder say with look at ranges.

 

[Office_Shredder]

I don't think the function 1/x is not continuous at x=0. Since it is not definite at this point, it does not need to be, i think. But yes, i think there are more functions, but i am not able to find them.

The function is continuous and, certain values, we can see that, vary between 2019 (which is f(2020)) and 1/2019, as Fresh showed. So, to be more specified, the intermediate theorem allow us to search for values between 2019 and 1/2019. 2018 need to be image of some x, f(x). I am just seeing what it implies. I think this is what @Office_Shredder say with look at ranges.

Right, so 2018 is in the range, which means f(2018) = 1/2018 as you figured out.

Now what are the possible graphs of this function?

Well we know between 1/2019 and 2019 it's just 1/x. One possible choice is that for x < 1/2019 Ave x > 2019, it wiggles around however you want so that f(x) is always between 1/2019 and 2019, and so f(2020)=2019.

You can expand it a bit, if you make it 1/x for x between 1/m and m for any m<2020 then you also get a valid function.

 

[fresh_42]

We have $\operatorname{range} f = 1/\operatorname{range}f \subseteq \operatorname{dom}f$ so $M:=\operatorname{range}f$ is closed under inversion, but not necessarily under multiplication. And we have $0\notin M$.

If $f(x\cdot y)=g(x)\cdot f(y)$ then $g(x)$ satisfies $g(x)\cdot g(g(x))=1$ as well.

It would be so a nice function if $f$ was the ordinary inversion or $2020$ was the golden ration instead.

Could it be, that such a function with $f(2020)=2020 - 1$ doesn't exist?

 

[Vanadium 50]

Also, they didn't tell us this function is continuous just for fun.

It does preclude the trivial solution: f(x) is usually equal to 1.

 

[Fred Wright]

How to find $f(x)$ in three easy steps:
1. differentiate
2. integrate
3. solve for the constant of integration