## Spring 2017, problem 38

Let $\mathbb{R}^+$ denote the set of positive real numbers. Find all functions $f : \mathbb{R}^+\to\mathbb{R}^+$ that satisfy $$ \left(1+y~f(x)\right)\left(1-y~f(x+y)\right)=1$$ for all $x,y\in\mathbb{R}^+$.

### Comments

The proof is a little bit longer . Seemingly you took x = 1 + y which works for x greater than 1 .

I may be wrong but after the second implication you find that the function has necssarily

some form for every x grater than one. Then you verify that this form works for every x. I think that this does not imply that there are no other solutions which are identical to the one you found only for every x greater than one. The fact that there are no other solutions follows by also assuming y to be one for every x and use f(1+x) with y in place of x in your solution

I am not enterely satisfied either with Hubert's proof and I would offer the folowing:

First let us assume that the function is continuous, then the original functional equation is equivalent to: f(x+y) - f(x) = -y f(x) f(x+y) This can be written as: (f(x+y) -f(x))/y = -f(x)f(x+y)

Now when y tends to 0 the left hand side is simply df/dx (the derivative and the right hand side is -f(x)^2 We now have a non linear differential equation whose general solution is 1/( a+x) So we have at least captured all continuous functions which satisfy thie original functional equation.

Hello Michel:

I too obtained the hyperbola $f(x) = \frac{1}{x + c}$ as the set of functions that satisfy the original equation. However, we need to be careful since we are concerned with mapping $\mathbb{R^{+}} \rightarrow \mathbb{R^{+}}$ only in the problem statement. This can be realized when $c \ge 0$ (i.e. a negative-valued vertical asymptote or the $y$-axis). If $c$ is negative (i.e. a positive-valued vertical asymptote), then we get a discontinuity at $x = c$ and $f(x)$ is negative for $x \in (0,c)$.

The best function that does this mapping for $x, y \in \mathbb{R^{+}}$ is $f(x) = \frac{1}{x}$ since it encompasses ALL positive reals in its domain and range. Otherwise, $f(x) = \frac{1}{x + c}$ has $y \in (0, \frac{1}{c})$, $x > 0$.

Let me know if you or anyone else here has questions/concerns.

Thanks, Tom Engelsman

**A remark**: I have been sending solutions here for years (**o**), before the new interactive presentation of 2015, about 50 % of the PDF problems were from analysis field (functions, integral, sequences, limits, ...), now it is close to 0%,

Why ?? analysis is so important, even in the us maths competition (see the Putnam or IMC problems ,,,), I deeply regret this lack, a real pity,, a well balanced variety of problems would emphasize solvers interest,, the idea of Purdue pow is so good,, meanwhile, I don't think I will often come back here.

(**o**) see the archive with my name Hubert Desprez in almost every solution since fall 2011

Hello Hubert:

I do recall you being on a number of past POW solutions.....nice to see a face to go with the name! I've also collaborated with Tairan Yuwen, Sorin Rubinstein, and Craig Schroeder often (before the Interactive Format started in Purdue's 2015-16 academic year).

Keep on solving & adieu, Tom Engelsman

Thank you Tom for remembering. That noninteractive period was kind of fun .

Since we started an exchange of ideas on the new format of the problem of the week I would also share with you some thoughts I have had for the past year. The interactive format is indeed interesting as it allows contributors to have an exchange and sometimes produce different solutions for the same problem. However I see two drawbacks with the present format: 1) Quite often it requires a mastery of LATEX software which I have not (and I would suspect a large part of amateur mathematicians like myself has not) and this restricts our contibutions to problems which do not involve complicated mathematical language. 2) More serioulsy it would be desirable to have,after a certain amount of time say one month, an official solution from the Department of mathematics. There is a quite large number of problems where there has not been any contibution and no answer and a fair number of problems which require some kind of statement from the Department (as an example take the preceeding problem # 37 which is still unclear to me).

Now on the first point if anyone knows a good way to learn LATEX....

Hi Michel:

Here's a link I've used to help teach myself lAtEx codes: http://artofproblemsolving.com/wiki/index.php?title=LaTeX:Symbols#Arrows

and one for syntax structure:
https://www.um.edu.mt/*_data/assets/pdf*file/0004/171373/LaTeX_Tutorial.pdf

Hate to say it, but the best way to learn is to practice!

Have a good one, Tom

Thanks Tom I will try it !

$x=1, \; a=f(1)\Rightarrow~f(1+y)=\frac a{1+ay}\Rightarrow~~f(x)=\frac a{1-a+ax}$ $f(x)\ge0\Rightarrow~~a\in(0,1] \Rightarrow~~$

$f(x)=\frac 1{\alpha+x}, \; \alpha\ge 0$ which works.