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Find $f$ if $$ f(x)+f\left(\frac{1}{1-x}\right)=x $$ I think, that I have to find x that $f(x) = f\left(\frac{1}{1-x}\right)$ I've tried to put x which make $x = \frac{1}{1 - x}$, but this equation has no roots in real numbers.

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    $\begingroup$ We are told that for all $x$, except presumably $x=1$, we have $f(x)+f\left(\frac{1}{1-x}\right)=x$. We are asked to identify all the functions $f$ for which this is true. Solving this kind of functional equation often breaks up into two parts: (i) by exploration, decide on what the $f$ that work might look like and (ii) show there are no others. $\endgroup$ – André Nicolas Nov 22 '15 at 16:39
  • $\begingroup$ You probably want to use that if $g(x)=\frac{1}{1-x}$ then $g(g(g(x)))=x$. $\endgroup$ – Thomas Andrews Nov 22 '15 at 16:55
  • $\begingroup$ This is a Putnam problem no? $\endgroup$ – anon Nov 22 '15 at 17:18
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    $\begingroup$ @YuiToCheng This post is older and the content is arguably equally as good. Why not link #2265891 as a duplicate to this one? I see that you just edited one of the answers there, but the whole situation is kind of toss-up isn't it? $\endgroup$ – Lee David Chung Lin May 27 '19 at 12:17
  • $\begingroup$ @LeeDavidChungLin I incline to close it the other way around because I like Jean Marie's spectacular answer there. $\endgroup$ – YuiTo Cheng May 27 '19 at 12:29
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You didn't mention the functional equation holds for all $x\in\Bbb R_{\neq 1}$. I'm assuming it.

Let $y=\frac{1}{1-x}$. Then $y$ reaches all real values except $0$, so:$$f\left(\frac{y-1}{y}\right)+f(y)=\frac{y-1}{y},\, \forall y\in\Bbb R_{\neq0 }\tag{1}$$

Let $y=\frac{t-1}{t}$. Then $t=\frac{1}{1-y}$ reaches all real values except $0,1$ (because $y$ can't reach $0$):

$$f\left(\frac{1}{1-t}\right)+f\left(\frac{t-1}{t}\right)=\frac{1}{1-t},\,\forall t\in\Bbb R_{\neq 0,1}$$

$$\iff f\left(\frac{t-1}{t}\right)=\frac{1}{1-t}-f\left(\frac{1}{1-t}\right),\, \forall t\in\Bbb R_{\neq 0,1}$$

Substitute this into $(1)$:

$$\frac{1}{1-t}-f\left(\frac{1}{1-t}\right)+f(t)=\frac{t-1}{t},\, \forall t\in\Bbb R_{\neq 0,1}$$

$$\iff f\left(\frac{1}{1-t}\right)=\frac{1}{1-t}+f(t)-\frac{t-1}{t},\, \forall t\in\Bbb R_{\neq 0,1}\tag{2}$$

Your original equation says:

$$f(t)+f\left(\frac{1}{1-t}\right)=t,\, \forall t\in\Bbb R_{\neq 1}$$

Subsitute $(2)$ into it:

$$2f(t)+\frac{1}{1-t}-\frac{t-1}{t}=t,\,\forall t\in\Bbb R_{\neq 0,1}$$

$$\iff f(t)=\frac{t+\frac{t-1}{t}-\frac{1}{1-t}}{2}=\frac{t^3-t+1}{2(t-1) t},\,\forall t\in\Bbb R_{\neq 0,1}$$

(WolframAlpha agrees this is a solution).

Now to find $f(0), f(1)$, note that the original equation only cares about $f(0), f(1)$ when $x=0$, so $f(0),f(1)$ can be any real numbers such that $f(1)=-f(0)$.

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If $g(x)=\frac{1}{1-x}$ then verify that $g(g(g(x)))=x$.

Then:

$$\begin{align} f(x)+f(g(x))&=x\\ f(g(x)) + f(g(g(x))) &= g(x)=\frac{1}{1-x}\\ f(g(g(x))) + f(x) &= g(g(x))=\frac{x-1}{x} \end{align}$$

So this gives three linear equations, so solve for $f(x)$.

Subtract the second row from the first: $$f(x)-f(g(g(x)))=x-\frac{1}{1-x}$$

Then add this to the third:

$$2f(x)=x-\frac{1}{1-x}+\frac{x-1}{x}$$

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  • $\begingroup$ That is indeed a very elegant solution. $\endgroup$ – mickep Nov 22 '15 at 17:00
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    $\begingroup$ It requires that "magic" about $g(x)$, which I just happened to remember from seeing problems like this before. Not sure how one is supposed to know this if one hasn't seen it before. $\endgroup$ – Thomas Andrews Nov 22 '15 at 17:03
  • $\begingroup$ Would it be possible to use this method in a systematic fashion for any sum of rational expressions of functions? $f(h_1(x)) + f(h_2(x)) + \cdots = l(x)$ where the $h_k$ are rational functions in x? $\endgroup$ – mathreadler Nov 22 '15 at 17:03
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    $\begingroup$ Not likely, no. @mathreadler This is a particular property of $g(x)=\frac{1}{1-x}$. It might be useable in a few other cases of Mobius transformations $\frac{ax+b}{cx+d}$ of finite degree. $\endgroup$ – Thomas Andrews Nov 22 '15 at 17:05
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    $\begingroup$ This assumes $x\neq 0,1$ and only finds $f(x)$ at $\Bbb R_{\neq 0,1}$. The values $f(0),f(1)$ can be any real numbers such that $f(1)=-f(0)$ (the original equation only concerns $f(0), f(1)$ when $x=0$). $\endgroup$ – user236182 Nov 22 '15 at 17:50
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In case you're a bit miffed by the "trickiness" of the solution, here's how I came upon the answer by playing around with the definition of $f$. The key, as in both answers above, is to realize there's some cyclicality induced by $\frac{1}{1-x}$.

My first instinct was to play around with $x=0$, from which we get $f(0)+f(1)=0$. In order to get something out of this, though, we look for another way to express $f(0)$ or $f(1)$. However, to get the left-hand term to be $f(1)$ requires plugging an undefined point ($x=1$) in the right-hand term, and getting $f(0)$ out of the right-hand term requires taking an infinite limit of the left-hand term. So that seems like a dead end.

I moved on to try $x=-1$ since this is also simple to play with. We get

$$f(-1)+f(\frac{1}{2})=-1$$

Just for kicks, I tried plugging in $x=\frac{1}{2}$ since the numbers in the first equation were pretty palatable:

$$f(\frac{1}{2}) + f(2)=\frac{1}{2}$$

Again, not bad -- nice, pretty numbers. So I tried once more:

$$f(2)+f(-1) = 2$$

And now we're on to something. We note that we got $f(-1)$ to show up again, and that in the three equations we've written, there are only three unknowns -- namely, $f(-1), f(\frac{1}{2})$, and $f(2)$. This means we can solve for these three values (and luckily it's a simple linear form).

Now we begin to wonder whether we happened upon this by accident or whether we've found a more general phenomenon -- it appears that churning $-1$ through $\frac{1}{1-x}$ recursively brought us back to $-1$; is this true for other starting values?

Indeed it is -- $\frac{1}{1-\frac{1}{1-x}}=-\frac{1-x}{x}$ and $\frac{1}{1-(-\frac{1-x}{x})}=x$, so we'll get the system of three unknowns we saw in our concrete example for any (valid) starting value of $x$.

The rest is details, which are sufficiently covered in the other answers.

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