Take the 2-minute tour ×

Can we find $f(x)$ given that $1-f(x) = f(-x)$ for all real $x$?

I start by rearranging to: $f(-x) + f(x) = 1$. I can find an example such as $f(x) = |x|$ that works for some values of $x$, but not all. Is there a method here? Is this possible?

share|improve this question
Homework? . . . –  Charles Stewart Jul 30 '10 at 8:33
Nope, just a problem posed to me - what is the line on homework, by the way? If I show I've sufficiently worked on a problem (i.e. not just posting the problem), then that's okay? –  Danish Kaneria Jul 30 '10 at 9:03
@Danish: That's correct. It's best to also tag the question as homework. –  Larry Wang Jul 30 '10 at 9:21
@Danish, @Kaestur: As a rule, it's good to say why you care about getting the answer to your question. –  Charles Stewart Jul 30 '10 at 9:31
@Danish Nice cricket name. –  A B Jul 30 '10 at 13:04

6 Answers 6

$$f(x)=\frac{1}{2}+\text{(any odd function)}.$$ For example, $f(x)=\frac{1}{2}+x$ or, say, $f(x)=\frac{1}{2}+99x^3+7x^5$.

share|improve this answer
Or in other words if $g(x) = f(x) - 1/2$ then $ f(x) + f(-x) = 1 \iff g(x) + g(-x) = 0$. –  Aryabhata Aug 26 '10 at 18:55
Hum, the full answer... –  awllower Apr 5 '11 at 10:23

Clearly, we only have relations between f(x) and f(-x). The relation means that 0 has to have value 1/2. We can divide all non-zero real numbers into disjoint pairs of x and -x and define the function f on each pair separately. For each pair, f(x) can be given any value and then f(-x) has a single valid value. As mentioned by Grigory, the valid functions can be characterised as any odd function plus 1/2.

share|improve this answer
I wonder if there is a more precise wording than 'define the function f on each pair separately" –  Casebash Jul 30 '10 at 8:57
One may also say that f(0)=1/2, for x>0 we define f in any way we like, and for x<0 it's defined by the condition f(-x)=1-f(x). –  Grigory M Jul 30 '10 at 9:40

Consider instead the functions $g$ that satisfy the identity $-g(x)=g(-x)$ for all $x$. If $(x,g(x))$ is a point of the function $g$, then $(-x,-g(x))$ is also a point (since $g(-x)=-g(x)$). Therefore, every function $g$ is symmetric when rotated by $180$ degrees about the point $(0,0)$.

How do things change for the identity $1-f(x)=f(-x)$? We merely shift the point of symmetry to $(0,1/2)$. Here the point $(x,f(x))$ implies the point $(-x,1-f(x))$.

The function $f$ satisfies the identity $1-f(x)=f(-x)$ for all real numbers $x$ if and only if it is symmetric when rotated about the point $(0,1/2)$ by $180$ degrees.

There's going to be many of these functions; some of which will be polynomials, some of which will not.

share|improve this answer

Usually simple problems like this ask you to find a function that respects the condition, not all of them. And (again) usually you start by checking if a simple polynomial function of the first degree could be a solution.

So, if

$$f(x) = ax + b$$


$$f(x) + f(-x) = 1 \implies ax + b + a \cdot (-x) + b = 1 \implies 2b = 1 \implies b = 1/2$$

So the condition is satisfied by any function of the type:

$$f(x) = ax + 1/2$$

share|improve this answer

WolframAlpha provides a solution to this (and many other) recurrence equations:

share|improve this answer
It doesn't seem to have found all solutions in this case. –  ShreevatsaR Jul 30 '10 at 17:42
@ShreevatsaR, WolframAlpha can only regard functional equations as recurrence equations to solve. –  doraemonpaul Sep 19 '12 at 0:47

Let $$f(x) = \begin{cases} 1 \quad x>0, \\ 1/2 \quad x=0, \\ 0 \quad x<0\end{cases}.$$

If $x > 0$, then $-x < 0$ so that $f(x)+f(-x)=1+0=1$

Likewise with $x<0$.

If $x=0$, then $f(x)+f(-x)=(1/2)+(1/2)=1$.

share|improve this answer

Your Answer

