Solution to $1-f(x) = f(-x)$

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?

Solution 1:

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

Solution 2:

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.

Solution 3:

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.

Solution 4:

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

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

Then

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

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

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

Solution 5:

WolframAlpha provides a solution to this (and many other) recurrence equations:
http://www.wolframalpha.com/input/?i=1-f(x)+%3D+f(-x)