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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?

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    $\begingroup$ Homework? . . . $\endgroup$ Jul 30, 2010 at 8:33
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    $\begingroup$ 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? $\endgroup$
    – Danish Kaneria
    Jul 30, 2010 at 9:03
  • $\begingroup$ @Danish: That's correct. $\endgroup$
    – Larry Wang
    Jul 30, 2010 at 9:21
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    $\begingroup$ @Danish, @Kaestur: As a rule, it's good to say why you care about getting the answer to your question. $\endgroup$ Jul 30, 2010 at 9:31
  • $\begingroup$ @Danish Nice cricket name. $\endgroup$
    – A B
    Jul 30, 2010 at 13:04

7 Answers 7

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

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    $\begingroup$ Or in other words if $g(x) = f(x) - 1/2$ then $ f(x) + f(-x) = 1 \iff g(x) + g(-x) = 0$. $\endgroup$
    – Aryabhata
    Aug 26, 2010 at 18:55
  • $\begingroup$ Hum, the full answer... $\endgroup$
    – awllower
    Apr 5, 2011 at 10:23
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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.

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  • $\begingroup$ I wonder if there is a more precise wording than 'define the function f on each pair separately" $\endgroup$
    – Casebash
    Jul 30, 2010 at 8:57
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    $\begingroup$ 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). $\endgroup$
    – Grigory M
    Jul 30, 2010 at 9:40
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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.

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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:

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

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

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  • $\begingroup$ It doesn't seem to have found all solutions in this case. $\endgroup$ Jul 30, 2010 at 17:42
  • $\begingroup$ @ShreevatsaR, WolframAlpha can only regard functional equations as recurrence equations to solve. $\endgroup$ Sep 19, 2012 at 0:47
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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 $f(x)+f(-x)=(1/2)+(1/2)=1$.

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We have $f(x)-1/2=-(1-f(x))+1/2=-f(-x)+1/2=-(f(-x)-1/2)$; hence $f(x)-1/2$ is an odd function. So, $f(x)=1/2+\phi(x)$ for some odd function $\phi$. Clearly, any such function satisfies the original equation.

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