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f(x) + f(x+T) = 2 ; where T is a fixed positive number.

The solution is given as:

put x = x+T

then given equation becomes

f(x+T) + f(x+2T) = 2

subtract given equation from above. You'll get: f(x) = f(x+2T).

Hence 2T is the period of f(x).

I don't get it. wouldn't putting x = x+T change the value of the function? How come we are still equating it to 2? If the function value doesn't change then we are implicitly assuming that T is the period right?

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To nitpick, this only shows that $2T$ is a period of $f$ (rather than the period). It is possible that there is a smaller period. E.g., this would hold for $f(x)=1+\cos(x)$, and $T=3\pi$. Many would regard it as incorrect to say that $6\pi$ is the period of $f$, even though it is true that $f(x+6\pi)=f(x)$ for all $x$. – Jonas Meyer Apr 3 '11 at 3:02
@Jonas: nitpicking encourages precision. Yours is a good example. Is the definition of period the smallest period or a period? – Ross Millikan Apr 3 '11 at 3:28
up vote 8 down vote accepted

The condition

$f(x)+f(x+T) = 2$

means that for every value of $x$, no matter what it is, if you evaluate $f$ at $x$ and at $x+T$, and add them, then you get $2$. So $f(1)+f(1+T) = 2$, $f(3.5) + f(3.5+T) = 2$, $f(0) + f(0+T) = 2$, $f(y) + f(y+T) = 2$, etc.

In particular, $f((x+T)) + f((x+T)+T) = 2$, by picking as our value of $x$ the value $x+T$.

This means that for every value of $x$, $f(x) + f(x+T) = f(x+T) + f(x+2T)$ (by substituting the $2$ in "$f(x)+f(x+T) = 2$" by $f(x+T) + f(x+2T)$, which we know is also equal to $2$). This says that for every value of $x$, $f(x)=f(x+2T)$.

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For a slightly different solution, you could consider the function $g(x)=f(x)-1$. Then

$\begin{align*} g(x)+g(x+T) &=(f(x)-1)+(f(x+T)-1)\\ &=(f(x)+f(x+T)) -2\\ &= 2- 2 \\ &= 0 \end{align*}$

for all $x$. This implies that $g(x)=-g(x+T)$ for all $x$, so $$g(x+2T)=g((x+T)+T)=-g(x+T)=-(-g(x))=g(x).$$

Then $f(x+2T)=g(x+2T)+1=g(x)+1=f(x)$.

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