Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Given:

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?

share|improve this question
4  
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
add comment

2 Answers

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

share|improve this answer
add comment

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

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.