# Period of $f(2x+3)+f(2x+7)=2$

I have a problem in finding the period of functions given in the form of functional equations.

Q. If $f(x)$ is periodic with period $t$ such that $f(2x+3)+f(2x+7)=2$. Find $t$. ($x\in \mathbb R$)

What I did:

$$f(2x+3)+f(2x+7)=2.........(1)$$

Replacing $x$ with $x-1$ in $(1)$,

$$f(2x+1)+f(2x+5)=2.........(2)$$

And replacing $x$ with $x+1$ in $(1)$,

$$f(2x+5)+f(2x+9)=2.........(3)$$

Subtracting $(2)$ and $(3)$, I get $$f(2x+1)=f(2x+9)$$

Since $x \in \mathbb R \iff 2x \in \mathbb R$, replace $2x$ with $x$ to get$$f(x)=f(x+8)\implies t=8$$

But sadly, my textbook's answer is $t=4$.

Is my method correct? How can I be sure that the $t$ so found is the least?

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As $2x+9=2(x+4)+1$ –  lab bhattacharjee Apr 16 '14 at 4:23
The period cannot be 4: It forces $f = 1$, a constant. Your argument shows that $f$ must have a period which is a rational multiple of $8$, but I am not really sure where to go from there. –  Braindead Apr 16 '14 at 4:38

I found this tricky and I may have got it wrong, but I think that you are correct and the book is not.

If $f(2x+3)+f(2x+7)=2$ for all real $x$, then we can take $x=(y-3)/2$ to give $$f(y)+f(y+4)=2\tag{*}$$ and then, essentially following your argument, $$f(y)=f(y+8)$$ for all real $y$. Now suppose that $f$ has period $t$: recall that this means $t>0$ and $f(y)=f(y+t)$ for all real $y$, and that no smaller positive $t$ has the same property. Suppose that there is an integer $k$ such that $$kt<8<(k+1)t\ ;$$ this can be rewritten as $$8=kt+t'\quad\hbox{with}\quad 0<t'<t\ .$$ Then we have $$f(y)=f(y+8)=f(y+t'+kt)=f(y+t')\quad\hbox{for all y},$$ which contradicts the fact that $f$ has period $t$. Therefore we must have $8=kt$ for some positive integer $k$, that is, $t=8/k$.

Now $k$ cannot be even, for if $k=2m$ then for all $y$ we have $f(y)=f(y+mt)=f(y+4)$; in conjunction with $(*)$ this shows that $f(y)=1$ for all $y$, which is not (strictly speaking) periodic.

However it is possible that the period could be $t=8/k$ with $k$ odd. For example, take $$f(y)=1+\sin\Bigl(\frac{2\pi y}{t}\Bigr)\ .$$ Then, as is well known, $f$ has period $t$; also $$f(y)+f(y+4) =1+\sin\Bigl(\frac{2\pi y}{t}\Bigr)+1+\sin\Bigl(\frac{2\pi y}{t}+k\pi\Bigr)=2$$ for all $y$, as required.

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Here $t$ is the period. So $f(y)=f(y+t)=f(y+2t)=f(y+12345t)=f(y+nt)$, where $n$ is any integer you like. BTW I found a better example for the last part of the question and have edited it - please re-read. –  David Apr 16 '14 at 7:58
Thanks.. So u have proven that the period is $8/k$ where $k$ is odd... How do u prove $k=1$?? –  Apurv Apr 16 '14 at 8:13
It is certainly possible to have $k=1$, just take my final example. If you mean how do you prove $k$ must be $1$, I don't think you can. In other words, your question does not have just one answer: the period might be $8$, or $\frac{8}{3}$, or $\frac{8}{5}$, etc. –  David Apr 16 '14 at 8:21
Perhaps, the original problem was meant to ask for integer periods. –  Braindead Apr 16 '14 at 12:28

"How can I be sure that the t so found is the least?"

Answer : by finding an example where $8$ is the smallest period. For example, take $f$ such that $f(x)=\frac{x-8k}{2}$ when $x\in[8k,8k+4[$ and $f(x)=\frac{8k+8-x}{2}$ when $x\in[8k+4,8k+8[$, for every $k\in{\mathbb Z}$.

This creates a "sawtooth" function and only multiples of $8$ are periods.

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Err... but this surely doesn't mean that 8 is the only possible period for $f$. –  Braindead Apr 16 '14 at 4:43

Since we have $f(2x) = f(2x + 8)$ for all $x$, If we assume $g(x) = f(2x)$ then we have $g(x + 4) = g(x)$, and hence $g(x)$ is periodic with period $4$, not $f(x)$.

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