Finding the period of $f$ if ${f}(x+7)+{f}(x+4)+{f}(x+3)+{f}(x)=1$

Question

If function $f: \mathbb{R} \to \mathbb{R}$ satisfies the relation ${f}(x+7)+{f}(x+4)+{f}(x+3)+{f}(x)=1$, then period of $f(x)$ is?

I have supposed $f(x)+f(x+3)=g(x)$ and period of $g(x)$ will be $6$.

Answer 1

Using $g(x):=f(x)+f(x+3)$ is a good idea. We can then simplify to get $g(x) + g(x+4) = 1$.

What is the period of this function? We need some $\lambda$ s.t. $g(x + \lambda) = g(x)$ using above equality:

$g(x) + g(x+4) = 1 = g(x+4) + g(x+8)$

We can simplify to get: $g(x) = g(x+8)$ which means that $g$ has a period at most $8$, it could have a period that is a divisor of $8$ as well.

What does this mean for $f$? We know that \begin{align} f(x) + f(x+3) &= f(x+8) + f(x+11) \\ f(x) + f(x+3) + f(x+4) + f(x+7) &= 1 \\ f(x+3) + f(x+6) + f(x+7) + f(x+10) &= 1 \end{align}

Setting equation 2 and 3 equal gives

\begin{align} f(x)+f(x+3)+f(x+4)+f(x+7) &= f(x+3)+f(x+6) +f(x+7)+f(x+10) \\ \Leftrightarrow f(x) + f(x+4) &=f(x+6) + f(x+10) \end{align}

which means that $h(x) = f(x)+f(x+4)$ has a period at most $6$. Using the given formula starting with $x+7$ gives that $j(x) = f(x) - f(x+6)$ has a period of $8$ as well.

However, I tried some combinations as well and I don't get a final period of $f$. Maybe these $3$ periodic functions $g,h,j$ are helpful and you can solve the problem now.

Answer 2

Let $f(x) - 1/4 = g(x)$, then $g(x+7) + g(x+4) + g(x+3) + g(x) = 0$.

Let $G(k) = g(x+k) + g(x+k+3) + g(x+k+4) + g(x+k+7) = 0$.

Applying the idea that LegNaiB had, (but this ended up being more complicated,) observe that $G(0) - G(3) - G(4) + G(6) + G(7) + G(8) - G(9) - G(10) - G(11) +G(13) + G(14) - G(17)$ when expanded out gives us:

$$g(x) + g(x+3) + g(x+4) + g(x+7) \\ - [ g(x+3) + g(x+6) + g(x+7) + g(x+10) ] \\ - [ g(x+4) + g(x+7) + g(x+8) + g(x+11) ] \\ g(x+6) + g(x+9) + g(x+10) + g(x+11) \\ g(x+7) + g(x+10) + g(x+11) + g(x+14) \\ g(x+8) + g(x+11) + g(x+12) + g(x+15) \\ - [ g(x+9) + g(x+12) + g(x+13) + g(x+16) ] \\ - [ g(x+10) + g(x+13) + g(x+14) + g(x+17) ] \\ - [ g(x+11) + g(x+14) + g(x+15) + g(x+18) ] \\ g(x+13) + g(x+16) + g(x+17) + g(x+20) \\ g(x+14) + g(x+17) + g(x+18) + g(x+21) \\ - [ g(x+17) + g(x+20) + g(x+21) + g(x+24) ] \\ = 0, $$

where all of the terms cancel out leaving us $ 0 = g(x) - g(x+24)$.
So $g$ must have period 24 (and it's multiples), and thus so must $f$.

Note that a particular solution could have a smaller fundamental period.

Note that we haven't yet shown that all solutions couldn't share a smaller period.

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How did we get at this? Apart from blind trial and error (which was my immediate approach, and things just happened to cancel out), if we express $ g(x+k) $ as the term $y^k$, we're then asking:

What is a (ideally smallest) value of $K$ such that $(1+y^3)(1+y^4) \mid y^K -1$?

Clearly (or we can check that), $K= 24$ works.
By appealing to the theory of Cyclotomic polynomials (esp in the general case), or otherwise (EG By checking all smaller values), this is indeed the smallest value.

In fact, $\frac{ y^{24} -1 } { (1 + y^3 )(1+y^4)} = y^{17} - y^{14} - y^{13} + y^{11} + y^{10} + y^9 - y^8 - y^7 - y^6 + y^4 + y^3 - 1$ tells us which $g(x+k)$ values to take. (Focus on the $\sum G(i)$ expression, or on the first column in the above list).

How can we construct a function $f(x)$ with fundamental period 24 and $f(x) + f(x+3)+f(x+4)+f(x+7) = 1 $?