f(x+ 1/6)-f(x)= f(x+ 1/6+ 1/7)-f(x+ 1/7)

Using iteration and x'=x+ 1/7 six times, we can get:

( 1)f(x+ 1/6)-f(x)= f(x+ 1/6+ 1)-f(x+ 1)

Repeat the iteration and use x'=x+ 1/6 for five times to get the following formula:

(2)f(x+2/6)-f(x+ 1/6)= f(x+2/6+ 1)-f(x+ 1/6+ 1)

(3)f(x+3/6)-f(x+2/6)= f(x+3/6+ 1)-f(x+2/6+ 1)

(4) ...

(5) ...

(6)f(x+ 1)-f(x+5/6)= f(x+ 1+ 1)-f(x+5/6+ 1)

When these six formulas are added, all the intermediate terms are cancelled, leaving only the first term and the last term, so:

f(x+ 1)-f(x)= f(x+2)-f(x+ 1)

Proving the periodic function is simple. Suppose f(x)=a, f(x+ 1)-f(x)=b, then:

f(x+ 1)-a=b

f(x+2)-f(x+ 1)= f(x+ 1)-f(x)= b

...

f(x+n)-f(x+n- 1)=b

All the above formulas add up to:

f(x+n)-a = n * b = & gt； f(x+n)=a+n*b

Reduction to absurdity, assuming that b is not equal to 0 and n is large enough to make |a+n*b| > 1 contradict the topic, so b=0.

So f(x+ 1)=f(x), and the period is 1.