Problem:
Consider those functions f that satisfy f(x+4)+f(x−4)=f(x) for all real x. Any such function is periodic, and there is a least common positive period p for all of them. Find p.
Answer Choices:
A. 8
B. 12
C. 16
D. 24
E. 32
Solution:
We may replace x with x+4 in
f(x+4)+f(x−4)=f(x) (1)
to get
f(x+8)+f(x)=f(x+4) (2)
From (1) and (2), we deduce that f(x+8)=−f(x−4). Replacing x with x+4, the latter equation yields f(x+12)=−f(x). Now replacing x in this last equation with x+12 yields f(x+24)=−f(x+12). Consequently, f(x+24)=f(x) for all x, so that a least period p exists and is at most 24. On the other hand, the function f(x)=sin(12πx) has fundamental period 24, and satisfies (1), so p≥24. Hence p=24.
OR
Let x0 be arbitrary, and let yk=f(x0+4k) for k=0,1,2,…. Then f(x+4)= f(x)−f(x−4) for all x implies yk+1=yk−yk−1, so if y0=a and y1=b, then y2=b−a,y3=−a,y4=−b,y5=a−b,y6=a, and y7=b. It follows that the sequence (yk) is periodic with period 6 and, since x0 was arbitrary, f is periodic with period 24. Since f(x)=sin(12πx) has fundamental period 24 and satisfies f(x+4)+f(x−4)=f(x), it follows that p≥24. Hence p=24.