How to find period of a real function f given the functional equation 3 f ( x ) = f ( x − 1 ) + f ( x + 1 )?If a periodic function satisfies the equation 3 f ( x ) = f ( x − 1 ) + f ( x + 1 ) for all real x then prove that fundamental period of the function is 12.Here fundamental period means the smallest positive real for which function repeats its value for all x.I tried replacing x by x ± 1 then try to find f(x) in terms of other but always end up with it in terms of sum of other two arguments in the function eg f ( x + 2 ) etc.Please provide a general method and also especially do give the thought process or reasoning for all the steps ie why you are doing these particular steps or what led you to thinking that doing these steps would give you the period of f.

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How to find period of a real function f given the functional equation       3    f  (  x  )  =  f  (  x  −  1  )  +  f  (  x  +  1  )?If a periodic function satisfies the equation       3    f  (  x  )  =  f  (  x  −  1  )  +  f  (  x  +  1  ) for all real x then prove that fundamental period of the function is 12.Here fundamental period means the smallest positive real for which function repeats its value for all x.I tried replacing x by   x  ±  1 then try to find f(x) in terms of other but always end up with it in terms of sum of other two arguments in the function eg   f  (  x  +  2  ) etc.Please provide a general method and also especially do give the thought process or reasoning for all the steps ie why you are doing these particular steps or what led you to thinking that doing these steps would give you the period of f.

vihralV5x · Accepted Answer

let   x  ∈      R   be fixed. Then   a  (  x  ,  1  )  :=  f  (  x  +  1  )and   a  (  x  ,  n  )  :=      3    a  (  x  ,  n  −  1  )  −  a  (  x  ,  n  −  2  ) for   n  ≥  2The roots of the characheristic polynomial       z    2    −      3    z  +  1 are                     3            +      i        2    =      e                            π          i                6                                 3            −      i        2    =      e          −                        π          i                6            . So   a  (  x  ,  n  )  =      c    1        e                            n          π          i                6              +      c    2        e          −                        n          π          i                6             where       c    1   and       c    2   are determined from a(x,0) and a(x,1). Since   f  (  x  +  n  )  =  a  (  x  ,  n  ) it is obvious that 12 is the period of f.f  is not identically 0, 12 is the fundamental period. is not identically 0, one can find   x  ∈      R   for which   f  (  x  )  ≠  f  (  x  +  1  )Replacing f by f−g, we can assume f(x)=0. As a resulut of this and scalar multiplication we can assume that f(x)=0 and   f  (  x  +  1  )  =  1. Then   f  (  x  +  n  )  =      1          α      −      β        (      α    n    −      β    n    ) where   α  =      e                            π          i                6              ,  β  =      e          −                        π          i                6            . From this   f  (  x  )  =  f  (  x  +  n  )  =  0 if and only if n is a mutiple of 6. But 6 cannot be a period because   f  (  x  +  1  )  ≠  f  (  x  +  7  )

How to find period of a real function f given the functional equation sqrt(3)f(x)=f(x−1)+f(x+1)? If a periodic function satisfies the equation sqrt(3)f(x)=f(x−1)+f(x+1) for all real x then prove that fundamental period of the function is 12.

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