Help with this? Suppose a function f is 26 times differentiable and f^(26)=f. Then, it turns out that f is infinitely differentiable and that for any positive number n, f^(n) equals one of the functions f,f′,f′′,...,f^(25).

Suppose a function f is 26 times differentiable and f^(26)=f. Then, it turns out that f is infinitely differentiable and that for any positive number n, f^(n) equals one of the functions f,f′,f′′,...,f^(25).
So, find a positive number k so that f(^134)=f^(k) and 0≤k≤25.

1 Answer
Feb 21, 2018

Ohh this looks fun. I'll be using #f^(3)# to represent #f'''#, not exponents.

What we know:
#f = f#
#f' = f'#
#f^(3) = f^(3)#
#f^26 = f#

Then, note that
#(f^26)' = f' = f^(26+1) = f^27#

Therefore,
#f^(n) = f^(n mod 26)#

Finally, the answer:
#f^134 = f^(134 mod 26) = f^4#
#k = 4#
#square#