Question

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.

Answer 1

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`