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).

Question

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.

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Answer 1 · 2018-02-21T23:09:28

Ohh this looks fun. I'll be using $f^{3}$ $f^(3)$ 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:
$f^134 = f^(134 mod 26) = f^4$
$k = 4$
$square$