The smallest n making |second derivative| greater than 2024 is 18

For this function the second derivative at zero is always negative and we want the absolute value, so just find -f"(x)

Method 1:

from Desmos; see   https://www.desmos.com/calculator/l65vpj1t0g

n  -fn''(0)

1  1

2  5

3  14

4  30

5  55

6  91

This is in oeis as A000330, and by counting terms, the 17th is 1785 and the 18th is 2109

Method 2:

Using Wolfram Alpha to take 2nd derivatives of the first several f_i(x) functions (multiplied by -1) shows

f''(x) = cos(x)

f''(x) = 5 cos(x) cos(2 x) - 4 sin(x) sin(2 x)

f''(x) = 2 (7 cos(x) cos(2 x) cos(3 x) - 2 sin(x) sin(2 x) cos(3 x) - 3 sin(x) sin(3 x) cos(2 x) - 6 sin(2 x) sin(3 x) cos(x))

f''(x) = 2 (15 cos(x) cos(2 x) cos(3 x) cos(4 x) - 2 sin(x) sin(2 x) cos(3 x) cos(4 x) - 3 sin(x) sin(3 x) cos(2 x) cos(4 x) - 6 sin(2 x) sin(3 x) cos(x) cos(4 x) - 4 sin(x) sin(4 x) cos(2 x) cos(3 x) - 8 sin(2 x) sin(4 x) cos(x) cos(3 x) - 12 sin(3 x) sin(4 x) cos(x) cos(2 x))

Replace all cos terms with 1, all sin terms with 0:

The sequence is:  1, 5, 14, 30

Method 3:

Numerically calculate second derivative.

f'(x) = (f(x+ε) - f(x))/ε

f"(x) = (f'x+ε) - f'x))/ε

      = (f(x+2ε) - 2f(x+ε) + f(x))/ε^2

Program output:

1 1

2 5

3 14

4 30

5 55

6 91

7 140

8 204

9 285

10 385

11 506

12 650

13 819

14 1015

15 1240

16 1496

17 1785

18 2109

The smallest n making |second derivative| greater than 2024 is 18

------------------

import math

epsilon = .0000001

def f(x,n):

    ans = 1

    for i in range(1,n+1):

        ans *= math.cos(i*x)

    return ans

def d2(x,n):

    """ Second derivative

    f'(x) = (f(x+ε) - f(x))/ε

    f"(x) = (f'x+ε) - f'x))/ε

          = (f(x+2ε) - 2f(x+ε) + f(x))/ε^2 """

    ans = (f(x+2*epsilon,n) - 2*f(x+epsilon,n) + f(x,n) ) / (epsilon**2)

    return - ans

for i in range(1,100):

    print(i, round(d2(0,i)))

    if d2(0,i) > 2024:

        print('The smallest n making |second derivative| greater than 2024 is', i)

        break