For natural number n, define f_n(x) as the product cos(x) * cos(2x) * cos(3x) * .... * cos(nx).
What is the smallest n such that |f_n''(0)| > 2024?
This is actually pretty easy to work out using the extended product rule for derivatives. In fact, the structure is so simple I don't feel it warrants a high difficulty rating.
The first few terms of the first derivative are
-sin(x)cos(2x)cos(3x)...
-2cos(x)sin(2x)cos(3x)...
-3cos(x)cos(2x)sin(3x)...
...
Note the sine term just shifts. Now each of these need a separate derivative for the second derivative.
-cos(x)cos(2x)cos(3x)... + 2sin(x)sin(2x)cos(3x)... + 3sin(x)cos(2x)sin(3x)...
+2sin(x)sin(2x)cos(3x)... - 4cos(x)cos(2x)cos(3x)... + 6cos(x)sin(2x)sin(3x)...
+3sin(x)cos(2x)sin(3x)... + 6cos(x)sin(2x)sin(3x)... - 9cos(x)cos(2x)cos(3x)...
...
So we have a sort-of nxn matrix each of which is a product of n sines and cosines. However any of these that contains a sine term is going to be zero. The terms that have only cosines are the main diagonal that I put in bold. They happen when the chained derivative cosine hits the second derivative sine. This explains why they are all negative and the leading coefficients are the squares.
So as a function of n, |f_n"(0)| is just the sum of the first n squares. The well known formula is n(n+1)(2n+1)/6.
Rather than solve the cubic, I used a table to show n=18 gives 2109.
Edited on February 28, 2024, 1:54 pm
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Posted by Jer
on 2024-02-28 13:50:38 |
Analytic solution
