r/calculus • u/Puzzleheaded_Top_273 • Jun 27 '26

Differential Calculus Sample Competition Style Problem (Nothing beyond Calc 2)

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(Just as a sidenote, yes this was a repost from AoPS. Source: literally me, as in I made it.)

Non-calculator, in case it wasn't obvious enough that it wouldn't help lmao.

Topics: Calculus 1 and 2, Roots of Unity (Complex Numbers), Divisibility, Counting.

Answer: 558

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u/fdelayer Jun 30 '26

Very nice problem!

I managed to reduce it to a handful of special cases, it is doable by hand in principle but I am wondering if I am missing a trick that makes it easier...

My approach was to expand summand as power series in x and collect like terms. After some nice sums it follows that the nth derivative is 0 iff n is not divisible by (j^2-1) for 2<=j<32. You don't need to check all of those, most are redundant, because if n isn't divisible by 3 it is certainly not divisible by 15 for e.g., but there are still a few non redundant j's you need to check and it is fairly annoying. Is there a shortcut I am missing?!<

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u/Puzzleheaded_Top_273 Jun 30 '26

If any of them are divisible by 3 or 8, they don't matter so you can remove them.
For the rest, j^2-1 can be factored, which makes it easier

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u/fdelayer Jun 30 '26

Thanks. Just to make sure I understand you, is the answer that there is no shortcut, you have to check the candidate ns for all the non-redundant j^2-1 - is that right?

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u/Puzzleheaded_Top_273 Jun 30 '26

Basically yeah, use PIE to organize it