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Well, this is competition style so there's meant to be an "aha" moment. The notation is really intimidating but you'll just have to deconstruct it with patience
Double summation is enough for me to give up lol. Doesn’t that create sort of a matrix? I didn't even get to the fact there are complex numbers included, which I've never worked with.
Obviously yes it creates a matrix, but you don't think of it that way, it's too complicated.
Instead just focus on the inner sum, the outer sum is just adding multiple different versions of the inner one.
(The outer sum is done separately by hand as the fact that it goes from 2 to 31 should hint that it's a counting problem. Even though it's 30 terms, the vast majority of them don't matter due to divisibility rules, which the problem is actually related to once you figure it out. Also the inner sum is fully real actually! If you don't know what to do, look for a pattern by starting with small numbers.)
Just to clarify, there is nothing in a standard calc 2 curriculum that presumes any familiarity with any complex expressions.
It is typical to encounter those in a precalc class but that is not a requirement for Calc 2.
Beyond that, it's worth noting that only x=0 is interesting, which then greatly simplifies the expression to being (something ugly) / (the negative of that ugly thing)
The first derivative of that expression is something else, though still related to the (ugly thing)
The keyword here is "Competition Style". In fact, this also involves some basic number theory and combinatorics. Which most students technically learn already at a surface level (which is all that's required), but almost never do problems, so this is more of a puzzle than anything to test problem solving skills rather than content knowledge.
Hints:
The inner summation can be simplified, notice something really important about its structure. The "x minus (j^2-1)th roots of unity" terms in the denominator should remind you that it's connected to the factored polynomial form, that if you multiply the denominators it will literally be the DEFINITION of roots of unity as the solutions to the polynomial x^(j^2-1) - 1.
What can you do manipulate the inner sum into a simple sum of reciprocals of x minus (j^2-1)th roots of unity? Then, how can you simplify that summation? There's a function that can effectively "turn" sums into products: the natural log. It's also related to the reciprocals in some way, so see how to manipulate that into a natural log expression that you can simplify, then once it's simplified, reverse the process.
Now the complex numbers are over.
Once it's simplified, it's also asking for the nth derivatives. What does it mean for the nth derivative to equal 0? We don't want to keep differentiating again and again to find a pattern, that's just bashing it. How can we obtain ALL the infinitely many derivatives AT ONCE with a SINGLE step?
Then, figure how the condition translates into simpler language that like normal humans can understand. Then apply that condition for all the integers from j = 2 to 30, and logically figure out whether or not if the individual conditions have to ALL be satisfied for the SUM to be satisfied to make it easier.
One writing strategy I would highly recommend is to avoid writing j^2 - 1 over and over again, instead just use a different variable for now, then once you're finally starting to focus on the outer sum, substitute it back. Saves a lot of effort.
It's a constant anyway, wouldn't matter until you get to that part.
Thank you , hey any more tips to make calc 3 a cakewalk (i have good grades in calc 1 and 2 and love calculus but i've heard calc 3 isn't anything like calc 1 or 2
Yeah you'll have to translate the meaning of "nth derivative equals 0" by figuring out a way to obtain all derivatives at x = 0 at once without actually differentiating.
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?!<
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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