>>10877130the absolute value of abs(f(x^k)) is differentiable almost everywhere, so g(x) should be differentiable at any point such that all of the f(x^k)s are differentiable at the point, with the only possible non-differentiable points such that f(x^k) = 0. In fact, when k is even, f(x^k) = 0 is impossible, and by its continuity, abs(f(x^k)) = f(x^k), so the even terms won’t make a difference. Additionally, for the odd terms, only x = -1 can make f(x^k) = 0, so g(x) is differentiable everywhere except possibly x = -1. To be differentiable, the derivative from the right should be equal to the derivative coming from the left. Suppose the derivative of f(x^k) at x = -1 is 0, then the derivative of abs(f(x^k)) at x = -1 is also 0 because the left and right derivatives have to be both 0, no matter if anything’s sign was flipped. Now, suppose the derivative of f(x^k) at x = -1 is non-zero, then, because f(x^k) is a monotonically increasing function that crosses 0 at x = -1 (we are only considering k is odd), exactly one of the sides will be flipped, and the other will star the same, so one of the left or right derivative will be the same, and the other will be flipped. Suppose we add up all of the cusp derivatives on the right hand side, and their sum is non-zero. Then, the left hand cusp sum will be the same sum, but the sign is flipped, resulting in the left and and right hand derivatives at x = -1 (for g) being unequal. Suppose the cusp sum for the right hand derivatives is 0, then the left hand sum is 0, so the sum of derivatives on the right and left will be the same, so g will be differentiable at x = -1 and everywhere else. The only functions abs(f(x^k)) with cusp derivatives are the ones such that k is odd, and the derivative of f(x^k) at x = -1 is non-zero, so, all of the natural, odd ks have cusp derivatives, and the sum of the cusps happens to be a sum of odd numbers, and that sum must be 100. N = 19 and N = 20 are the only Ns that work, so the sum is 39
