You own one-month calls struck at 110 with the share at 100, and you short 0.1452 shares against each one. If the share rallies to exactly 110 and stops, the calls expire worthless while the short has lost ten dollars a share, so the hedged position is down 2.074208 where the unhedged one would have lost only its 0.622212 premium. The worst case sits at the strike because the profit is piecewise linear with slopes of -0.1452 and +0.8548, and a rebalanced hedge on the same path loses 3.058738.
Sorting the list finds the gap and is merely wasteful, which is why the article says so rather than striking it out. The subtraction works because 5050 is a closed form available before the list is read, and the two conditions carrying it are distinctness and a known range. The article adds the duplicate-hunting mirror image, the sum-of-squares route when two values are absent, and the exclusive-or accumulator for when the total would overflow.
Stack i + j - 1 cubes on every square of a 20 by 20 board and the total is 8000, which is 20 cubed. Folding the board across the squares that are exactly 20 deep pairs every stack with a mirror stack, and each pair adds to 40, so the average depth is 20. The double sum gets the same answer and is merely slow, and the main diagonal is the fold that proves nothing.
A driver who eats one apple per loaded mile delivers 833 of 3000 across a thousand miles, because the price of a mile is the ceiling of the stock over the truck's capacity: three apples, then two, then one. The continuous optimum is 2500/3, but rounding the switch point down to mile 333 leaves 2001 apples needing three passes and produces a fake 834. A lower bound on loaded traversals proves 833 optimal without a dynamic programme, and the free return legs are the clause that separates 833 from 533.
Three feet up each day, one foot back each night, ten feet to climb. Dividing ten by the net two feet a day gives five days, and it charges the snail for a night it never spends. The dawn heights settle the day, the last climb settles the moment, and the closed form the video had no room to voice is a single ceiling function.
On the unit interval the indicator of the rationals has upper sum 1 and lower sum 0 for every partition ever written, so the two never meet and Riemann returns nothing at all. Give the k-th rational an interval of width ε/2^k and the whole countable set sits inside a total length of ε, which puts its measure at 0 and its Lebesgue integral at 0. Lebesgue's criterion turns that into the general law — a bounded function on a compact interval is Riemann integrable exactly when its discontinuities have measure zero — which is why Thomae's function, discontinuous on the same dense set, is integrable and this one is not. The trade is not free: sin x over x on the half line has an improper Riemann value of π/2 and no Lebesgue integral at all.