Sine is differentiable everywhere and never leaves the band from minus one to one on the real axis, which makes it the counterexample everyone reaches for, and the modulus of sine at 10i is already 11013.23. The Cauchy estimate caps every Taylor coefficient by M over r to the n on a circle of radius r, so growing r kills every coefficient above the constant one and nothing but a constant survives. The same estimate with a polynomial growth bound gives more: an entire function bounded by C times one plus the modulus of z, all to the k, is a polynomial of degree at most k, and one corollary of that is the fundamental theorem of algebra.
A plain t sitting next to the t squared in a Gaussian exponent looks like a new function and is only a shift. Completing the square turns the integral of e to the minus a t squared over two plus b t, from x to infinity, into e to the b squared over 2a times the root of 2 pi over a times the standard normal at a rescaled and shifted argument, never at x itself. The worked case comes out as exactly half a bell, e root pi over two or 2.40901455, but only because its lower limit happens to land on the centre b over a.
The characteristic roots of u'' + u' + u are the primitive cube roots of unity, so the homogeneous part decays with envelope e to the minus x over two and oscillates with period 4 pi over root 3, which is 7.2552. Substituting that homogeneous solution back into the equation leaves a residual of exactly minus one, and that residual is the whole distance between the common wrong answer and the right one. The constant u = 1 solves the equation by itself, so every solution settles on 1, and the constant trial works only because the coefficient on u is not zero.
The ratio and root tests both return 1 on the sum of e to the minus root n, which settles nothing, and the usual write-up of the problem then quotes 4 over e as the answer. That number is the floor rather than the cap: the sum is 1.6704068, which is 13.52 percent above it, and the usable bound comes from integrating from 0 instead of from 1, giving exactly 2. Where each bar of width one sits relative to its index is the single step that decides which way the inequality points.
Buying two-year notes because you expect the curve to steepen is a position on the level of rates: the same correct view loses two dollars if the steepening arrives with everything rising. Matching the two legs on dollar duration removes the parallel part of the move as an algebraic identity, so half a point of widening pays four dollars whatever the level does. Matching market value as well is impossible with only two bonds and needs a third leg carrying no duration, and on a full cash-flow reprice the level survives at second order, worth 0.07 against a four-dollar profit at fifty basis points.
A share at 100 that jumps to either 80 or 130 gives a call an exact price of 12, from two equations in two unknowns and no probability at all. Let the jump size be random, so 110 is also reachable, and that same hedge pays 18 where the option pays 10 while no other portfolio does better. The arbitrage-free prices then fill the whole interval from 20/3 to 12, and the obstruction turns out to be the kink in the payoff rather than the number of states.
A riskless zero-coupon bond at 100 has a six-month forward of 102.531512, a premium. Give the same bond an 8% coupon and the forward drops to 98.511194, a discount, because the sign of the premium is the sign of the rate minus the coupon and nothing else. Quoting the forward at spot when the coupon is rich hands the other side a riskless 1.518802 per hundred, and the discrete-coupon version shows the answer also turns on whether a payment date falls before delivery.
A top-rated issuer picks the coupon that prices its ten-year bond at exactly 100 off its own flat 5 percent curve, and the same cash flows discounted off a swap curve 25 basis points lower come to 101.954087. Two facts do the work: a present value is strictly decreasing in every rate it is discounted at, and for a top-rated name the swap curve sits below its own bond curve because a swap risks no principal and is margined daily. A modified duration of 7.7217 times the spread accounts for 1.9304 of the lift, and a convexity of 74.9977 supplies the last two cents.
Give a pulled-back log price the same 20 percent instantaneous swing as a free-wandering one and its horizon variance stops being sigma squared times T: at one reversion time only 0.432332 of it survives, the volatility that prices a one-year call is 13.1504 percent, and the call falls from 7.9656 to 5.2425. The same pull makes consecutive returns fight each other, with a first-order autocorrelation of exactly minus half of one minus phi, and that is the independence the pricing model rests on. The formula still returns the right European price and has lost the hedging argument that justified it.
A six-month at-the-money call on a 50 dollar share sheds a cent and a half a night to time decay, and its expected price tomorrow is higher anyway. The deterministic total differential gives minus 0.66 cents and predicts the opposite of the truth, while Ito's third term, half the gamma times the squared move, adds plus 1.13 and runs on variance rather than direction. Substituting the pricing equation for theta cancels that term exactly and leaves an expected return of the riskless rate plus elasticity times the premium, which is 35.54 percent a year here and turns negative below a real drift of 4.18.
Sketch a one-year call struck at 100 with a five percent rate. Deep in the money the curve straightens into a line of slope one, and that line crosses at 95.122942 rather than at 100, so drawing it through the strike is out by 4.877058 for ever. That gap is the interest saved on the strike, and it is also why an American call on a share paying no dividends is never exercised early. Plot the same option against the futures price and the crossing returns to 100 while the slope drops to 0.951229.
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.
A bond paying 100 in ten years costs 67.5564 at a four percent yield. The first two points of yield cost 11.7169 and the next two only 9.5201, so the curve bends. The usual explanation blames duration falling as yields rise, and this bond refutes it: with a single cash flow its Macaulay duration is exactly ten at every yield. The slope is minus duration times price over one plus the yield, and the general statement needs no duration at all, only that every discount factor is convex.
A 20-year 7 percent bond on a flat 10 percent curve prices at 744.5931, and a one-point rise costs exactly 63.1262. The tangent alone says 67.7028, and adding the second-order term of 4.8416 lands at 62.8612, inside 27 cents of the truth. Note that the correction and the error it corrects are two different numbers, which is why the estimate ends up on the wrong side of the answer.
Whether an up move followed by a down move lands where a down move followed by an up move lands decides between a quadratic node count and an exponential one, and at twenty steps the gap is a factor of 9,078.6. Both sums carry N+1 terms rather than N, because a twenty-step tree has twenty-one dates on it, and the off-by-one costs the entire final row. The article also states the recombination hypothesis exactly, which is weaker than the usual ud = 1.
Cutting a hundred million of a thirty-year bond down to fifty takes five hundred futures, and the usual arithmetic of fifty million over a hundred thousand lands there only because the contract's duration per dollar of face happens to match the bond's. What a hedge matches is dollars per basis point: 56,288.92 against 112.5778. Hold a thirty-year zero instead and the same job needs 1,234 contracts, while five hundred three-month contracts would cover 22.2 per cent of it.
An eight per cent thirty-year bond at par loses 27.49 dollars when its yield rises 25 basis points, and its yield moves because the principal is collateralised in United States Treasuries. The answer that circulates, about thirty-five dollars, needs a duration of fifteen, and a par bond at an eight per cent yield cannot have one: its modified duration is its own annuity factor, capped at 12.5 at any maturity whatsoever. The pass-through, the only soft number in the chain, is swept from an eighth to a half.
Heads pays $7 in eighteen months, tails costs $2 today, and the curve gives 12% for one year and 18% for two. Averaging the amounts gives $2.50, which is 38.68% too high, because expectation and discounting only commute when every cash flow lands on the same date. The answer is about $1.80, and four defensible compounding conventions spread it from 1.7862 to 1.8381, so one decimal is honest and two are not.
Shade the region between a diffusing particle and the time axis over one second. The box is one wide and about one tall, so the eye guesses a variance of one, and the answer is one third because each increment counts only for the time remaining after it. No stochastic integration is needed to define the object, only continuity of the path, and the constant is pinned twice over: once by the weight (T minus t) and once by integrating the covariance min(s,t) across the square.
The integral of x dx is x squared over two, so the integral of W dW ought to be W(T) squared over two, and the only false step in that chain is the conclusion. A dissected square turns the Riemann sum into an identity exact at every partition, and the term that refuses to vanish is the total of the squared steps, which equals T rather than zero. The reflex answer is the exact value of the midpoint sum over the same partition, which is why it feels so solid.
The anchor is 45 squared, the shortfall is 1500, and one division by 900 lands on 1355/3 = 451.6667 against a true 451.66359. The estimate overshoots by exactly h squared over four a squared, which is 25/9 in the square here, so the error has a known sign as well as a known size. The article carries the bracket that names 452 as the nearest integer, one Newton step to nine figures, and what happens when the anchor is chosen too far away.
Four settlements in five come back below the 1.50 outlay, and the average payoff is still 1.80, an edge of 0.30 a contract or twenty percent of the money at risk. The reflex is not bad arithmetic, it is the mode standing in for the mean. The article carries the tally over one full cycle, the threshold saying you need the large outcome more often than one time in eight, and the reason waiting longer can leave you less likely to be ahead.
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.
Because the horizons are ten and five, both sides of the no-arbitrage equation are fifth powers and the root disappears, leaving 1 + f = 1.15 squared over 1.10 = 529/440, so f = 89/440 exactly. Reflecting 10 percent around 15 to get 20 is low by exactly (b - a) squared over (1 + a), a square over a positive number, which is why the reflection can never overshoot for any pair of rates. Under continuous compounding the same problem is linear and 20 percent is exactly right, so the instinct is correct machinery pointed at the wrong convention.
Logarithmic differentiation turns the exponent into a factor and gives x^x times (1 + ln x), which is exactly the sum of the power-rule answer x^x and the exponential-rule answer x^x ln x. That is a theorem rather than a coincidence: the two rules are the partial derivatives of u^v, and walking the diagonal u = v = x adds both partial effects. The power rule accidentally returns the correct slope at x = 1, which is the one point nobody should use to test a rule.
There is no elementary antiderivative to evaluate, and the checkable slice of that is one line: if p is a polynomial then p' - 2xp has degree deg p + 1, which can never equal the degree of 1. Squaring the integral turns it into a rotationally symmetric integral over the plane, where the polar area element supplies the factor r that makes the radial integral elementary, so I squared equals 2 pi times one half. The same idea survives without polar coordinates via the substitution y = xt, and it fails for e to the minus x to the fourth because x^4 + y^4 is not a function of the radius.
The fraction that maximises long-run growth is exactly the edge, 2p-1, which is 0.2 on this coin, and one derivative gets you there. Double it and the growth rate is -0.0024469 a flip, negative on a game that leans your way three hundred times in a row, and the crossing happens at 0.3894 rather than at 0.4. The article carries the exact median over 300 flips, 25 dollars to 10504.19 at the optimum and to 12.00 at double, the reason about 48 percent of overbettors still finish ahead anyway, and the place where the textbook approximation mean minus half the variance returns the opposite sign.
A contract struck at the current share price is worth roughly 0.3989 S sigma root T, so doubling the time to expiry multiplies the price by root two and takes 100 dollars to about 141 rather than 200. The exact ratio erf(s/2) over erf(s/(2 root 2)) is always strictly below root two because erf is concave, so 141.42 is a ceiling never reached. Strip out the strike condition and the rule collapses: the same doubling multiplies a strike 30 percent above spot by 4.19 and one 30 percent below by 1.01.
Differentiating y = L tan(wt) gives a spot speed of w R squared over L, so the footprint accelerates with the square of its distance from the lamp: a tenth of pi directly opposite, and exactly pi miles per second nine miles along. The 9 is the along-shore leg, which makes 90 the squared hypotenuse rather than the square of nine, and that misreading is the usual failure. Nothing physical moves at that speed, and a straight coast running 2310 miles would carry a nominally faster-than-light spot carrying no information at all.
Six months of a sixty dollar year carries 60 over root two, which is 42.43 rather than 30, because variances add over disjoint intervals and standard deviations do not, so the digital is worth exactly $239,750. The figure of $250,000 in circulation comes from rounding the z score 0.7071 up to 0.75 and then reading the tail at 0.75 as 0.25, but Phi(0.75) = 0.773373, so even the rounded chain gives 0.2266. Rounding z upward has to make the tail smaller, and 0.25 is larger, which is the tell that a symbol changed meaning mid-calculation.
A share sits at 75, the rate is zero, and a perpetual claim pays one dollar the first time the price ever touches 100. It is worth exactly 75 cents, and no volatility number is needed to say so. The reflex answer of a dollar assumes the barrier is always reached, which a price with a floor at zero never promises: a quarter of the paths fade away without paying anything.
A three-month at-the-money call on a stock at 100 with 40% volatility is worth about eight dollars, and you can get there in two multiplications. The constant four tenths turns out to be the height of the normal bell at its peak, and the whole error of the mental rule is one rounding plus one cubic term. Scaling volatility linearly with time instead of with its square root gives ten dollars, which is 25.5% too high.
The midpoint of p and q sits strictly between them, and consecutive means precisely that no prime lives in that interval, so the answer is never and the proof is two lines with no arithmetic in it. The pair 2 and 3 survives for a different reason, since five halves is not an integer, and it is the only such pair.
Multiplying by the conjugate turns the difference into x over the sum of the same two terms, exactly, with no series and no error term, and dividing through by x reads off the limit one half. Completing the square gives a constant excess of a quarter, so the gap climbs toward a half strictly from below and never reaches it.
Unfolding two faces into a 2 by 1 rectangle turns the walk into a straight segment of length root five, crossing the shared edge at half height. The reflex answer of one plus root two is the same one-parameter family evaluated at the end of that edge instead of its middle, so the trap and the answer are two points on one curve.
Halving the length and timing the flame is not a biased estimator of half the time, it is unrelated to it: across four thousand random cords the midpoint method scattered from under fifteen seconds to over forty-five. Lighting both ends gives exactly thirty on every cord, by an argument that never evaluates the burn rate.
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.
Draw X and Y uniformly from the unit interval and their product beats a half with probability (1 - ln 2)/2, about 15.3 percent. The reflex answer of a quarter counts a condition that is genuinely necessary and treats it as sufficient, which is why 0.8 times 0.6 sits inside the quarter square and still loses. The hyperbola y = 1/(2x) cuts the winners down to a sliver, and one integral measures it.
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.
Two riders twenty-five miles apart close at fifty miles an hour, so a forty mile an hour fly shuttling between them flies exactly twenty miles. The series of shuttle legs gives the same twenty, with a first leg of 100/7 and a round-trip ratio of 1/21, but no finite number of legs ever reaches it: after eighty legs the exact total is twenty minus about 2.6e-52. The general law is wD/(a+b), and it holds identically in the rider speeds rather than by luck at 20 and 30.
One lily doubling daily covers the pond on day thirty, so eight lilies must finish in 30/8 = 3.75 days. They finish on day twenty-seven, because eight is two cubed and that slides the whole schedule exactly three days earlier. The article carries the general rule that k lilies save the floor of log base two of k, the case where five lilies save only two, and the non-overlap assumption the answer quietly rests on.
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.
Fifty white marbles, fifty black, two jars, and a fair coin choosing which jar gets drawn from. The even split gives exactly one half, and so does every other split where the jars are the same size. Isolating a single white marble reaches 74/99, an exchange argument proves nothing beats it, and three quarters turns out to be a ceiling no arrangement ever touches.
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.