Every solution of y′ = k·y doubles in the same time, T = ln 2/k, whatever its starting amount, and a decaying one halves in ln 2/|k|. The time is irrational, so the lab prints it rounded. The exact Euler sequence passes the target at a particular step, and the lab prints that step and its time as exact values: they sit on the grid of steps, not at the true time.
Core ideas
Three ideas: the derivation, the independence from the start, and the difference between the true time and the step on which the exact sequence crosses.
The solution y₀·e^(kt) has doubled when e^(kT) = 2. Taking the natural logarithm of both sides gives kT = ln 2, so T = ln 2/k. For k < 0 the quantity halves when e^(kT) = 1/2, and T = ln 2/|k|. The number ln 2 is irrational: ≈ 0.693147 to six figures.
The amount at time t + T is y₀·e^(k·(t + T)) = y₀·e^(kt)·e^(kT) = 2·y(t). Whatever y(t) is, doubling it takes T. A culture of a thousand and a culture of a million, with the same k, have the same doubling time.
The exact sequence y₀·(1 + kh)ⁿ exists only at multiples of h. The first step n with yₙ at least double the start is an exact statement about the sequence, and the lab prints it with its time. For growth it is never earlier than T, and how much later it falls depends on the step: within a step for a fine one, more than a step for a coarse one.
The material
The derivation, the independence from the start, and the exact crossing step of the sequence from “Exponential Growth”.
Take y′ = (1/2)·y, so k = 1/2. The solution is y₀·e^(t/2), and it doubles when e^(T/2) = 2, which is T/2 = ln 2 and T = 2·ln 2. The lab prints this as ln 2/(1/2) and then ≈ 1.38629, rounded; the irrational factor is ln 2 ≈ 0.693147.
A bigger rate constant gives a shorter doubling time, and the product k·T = ln 2 is the same for every equation of this kind. A useful consequence: T ≈ 0.693/k, so a rate of k = 0.07 per year, about seven percent, doubles in about ten years. That is rounded, and the lab prints it as rounded whenever it appears.
Begin with y₀ = 1 and the target 2, and again with y₀ = 8 and the target 4 and k = −1/2: the lab prints the same time, ≈ 1.38629, because k is the only thing the formula uses. Change the start to 5 and the target to 10 on the first preset to see the same step and the same time again. The amount scales the whole curve and moves nowhere along the time axis.
With k = 1/2 and h = 1/4 the factor is 9/8. The terms (9/8)⁵ = 59049/32768 ≈ 1.80203 and (9/8)⁶ = 531441/262144 ≈ 2.02729 straddle 2, so the exact sequence first reaches double at step 6, which is t = 6·(1/4) = 3/2. The lab prints step 6 (t = 3/2).
The true curve passes 2 at T ≈ 1.38629, between steps 5 and 6. For growth the exact Euler sequence lies below the curve, so it cannot cross earlier than the curve does: the crossing step is at or after the first grid point past T, and here it is that grid point. It need not be. A coarser step leaves the sequence further below the curve, and in the third preset the crossing comes a whole step after the grid point that follows T.
For k = 1/10 and h = 1 the factor is 11/10. The terms (11/10)⁷ = 19487171/10000000 ≈ 1.94872 and (11/10)⁸ ≈ 2.14359 straddle 2, so the lab prints step 8 (t = 8). The true doubling time is T ≈ 6.93147, so at t = 7 the closed form column already reads ≈ 2.01375, while the exact sequence at step 7 is still below 2: it is late by a whole step.
Interactive
A geometric sequence beside an exponential
Method
Five moves, and the last one is the one that stops a coarse step being mistaken for the answer.
Positive means doubling, negative means halving. The time uses |k| only.
y₀·e^(kT) = 2·y₀ for doubling, y₀·e^(kT) = y₀/2 for halving. The y₀ cancels, and that is why the answer does not depend on it.
kT = ln 2, so T = ln 2/|k|. Quote it rounded and marked ≈, since ln 2 is irrational.
Compute (1 + kh)ⁿ for increasing n until it passes 2 (or falls to 1/2). The first such n and the time n·h are exact.
The grid time is at least T for growth, with the gap set by h. Report the step size alongside it.
The true time first, rounded, and then the exact sequence's crossing, which is exact.
Euler underestimates growth, so it is late by the part of a step it needs to make up. Here that is part of a quarter-step, and with the step equal to 1 in the third preset it is more than a whole one. Neither is wrong arithmetic; the first is the exact sequence and the second is the true curve, and they are different objects.
The decay preset runs the same arithmetic downward: the factor is 7/8 and the exact sequence from 8 first reaches 4 or below at step 6, the same step, though not for the same reason. For decay the factor 7/8 is below e^(−1/8), so the sequence falls faster than the curve and could cross early; here both cross between steps 5 and 6. Its time is the half-life, which equals the doubling time for the opposite rate.
You should be able to turn a rate constant into a doubling time or a half-life, explain why the starting amount does not enter, compute powers of the Euler factor until one passes the target, and say how the grid time and the true time differ.
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