Money grows at 10 percent a year. Paid in one jump, in 12 monthly jumps, or smoothly all the time, you get slightly different amounts. One pair of rules handles jumps, smooth stretches and mixtures of the two.
Name ____________________Date ____________
Tonight's idea
A quantity \(y\) grows at rate \(r\): in a short time \(h\) it gains about \(rhy\).
Jump. Wait a time \(h\), then add \(rhy\) all at once. Smooth. Grow all the time, so that the slope is always \(r\) times the amount: \(y' = ry\). Then \(y\) is \(e^{rt}\) times its start (grade 10).
A jump across a gap \(h\) multiplies \(y\) by \(1 + rh\). A smooth stretch of length \(t\) multiplies \(y\) by \(e^{rt}\).
A calendar can mix the two: separate dots where \(y\) jumps to the next dot, solid stretches where it grows smoothly. Such a set of times is called a time scale. In 1988 Stefan Hilger introduced a calculus that works on all of them.
To follow \(y' = ry\) on a computer, step forward a time \(h\) using the slope at the start of the step: new \(y = y + h \cdot ry\). This is Euler's method, and here it is exactly the jump rule. Small steps stay close to the smooth answer. Steps that are too big can go badly wrong (exercise 4).
Watch one: 10 percent for one year
Start with 1000 dollars and \(r = 0.1\) per year. Split the year into \(n\) equal gaps of \(h = 1/n\) and jump across each one. Every jump multiplies by \(1 + 0.1/n\).
How it is paid
1 jump
2 jumps
12 jumps
smooth
Multiplier
\(1.1\)
\(1.05^2 = 1.1025\)
\((1 + 0.1/12)^{12} = 1.104713\)
\(e^{0.1} = 1.105171\)
Dollars
1100.00
1102.50
1104.71
1105.17
More jumps pay more, because each later jump works on a bigger pile. Smooth growth pays most. For \(r \gt 0\) the multiplier \((1 + r/n)^n\) climbs with every extra jump and closes in on \(e^r\) without reaching it.
Read the table as Euler's method too. The true answer to \(y' = 0.1y\) after one year is 1105.17. One big step misses by 5.17 dollars. Twelve steps miss by 46 cents.
The first column against the last is one case of a fact you will need: \(1 + x \lt e^x\) for every \(x \neq 0\). So for \(r \gt 0\), a jump across a gap always gives a little less than smooth growth across the same gap.
Your turn
Carry on the worked example. (a) Find the dollars after one year with 4 jumps, and with 365 jumps. (b) How many cents short of smooth growth are the 365 jumps?
(a) Start Euler's method on \(y' = ry\) at \(y = 1\) and take \(n\) steps of size \(h = 1/n\). Show that you land on \((1 + r/n)^n\). (b) Expand the case \(n = 2\) and show that two jumps beat one jump by exactly \(r^2/4\). Check this against the table.
One week as a time scale. Each label marks midnight at the start of that day.A made-up account grows at rate \(r\) per day. From Monday to Saturday it changes only at midnight, once a day: five jumps, each across a gap of one day, the last from Friday to Saturday. From then until the next Monday it grows smoothly. (a) Write the multiplier for the whole week. (b) Take \(r = 0.01\), far more than any real account pays (this is a model). Compute the week's multiplier, and compare it with 7 smooth days and with 7 daily jumps. (c) Put the three in order and explain why that order holds for every \(r \gt 0\).
Now a quantity that shrinks: \(y' = -2y\), starting at \(y = 1\). The true answer \(e^{-2t}\) stays positive and fades towards 0. Euler's method with step \(h\) multiplies \(y\) by \(1 - 2h\) each step. Write down \(y\) after each of the first four steps for (a) \(h = \tfrac14\), (b) \(h = 1\), (c) \(h = 1.5\). Say in words what each list does. For (a), compare the fourth value with the true value at \(t = 1\).
Stretch: how big a step is too big?
Keep \(y' = -2y\). (a) For exactly which steps \(h \gt 0\) do the Euler values shrink towards 0? Is there a largest such step? (b) For which steps do the values also stay positive, as the true answer does? What happens at \(h = \tfrac12\)? (c) Answer (a) again for \(y' = -ky\) with \(k \gt 0\). What does a faster decay demand of the step?
Answers
(a) \(1000 \times 1.025^4 = 1103.81\) dollars, and \(1000(1 + 0.1/365)^{365} = 1105.16\) dollars. (b) To more places the two amounts are 1105.1558 and 1105.1709, so daily jumps fall about 1.5 cents short. Going from 12 jumps to 365 gained only 44 cents.
(a) One step takes \(y\) to \(y + h \cdot ry = y(1 + rh) = y(1 + r/n)\). Each step multiplies by the same number, so \(n\) steps from 1 give \((1 + r/n)^n\). (b) \((1 + r/2)^2 = 1 + r + r^2/4\), which is \(r^2/4\) more than \(1 + r\). With \(r = 0.1\) that is 0.0025, and \(1.1025 - 1.1 = 0.0025\): the 2.50 dollars in the table.
(a) Five jumps across gaps of 1, then a smooth stretch of length 2: \((1 + r)^5 e^{2r}\). (b) \(1.01^5 \times e^{0.02} = 1.051010 \times 1.020201 = 1.072242\). Seven smooth days: \(e^{0.07} = 1.072508\). Seven jumps: \(1.01^7 = 1.072135\). (c) Seven jumps \(\lt\) mixed week \(\lt\) seven smooth days. Start from seven jumps, \((1 + r)^7\). Since \(1 + r \lt e^r\), swapping the two weekend jumps \((1 + r)^2\) for \(e^{2r}\) raises the total, and swapping the five weekday jumps for \(e^{5r}\) raises it again.
(a) The multiplier is \(\tfrac12\): \(\tfrac12, \tfrac14, \tfrac18, \tfrac1{16}\). It fades and stays positive, like the true answer, though too fast: \(\tfrac1{16} = 0.0625\) against \(e^{-2} = 0.1353\). (b) The multiplier is \(-1\): \(-1, 1, -1, 1\). It flips sign for ever and never fades. (c) The multiplier is \(-2\): \(-2, 4, -8, 16\). It flips sign and doubles in size each step. It blows up, although the true answer fades away.
Stretch. (a) After \(n\) steps \(y = (1 - 2h)^n\). This shrinks towards 0 exactly when \(-1 \lt 1 - 2h \lt 1\), that is \(0 \lt h \lt 1\). There is no largest such step: every \(h\) below 1 works (slowly, when \(h\) is near 1), and \(h = 1\) itself does not. (b) The values stay positive exactly when \(1 - 2h \gt 0\), that is \(h \lt \tfrac12\). At \(h = \tfrac12\) the multiplier is 0, so \(y\) drops to 0 in one step and stays there, which the true answer never does. For \(\tfrac12 \lt h \lt 1\) the values shrink but flip sign each step. (c) The multiplier is \(1 - kh\), so you need \(0 \lt h \lt 2/k\). The faster the decay, the smaller the step must be. On a time scale this is a warning about jumps: across a gap of \(h\), a rate below \(-1/h\) makes \(y\) change sign.