Unit 25 · Sequences, Series and Modeling

A stack of one plastic chair is 30 inches tall. Each chair you add raises the stack by 3 inches. So the heights go 30, 33, 36, 39, 42 and so on. An ordered list like this is a sequence. Each number in it is a term, and the position of a term is its index. We write a_n (read "a sub n") for the term in position n. Here a_1 = 30, a_2 = 33 and a_4 = 39. The index counts chairs; the term measures inches.

There are two ways to write the rule. An explicit rule gives any term straight from its index: a_n = 30 + 3(n − 1). Test it on the fourth term: 30 + 3(3) = 39, which matches. It jumps straight to the tenth term too: a_10 = 30 + 3(9) = 57 inches. A recursive rule gives each term from the one before it: a_1 = 30, and a_n = a_(n−1) + 3. It needs a starting term, because "add 3" means nothing until you know where to begin.

Each kind of rule has a job. The explicit rule is fast for far-off terms; the recursive rule matches how the stack actually grows, one chair at a time. To find a_10 recursively you would add 3 nine times. A common mistake is to write the explicit rule as a_n = 30 + 3n. That gives a_1 = 33, but the first stack is 30. Always test a rule on the first term before you trust it.

Not every sequence adds a fixed amount. The list 1, 4, 9, 16, 25 has the rule a_n = n^2. The list 2, 4, 8, 16, 32 doubles each time, so a_n = 2^n. The list 5, −5, 5, −5 flips sign, so a_n = 5 × (−1)^(n+1). Check that last one: for n = 1 the exponent is 2, and (−1)^2 = 1, so a_1 = 5. For n = 2 the exponent is 3, and a_2 = −5. A rule is right only if it produces every term you were given.

A gym charges $40 to join and $25 each month. The total you have paid after 1, 2, 3, 4 months is 65, 90, 115, 140. Every term is the one before plus 25. A sequence that grows by the same amount each step is an arithmetic sequence, and that fixed amount is the common difference, written d. Here a_1 = 65 and d = 25, so the explicit rule is a_n = 65 + 25(n − 1). After 12 months: 65 + 25(11) = 65 + 275 = 340. Check another way: the $40 fee plus 12 × $25 = 40 + 300 = 340. It matches.

You can build the rule from any two terms. Suppose a_3 = 14 and a_7 = 26. From position 3 to position 7 is 4 steps, and the terms rose by 26 − 14 = 12, so d = 12 ÷ 4 = 3. Walk back two steps from a_3 to find a_1 = 14 − 2(3) = 8. The rule is a_n = 8 + 3(n − 1), which simplifies to a_n = 3n + 5. Check: a_7 = 3(7) + 5 = 26. Good.

The common difference can be negative. A cup of water at 52 degrees cools 4 degrees each hour: 52, 48, 44, 40. The rule is a_n = 52 − 4(n − 1). When does it hit 20 degrees? Set 52 − 4(n − 1) = 20, so 4(n − 1) = 32, so n − 1 = 8 and n = 9. Check: a_9 = 52 − 4(8) = 52 − 32 = 20. The most common error is writing 4n instead of 4(n − 1); that shifts every term by one position.

If you plot an arithmetic sequence with the index on the horizontal axis, the points lie on a straight line whose slope is d. That is why a_n = 3n + 5 looks like the linear function y = 3x + 5. The difference is that a sequence uses only whole-number positions 1, 2, 3 and so on, so the graph is a row of separate dots, not a solid line.

On Monday three people hear a rumor. Each day, every person who knows it tells two new people, so the count of new hearers goes 3, 6, 12, 24, 48. Each term is the one before times 2. A sequence that multiplies by the same number each step is a geometric sequence, and that number is the common ratio, written r. The explicit rule is a_n = a_1 × r^(n − 1), so here a_n = 3 × 2^(n − 1). On day 7: 3 × 2^6 = 3 × 64 = 192. Doubling by hand, 3, 6, 12, 24, 48, 96, 192, agrees.

The ratio can be less than 1. A ball dropped from 200 cm bounces back to 3/5 of its previous height each time: 200, 120, 72, 43.2. The fourth height is 200 × 0.6^3 = 200 × 0.216 = 43.2 cm. The terms shrink toward 0 but never reach it. A negative ratio makes the signs alternate: 5, −10, 20, −40 has r = −2.

To find r, divide any term by the one before it: 12 ÷ 6 = 2 and 6 ÷ 3 = 2. To decide whether a list is geometric, check that every such ratio is the same. The most common error is using r^n instead of r^(n − 1). For the rumor that gives 3 × 2^7 = 384 on day 7, one day too far. Another error is multiplying a_1 by n × r, as if the sequence were arithmetic.

Arithmetic sequences add; geometric sequences multiply. The difference shows in the graph. An arithmetic sequence plots as dots on a line. A geometric sequence with r greater than 1 curves upward faster and faster, the shape of an exponential function. With r between 0 and 1 it curves down toward the axis without touching it.

A reading log shows 10 pages on day 1, then 15, 20, 25 and 30 on the next four days. The total is 10 + 15 + 20 + 25 + 30 = 100. Mathematicians write a sum like this with the Greek letter Σ (sigma): Σ from k = 1 to 5 of (5k + 5). The letter k is the index of summation. The 1 below the sigma is where k starts, the 5 above is where it stops, and the expression after it is the formula for each term. Check the ends: k = 1 gives 5(1) + 5 = 10 and k = 5 gives 5(5) + 5 = 30.

To expand a sigma expression, substitute each index value and add. Σ from k = 1 to 4 of k^2 means 1 + 4 + 9 + 16 = 30. The lower limit is not always 1. Σ from k = 3 to 6 of (2k − 1) means 5 + 7 + 9 + 11 = 32. The number of terms is the upper limit minus the lower limit plus 1, so from 3 to 6 there are 6 − 3 + 1 = 4 terms, not 3.

The sum of the terms of a sequence is called a series. The sum of just the first n terms is the partial sum, written S_n. For the reading log, S_3 = 10 + 15 + 20 = 45. Three errors show up often: starting k at 1 when the lower limit is something else, counting the terms as upper minus lower, and multiplying the formula by the number of terms instead of adding the separate values.

One useful fact: a constant factor can be pulled out front. Σ from k = 1 to 4 of 3k is 3 + 6 + 9 + 12 = 30, and 3 × (1 + 2 + 3 + 4) = 3 × 10 = 30. Sums of separate pieces also split: Σ of (k + 2) equals Σ of k plus Σ of 2. These rules turn a messy sigma into simpler ones you already know how to add.

Gauss’s pairing trick works for any arithmetic sequence. A theater has 20 seats in the front row, and each row back has 2 more seats. There are 15 rows. The last row has a_15 = 20 + 2(14) = 48 seats. Pair the first and last rows: 20 + 48 = 68. The second and second-to-last: 22 + 46 = 68. Every pair gives 68, the sum of the first and last terms. Fifteen terms make 7 pairs plus the middle term, a_8 = 20 + 2(7) = 34. So the total is 7 × 68 + 34 = 476 + 34 = 510 seats.

The formula packages this: S_n = n(a_1 + a_n)/2. For the theater, S_15 = 15 × (20 + 48)/2 = 15 × 34 = 510. Same answer. The fraction (a_1 + a_n)/2 is the average of the first and last term, and in an arithmetic sequence that is the average of all the terms. Multiply an average by a count and you get a total. If you do not know the last term, substitute a_n = a_1 + (n − 1)d to get S_n = (n/2)[2a_1 + (n − 1)d].

Try a sigma sum: Σ from k = 1 to 40 of (3k + 2). The first term is 5, the 40th is 3(40) + 2 = 122. So S_40 = 40 × (5 + 122)/2 = 40 × 63.5 = 2540. Check with the second form: (40/2)[2(5) + 39(3)] = 20 × (10 + 117) = 20 × 127 = 2540. Both routes agree.

A savings challenge: put away $5 in week 1, $10 in week 2, $15 in week 3, and so on for 52 weeks. Week 52’s deposit is 5 × 52 = $260. The total is 52 × (5 + 260)/2 = 52 × 132.5 = $6,890. The most common error is multiplying the count by the last term, 52 × 260 = 13,520, which is about twice the truth. The other is forgetting to divide by 2 after adding first and last.

Take S = 3 + 6 + 12 + 24 + 48. Multiply every term by the ratio 2: 2S = 6 + 12 + 24 + 48 + 96. Now subtract the first line from the second. Almost everything cancels: 2S − S = 96 − 3, so S = 93. Check by adding: 3 + 6 = 9, 9 + 12 = 21, 21 + 24 = 45, 45 + 48 = 93. The same subtraction on a general geometric series gives the formula S_n = a_1(r^n − 1)/(r − 1). For this one: 3 × (2^5 − 1)/(2 − 1) = 3 × 31 = 93.

The formula handles ratios less than 1 too, where it is often written S_n = a_1(1 − r^n)/(1 − r). Add 100 + 50 + 25 + ... for 8 terms: 100 × (1 − 0.5^8)/(1 − 0.5) = 100 × (1 − 1/256)/0.5 = 200 × 255/256 = 199.21875. Adding the eight terms by hand, 100 + 50 + 25 + 12.5 + 6.25 + 3.125 + 1.5625 + 0.78125, gives the same 199.21875. Notice how close the total is to 200 already.

An old puzzle asks for one penny on day 1, two on day 2, four on day 3, and so on for 30 days. Day 30 alone is 2^29 pennies. The total is 1 × (2^30 − 1)/(2 − 1) = 2^30 − 1 = 1,073,741,823 pennies, which is a little over $10.7 million. Doubling is slow at first and then overwhelming. The most common errors are using r^(n − 1) in the formula instead of r^n, and dropping the −1 in the numerator.

A quick sanity check: the last term of a doubling series is always one more than the sum of all the earlier terms. In 3 + 6 + 12 + 24 + 48, the earlier terms add to 45 and 48 = 45 + 3, one first term more. For the pennies, the day-30 pile of 2^29 pennies is one penny more than days 1 through 29 combined. If your sum is far from twice the last term, look for a slip.

Walk toward a wall, covering half the remaining distance with each step. Your first step covers 1/2 of the way, the next 1/4, then 1/8, then 1/16. The partial sums are 1/2, 3/4, 7/8, 15/16. They creep toward 1 and never pass it; the gap to the wall halves each time. An endless sum whose partial sums settle on one number is said to converge, and that number is its sum. Here 1/2 + 1/4 + 1/8 + ... = 1.

Why? Start from S_n = a_1(1 − r^n)/(1 − r). When r is between −1 and 1, the power r^n shrinks toward 0 as n grows; 0.5^10 is already under 0.001. Dropping that vanishing piece leaves S = a_1/(1 − r). For the walk, (1/2)/(1 − 1/2) = 1. Another example: 6 + 2 + 2/3 + ... has r = 1/3, so S = 6/(1 − 1/3) = 6/(2/3) = 9. The partial sums 6, 8, 8.67, 8.89 are closing in on 9.

When r is 1 or more, or −1 or less, the series diverges: 1 + 2 + 4 + 8 + ... grows without bound, and 1 − 1 + 1 − 1 + ... bounces between 1 and 0 forever. Plugging r = 2 into the formula gives 1/(1 − 2) = −1, a nonsense answer, so always check the ratio before using it. Here is the formula at work: a ball dropped from 10 m rebounds to 0.6 of each height. Its total up-and-down travel is 10 + 2 × (6 + 3.6 + 2.16 + ...) = 10 + 2 × 6/(1 − 0.6) = 10 + 2 × 15 = 40 m.

The decimal 0.7777... means 0.7 + 0.07 + 0.007 + ..., a geometric series with a_1 = 0.7 and r = 0.1. Its sum is 0.7/(1 − 0.1) = 0.7/0.9 = 7/9. Divide 7 by 9 and you get 0.777..., so the formula is right. When the repeating block has two digits, the ratio is 0.01. So 0.363636... = 0.36/(1 − 0.01) = 0.36/0.99 = 36/99 = 4/11. Check: 4 ÷ 11 = 0.3636...

A famous case: 0.9999... = 0.9/(1 − 0.1) = 0.9/0.9 = 1. That surprises people, but the partial sums 0.9, 0.99, 0.999 close in on 1 and nothing else. A decimal with a non-repeating start splits into pieces: 2.1555... = 2.1 + (0.05 + 0.005 + ...) = 2.1 + 0.05/0.9 = 2.1 + 1/18 = 21/10 + 1/18. With the common denominator 90 that is 189/90 + 5/90 = 194/90 = 97/45. Check: 97 ÷ 45 = 2.1555...

The same idea runs a medicine schedule. A patient takes 100 mg every morning, and 40% of the drug is still in the body 24 hours later. Right after the second dose the body holds 100 + 40 = 140 mg. After the third: 100 + 0.4 × 140 = 156. After the fourth: 100 + 0.4 × 156 = 162.4. In the long run the amount just after a dose settles at a steady state: 100 + 40 + 16 + 6.4 + ... = 100/(1 − 0.4) = 166.67 mg. The most common error is using the wrong ratio, such as 0.1 for a two-digit repeating block.

In 1202 Leonardo of Pisa, known as Fibonacci, published Liber Abaci, a book that helped bring Hindu-Arabic numerals to Europe. One of its puzzles asks how rabbit pairs multiply if each grown pair produces a new pair every month and a new pair takes a month to grow up. The pair counts follow a rule with two starting terms: F_1 = 1, F_2 = 1, and F_n = F_(n−1) + F_(n−2). That gives 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144. This is the Fibonacci sequence, the best-known example of a rule that looks back two steps.

Divide each term by the one before it: 8/5 = 1.6, 13/8 = 1.625, 21/13 ≈ 1.615, 34/21 ≈ 1.619, 55/34 ≈ 1.618, 89/55 ≈ 1.618. The ratios bounce above and below a fixed number and close in on it. That number is the golden ratio, φ = (1 + √5)/2 ≈ 1.618. It is the positive solution of x^2 = x + 1; check that 1.618^2 ≈ 2.618 ≈ 1.618 + 1. A rectangle whose long side is φ times its short side is a golden rectangle: cut off a square and the leftover piece is another golden rectangle.

Fibonacci numbers turn up when counting spirals on a pineapple or the seed head of a sunflower, which often number 8, 13, 21 or 34. The sequence is not geometric, since the ratio is only approximately constant, but it gets closer to geometric with every term. A useful shortcut: F_n is φ^n/√5 rounded to the nearest whole number. For n = 10 that is about 122.99/2.236 ≈ 55.0, and F_10 = 55. Any sequence with the same add-the-last-two rule, such as 2, 1, 3, 4, 7, 11, has ratios that approach φ too.

You borrow $1,000 at 1% interest per month and pay $100 each month. Each month the lender adds interest first, then subtracts your payment. The rule is B_0 = 1000 and B_n = 1.01 × B_(n−1) − 100. Step it: B_1 = 1010 − 100 = 910. B_2 = 1.01 × 910 − 100 = 919.10 − 100 = 819.10. B_3 = 1.01 × 819.10 − 100 = 827.29 − 100 = 727.29. The debt falls a little faster each month because the interest charge shrinks as the balance shrinks.

A deer herd of 500 grows 8% a year, and wildlife managers remove 30 deer each year. The rule is P_n = 1.08 × P_(n−1) − 30. Then P_1 = 540 − 30 = 510, P_2 = 550.8 − 30 = 520.8, and P_3 = 562.46 − 30 = 532.46. Is there a size where the herd holds steady? Set P = 1.08P − 30, so 0.08P = 30 and P = 375. That is the equilibrium. Check: 1.08 × 375 = 405, and 405 − 30 = 375. A herd of exactly 375 stays at 375. Our herd started above 375, so growth outruns the removals and it keeps rising.

Recursive rules are how spreadsheets and calculators handle these problems: fill in the rule once and drag it down the column. An explicit formula exists but is messier, so people usually step the rule. Two errors to avoid: subtracting the payment before the interest is added, which understates the charge, and applying the growth rate to the original amount every time instead of to the new total. Growth on growth is called compounding, and it is why the herd’s yearly increase gets bigger each year.

Start with a straight segment of length 1. Replace its middle third with two sides of a small triangle that points outward. Now you have 4 segments, each 1/3 long, total length 4/3. Do the same to each of those 4 segments: 16 segments, each 1/9 long, total length 16/9. This is the Koch curve. At stage n it has 4^n segments, each of length (1/3)^n, so its total length is (4/3)^n. At stage 3 that is 64 segments of length 1/27, total 64/27 ≈ 2.37. The length forms a geometric sequence with ratio 4/3, greater than 1, so it grows without bound, even though the whole curve fits in a small box.

The Sierpinski triangle goes the other way. Take a shaded triangle and remove the middle triangle formed by the midpoints of its sides, leaving 3 shaded triangles, each with 1/4 of the area. Repeat on each of those. At stage n there are 3^n shaded triangles, each with (1/4)^n of the original area, so the shaded area is (3/4)^n. At stage 4 that is 81 triangles covering 81/256 ≈ 0.316 of the original. The shaded area shrinks toward 0 while the count of triangles explodes.

Shapes like these are fractals: the word was coined by Benoit Mandelbrot in 1975 for shapes that look similar at every scale. Zoom in on a piece of the Koch curve and you see the same crinkled shape again; that property is called self-similarity. Coastlines, fern leaves, river networks and the branching airways of the lungs all show it in rough form. The common error is mixing up the two sequences: the count of pieces grows (ratio 4, or 3), while the size of each piece shrinks (ratio 1/3, or 1/4). Both are geometric, but with different ratios.

A school installs a water-bottle filling station. "Was it worth it?" is too vague to compute. A sharper question is: how many disposable bottles will students skip in one school year? To answer it you need assumptions, statements you accept as true for now so the math can begin. Suppose the school has 800 students, each uses the station 0.5 times per day instead of buying a bottle, and the year has 180 school days. Then bottles skipped = 800 × 0.5 × 180 = 72,000.

That calculation is a model: a simplified version of the situation written in math. Give its parts names, and it becomes a function you can reuse. Let S be the number of students, u the uses per student per day and D the school days; then B = S × u × D. Every assumption is a place the model could be wrong, so write them down in a list. Perhaps not every fill replaces a bottle; if only 1 in 4 does, the estimate drops to 72,000 ÷ 4 = 18,000. Perhaps attendance is not 100%.

The modeling cycle is the loop you walk around: ask a clear question, list assumptions, build the model, compute, check the result against reality, revise, and report. You rarely get it right the first time, and that is the point of the loop. A units check catches many errors early: students × (uses per student per day) × days = uses. If your formula ends in "students × days" with nothing per day, a factor is missing.

A cup of coffee starts at 90 °C in a 20 °C room. First model: it cools 10 degrees per minute, so T(t) = 90 − 10t. After 5 minutes it predicts 40 °C. After 10 minutes it predicts 90 − 100 = −10 °C. That is colder than the room, which cannot happen. The model has failed a reality check. The wrong assumption is "constant rate"; coffee cools fast when it is hot and slowly when it is nearly room temperature.

Revised model: the gap between coffee and room shrinks by the same percent each minute. The gap starts at 70 degrees. If it loses 10% a minute, T(t) = 20 + 70 × 0.9^t. At t = 10: 0.9^10 ≈ 0.349, so 70 × 0.349 ≈ 24.4, and T ≈ 44.4 °C. A thermometer reading of 45 °C at 10 minutes says the revised model is close. Notice that you did not fudge the answer; you changed an assumption and recomputed. That is what revise means.

Three checks catch most bad models. Check the boundary cases: at t = 0 the model gives 20 + 70 = 90, correct, and for very large t it approaches 20, the room, also correct. Check against a measured point, as we did at 10 minutes. Check the direction and size of change: cooling should slow down, and it does. A model that matches one data point can still be wrong, so use several checks, not one.

A table with equally spaced inputs tells you its family if you look at how the outputs change. For x = 0, 1, 2, 3, 4 with y = 3, 7, 11, 15, 19, the first differences are 4, 4, 4, 4. Constant first differences mean a linear model, y = 4x + 3, with the difference as the slope. Check x = 4: 4(4) + 3 = 19.

For x = 1, 2, 3, 4, 5 with y = 2, 5, 10, 17, 26, the first differences are 3, 5, 7, 9. Not constant. Take the differences of those: 2, 2, 2. Constant second differences mean a quadratic, and the second difference equals 2a, so a = 1. Here y = x^2 + 1 fits every row: 1 + 1 = 2, 4 + 1 = 5, 9 + 1 = 10, 16 + 1 = 17, 25 + 1 = 26.

For x = 0, 1, 2, 3, 4 with y = 5, 10, 20, 40, 80, the first differences 5, 10, 20, 40 grow, but the ratios 10/5, 20/10, 40/20, 80/40 are all 2. A constant ratio means an exponential, y = 5 × 2^x. Real data are never this clean, so look for differences or ratios that are roughly constant. The common error is to see growing differences and shout "exponential" without checking ratios; a quadratic’s differences also grow, but in a straight-line way.

One more caution: the inputs must be equally spaced for these tests to work. If x jumps 1, 2, 4, 8, the differences mean nothing, because each step covers a different distance. When a table has uneven spacing, either compute the change per unit of x or plot the points and look at the shape of the graph.

A taxi charges $3 plus $2.25 per mile. "Plus a fixed amount per mile" is the signature of a linear model: C(m) = 3 + 2.25m, so 10 miles cost 3 + 22.5 = $25.50. A bacteria culture doubles every 3 hours. "Doubles every" or "grows 5% each" is the signature of an exponential: N(t) = N_0 × 2^(t/3), so after 12 hours the count is 2^4 = 16 times the start.

A ball thrown upward at 40 feet per second from a height of 5 feet follows h(t) = −16t^2 + 40t + 5, in feet. Anything thrown, and any area that depends on a length, is quadratic. The ball peaks at t = 40/32 = 1.25 seconds, where h = −16(1.5625) + 50 + 5 = −25 + 55 = 30 feet. Hours of daylight in Chicago rise and fall every year, so daylight is periodic and calls for a sine or cosine. A rumor in a school of fixed size grows fast, then levels off as it runs out of people: that is logistic.

The trap is linear versus exponential when the word "percent" is nearby. "Grows by 5 a year" is linear; "grows by 5% a year" is exponential. Start both at 100 and run 20 years. Linear: 100 + 5 × 20 = 200. Exponential: 100 × 1.05^20 ≈ 100 × 2.653 ≈ 265. The gap only widens with time. When a situation mentions a rate, ask whether it is an amount per unit or a percent per unit before you choose.

Five students record hours studied and quiz scores: (1, 62), (2, 70), (3, 74), (4, 83), (5, 88). A calculator or spreadsheet computes the line of best fit, also called the regression line: y = 6.5x + 55.9. The slope says each extra hour goes with about 6.5 more points; the intercept, 55.9, is the model’s score for zero hours. Predict 6 hours: 6.5(6) + 55.9 = 39 + 55.9 = 94.9.

How good is the fit? For each point, compute the residual: actual minus predicted. At x = 3 the line predicts 6.5(3) + 55.9 = 75.4, the actual score is 74, so the residual is 74 − 75.4 = −1.4. At x = 1: predicted 62.4, residual −0.4. At x = 4: predicted 81.9, residual 1.1. The residuals are small and scattered above and below zero, which is what a good fit looks like. If the residuals form a pattern, such as all negative in the middle and positive at the ends, the data are curving and you should try another family.

Technology also fits exponential and quadratic models, and reports the correlation coefficient r, which runs from −1 to 1. Values near 1 or −1 mean the points hug a line; values near 0 mean no linear pattern. Two cautions. First, a strong r does not prove that one thing causes the other. Second, do not extrapolate far beyond the data: 12 hours gives 6.5(12) + 55.9 = 133.9, more than 100 points, which is impossible. The model is only trusted where the data live.

A club sells T-shirts. Making them costs $200 for the design plus $4 per shirt, so C(x) = 200 + 4x. Each shirt sells for $12, so revenue is R(x) = 12x. Profit is what is left: P(x) = R(x) − C(x) = 12x − (200 + 4x) = 8x − 200. The break-even point is where profit is zero: 8x = 200, so x = 25 shirts. Check: 25 shirts bring in 12 × 25 = $300 and cost 200 + 100 = $300. Sell 40 and the profit is 8(40) − 200 = $120.

Sometimes one function feeds another. A store takes 20% off, so D(p) = 0.8p, and then the register adds 8% tax, so T(p) = 1.08p. Doing the discount first and then the tax is the composition T(D(p)) = 1.08 × 0.8p = 0.864p. Order does not matter for two multiplications. It matters when a $10 coupon is involved. Coupon then tax: 1.08(p − 10) = 1.08p − 10.8. Tax then coupon: 1.08p − 10. On a $50 item that is $43.20 versus $44.00.

Multiplying functions builds a model too. Suppose the number of shirts you can sell at price p is q(p) = 500 − 20p; raise the price and fewer people buy. Revenue is price times quantity: R(p) = p(500 − 20p) = 500p − 20p^2, a quadratic that opens downward. Its peak is at p = 500/40 = $12.50, where R = 500(12.5) − 20(156.25) = 6250 − 3125 = $3,125. Two linear pieces combined to make a quadratic with a best price, something neither piece showed alone.

Chicago gets about 15 hours of daylight in late June and about 9 in late December, so the average is about 12 and the swing above or below it is about 3 hours. A sine wave with those numbers, with t the day of the year, is D(t) = 12 + 3 sin(2π(t − 80)/365). The 12 is a vertical shift: it lifts the wave from a middle of 0 to a middle of 12. The 3 is a vertical stretch: it makes the wave rise and fall 3 hours instead of 1. The (t − 80) is a horizontal shift that puts the upward crossing near day 80, around the spring equinox, and the 365 sets the period at one year.

The same moves adjust any parent function. A ball thrown from a 5-foot height follows h(t) = −16t^2 + 40t + 5. Throw it the same way from a balcony 20 feet higher and every height goes up 20: h(t) = −16t^2 + 40t + 25. Throw it 2 seconds later and the clock is shifted: h(t) = −16(t − 2)^2 + 40(t − 2) + 5. In general, f(x) + k moves a graph up k, f(x − h) moves it right h, a × f(x) stretches it vertically by a, and −f(x) flips it.

Horizontal shifts matter most with exponentials. A town of 1,200 people in 2010 grows 3% a year. The model P(t) = 1200 × 1.03^(t − 2010) uses the year directly; the shift makes the exponent count years since 2010. For 2025: 1200 × 1.03^15 ≈ 1200 × 1.558 ≈ 1,870. Forget the shift and you compute 1.03^2025, a number with 26 digits. Always ask what value of the input makes the exponent zero, and check that the model gives the starting value there.

A garage charges $5 for the first hour and $3 for each additional hour or any part of an hour. For 3.5 hours you pay $5 for hour one, then 2.5 extra hours round up to 3, so 5 + 3 × 3 = $14. A rule like this is a piecewise function: one formula on one interval of inputs, another formula on the next. Because the price jumps at each whole hour and stays flat in between, this piece is also a step function.

Income taxes use thresholds too. Take an example schedule: no tax on the first $10,000, 10% on the part from $10,000 to $40,000, and 20% on the part above $40,000. On $55,000 of income the tax is 0 + 0.10 × 30,000 + 0.20 × 15,000 = 3,000 + 3,000 = $6,000. The common error is charging 20% on all $55,000, which gives $11,000, nearly double. Written as a function: T(x) = 0 for x ≤ 10,000; T(x) = 0.10(x − 10,000) for 10,000 < x ≤ 40,000; T(x) = 3,000 + 0.20(x − 40,000) for x > 40,000. Check the seam at 40,000: the middle piece gives 0.10 × 30,000 = 3,000, and the top piece starts at 3,000, so the pieces meet.

The Soldier Field model from the chapter opening is piecewise as well. Let F(t) be the flow in people per minute at minute t. For 0 ≤ t ≤ 2 the flow ramps: F(t) = 3000t. For t > 2 it holds: F(t) = 6000. The number of people out by minute 2 is the average flow times the time, 3000 × 2 = 6,000, and after that 6,000 leave each minute, so by minute 10 the count is 6,000 + 8 × 6,000 = 54,000. A single formula could not capture the ramp and the plateau; two pieces do.

In the Soldier Field model, 61,000 people at 100 per second gives 610 seconds. But the rate was a guess. Try 80 per second: 61,000 ÷ 80 = 762.5 seconds, about 12.7 minutes. Try 120: 61,000 ÷ 120 ≈ 508 seconds, about 8.5 minutes. Moving one input by 20% moved the answer by about 20% as well. That test is called sensitivity analysis, and it tells you which assumptions deserve the most care. An honest report gives the range: about 8.5 to 12.7 minutes.

When you can compare a model to a measurement, compute the percent error: |predicted − actual| ÷ actual × 100. The coffee model predicted 44.4 °C at 10 minutes and the thermometer read 45 °C, so the error is 0.6 ÷ 45 × 100 ≈ 1.3%. The absolute error is the plain difference, 0.6 degrees. Measurements carry their own error too: a length recorded as 12.5 cm to the nearest 0.1 cm could really be anywhere from 12.45 to 12.55 cm.

Errors spread through a calculation. A rectangle measured at 12.5 cm by 8.0 cm has an area of 100 cm^2 by the numbers. But the true sides could be as small as 12.45 and 7.95, giving 12.45 × 7.95 ≈ 98.98 cm^2, or as large as 12.55 and 8.05, giving 12.55 × 8.05 ≈ 101.03 cm^2. Reporting "100.0000" would claim a precision the measurements never had. Report the area as about 99 to 101 cm^2, an interval estimate.

An 8-year-old is 128 cm tall and has grown 6 cm a year. The linear model h = 128 + 6(a − 8) fits ages 6 to 10 well. Push it to age 30 and it predicts 128 + 6 × 22 = 260 cm, taller than any person who has ever lived. The model is not wrong; it was used outside its domain of validity, the range of inputs where its assumptions hold. Growth stops in the late teens, and the model never knew that.

Exponential growth has the same problem. In 1798 Thomas Malthus argued that populations tend to grow geometrically while food supplies grow arithmetically. In 1838 Pierre Verhulst proposed a fix, the logistic model, in which growth slows as the population nears a ceiling called the carrying capacity. One form is P(t) = 1000/(1 + 9 × 0.5^t), with t in days. At t = 0: 1000/(1 + 9) = 100. At t = 3: 0.5^3 = 0.125, so P = 1000/(1 + 1.125) = 1000/2.125 ≈ 471. At t = 10: 0.5^10 ≈ 0.00098, so P ≈ 1000/1.0088 ≈ 991. The curve starts like an exponential and flattens at 1000.

Other limits hide in plain sight. Models ignore randomness, so a real crowd will not empty on the same second twice. Inputs were measured once, under one set of conditions; a rainy day changes the gate rate. Assumptions can be unspoken: the Soldier Field model assumes everyone leaves at the final whistle and no gate is blocked. And precision can lie: reporting 10.1667 minutes when the rate is known only to ±20% pretends to a certainty you do not have. When a model guides a real decision, like how many gates to staff, it should be checked against a drill.

A model nobody can follow is not finished. A good report has five parts. The question: how long to empty Soldier Field? The assumptions: 61,000 people, a combined gate rate near 100 per second, a two-minute ramp-up. The method: a piecewise flow model, ramp then plateau. The answer with its range: about 11 minutes, likely between 9 and 13. The limitations: the rate is a guess, weather and blocked gates were ignored, and a drill should confirm it. A reader can now check every step and reproduce the number.

Show the model and the data together on one graph, with both axes labeled and units named. Mark the range where the model applies, and stop drawing it where it does not. In tables, round to a precision the inputs support: "about 10 minutes," not "10.1667 minutes." Then say what the number means for the decision. "Ten to thirteen minutes" matters only if the reader learns that the goal was ten, so the current gates are borderline.

The common failures are the same everywhere: hiding an assumption, reporting more digits than the data allow, leaving out the question so the number floats free, and choosing only the data that agree with you. A simple test is to hand your report to a classmate. If they can reproduce your answer from what you wrote, and can name the assumption they would attack first, the report is doing its job. That is the standard a transit planner, an engineer or a city budget analyst is held to as well.