Study Companion: Programme F.11 — Trigonometric and Exponential Functions

Engineers meet the same few shapes again and again: the wave of a wheel going round, the curve of something growing or fading by a steady fraction, and the question of where a value settles as you creep up on it. Programme F.11 builds those shapes. The one idea behind them is that each is the record of a simple repeated action. Turn at a steady rate and the height you reach is a sine wave. Grow by a steady fraction and the quantity is an exponential. Ask what the action settles towards as the steps shrink and you have a limit. Every rule in the programme, from the period of \tan5\theta to the shape of \cosh x, follows from one of those three actions, so none of it needs to be memorised as a bare fact.

The rules, procedures and drill are in the companion reference sheet, which follows the book's topic order. This file tells the story in 7 units instead, because the programme's many topics are really 7 ideas.

Notation, as in the book: \sin^{-1}x for the inverse sine (never 1/\sin x), \operatorname{cosec} for 1/\sin, \equiv for identities, and phase difference measured against a reference function, positive when the wave leads. Angles are in radians unless a degree sign says otherwise, and every calculator value below was taken in the mode that matches.

How to use this companion

Each unit is an argument: a real situation and a question, the attempt a sensible person would make first, the picture that settles it, a check with numbers, and only then the rule. Read it once front to back. The Narration spine at the end of each unit summarises what the video shows.

Unit 01: Rotation Defines Sine and Cosine for Every Angle

The problem

A Ferris wheel of radius 10 m turns anticlockwise about an axle 12 m above the ground. A rider starts level with the axle, at the three o'clock position. When the wheel has turned through 153^\circ, how high is she? Your calculator will say \sin153^\circ=0.4540, but Programme F.9 defined sine as "opposite over hypotenuse" in a right-angled triangle, and no triangle has a 153^\circ angle and a right angle as well. What is the calculator computing?

First attempt

Use what F.9 gave us: the rider's height above the axle is 10\sin\theta. For \theta=30^\circ that is a real triangle, the radius as hypotenuse, and 10\sin30^\circ=5 m, which is right. But push on to 153^\circ: a triangle needs its three angles to add to 180^\circ, and 153^\circ+90^\circ=243^\circ. The recipe has nothing to measure. Worse, at \theta=200^\circ the rider is below the axle, and a triangle's side is a length and cannot be negative. The triangle definition simply stops working at 90^\circ, while the wheel keeps turning.

The picture

Keep what the triangle was really doing and drop the triangle. Let a line OA of length 1 turn anticlockwise about O, and let B be the foot of the perpendicular from A to the horizontal axis. For an acute angle \theta the triangle OAB is exactly the F.9 triangle, with hypotenuse OA=1, so its two ratios are plain measurements:

Now let the line go on turning and keep the measurements:

• \sin\theta is the height of A above the horizontal axis, for any angle. Past 180^\circ the point is below the axis and the height is negative.
• \cos\theta is the signed distance of B from O, for any angle. Past 90^\circ the foot B is to the left of O and the distance is negative.
• A negative angle is a clockwise turn.

Now carry the height of A across to a graph of height against angle and let the line turn. The point's height rises to 1, falls through 0, bottoms out at -1, and the trace is the sine wave:

Cosine is the same wave measured sideways: the distance OB is 1 at \theta=0, falls to 0 at \pi/2 and reaches -1 at \pi.

For the tangent, divide:

which is the slope of the turning line, rise over run. When the line stands upright, B sits on O, the run is 0 and the slope has no value: the graph has a vertical asymptote at every odd multiple of \pi/2. After half a turn the line points the opposite way, so \sin and \cos both change sign and their ratio does not: \tan(\theta+\pi)=\tan\theta.

This agrees with F.9, as it must: for acute angles the measurements are the triangle's ratios, so the new definition extends the old one rather than replacing it.

Check it with numbers

At \theta=153^\circ the line points into the upper left, 27^\circ above the negative horizontal axis. The perpendicular has the height of the 27^\circ triangle, so \sin153^\circ=\sin27^\circ=0.4540, positive because A is above the axis, and \cos153^\circ=-\cos27^\circ=-0.8910, negative because B is left of O. The rider is 12+10\times0.4540=16.54 m above the ground.

At \theta=-\pi/4 (the wheel turned clockwise through 45^\circ) the line points into the lower right: \cos(-\pi/4)=0.7071 and \sin(-\pi/4)=-0.7071, so the rider is at 12-7.071=4.93 m.

Further calculator values, each readable from the picture: \cos(2\pi/3)=-0.5 (120^\circ, left of O, a 60^\circ triangle); \cos(-272^\circ)=0.0349 (-272^\circ is the line at 88^\circ, just short of straight up, so B is just right of O); \tan333^\circ=-0.5095 (pointing down and to the right, so the slope is negative); \tan(-6\pi/5)=-0.7265.

The rule it gives

The sine of an angle \theta, of any size and either sign, is the signed height of the end of the unit line turned through \theta. The cosine is the signed distance of its foot from O. Both lie between -1 and +1 and both repeat after one full turn:

The tangent is \tan\theta=\sin\theta/\cos\theta, undefined wherever \cos\theta=0, that is at \theta=\pi/2+n\pi, and \tan(\theta+n\pi)=\tan\theta for every whole number n. Three facts to carry into a sketch: the sine crosses the axis at every multiple of \pi; the cosine crosses it at every odd multiple of \pi/2; each branch of the tangent climbs from -\infty to +\infty between two asymptotes. Reciprocals keep the book's names: \sec\theta=1/\cos\theta, \operatorname{cosec}\theta=1/\sin\theta, \cot\theta=1/\tan\theta.

To evaluate by hand, draw the line, find the acute angle it makes with the horizontal axis, and give the answer the sign of its quadrant.

Set the calculator to the unit the angle is written in. \sin30 is 0.5 in degree mode and -0.9880 in radian mode.

Worked example

Find \sin153^\circ and \tan333^\circ without a calculator, then check. 153^\circ is 27^\circ short of a half turn. The line is the mirror image of the 27^\circ line in the vertical axis, so the height is unchanged: \sin153^\circ=\sin27^\circ. For 333^\circ, the line is 27^\circ below the positive horizontal axis, a mirror image in the horizontal axis of the 27^\circ line, so the rise flips sign and the run does not: \tan333^\circ=-\tan27^\circ=-0.5095. Both agree with the calculator.

Where this shows up

A Ferris wheel, as above, and any crank, rotor or alternator coil: its projected height or sideways reach is R\sin\theta or R\cos\theta for whatever angle it has reached. You can watch it. Stick a strip of tape on the side of a bicycle tyre, spin the wheel slowly and film it from the side. The tape's height against time, read off frame by frame, is the sine curve of the unit above, scaled by the wheel's radius.

Narration spine. A Ferris wheel has turned through 153^\circ and the triangle recipe has nothing to say. A unit line turns about a centre, its foot dropped onto the axis: for acute angles the old triangle appears. The line swings on past vertical, and the two measurements, the height and the foot's distance, keep going, with the foot crossing to the left and the height dipping below the axis. A dot carries the height across to a graph and the sine wave is drawn; the foot's distance draws the cosine. The slope of the line draws the tangent, which tears away at the vertical, and half a turn later it repeats. Then the numbers go onto the picture.

Unit 02: Period, Amplitude and Phase Are the Three Dials of a Wave

The problem

British mains electricity is 230 V, yet an oscilloscope on it shows a wave that climbs to 325 V. The voltage is

and a motor on the circuit draws a current i=10\sin(100\pi t-\pi/6) amperes. How long is one cycle? What does 325 have to do with 230? And by how much time does the current trail the voltage? Each answer should come from reading the formula, not from plotting it.

First attempt

Sine repeats every 2\pi, so any \sin(\text{something}) must repeat after 2\pi. Try it on a gentler example, y=\sin(\theta/3), with 2\pi as its period: \sin(0)=0 but \sin(2\pi/3)=0.8660. The wave is nowhere near repeating. The input to the sine was \theta/3, and it only reaches 2\pi when \theta reaches 6\pi. Reading the period straight off "sine" is wrong whenever the angle is scaled.

The picture

Write the wave as y=A\sin(k\theta+\varphi) and treat A, k and \varphi as three dials:

• Amplitude. Turning A stretches the wave up and down. The book defines amplitude as the difference between the maximum value and the average value over one period. For \sin that is 1-0=1; for 4\cos(2\theta-3) it is 4-0=4. Because it is max minus average, it works for waves that are not sinusoidal too.

• Period. Turning k squeezes the wave sideways. To find how far to move along before the wave repeats, add the basic period inside the argument and factor out k:

  \sin\!\big(k\theta+2\pi\big)=\sin\!\Big(k\Big[\theta+\frac{2\pi}k\Big]\Big),

  so moving the input on by 2\pi/k gives the same value again. The wave repeats every 2\pi/k. The same working for the tangent, whose basic period is \pi, gives \pi/k for \tan k\theta. A constant added inside, the \varphi dial, never changes the period.

• Phase. Turning \varphi slides the wave sideways without changing its shape. Measured against a reference function \sin k\theta, the phase difference is the interval of the input by which the new wave leads or lags. Factor out k to read it: \sin(k\theta+\varphi)=\sin k\big(\theta+\frac\varphi k\big), so the phase difference is \varphi/k. A positive value means leads.

  Why is a shift to the left a lead? At \theta=0 the wave \sin(\theta+\pi/4) already has the value \sin\frac\pi4=0.7071 that the reference \sin\theta does not reach until \theta=\pi/4. It gets there first. Read the other way, \sin(\theta-\pi/4) is the reference shifted right by \pi/4 and lags it.

A wave need not be a sinusoid. A periodic function of period P is described by giving one stretch and saying it repeats: the sawtooth

rises from 0 to 1 and drops back, over and over. Its period is 1. Its largest value is 1 and its average over a period is \frac12, so its amplitude, max minus average, is \frac12. To evaluate far along, subtract whole periods: f(2.5)=f(0.5+2)=f(0.5)=0.5.

Check it with numbers

Mains. The period is 2\pi/100\pi=0.02 s, so the supply completes 50 cycles a second. At t=0.0025 s the angle 100\pi t equals \pi/4 and v=325\times0.7071=229.8 V, which is the quoted 230 V: that figure is the peak divided by \sqrt2, the steady voltage that would heat a resistor equally (the root-mean-square value). The current i=10\sin100\pi\big(t-\frac1{600}\big) has phase difference -\frac1{600} s relative to \sin100\pi t: it lags by \frac{\pi/6}{100\pi}=\frac1{600} s =1.667 ms.

The surprise. \sin(\theta/3)=\sin\frac13(\theta+6\pi), period 6\pi, and a numerical check agrees: \sin(1.234) and \sin(1.234+6\pi) are equal, while \sin(1.234+2\pi) differs by 0.194.

Add and factor. \cos3\theta: period \frac{2\pi}3. \tan5\theta=\tan5(\theta+\frac\pi5): period \frac\pi5 (check: \tan5\theta at 0.31 and at 0.31+\frac\pi5 agree). \cos\big(\frac\theta2+\frac\pi3\big)=\cos\frac12\big(\theta+\frac{2\pi}3\big): period 4\pi, and the shift is \frac{2\pi}3, not \frac\pi3. Check: at \theta=-\frac{2\pi}3 the argument is -\frac\pi3+\frac\pi3=0 and the cosine is at its crest, 1.000; at \theta=-\frac\pi3 it is only 0.866.

The rule it gives

For y=A\sin(k\theta+\varphi) or y=A\cos(k\theta+\varphi):

Always factor out k before reading the shift; the \varphi in the bracket alone is not the shift. A periodic function with period P satisfies f(x+nP)=f(x) for whole n.

Worked example

For y=4\cos(2\theta-3): amplitude 4; period \frac{2\pi}2=\pi; \cos(2\theta-3)=\cos2\big(\theta-\frac32\big), so the phase difference from \cos2\theta is -\frac32 rad: the wave lags, shifted 1.5 units to the right. A second: \cos\theta=\sin\big(\theta+\frac\pi2\big), so cosine leads sine by \frac\pi2 rad. A third: \sin\big(3x+\frac\pi8\big)=\sin3\big(x+\frac\pi{24}\big) leads \sin3x by \frac\pi{24}.

Where this shows up

The mains example above works a real quantity to a number: period 20 ms, peak 325 V, a lag of 1.667 ms. A lagging current, one that trails the voltage, is what an induction motor draws. You can hear it. The 50 Hz hum from an unshielded guitar lead or an old amplifier is this wave, and doubling k halves the period and doubles the pitch: k is a squeeze.

Narration spine. A scope trace of mains: the formula 325\sin(100\pi t) and the question of how long one cycle takes. A guess that sine always repeats after 2\pi is tested on \sin(\theta/3) and fails. Three dials appear on the wave: amplitude stretches it up and down, k squeezes it sideways, \varphi slides it. To find the period, 2\pi is added inside the brackets and k is factored out, and a tracker on k confirms the repeat. Beside a reference wave, the shifted copy is seen to reach each value first: leading is shifting left. A sawtooth shows that a wave need not be a sinusoid, and its average and peak give its amplitude. The mains numbers are read off.

Unit 03: Inverse Trigonometric Functions Need a Cut

The problem

A wheelchair ramp rises 1 m over a horizontal run of 12 m, so its slope is \tan\theta=\frac1{12}. Press \tan^{-1} on a calculator and it shows 4.764^\circ. But \tan184.764^\circ is also \frac1{12}, and so is \tan364.764^\circ. Which angle does the calculator mean, and why does it choose only one when infinitely many angles share the same slope?

First attempt

Run the sine machine backwards, as F.10 taught for any function: the graph of the inverse is the mirror image of the graph in the line y=x (F.10 Unit 04). So reflect the whole of y=\sin x in y=x. The result is a wave on its side, and it fails at once. Draw the vertical line x=0.5: it meets the sideways wave at 30^\circ, 150^\circ, 390^\circ, 510^\circ, and on forever. One input has many outputs. The inverse of the sine function is a relation, not a function, so "the inverse sine" has no single value to return.

The picture

A function needs one output per input, so keep only a piece of the sine curve that is one-to-one, one that rises through every value from -1 to 1 exactly once, and reflect that:

The piece between -\pi/2 and \pi/2 goes through every output value once, and its mirror image is a curve with one height for each x between -1 and 1. That is the inverse sine function, \sin^{-1}x.

• \cos is cut to 0\le x\le\pi, which gives \cos^{-1}x, with outputs from 0 to \pi.
• \tan is cut to -\frac\pi2<x<\frac\pi2 and reflects to \tan^{-1}x, a curve squeezed between horizontal asymptotes at \pm\frac\pi2.

The cut interval has become the range of the inverse: \sin^{-1} takes an input in [-1,1] and returns an angle in [-\frac\pi2,\frac\pi2].

The three reciprocal inverses are defined by flipping the input:

Check it with numbers

\sin^{-1}0.5=\frac\pi6=30^\circ: the calculator returns the angle from the cut piece and not 150^\circ (which is \sin150^\circ=0.5 too, but lies outside the cut). \tan^{-1}(-3.5)=-1.2925 rad =-74.05^\circ, a negative angle because the cut piece of the tangent runs through negative angles. For \sec^{-1}10: \sec\theta=10 means \cos\theta=\frac1{10}, so \theta=\cos^{-1}0.1=84.26^\circ=1.4706 rad. The ramp: \tan^{-1}\frac1{12}=4.764^\circ.

The rule it gives

The inverse of a trigonometric function (the whole reflected curve) is many-valued. The inverse trigonometric function \sin^{-1}, \cos^{-1}, \tan^{-1} (the reflected cut piece) is single valued, and is the one the calculator keys give. The notation \sin^{-1}x means this inverse and does not mean \frac1{\sin x}: \sin^{-1}0.5=0.5236 but 1/\sin0.5=2.0858. (Many texts and calculators write \arcsin, \arccos, \arctan for the same functions.) To recover every angle with a given value, which the calculator does not return, find the principal angle and add the partner and the repeats; Unit 04 does exactly that.

Worked example

Find the angle of the ramp, and name what the calculator has left out. \theta=\tan^{-1}\frac1{12}=4.764^\circ. The same slope belongs to 4.764^\circ+180^\circ and every other angle 4.764^\circ+n\times180^\circ, since \tan has period 180^\circ; none of these is a ramp, because a ramp's angle lies between 0^\circ and 90^\circ. The cut piece of the tangent holds exactly the angles a sloping line can make with the horizontal, which is why the key gives the right answer here.

Where this shows up

Building rules in many countries cap an accessible ramp at a slope of 1 in 12, which is \tan^{-1}\frac1{12}=4.764^\circ: a number you can confirm with a tape measure and a spirit level against any ramp, or on a phone's level app. The same step, turning a measured ratio back into an angle, recovers a phase from the ratio of two voltages and a bearing from two distances. In each case the calculator's single answer is the principal one, and any quadrant correction is the user's job.

Narration spine. A ramp rises 1 m in 12 and the calculator offers one angle although infinitely many share the slope. The whole sine curve is reflected in y=x and a vertical line cuts the sideways wave again and again: not a function. The curve is cut down to the piece between -\frac\pi2 and \frac\pi2; only that piece is reflected, and the result is single valued. Cosine and tangent are cut the same way. \sin^{-1}0.5 lands on \frac\pi6 and the other angle with that sine is left behind. The ramp gives 4.764^\circ.

Unit 04: Trigonometric Equations Have Infinitely Many Solutions

The problem

The Ferris wheel from Unit 01 turns once every minute, and the rider's height above the axle is 10\sin\theta metres at angle \theta. When is she 5 m above the axle? A calculator gives \sin^{-1}0.5=\frac\pi6 and stops. But the wheel keeps turning, so she is surely at 5 m more than once. How many times, and when?

First attempt

Take the calculator at its word: the solution of \sin\theta=\frac12 is \theta=\frac\pi6, one angle. Now watch the wheel. The rider is 5 m above the axle on the way up at \frac\pi6. She keeps climbing to 10 m, comes down, and is 5 m above the axle again on the way down, and then again on the next lap, and the next. One angle is not the answer; it is the first of many.

The picture

Draw the sine wave and the horizontal line y=\frac12. The line meets the wave twice per period and then the pattern repeats:

The first hit is the calculator's, x=\frac\pi6. By the symmetry of the wave about its crest at \frac\pi2, the second is its mirror image, \pi-\frac\pi6=\frac{5\pi}6. Every later hit is 2\pi further along. So

The cosine behaves alike but its two hits are mirror images about x=0, so they come as a pair \pm: \cos x=\frac12 gives x=\pm\frac\pi3+2n\pi. The tangent takes each value once per period \pi, so \tan x=t gives x=\tan^{-1}t+n\pi.

When the angle is scaled, the same line meets a squeezed wave. For 2\sin3x=1, put u=3x: \sin u=\frac12 gives u=\frac\pi6+2n\pi or u=\frac{5\pi}6+2n\pi. Divide by 3 at the end, including the 2n\pi:

The sum of a cosine and a sine. Now the harder case, a\cos x+b\sin x=c. Two 50 Hz sources are wired in series, one a cosine of 3 V and the other a sine of 4 V. The total is 3\cos x+4\sin x with x=100\pi t. Plot it and it is visibly one wave: a smooth sinusoid, higher than either, and shifted. A sinusoid of that frequency has the form R\sin(x+\theta). To find R and \theta expand it with the compound-angle formula from F.9 Unit 06:

Matching this with a\cos x+b\sin x term by term gives R\sin\theta=a and R\cos\theta=b. Squaring and adding, R^2(\sin^2\theta+\cos^2\theta)=R^2=a^2+b^2, and dividing, \tan\theta=a/b. For 3\cos x+4\sin x: R=5 and \theta=\tan^{-1}\frac34=0.6435. The two sources together act as one source of peak 5 V:

That one wave can be written equally as R\cos(x-\alpha) with R\cos\alpha=a, R\sin\alpha=b, here \alpha=\tan^{-1}\frac43=0.9273. They describe the same curve: \alpha=\frac\pi2-\theta.

Check it with numbers

A single family. Solve 3\cos x+4\sin x=5: the line y=5 just touches the crest of the wave of amplitude 5, once per period. \sin(x+\theta)=1 gives x+\theta=\frac\pi2+2n\pi, so x=\frac\pi2-0.6435+2n\pi=0.9273+2n\pi. Test x=0.9273: \cos=0.6, \sin=0.8, so 3(0.6)+4(0.8)=1.8+3.2=5. ✓

Two families. Solve 3\cos x+4\sin x=2.5. In the cosine form 5\cos(x-\alpha)=2.5 gives \cos(x-\alpha)=\frac12, so x-\alpha=\pm\frac\pi3+2n\pi and x=0.9273\pm1.0472+2n\pi, that is x=1.9745+2n\pi or x=-0.1199+2n\pi. Both satisfy the equation: each gives 3\cos x+4\sin x=2.500.

Do not stop at the first family. Solve \sin x-\sqrt2\cos x=1. Here a=-\sqrt2 and b=1, so R=\sqrt3 and \theta=\tan^{-1}(-\sqrt2)=-0.9553. Then \sin(x+\theta)=\frac1{\sqrt3}, with principal value \sin^{-1}\frac1{\sqrt3}=0.6155 and partner \pi-0.6155=2.5261. So

Check both: at x=\frac\pi2, \sin=1 and \cos=0, so the left side is 1. ✓ At x=3.4814, \sin=-0.3333 and \cos=-0.9428, so -0.3333+\sqrt2(0.9428)=-0.3333+1.3333=1. ✓ A solution that gives only 1.5708+2n\pi misses the second family, because the equation \sin(x+\theta)=\frac1{\sqrt3} has the principal value and its partner, as in the opening picture.

The rule it gives

To solve \sin x=s, \cos x=s or \tan x=s: find the principal value with the inverse key, add the partner (for sine, \pi-x_0; for cosine, -x_0; the tangent has none), then add the repeats, 2n\pi for sine and cosine and n\pi for tangent, with n=0,\pm1,\pm2,\dots. For a scaled angle, solve for the whole angle kx first, then divide everything by k.

To solve a\cos x+b\sin x=c: write the left side as R\sin(x+\theta) with R=\sqrt{a^2+b^2}, R\sin\theta=a, R\cos\theta=b; solve \sin(x+\theta)=c/R (possible only if \lvert c\rvert\le R); choose \theta in the quadrant that matches the signs of R\sin\theta=a and R\cos\theta=b; and subtract \theta from both families. R\cos(x-\alpha) is the same wave, with \alpha=\frac\pi2-\theta.

Worked example

Solve 2\sin3x=1 and give three solutions. \sin3x=\frac12, so 3x=\frac\pi6+2n\pi or \frac{5\pi}6+2n\pi, and x=\frac\pi{18}+\frac{2n\pi}3 or \frac{5\pi}{18}+\frac{2n\pi}3. With n=0: 0.1745 and 0.8727; with n=1: 2.2689 and 2.9671. Check one: 2\sin(3\times0.1745)=2\sin(0.5236)=1.000. ✓

Where this shows up

The two-source circuit is a real calculation: series sources of 3\cos x V and 4\sin x V with x=100\pi t peak at 5 V, reached when x=0.9273, which is at t=0.9273/100\pi=2.95 ms after the cosine's own crest. The rider on the wheel is a second example, with the repeats made physical: she is at 5 m above the axle 2 times per minute, at the two angles \frac\pi6 and \frac{5\pi}6, and the 2n\pi is simply the next lap. You can watch it: on any Ferris wheel, note when a particular gondola is level with the hub, once going up and once coming down.

Narration spine. A Ferris wheel and the question of when the rider is 5 m above the axle. The calculator offers \frac\pi6 and the wheel keeps turning. On the sine graph a horizontal line at \frac12 meets the wave twice per period: the calculator's hit, its mirror image, then each shifted by whole periods. The family is written out with n. A scaled angle squeezes the wave and the hits crowd together; dividing by 3 crowds the 2n\pi too. Two waves of the same frequency, 3\cos x and 4\sin x, are added point by point and the sum is one wave of height 5; the line at 2.5 crosses it twice per period. A second equation shows the partner family that a careless solution drops.

Unit 05: Exponentials and Logarithms Undo Each Other, and That Solves Indicial Equations

The problem

A 1000\ \muF capacitor is charged to 12 V and left to drain through a 2\ \text{k}\Omega resistor. Its voltage is v=12e^{-t/2} volts (t in seconds). A relay drops out when the voltage falls to 5 V. How long does that take? The unknown t sits in an index, and the algebra of the earlier programmes has no operation that takes it down.

First attempt

Guess and refine. At t=1 s, v=12e^{-0.5}=7.28 V, too high. At t=2 s, v=12e^{-1}=4.42 V, too low. Try 1.75 s: 5.00 V, close. It works, but each digit costs another trial, and a different capacitor means starting again. Trial and error finds one answer; it gives no way to write the answer as a formula in the numbers of the problem.

The picture

The curve y=e^x is the function that grows at a rate proportional to its own size (the same property that produced e in F.7 Unit 05). Its graph lies wholly above the x-axis and climbs ever faster. Its reflection in the line y=x is its inverse, the natural logarithm y=\ln x (F.10 Unit 04):

Any other base is the same curve with the x-axis rescaled. Since a=e^{\ln a} (the logarithm undoes the exponential), raising to the power x gives

So y=a^x is y=e^x with the horizontal scale stretched by the factor \ln a. For a<e we have \ln a<1 and the curve climbs more slowly than e^x; for a>e it climbs faster. The inverse of a^x is \log_ax, and the exponential and the logarithm are mutual inverses: each undoes the other.

The logarithm is what pulls an index down. A logarithm turns multiplication into addition (\log AB=\log A+\log B), and a power is repeated multiplication, so

An equation with the unknown in an index is solved by taking the logarithm of both sides, which brings the unknown down beside a number. Any base works, and the book uses base 10 with \log.

Indicial equations quadratic in a^x. Sometimes the unknown appears as both a^x and a^{2x}=(a^x)^2. Treat y=a^x as a new unknown and the equation is an ordinary quadratic in y, solved by the methods of F.6, after which each root is a separate simple indicial equation a^x=\text{root}. A root that is zero or negative is rejected, since a^x>0 for every x: the curve never reaches y\le0.

Check it with numbers

The capacitor. 12e^{-t/2}=5, so e^{-t/2}=\frac5{12} and -\frac t2=\ln\frac5{12}, giving t=2\ln\frac{12}5=2\ln2.4=1.751 s. Check: 12e^{-1.751/2}=12\times0.4167=5.00 V. ✓

A simple indicial equation. 12^{2x}=35.4. Taking logs: 2x\log12=\log35.4, so x=\frac{1.5490}{2\times1.0792}=0.7177. Check: 12^{2\times0.7177}=12^{1.4354}=35.40. ✓ The equation is also e^{2x\ln12}=35.4, so the same answer comes from natural logarithms: x=\frac{\ln35.4}{2\ln12}=\frac{3.5667}{4.9698}=0.7177.

Quadratic in a^x. 7^{2x}-9\cdot7^x+14=0. Put y=7^x: y^2-9y+14=(y-2)(y-7)=0, so 7^x=2 or 7^x=7. Then x=\frac{\ln2}{\ln7}=0.3562 or x=1. Check x=1: 49-63+14=0. ✓ Check x=0.3562: 7^{0.3562}=2.000, so 4-18+14=0. ✓

The rule it gives

e^x and \ln x are inverse functions, and so are a^x and \log_ax: e^{\ln x}=x and \ln(e^x)=x. Also a^x=e^{x\ln a}, and \log(A^n)=n\log A. To solve an indicial equation, take logs and bring the index down. If it is quadratic in a^x, substitute y=a^x, solve for y, reject any y\le0, then solve a^x=y.

To compute e^x itself, the book gives the series e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots which agrees with the value e=2.7182818\ldots at x=1 to seven places after eleven terms.

Worked example

Solve \log_x49=2. By the definition of a logarithm this says x^2=49, so x=7 or x=-7. A base must be positive and not 1, so the answer is x=7. Then solve 5^{2x}-6\cdot5^x+5=0: with y=5^x, y^2-6y+5=(y-1)(y-5)=0, so 5^x=1 or 5, giving x=0 or x=1. Check x=0: 1-6+5=0. ✓

Where this shows up

The capacitor above is Level 2 and Level 4 at once: the voltages at t=0,1,2 s are 12,\ 7.28,\ 4.42 V, which you can measure by charging a 1000\ \muF electrolytic capacitor from a 12 V supply and draining it through a 2\ \text{k}\Omega resistor with a voltmeter across it. The relay's 1.751 s is the answer to 12e^{-t/2}=5. A cooling cup of coffee follows the same curve, its temperature above the room's falling by the same fraction in each equal interval, and a cut-off time follows from the same logarithm.

Narration spine. A capacitor drains through a resistor and its voltage follows 12e^{-t/2}; the question is when it reaches 5 V, with t stuck in the index. Guessing narrows in but never ends. The graph of e^x is drawn and reflected in y=x to give \ln x. A tracker on the base a makes a^x steepen or flatten about e^x, and the curve is shown to be e^x stretched sideways by \ln a. The logarithm pulls the index down and the capacitor time falls out. A quadratic in 7^x is solved as a quadratic in y, and each root becomes a height at which the curve 7^x is cut.

Unit 06: Odd and Even Parts Split a Function by Symmetry

The problem

Two children sit on a see-saw 1 m either side of the pivot: a 30 kg child on the left and a 10 kg child on the right. Treat the load as a function of position, w(x) in kg at distance x metres to the right of the pivot, so w(-1)=30 and w(1)=10. Which part of this loading is the "balanced" part that would press straight down on the pivot, and which part is the "tilting" part? A total of 40 kg is pressing down, yet the see-saw is far from level.

First attempt

Replace the pair by their average, 20 kg on each side. That loading is perfectly balanced, and the total is right, 40 kg. But it is also completely level, and the actual see-saw tips. The average has thrown away exactly what makes it tip, and a description of the see-saw has to keep it.

The picture

Take any function f and reflect its graph in the vertical axis; the mirror copy is f(-x). At each x the two curves sit at different heights:

Two numbers at each x can be rebuilt from their midpoint and their half gap:

Three one-line checks say this does what it claims.

• f_e is even: f_e(-x)=\frac{f(-x)+f(x)}2=f_e(x). Its graph is its own mirror image in the vertical axis.
• f_o is odd: f_o(-x)=\frac{f(-x)-f(x)}2=-f_o(x). Its graph maps onto itself under a half turn about the origin.
• They add back: f_e+f_o=\frac{f(x)+f(-x)}2+\frac{f(x)-f(-x)}2=f(x).

So every function whose f(-x) is also defined is the sum of an even part and an odd part. This also explains why sine is odd and cosine even, which Unit 01 already showed: turning \theta\to-\theta reflects the unit line in the horizontal axis, which flips the height (\sin(-\theta)=-\sin\theta) and keeps the foot (\cos(-\theta)=\cos\theta).

For polynomials the parts can be read off at sight: even powers make the even part and odd powers make the odd part.

Check it with numbers

The see-saw: f_e(1)=\frac{10+30}2=20 kg and f_o(1)=\frac{10-30}2=-10 kg, and at x=-1 the even part is again 20 and the odd part +10. Rebuilding: 20+(-10)=10 at x=+1 and 20+10=30 at x=-1. ✓

A polynomial: f(x)=x^3-2x^2-3x+4. The even powers are -2x^2+4 and the odd ones are x^3-3x, so f_e=-2x^2+4 and f_o=x^3-3x. At x=1: f(1)=0, f(-1)=4, f_e(1)=2, f_o(1)=-2, and 2-2=0. ✓ At x=-1: 2+2=4. ✓

The rule it gives

A function is even if f(-x)=f(x) (x^2, \cos x) and odd if f(-x)=-f(x) (x^3, \sin x, \tan x). The even part and odd part of any f are

More examples: 3x^2-2x+1 has f_e=3x^2+1 and f_o=-2x; x^3(x^2-3x+5)=x^5-3x^4+5x^3 has f_e=-3x^4 and f_o=x^5+5x^3; \frac1{x-1} has f_e=\frac1{(x-1)(x+1)} and f_o=\frac x{(x-1)(x+1)}.

Some functions have no parts. \ln x is defined only for x>0, so \ln(-x) does not exist and the formulas have nothing to work with: \ln x is neither odd nor even and has no even or odd part, and neither does x^4\ln x.

The hyperbolic functions. The even and odd parts of the exponential are named: \cosh x=\frac{e^x+e^{-x}}2 and \sinh x=\frac{e^x-e^{-x}}2, so that e^x=\cosh x+\sinh x, and \tanh x=\frac{\sinh x}{\cosh x}. At x=1: \cosh1=1.5431 and \sinh1=1.1752, and their sum is 2.7183=e. ✓

Beyond the book: \cosh^2x-\sinh^2x=1, which follows from expanding \big(\frac{e^x+e^{-x}}2\big)^2-\big(\frac{e^x-e^{-x}}2\big)^2=\frac14\cdot4e^xe^{-x}=1. At x=1: 1.5431^2-1.1752^2=1.0000.

Worked example

Split \frac1{x-1} into its parts. Here f(-x)=\frac1{-x-1}=-\frac1{x+1}. So f_e=\frac12\Big(\frac1{x-1}-\frac1{x+1}\Big)=\frac1{(x-1)(x+1)} and f_o=\frac12\Big(\frac1{x-1}+\frac1{x+1}\Big)=\frac x{(x-1)(x+1)}. Check at x=2: f_e=\frac13, f_o=\frac23, and \frac13+\frac23=1=\frac1{2-1}. ✓

Where this shows up

The see-saw is the problem worked in the open: it carries a total load of 40 kg, and all of it comes from the even part (20+20), since the odd parts cancel in the sum. The turning effect about the pivot is the opposite: the moment is 30\times1-10\times1=20 kg m, and all of it comes from the odd part (\pm10 kg at \pm1 m), because the even part's 20 kg either side produces a moment of 20-20=0. In \text{N m} that is 20\times9.81=196 N m. You can watch the split by sitting two people on a see-saw and moving one: shifting weight makes the see-saw tip by the odd part alone, while the pivot always carries the even part.

Narration spine. A see-saw with 30 kg on one side and 10 kg on the other: averaging the two to 20 and 20 balances it, which is wrong because the see-saw tips. A curve is reflected in the vertical axis; between the curve and its mirror image each point has a midpoint and a half gap. The midpoints trace a curve that is its own mirror image and the half gaps trace one that turns over; adding them restores the original. Sine and cosine fall out of the reflection of the turning line. The exponential splits into \cosh and \sinh. A logarithm, which lives only on one side of the axis, cannot be reflected.

Unit 07: Limits Make Getting Arbitrarily Close Precise

The problem

A car pulls away so that its distance from the start is s=t^2 metres after t seconds. What does the speedometer read at exactly t=1 s? Speed is distance over time, so over the interval from 1 s to t s the average speed is

Put in t=1 to get the speed at that instant and the formula returns \frac00. A speedometer reads something sensible, so what has the formula failed to say?

First attempt

Cancel the (t-1): \frac{t^2-1}{t-1}=\frac{(t-1)(t+1)}{t-1}=t+1, which at t=1 gives 2. But cancelling a factor of t-1 divides by zero when t=1, the very case being asked about. The cancellation is only licensed for t\ne1, so it has not yet shown that the answer at t=1 is 2. It has only shown what the formula does everywhere else.

The picture

Look at what the formula does near t=1, without ever reaching it. The graph of y=\frac{x^2-1}{x-1} is the straight line y=x+1 with a single hole at x=1:

Squeeze in on x=1 from both sides and the height squeezes in on 2:

The function never equals 2 at x=1, yet it comes as near as you please. That is a limit, written with the book's \mathop{Lim}:

"As near as you please" can be made exact with a band and a window. Choose a band of any width \varepsilon around the height 2. Then there is a window of width \delta around x=1 (not including x=1 itself) such that the graph inside the window stays inside the band. Here \delta=\varepsilon works, because \Big\lvert\frac{x^2-1}{x-1}-2\Big\rvert=\lvert x-1\rvert is less than \varepsilon exactly when x is within \varepsilon of 1. Shrink the band and the window shrinks with it, and there is always a window.

The rules of limits. Suppose \mathop{Lim}_{x\to x_0}f(x)=A and \mathop{Lim}_{x\to x_0}g(x)=B. Then:

• Sum and difference. Use two bands of width \varepsilon, one round A and one round B. In a window small enough for both, f is within \varepsilon of A and g within \varepsilon of B, so f+g is within 2\varepsilon of A+B. Since \varepsilon is arbitrary, so is 2\varepsilon, and \mathop{Lim}(f\pm g)=A\pm B. This is a proof.
• Product. Write f=A+e_1 and g=B+e_2 with e_1,e_2 as small as you like. Then fg=AB+Ae_2+Be_1+e_1e_2, and every term after AB is small. So \mathop{Lim}(fg)=AB. This is an argument rather than a full proof: it takes for granted that a sum of small terms is small.
• Quotient. \mathop{Lim}\frac fg=\frac AB, provided B\ne0. If B=0 the rule is silent: the function \frac{x^2-1}{x-1} itself has a numerator and a denominator that both go to 0, and its limit 2 had to be found by cancelling for x\ne1.
• Composition. \mathop{Lim}f(g(x))=f(B), provided f is continuous at B. The proviso matters. Let f(y)=0 for y\ne0 and f(0)=1, and let g(x)=x^2. For every x\ne0, f(g(x))=0, so \mathop{Lim}_{x\to0}f(g(x))=0, but f(\mathop{Lim}g)=f(0)=1. The function f jumps at 0, so the limit cannot be moved inside.

Check it with numbers

\mathop{Lim}_{x\to-3}\frac{x^2-9}{x+3}: for x\ne-3 the quotient is x-3, which heads for -6. Numerically, at x=-3.01 it is -6.01 and at x=-2.99 it is -5.99. ✓ Rule examples: \mathop{Lim}_{x\to\pi}(x^2-\sin x)=\pi^2-0=9.8696; \mathop{Lim}_{x\to\pi}(x^2\sin x)=\pi^2\times0=0; \mathop{Lim}_{x\to\pi/4}\frac{\tan x}{\sin x}=\frac1{1/\sqrt2}=\sqrt2 (the denominator's limit is \ne0); \mathop{Lim}_{x\to1}\cos(x^2-1)=\cos0=1 (cosine is continuous). Numerically \cos(1.001^2-1)=0.999998. ✓

The rule it gives

\mathop{Lim}_{x\to x_0}f(x)=A means that f(x) can be made as close to A as you like by taking x close enough to x_0, without needing f to be defined at x_0 itself. The limit laws for sums, products and quotients, and for compositions with a continuous outer function, are as listed above, each with its proviso. When numerator and denominator both vanish at the limit point, cancel the common factor for x\ne x_0 first.

Worked example

Find the speedometer reading at t=1 s. The average speed over [1,1+h] is \frac{(1+h)^2-1}h=\frac{2h+h^2}h=2+h for h\ne0. As h shrinks the average heads for 2: it is 2.1 for h=0.1, 2.01 for h=0.01 and 2.001 for h=0.001. The limit is 2 m/s, which is 7.2 km/h. This limit, the one a speedometer needs, is what F.12 will call the derivative (F.12 Unit 01).

Where this shows up

A speedometer is the observable case: it reads a limit, the distance covered in an ever-shorter time divided by that time. On a stopwatch and a measured track you can see the averages settle as the interval shrinks, the way the table of 2.1,\ 2.01,\ 2.001 does. The same step underlies every instantaneous rate: the current in a capacitor, the slope of a ramp, the speed of the Ferris wheel's rider at the moment she passes the axle.

Narration spine. A car's speedometer and a formula that returns \frac00 at the instant of interest. Cancelling is refused because it divides by zero. The graph appears as a straight line with a hole, and a dot creeps in on the hole from each side with its height read off, closing in on 2. A band of width \varepsilon is drawn round 2 and a window round 1; as the band shrinks the window follows. Two bands of width \varepsilon add to 2\varepsilon to justify the sum rule, a function with a jump shows why the composition rule needs continuity, and the speedometer reads 2 m/s.