Answer key · Unit 10

Complex numbers.

Each lesson’s practice questions, their correct answers and the mistakes they spot. Answers read off the Scene use the same setting the learner sees.

Misconception catalogue

Answer key 3

Lesson 1 of 3 · About 25 minutes

Multiplying by i

The number i, whose square is minus 1, and the complex plane: multiplying by i as a quarter turn, the powers of i, a complex number by length and angle, and a number that three times over makes minus 1.

Open the lesson

The 6 steps and their answers

Step 1 of 6
Open this step
The square of i
Practice · Predict, then look
Question
Turn per step is $90^\circ$ ⁠ and Stretch per step is $1$ ⁠: each step multiplies by $i$ ⁠. Start from $1$ ⁠. Where will the point be after two steps, $1 \cdot i \cdot i$ ⁠?

Correct answer
- Correct $-1$ ⁠
Setup, choices and feedback
The Scene at this step
Turn per step
90 °fixed
Stretch per step
1fixed
Steps
0fixed
Build Euler’s formula
Steps
Choices
- Correct $-1$ ⁠
- Wrong $1$ ⁠
  Misconception: Thinks i² = 1, as if i were an ordinary number whose square is positive.
  Feedback from the catalogue
  $i$ ⁠ is defined so that $i^2 = -1$ ⁠; that is why it is no real number, whose square is always positive or zero. On the complex plane, multiplying by $i$ ⁠ is a quarter turn, and multiplying by $i$ ⁠ twice is a half turn: from $1$ ⁠ to $-1$ ⁠.
- Wrong $i$ ⁠
  Feedback
  $i$ ⁠ is where it is after one step: a quarter turn, to the top of the imaginary axis.
- Wrong $2i$ ⁠
  Feedback
  Multiplying by $i$ ⁠ changes no length: the point stays on the unit circle. It only turns.
Once the prediction is locked in, the Scene changes
Steps
2fixed
The Scene then shows: -1 · Complex: the real part of the last product zₙ/nodes/board/outputs/complexRe
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
$-1$ ⁠: two quarter turns make a half turn. That is the whole secret of $i$ ⁠: $i^2 = -1$ ⁠.

Worked solution
1. Multiplying by $i$ ⁠ turns by $90^\circ$ ⁠ without changing length.
2. $1 \cdot i = i$ ⁠: a quarter turn, to the top of the imaginary axis.
3. $i \cdot i$ ⁠: another quarter turn, to the negative end of the real axis, $-1$ ⁠.
If you chose $1$ ⁠, you treated $i$ ⁠ as an ordinary number, whose square is positive.
Step 2 of 6
Open this step
The powers of i
Practice · Reach the target
Question
Each step multiplies by $i$ ⁠. Change Steps so that the last point is $-i$ ⁠, then press Check.

Goal
- 0 ± 0.01
  At the start of the step: 0
- -1 ± 0.01
  At the start of the step: 1
Setup, choices and feedback
The Scene at this step
Turn per step
90 °fixed
Stretch per step
1fixed
Steps
1
Build Euler’s formula
Steps
Read from the Scene at
Complex: the real part of the last product zₙ/nodes/board/outputs/complexRe
Read from the Scene at
Complex: the imaginary part of zₙ/nodes/board/outputs/complexIm
One setting that reaches it
Steps
3
Feedback for a right answer
$i^3 = i^2 \cdot i = -i$ ⁠: three quarter turns, to the bottom of the imaginary axis. $7$ ⁠ and $11$ ⁠ steps get there too: the powers of $i$ ⁠ repeat every four, $1, i, -1, -i, 1, \dots$ ⁠

Feedback for a wrong answer
Not yet. $-i$ ⁠ is at the bottom of the imaginary axis. How many quarter turns does that take?

Worked solution
1. Each step adds $90^\circ$ ⁠.
2. $-i$ ⁠ is at $270^\circ$ ⁠.
3. $\frac{270^\circ}{90^\circ} = 3$ ⁠ steps: $i^3 = -i$ ⁠.
4. Set Steps to $3$ ⁠.
Step 3 of 6
Open this step
Length and angle
Practice · Calculate
Question
$w$ ⁠ turns by $30^\circ$ ⁠ and has length $1$ ⁠: $w = \cos 30^\circ + i \sin 30^\circ$ ⁠. What is the real part of $w$ ⁠?

Correct answer
0.866 ± 0.005
Setup, choices and feedback
The Scene at this step
Turn per step
30 °fixed
Stretch per step
1fixed
Steps
0fixed
Build Euler’s formula
Steps
Accepted within
± 0.005
Computed from the Scene at this step’s setting
Complex: the real part of w, r·cos φ/nodes/board/outputs/stepRe
Wrong answers it recognises
- 0.5 ± 0.005
  Misconception: Takes the real part of r(cos φ + i sin φ) from sin φ.
  Feedback from the catalogue
  A number of length $r$ ⁠ at angle $\varphi$ ⁠ is $r\cos\varphi + i \, r\sin\varphi$ ⁠: the real, horizontal part is the cosine, like a vector’s horizontal component, and the imaginary, vertical part is the sine.
Feedback for a right answer
$\cos 30^\circ \approx 0.866$ ⁠, and $w \approx 0.866 + 0.5i$ ⁠. This is the same split into components as for a vector of length $1$ ⁠.

Worked solution
1. A number of length $r$ ⁠ at angle $\varphi$ ⁠ is $r(\cos\varphi + i\sin\varphi)$ ⁠.
2. The real part is $r\cos\varphi = 1 \cdot \cos 30^\circ$ ⁠.
3. $\cos 30^\circ = \frac{\sqrt3}{2} \approx 0.866$ ⁠.
If you answered $0.5$ ⁠, that is the imaginary part, $\sin 30^\circ$ ⁠.
Step 4 of 6
Open this step
Roots of minus 1
Practice · Reach the target
Question
Steps is locked at $3$ ⁠ and Stretch per step at $1$ ⁠. Change Turn per step so that $w^3 = -1$ ⁠, then press Check.

Goal
- -1 ± 0.01
  At the start of the step: -0.7071
- 0 ± 0.01
  At the start of the step: 0.7071
Setup, choices and feedback
The Scene at this step
Turn per step
45 °
Stretch per step
1fixed
Steps
3fixed
Build Euler’s formula
Steps
Read from the Scene at
Complex: the real part of the last product zₙ/nodes/board/outputs/complexRe
Read from the Scene at
Complex: the imaginary part of zₙ/nodes/board/outputs/complexIm
One setting that reaches it
Turn per step
60 °
Feedback for a right answer
Three turns must come to a half turn: $3\varphi = 180^\circ$ ⁠, so $\varphi = 60^\circ$ ⁠. $\varphi = 180^\circ$ ⁠ works too, $3 \cdot 180^\circ = 540^\circ$ ⁠, one and a half turns: $(-1)^3 = -1$ ⁠. $-1$ ⁠ has three cube roots, and two of them are not real.

Feedback for a wrong answer
Not yet. $-1$ ⁠ is a half turn round. Which turn, three times over, gets there?

Worked solution
1. The length stays $1$ ⁠, so only the angle matters: $3\varphi$ ⁠ must be $180^\circ$ ⁠, or $180^\circ$ ⁠ plus whole turns.
2. $3\varphi = 180^\circ$ ⁠ gives $\varphi = 60^\circ$ ⁠.
3. Set Turn per step to $60^\circ$ ⁠.
Step 5 of 6
Open this step
Squaring
Practice · Choose
Question
$w = \cos 30^\circ + i \sin 30^\circ$ ⁠. What is $w^2$ ⁠?

Correct answer
- Correct $\cos 60^\circ + i \sin 60^\circ$ ⁠
Setup, choices and feedback
The Scene at this step
Turn per step
30 °
Stretch per step
1
Steps
1
Build Euler’s formula
Steps
Choices
Shown in a random order.
- Correct $\cos 60^\circ + i \sin 60^\circ$ ⁠
- Wrong $\cos^2 30^\circ + i \sin^2 30^\circ$ ⁠
  Misconception: Squares a complex number part by part, (a + bi)² = a² + b²i, forgetting the cross term.
  Feedback from the catalogue
  $(a + bi)^2 = a^2 + 2abi + b^2 i^2 = (a^2 - b^2) + 2ab \, i$ ⁠, like $(x + y)^2$ ⁠ in the algebraic expressions Unit. On the complex plane it is easier: squaring multiplies the length by itself and doubles the angle.
- Wrong $2(\cos 30^\circ + i \sin 30^\circ)$ ⁠
  Feedback
  That is $2w$ ⁠, not $w^2$ ⁠. $w$ ⁠ has length $1$ ⁠, and $1 \cdot 1 = 1$ ⁠.
- Wrong $-1$ ⁠
  Feedback
  $-1$ ⁠ is $w^6$ ⁠: six turns of $30^\circ$ ⁠.
Feedback for a right answer
$w^2 = \cos 60^\circ + i \sin 60^\circ$ ⁠: length $1 \cdot 1 = 1$ ⁠, angle $30^\circ + 30^\circ$ ⁠. Worked out directly: $(0.866 + 0.5i)^2 = 0.75 - 0.25 + 2 \cdot 0.866 \cdot 0.5 \, i = 0.5 + 0.866i$ ⁠.

Worked solution
1. When multiplying, the lengths multiply: $1 \cdot 1 = 1$ ⁠.
2. And the angles add: $30^\circ + 30^\circ = 60^\circ$ ⁠.
3. So $w^2 = \cos 60^\circ + i \sin 60^\circ = 0.5 + 0.866i$ ⁠.
4. Check on the board: set Turn per step to $30^\circ$ ⁠ and Steps to $2$ ⁠.
Step 6 of 6
Open this step
The imaginary part
Practice · Calculate
Question
Turn per step $60^\circ$ ⁠ and Stretch per step $1.04$ ⁠: $w$ ⁠ has length $1.04$ ⁠ and angle $60^\circ$ ⁠. What is the imaginary part of $w$ ⁠?

Correct answer
0.9007 ± 0.005
Setup, choices and feedback
The Scene at this step
Turn per step
60 °fixed
Stretch per step
1.04fixed
Steps
1fixed
Build Euler’s formula
Steps
Accepted within
± 0.005
Computed from the Scene at this step’s setting
Complex: the imaginary part of w, r·sin φ/nodes/board/outputs/stepIm
Wrong answers it recognises
- 0.52 ± 0.005
  Misconception: Takes the real part of r(cos φ + i sin φ) from sin φ.
  Feedback from the catalogue
  A number of length $r$ ⁠ at angle $\varphi$ ⁠ is $r\cos\varphi + i \, r\sin\varphi$ ⁠: the real, horizontal part is the cosine, like a vector’s horizontal component, and the imaginary, vertical part is the sine.
Feedback for a right answer
About $0.901$ ⁠: $w \approx 0.52 + 0.901i$ ⁠.

Hint
$w = r(\cos\varphi + i\sin\varphi)$ ⁠: the imaginary part is $r\sin\varphi$ ⁠.

Worked solution
1. $r = 1.04$ ⁠ and $\varphi = 60^\circ$ ⁠.
2. $r\sin\varphi = 1.04 \times 0.866 \approx 0.901$ ⁠.
3. The real part: $1.04 \times 0.5 = 0.52$ ⁠. So $w \approx 0.52 + 0.901i$ ⁠.
If you answered $0.52$ ⁠, you gave the real part. The imaginary part is the point’s height, so it comes with the sine.

Lesson 2 of 3 · About 25 minutes

Turns, stretches and Euler’s formula

Multiplying complex numbers multiplies lengths and adds angles: powers as spirals, doubling time on the complex plane, a spiral that closes in on zero, and Euler’s formula, in which e to the iπ equals minus 1.

Open the lesson

The 5 steps and their answers

Step 1 of 5
Open this step
Angles add
Practice · Predict, then look
Question
Turn per step is $40^\circ$ ⁠ and Stretch per step is $1$ ⁠. What will the point’s angle be after $5$ ⁠ steps, in degrees?

Correct answer
200 ° ± 1 °
A number prediction counts as right within ± 1 °
Setup, choices and feedback
The Scene at this step
Turn per step
40 °fixed
Stretch per step
1fixed
Steps
0fixed
Build Euler’s formula
Steps
Once the prediction is locked in, the Scene changes
Steps
5fixed
The Scene then shows: 200 ° · Complex: the total turn n·φ, as the board prints it/nodes/board/outputs/complexTurn
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
$5 \cdot 40^\circ = 200^\circ$ ⁠: each multiplication adds $40^\circ$ ⁠, and the point has passed the negative real axis.

Worked solution
1. Each multiplication by $w$ ⁠ adds $w$ ⁠’s angle, $40^\circ$ ⁠.
2. After $5$ ⁠ multiplications: $w^5$ ⁠, at $5 \cdot 40^\circ = 200^\circ$ ⁠.
3. $200^\circ$ ⁠ is just past $180^\circ$ ⁠: below the real axis, on the left.
Step 2 of 5
Open this step
Lengths multiply
Practice · Reach the target
Question
Stretch per step is locked at $1.08$ ⁠: each multiplication lengthens by $8\%$ ⁠. Change Steps so that the last point’s length is $2$ ⁠, then press Check.

Goal
- 2 ± 0.01
  At the start of the step: 1.08
Setup, choices and feedback
The Scene at this step
Turn per step
30 °fixed
Stretch per step
1.08fixed
Steps
1
Build Euler’s formula
Steps
Read from the Scene at
Complex: |zₙ| = rⁿ/nodes/board/outputs/complexLength
One setting that reaches it
Steps
9
Feedback for a right answer
$1.08^9 \approx 2.00$ ⁠: nine steps double the length. This is the doubling time from the exponents Unit: the lengths multiply like a geometric sequence.

Feedback for a wrong answer
Not yet. The length after $n$ ⁠ steps is $1.08^n$ ⁠. When is that $2$ ⁠?

Worked solution
1. The lengths multiply: after $n$ ⁠ steps the length is $1.08^n$ ⁠.
2. $1.08^n = 2$ ⁠, so $n = \frac{\log 2}{\log 1.08} \approx 9.0$ ⁠.
3. Set Steps to $9$ ⁠.
Step 3 of 5
Open this step
The rule
Practice · Choose
Question
$z$ ⁠ has length $3$ ⁠ and angle $50^\circ$ ⁠, and $w$ ⁠ has length $2$ ⁠ and angle $30^\circ$ ⁠. What is $z \cdot w$ ⁠?

Correct answer
- Correct Length $6$ ⁠, angle $80^\circ$ ⁠
Setup, choices and feedback
The Scene at this step
Turn per step
30 °
Stretch per step
1.08
Steps
4
Build Euler’s formula
0.3 (Steps → Euler's formula)
Choices
Shown in a random order.
- Correct Length $6$ ⁠, angle $80^\circ$ ⁠
- Wrong Length $5$ ⁠, angle $80^\circ$ ⁠
  Misconception: Adds the lengths when multiplying complex numbers, instead of multiplying them.
  Feedback from the catalogue
  When complex numbers multiply, the lengths multiply and the angles add. That carries on ordinary multiplication: $2 \cdot 3 = 6$ ⁠, not $5$ ⁠. It is the angles that add, because a turn after a turn is a turn by the sum of the angles.
- Wrong Length $6$ ⁠, angle $1500^\circ$ ⁠
  Misconception: Multiplies the angles when multiplying complex numbers, instead of adding them.
  Feedback from the catalogue
  Multiplying by $w$ ⁠ turns by $w$ ⁠’s angle. Multiplying by $w$ ⁠ and then by $z$ ⁠ turns once and then again, so the angles add. Only a power, $w^n$ ⁠, multiplies the angle by a number: $n$ ⁠ equal turns.
- Wrong Length $6$ ⁠, angle $20^\circ$ ⁠
  Feedback
  The difference of the angles belongs to division: $\frac{z}{w}$ ⁠ has length $1.5$ ⁠ and angle $20^\circ$ ⁠.
Feedback for a right answer
Length $3 \cdot 2 = 6$ ⁠, and angle $50^\circ + 30^\circ = 80^\circ$ ⁠. Multiplying by $w$ ⁠ stretches by $2$ ⁠ and turns by $30^\circ$ ⁠, like one step on the board.

Worked solution
1. When complex numbers multiply, the lengths multiply: $3 \cdot 2 = 6$ ⁠.
2. And the angles add: $50^\circ + 30^\circ = 80^\circ$ ⁠.
3. The reason: $z = 3(\cos 50^\circ + i \sin 50^\circ)$ ⁠, and on multiplying, the sine and cosine of a sum of angles give $6(\cos 80^\circ + i \sin 80^\circ)$ ⁠.
Step 4 of 5
Open this step
Spiralling in
Practice · Reach the target
Question
Steps is locked at $12$ ⁠. Change Turn per step and Stretch per step so that the last point lies on the negative real axis, at angle $180^\circ$ ⁠, with a length of about a half, between $0.45$ ⁠ and $0.55$ ⁠. Press Check.

Goal
- 180 ° ± 0.5 °
  At the start of the step: 0 °
- 0.5 ± 0.05
  At the start of the step: 1
Setup, choices and feedback
The Scene at this step
Turn per step
30 °
Stretch per step
1
Steps
12fixed
Build Euler’s formula
0.5 (Steps → Euler's formula)
Read from the Scene at
Complex: the angle of zₙ, the total turn less whole turns, 0 up to 360/nodes/board/outputs/complexArgument
Read from the Scene at
Complex: |zₙ| = rⁿ/nodes/board/outputs/complexLength
One setting that reaches it
Turn per step
15 °
Stretch per step
0.94
Feedback for a right answer
The length $r^{12}$ ⁠ must be about $0.5$ ⁠, so $r \approx 0.5^{1/12} \approx 0.94$ ⁠ or $0.95$ ⁠. The angle $12\varphi$ ⁠ must be $180^\circ$ ⁠ plus whole turns: $\varphi = 15^\circ$ ⁠, $45^\circ$ ⁠, $75^\circ$ ⁠ and so on. The points spiral down towards $0$ ⁠.

Feedback for a wrong answer
Not yet. Work them out one at a time: which $r$ ⁠ gives $r^{12} \approx 0.5$ ⁠, and which $\varphi$ ⁠ gives $12\varphi = 180^\circ$ ⁠?

Worked solution
1. The length: $r^{12} = 0.5$ ⁠, so $r = 0.5^{1/12} \approx 0.944$ ⁠. On the slider’s steps, $0.94$ ⁠ gives $0.48$ ⁠ and $0.95$ ⁠ gives $0.54$ ⁠.
2. The angle: $12\varphi = 180^\circ$ ⁠ gives $\varphi = 15^\circ$ ⁠. $12\varphi = 540^\circ$ ⁠, so $\varphi = 45^\circ$ ⁠, works too.
3. Set Stretch per step to $0.94$ ⁠ and Turn per step to $15^\circ$ ⁠.
Step 5 of 5
Open this step
Euler’s formula
Practice · Predict, then look
Question
Turn per step is $180^\circ$ ⁠, that is $\pi$ ⁠ radians, with length $1$ ⁠. By Euler’s formula, $w = e^{i\pi} = \cos\pi + i\sin\pi$ ⁠. What is $e^{i\pi}$ ⁠?

Correct answer
- Correct $-1$ ⁠
Setup, choices and feedback
The Scene at this step
Turn per step
180 °fixed
Stretch per step
1fixed
Steps
1fixed
Build Euler’s formula
Steps
Choices
- Correct $-1$ ⁠
- Wrong $1$ ⁠
  Feedback
  $1$ ⁠ is $e^{i \cdot 0}$ ⁠, or $e^{2\pi i}$ ⁠: a whole turn. $\pi$ ⁠ is only half a turn.
- Wrong $i$ ⁠
  Feedback
  $i$ ⁠ is $e^{i\pi/2}$ ⁠, a quarter turn.
- Wrong A huge number, since $e^{\pi} \approx 23$ ⁠
  Feedback
  $e^{\pi}$ ⁠ is indeed about $23$ ⁠, but $e^{i\pi}$ ⁠ is a turn, not growth: its length is $1$ ⁠.
Once the prediction is locked in, the Scene changes
Build Euler’s formula
Euler's formula
The Scene then shows: -1 · Complex: the real part of w, r·cos φ/nodes/board/outputs/stepRe
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
$e^{i\pi} = \cos\pi + i\sin\pi = -1 + 0i = -1$ ⁠. So $e^{i\pi} + 1 = 0$ ⁠: the five numbers $e$ ⁠, $i$ ⁠, $\pi$ ⁠, $1$ ⁠ and $0$ ⁠ in one formula.

Worked solution
1. Euler’s formula: $e^{i\theta} = \cos\theta + i\sin\theta$ ⁠, the point on the unit circle at angle $\theta$ ⁠ radians.
2. $\theta = \pi$ ⁠ is half a turn: $\cos\pi = -1$ ⁠ and $\sin\pi = 0$ ⁠.
3. $e^{i\pi} = -1$ ⁠.

Lesson 3 of 3 · About 20 minutes

Adding sines

Every repeating wave is a sum of sines: circles spinning on circles, the square wave’s odd sines and their sizes, the triangle wave that converges fast, and the bump beside a jump that never goes away.

Open the lesson

The 6 steps and their answers

Step 1 of 6
Open this step
One circle
Practice · Calculate
Question
Terms is $1$ ⁠: one circle, and the square wave’s first sine, $b_1 \sin t$ ⁠ with $b_1 = \frac{4}{\pi}$ ⁠. What is the circle’s radius, $b_1$ ⁠, as a decimal?

Correct answer
1.273 ± 0.005
Setup, choices and feedback
The Scene at this step
Terms
1fixed
Wave
Squarefixed
Show the sines
Circles
Accepted within
± 0.005
Computed from the Scene at this step’s setting
Fourier: the radius of the first circle, |b₁|/nodes/board/outputs/fourierFirstAmplitude
Wrong answers it recognises
- 1 ± 0.005
  Misconception: Expects the first sine of a square wave to be exactly as tall as the wave.
  Feedback
  $1$ ⁠ is the square wave’s height. One sine that follows the square as closely as possible must rise above it in the middle to make up for its low shoulders.
Feedback for a right answer
$\frac{4}{\pi} \approx 1.273$ ⁠: the first sine is taller than the square wave, which reaches only $1$ ⁠. The next terms pull the peak down and lift the shoulders.

Worked solution
1. $b_1 = \frac{4}{\pi}$ ⁠.
2. $\pi \approx 3.1416$ ⁠, so $\frac{4}{3.1416} \approx 1.273$ ⁠.
If you answered $1$ ⁠, you assumed the first sine is as tall as the wave. To match the area under the square, it has to be taller.
Step 2 of 6
Open this step
More circles
Practice · Predict, then look
Question
There are $4$ ⁠ circles of the square wave: they spin $1$ ⁠, $3$ ⁠, $5$ ⁠ and $7$ ⁠ times as fast as the first. How many turns will the fifth circle make while the first turns once?

Correct answer
9 ± 0.1
A number prediction counts as right within ± 0.1
Setup, choices and feedback
The Scene at this step
Terms
4fixed
Wave
Squarefixed
Show the sines
0.3 (Circles → Sines inside)
Once the prediction is locked in, the Scene changes
Terms
5fixed
The Scene then shows: 9 · Fourier: k of the last term, how many times its circle turns while the first turns once/nodes/board/outputs/fourierLastHarmonic
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
$9$ ⁠: a square wave has only odd sines, $\sin t$ ⁠, $\sin 3t$ ⁠, $\sin 5t, \dots$ ⁠, and the fifth term is $\frac{4}{9\pi}\sin 9t$ ⁠.

Worked solution
1. The square wave uses only odd frequencies: $1, 3, 5, 7, 9, \dots$ ⁠
2. The fifth in the list is $9$ ⁠.
If you predicted $5$ ⁠, you counted every whole number. That is how a sawtooth is built, not a square wave.
Step 3 of 6
Open this step
The triangle wave
Practice · Calculate
Question
Wave is the triangle, with $3$ ⁠ terms. Term $k$ ⁠ has size $\frac{8}{\pi^2 k^2}$ ⁠, with $k$ ⁠ odd only. What is the radius of the third circle?

Correct answer
0.03242 ± 0.0005
Setup, choices and feedback
The Scene at this step
Terms
3fixed
Wave
Trianglefixed
Show the sines
0.3 (Circles → Sines inside)
Accepted within
± 0.0005
Computed from the Scene at this step’s setting
Fourier: the radius of the last (smallest) circle, |b| of its term/nodes/board/outputs/fourierLastAmplitude
Wrong answers it recognises
- 0.162 ± 0.002
  Misconception: Uses 1/k for the size of the triangle wave’s terms instead of 1/k².
  Feedback from the catalogue
  In the square and sawtooth waves term $k$ ⁠ shrinks like $\frac1k$ ⁠, but in the triangle wave like $\frac{1}{k^2}$ ⁠: that wave is smoother, with no jumps, so the fast sines are needed less. The term at frequency $5$ ⁠ is $\frac{1}{25}$ ⁠ of the first, not $\frac15$ ⁠.
- 0.0901 ± 0.001
  Misconception: Expects a square wave to use every whole-number harmonic, 1, 2, 3, …, not only the odd ones.
  Feedback
  $\frac{8}{9\pi^2} \approx 0.090$ ⁠ is the term $k = 3$ ⁠. In the triangle wave too $k$ ⁠ is odd only, so the third is $k = 5$ ⁠.
Feedback for a right answer
The third circle is $k = 5$ ⁠: $\frac{8}{25\pi^2} \approx 0.0324$ ⁠, less than $4\%$ ⁠ of the first. In the triangle wave the terms shrink fast, so even three terms look almost perfect.

Worked solution
1. The odd values of $k$ ⁠ are $1, 3, 5$ ⁠: the third circle is $k = 5$ ⁠.
2. $\frac{8}{\pi^2 \cdot 25} = \frac{8}{246.7} \approx 0.0324$ ⁠.
If you got $0.162$ ⁠, you divided by $k = 5$ ⁠ instead of $k^2 = 25$ ⁠.
Step 4 of 6
Open this step
Beside the jump
Practice · Predict, then look
Question
Beside each jump of the square wave the sum rises above the wave, by about $9\%$ ⁠ of the jump, with $5$ ⁠ terms. What will happen to this bump as you add terms, up to $15$ ⁠?

Correct answer
- Correct It will crowd in towards the jump but stay at about the same height
Setup, choices and feedback
The Scene at this step
Terms
5
Wave
Squarefixed
Show the sines
Sines insidefixed
Choices
- Correct It will crowd in towards the jump but stay at about the same height
- Wrong It will shrink until it disappears
  Misconception: Expects the overshoot beside a jump to vanish as more sines are added.
  Feedback from the catalogue
  Adding terms narrows the bump beside the jump and squeezes it closer, but does not lower it: it stays at about $9\%$ ⁠ of the jump however many terms are added. This is the Gibbs phenomenon. A finite sum of smooth sines cannot jump all at once.
- Wrong It will grow with every term
  Feedback
  Each added term is smaller than the last, so the bump cannot grow without end. It settles.
Once the prediction is locked in, the Scene changes
Terms
15
Show the sines
Sines insidefixed
The Scene then shows: 8.968 · Fourier: how far the peak rises above the top, in per cent of the jump of 2; 0 for the triangle, which has no jump/nodes/board/outputs/fourierOvershoot
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
The bump crowds in towards the jump but stays at about $9\%$ ⁠: the Gibbs phenomenon. A sum of smooth sines cannot jump all at once, so it always rises above the wave beside a jump.

Worked solution
1. With $5$ ⁠ terms the peak is about $1.182$ ⁠: $9.1\%$ ⁠ of the jump, which is $2$ ⁠ high.
2. With $15$ ⁠ terms the peak is about $1.179$ ⁠: $9.0\%$ ⁠.
3. The limit as terms are added for ever is $\frac{2}{\pi}\operatorname{Si}(\pi) \approx 1.179$ ⁠, about $9\%$ ⁠, not zero.
Step 5 of 6
Open this step
No jump
Practice · Choose
Question
Try all three Wave shapes with $15$ ⁠ terms, and drag Show the sines all the way. In which wave does the sum not rise above the wave?

Correct answer
- Correct The triangle wave
Setup, choices and feedback
The Scene at this step
Terms
15fixed
Wave
Square
Show the sines
0.5 (Circles → Sines inside)
Choices
Shown in a random order.
- Correct The triangle wave
- Wrong The square wave
  Misconception: Expects the overshoot beside a jump to vanish as more sines are added.
  Feedback from the catalogue
  Adding terms narrows the bump beside the jump and squeezes it closer, but does not lower it: it stays at about $9\%$ ⁠ of the jump however many terms are added. This is the Gibbs phenomenon. A finite sum of smooth sines cannot jump all at once.
- Wrong The sawtooth wave
  Feedback
  The sawtooth jumps once a cycle, from $1$ ⁠ to $-1$ ⁠, and beside the jump is the same bump of about $9\%$ ⁠.
Feedback for a right answer
The triangle wave: it has no jumps, only corners, so there is no bump. The Gibbs bump belongs to jumps, and the waves with a jump, square and sawtooth, keep it.

Worked solution
1. The square and sawtooth waves have jumps, and beside them the sum rises about $9\%$ ⁠ above the wave.
2. The triangle wave has no jumps: it is continuous and only changes direction at its corners.
3. So its sum closes in on the wave without rising above it, and the peak stays below $1$ ⁠.
Step 6 of 6
Open this step
The highest point
Practice · Calculate
Question
Wave: Square, and Terms $15$ ⁠. The wave jumps from $-1$ ⁠ to $1$ ⁠, and beside each jump the sum rises above the wave by about $9\%$ ⁠ of the size of the jump. What is the highest value the sum reaches?

Correct answer
1.179 ± 0.01
Setup, choices and feedback
The Scene at this step
Terms
15fixed
Wave
Squarefixed
Show the sines
Sines inside
Accepted within
± 0.01
Computed from the Scene at this step’s setting
Fourier: the highest value the sum reaches (the wave’s top is 1)/nodes/board/outputs/fourierPeak
Feedback for a right answer
About $1.18$ ⁠: $18\%$ ⁠ above the wave, even with $15$ ⁠ terms.

Hint
How big is the jump, from $-1$ ⁠ to $1$ ⁠?

Worked solution
1. The size of the jump: $1 - (-1) = 2$ ⁠.
2. The bump: $0.09 \times 2 = 0.18$ ⁠.
3. The highest value: $1 + 0.18 = 1.18$ ⁠.
This is the Gibbs phenomenon: with more terms the bump gets narrower and hugs the jump, but no lower. If you answered $1.09$ ⁠, you took $9\%$ ⁠ of the wave’s height, $1$ ⁠, not of the whole jump.

Across the unit · 9 misconceptions

Misconception catalogue

Each common mistake this unit recognises, in a few words. Open one for what the learner reads when an answer matches it, and where it is asked.

Index

Multiplying by i

The square of i

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

The powers of i

Question

Goal

The Scene at this step

One setting that reaches it

Worked solution

Length and angle

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Worked solution

Roots of minus 1

Question

Goal

The Scene at this step

One setting that reaches it

Worked solution

Squaring

Question

Correct answer

The Scene at this step

Choices

Worked solution

The imaginary part

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Turns, stretches and Euler’s formula

Angles add

Question

Correct answer

The Scene at this step

Once the prediction is locked in, the Scene changes

Worked solution

Lengths multiply

Question

Goal

The Scene at this step

One setting that reaches it

Worked solution

The rule

Question

Correct answer

The Scene at this step

Choices

Worked solution

Spiralling in

Question

Goal

The Scene at this step

One setting that reaches it

Worked solution

Euler’s formula

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

Adding sines

One circle

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Worked solution

More circles

Question