Everything in MTH603, rebuilt for your phone.
Lectures 19–45 in full — from the official handouts, the lecture slides, and your practice question banks. Every method starts with a plain-English explanation, then the formula, then a worked example you can step through one line at a time.
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00 Start here — how to use this in 4 hours ▶
This is your whole MTH603 final term syllabus (Lectures 19–45), rebuilt for a phone. Everything is here — you do not need the handouts open alongside it.
Every topic follows the same shape, so you always know what you are looking at:
- In plain words — the idea with no symbols at all. Read this first, always.
- Formula slip — the formula on ruled paper, exactly as you must write it in the exam.
- What each symbol means — every letter decoded.
- Worked example — tap Next step and watch it build one line at a time.
- Common mistake and Exam alert — where marks get lost.
- Practice questions — tap an answer, get told why.
Your 4-hour plan
| Time | Do this |
|---|---|
| 0:00–0:20 | Section 01 — Maths basics. Do not skip this if symbols confuse you. |
| 0:20–1:05 | Sections 02–04 — Operators and the two Newton formulas |
| 1:05–1:35 | Sections 05–07 — Lagrange, divided differences, error |
| 1:35–2:10 | Sections 08–09 — Differentiation and Richardson |
| 2:10–2:55 | Sections 10–11 — Integration (do one by hand!) |
| 2:55–3:35 | Sections 12–15 — ODEs (do one RK4 step by hand!) |
| 3:35–4:00 | Sections 17–18 — Formula sheet + Last-minute sheet |
The MTH603 final is mostly MCQs. Across the circulating past-paper collections, the same handful of question types come back every semester:
- Operator identities (
Δ = E − 1,δ = E^½ − E^−½) - "How many intervals?" for Simpson's rules
- Which formula was derived from which
- The four
kvalues in Runge–Kutta
So memorising formulas exactly beats deep understanding if you are short on time. This guide is weighted that way.
Features you should use
- Search box at the top — type "Simpson" or "k2" to jump straight there.
- Aa button — makes text bigger for phone reading.
- ☾ button — dark mode for late-night study.
- Mark studied at the end of each section — the progress bar tracks you.
- Contents button at the bottom — jump anywhere.
Your progress and settings are saved on this device. Close the page and come back — you will not lose your place.
01 Maths basics you need before anything else ▶
You said maths is hard for you. Fine — this section fixes that. Numerical Analysis looks terrifying because of its symbols, not because the ideas are hard. Learn these ten symbols and 80% of the fear disappears.
1. Subscripts — the little numbers below
y₀, y₁, y₂, y₃ just means first value, second value, third value, fourth value. Nothing more.
If a table gives you y = 2, 5, 10, 22 then y₀ = 2, y₁ = 5, y₂ = 10, y₃ = 22.
Read y₀ out loud as “y-zero” or “the first y”. Read yₙ as “the last y”. That is all n ever means in this course: the last one.
2. h — the step size
h is the gap between two x-values. If your table has x = 1, 2, 3, 4 then h = 1. If x = 0.1, 0.2, 0.3 then h = 0.1.
h = (b − a) / n ← when integrating from a to b with n strips
3. Δ, ∇, δ — the three difference symbols
These are just subtraction with a direction. That is genuinely all they are.
| Symbol | Name | Means | Example with y = 2, 5, 10 |
|---|---|---|---|
| Δ | delta (forward) | next − current | Δy₀ = 5 − 2 = 3 |
| ∇ | nabla (backward) | current − previous | ∇y₂ = 10 − 5 = 5 |
| δ | small delta (central) | half ahead − half behind | looks both ways |
← swipe the table sideways →
Δ points forward, ∇ points backward. Look at the shapes: Δ is a triangle sitting up (going up/forward), ∇ is upside down (going back). That is how you keep them straight under exam pressure.
4. E — the shift operator
E means “move one step to the right”.
E f(x) = f(x + h) — you gave it x, it hands you back the next value.
E² means move two steps. E⁻¹ means move backwards one step.
5. The ! symbol — factorial
n! means multiply every whole number from 1 up to n.
3! = 3 × 2 × 1 = 6
4! = 4 × 3 × 2 × 1 = 24
5! = 120
You will see p(p−1)/2!. That just means divide by 2. And p(p−1)(p−2)/3! means divide by 6. Write the number, not the exclamation mark, when you calculate.
6. Σ — sum everything up
Σ (sigma) means “add all of these together”. When you see Σ Lₖ(x)yₖ, it means: work out each piece, then add them.
7. What a derivative actually is
A derivative y′ (say: "y-prime") is just the steepness of a curve — how fast the value is climbing or falling at that exact point.
Driving a car: your position is y. Your speed is y′. Your acceleration is y″.
In this course you almost never differentiate by hand. You just plug table numbers into a formula that estimates the steepness. That is what "numerical differentiation" means.
8. What an integral actually is
An integral ∫ is the area underneath a curve.
Imagine the curve drawn on graph paper. Colour in everything between the curve and the bottom axis. That coloured area is the integral.
"Numerical integration" = chopping that area into strips and adding the strips up. Trapezoidal rule uses strips with slanted tops. Simpson's rule uses strips with curved tops. That is the entire chapter.
9. What an ODE and an "initial value problem" are
You are told the steepness rule (for example dy/dx = x + y) and one starting point (for example y(0) = 1). You must work out where the curve goes next.
It is like being told "you are at kilometre 0, and your speed at any moment equals your position plus the time" — and being asked where you are after 0.1 hours. You cannot jump there; you step forward a little at a time. Every ODE method in this course is a different rule for taking that step.
10. ξ — the mystery point in error formulas
ξ (say: "xi") appears in every error formula. It means “some unknown point somewhere inside the interval”. You never calculate it. You just write it down.
Two rules that cost students marks every single semester:
- All trigonometry in this course is in RADIANS, never degrees. Check your calculator mode before you start.
- Number of intervals
n= (number of points) − 1. Five values in a table means four intervals, not five.
Quick check
If a table gives x = 0.2, 0.4, 0.6, 0.8, what is h?
A table has 7 values. How many intervals n is that?
What does 3! equal?
02 Difference operators & their identities ▶
This is the highest-scoring section in the entire paper. These identities appear as MCQs in essentially every past paper. If you learn only one section, learn this one.
All six "operators" are just instructions written as a single symbol. They are shorthand, like how "+" is shorthand for "add".
Δ says subtract going forward. ∇ says subtract going backward. E says shift one step right. That is it. The scary-looking identities are just these three instructions rewritten in terms of each other.
The six operators
| Operator | Symbol | Definition | Plain meaning |
|---|---|---|---|
| Forward difference | Δ | Δyᵢ = yᵢ₊₁ − yᵢ | next minus current |
| Backward difference | ∇ | ∇yᵢ = yᵢ − yᵢ₋₁ | current minus previous |
| Central difference | δ | δyᵢ = yᵢ₊½ − yᵢ₋½ | half ahead minus half behind |
| Shift | E | E f(x) = f(x+h) | push forward one step |
| Average / mean | μ | μf(x) = ½[f(x+h/2) + f(x−h/2)] | average of half-step neighbours |
| Differential | D | Df(x) = f′(x) | ordinary derivative |
← swipe the table sideways →
Extra shift-operator facts that get asked directly
E² f(x) = f(x + 2h)
Eⁿ f(x) = f(x + nh)
E⁻¹ f(x) = f(x − h)
E⁻ⁿ f(x) = f(x − nh)
Ey₀ = y₁ E²y₀ = y₂ E²y₂ = y₄
⭐ The identities — memorise these exactly
∇ = 1 − E⁻¹ = (E − 1)/E
δ = E^½ − E^−½
μ = ½(E^½ + E^−½)
E = e^(hD) ⟺ hD = log E
hD = log(1 + Δ) = −log(1 − ∇)
Δ − ∇ = Δ∇ = ∇Δ = δ²
μ² = 1 + δ²/4
E^½ = μ + δ/2
Δ = E∇ = ∇E = δE^½
∇ = ΔE⁻¹
The one strategy that solves every operator proof: turn everything into E, simplify with ordinary algebra, then turn it back.
You do not need clever tricks. Every single proof in this topic is that one move.
Worked proof — Prove that Δ − ∇ = δ²
- Step 1 — Write both sides using E only
Replace Δ and ∇ with their E-forms from the identity sheet.
Δ = E − 1 and ∇ = 1 − E⁻¹ - Step 2 — Subtract them
Just remove the brackets carefully. Watch the double minus.
Δ − ∇ = (E − 1) − (1 − E⁻¹)
= E − 1 − 1 + E⁻¹
= E − 2 + E⁻¹ - Step 3 — Now expand δ² separately
δ = E^½ − E^−½, so square it like any bracket: (a − b)² = a² − 2ab + b².
δ² = (E^½ − E^−½)²
= E − 2(E^½)(E^−½) + E⁻¹
= E − 2 + E⁻¹ - Step 4 — Compare
Both sides came out identical. Proof complete.
Δ − ∇ = E − 2 + E⁻¹ = δ² ✔
Worked proof — Prove that μ² = 1 + δ²/4
- Step 1 — Square μ
μ = ½(E^½ + E^−½). Square the bracket: (a + b)² = a² + 2ab + b².
μ² = ¼(E^½ + E^−½)² = ¼(E + 2 + E⁻¹) - Step 2 — Work out δ²/4
We already know δ² = E − 2 + E⁻¹ from the previous proof.
δ²/4 = ¼(E − 2 + E⁻¹) - Step 3 — Add 1 to it
Write the 1 as 4/4 so you can combine the fractions.
1 + δ²/4 = ¼(4 + E − 2 + E⁻¹) = ¼(E + 2 + E⁻¹) - Step 4 — Compare
Identical to Step 1.
μ² = ¼(E + 2 + E⁻¹) = 1 + δ²/4 ✔
Where E = e^(hD) comes from
- Step 1 — Start with what E does
E shifts forward by h.
Ey_x = y_(x+h) = f(x + h) - Step 2 — Expand f(x+h) by Taylor's series
Standard Taylor expansion around x.
f(x+h) = f(x) + h f′(x) + (h²/2!) f″(x) + … - Step 3 — Rewrite derivatives using D
f′ = Df, f″ = D²f, and so on.
= f(x) + hD f(x) + (h²/2!) D²f(x) + … - Step 4 — Recognise the exponential series
1 + z + z²/2! + z³/3! + … is exactly e^z, with z = hD.
= [1 + hD + (hD)²/2! + …] f(x) = e^(hD) f(x) - Step 5 — Conclude
Take logs of both sides for the second form.
E = e^(hD) and hD = log E ✔
Building a difference table
A difference table is just a triangle of subtractions. Write the numbers in a column, subtract each pair to make the next column, keep going until you run out.
Always subtract lower minus upper. Go down the column, never up.
Example with x = 1, 2, 3, 4, 5 and y = 2, 5, 10, 22, 39:
| x | y | Δy | Δ²y | Δ³y | Δ⁴y |
|---|---|---|---|---|---|
| 1 | 2 | ||||
| 3 | |||||
| 2 | 5 | 2 | |||
| 5 | 5 | ||||
| 3 | 10 | 7 | −7 | ||
| 12 | −2 | ||||
| 4 | 22 | 5 | |||
| 17 | |||||
| 5 | 39 |
← swipe the table sideways →
How each column forms: Δy: 5−2=3, 10−5=5, 22−10=12, 39−22=17 → Δ²y: 5−3=2, 12−5=7, 17−12=5 → Δ³y: 7−2=5, 5−7=−2 → Δ⁴y: −2−5=−7.
Never build two tables. The forward, backward and central tables contain the exact same numbers — only the labels differ.
Build one table. For a forward question read the top diagonal. For a backward question read the bottom row. This saves you five minutes in the exam.
Properties that get asked
- For a polynomial of degree n, the nth differences are constant and the (n+1)th are zero.
- Difference operators need equally spaced data. Lagrange and divided differences do not.
- Δ, ∇, δ, E, μ are all linear operators.
- Δ of a constant = 0.
- With n+1 data points, the highest difference you can compute is Δⁿ.
- "Central difference gives a better approximation but requires more computation" → TRUE.
Subtracting in the wrong direction. For Δ it is always lower minus upper. Students under time pressure flip this and lose every mark in the table.
Practice questions
Symbol used for forward differences is:
The relationship between the central difference operator and the shift operator is:
Δ equals:
The average operator μ is defined as:
Eⁿ f(x) equals:
Δ∇ equals:
μ² equals:
If the third differences of a tabulated function are constant, the data represents a polynomial of degree:
The inverse shift operator E⁻¹ f(x) equals:
Likely subjective question
“Define shift operator, average operator and differential operator.” — from your practice bank, and a near-certain 3-mark question.
Model answer:
Shift operator E: E f(x) = f(x+h). It shifts the function forward by one step h. In general Eⁿf(x) = f(x+nh), and E⁻¹f(x) = f(x−h).
Average (mean) operator μ: μ f(x) = ½[f(x + h/2) + f(x − h/2)]. It gives the mean of the function values half a step either side.
Differential operator D: D f(x) = f′(x) = dy/dx, and D²f(x) = f″(x). It relates to E by E = e^(hD).
03 Newton's forward difference formula ▶
You have a table of values, and you need a value that is not in the table — something in between. That is interpolation.
Newton's forward formula builds an estimate by starting at the first value and adding correction terms: first the change, then the change-in-the-change, and so on. Each extra term makes the estimate more accurate.
Use it when the value you want is near the top of the table.
- x₀
- the first x-value in the table
- h
- the gap between x-values
- x
- the point you want the answer at
- p
- how many steps x is away from x₀ — no units, just a number
- Δᵏy₀
- the kth forward difference, taken from the top diagonal of your table
Where it comes from (one line)
f(x₀ + ph) = E^p y₀ = (1 + Δ)^p y₀, then binomial-expand (1+Δ)^p. Write that line if a derivation is asked.
Worked example — estimate f(0.36)
| x | 0.1 | 0.2 | 0.3 | 0.4 |
|---|---|---|---|---|
| y | 0.1003 | 0.2027 | 0.3093 | 0.4228 |
- Step 1 — Build the difference table
Subtract downwards to get each column.
Δy: 0.1024, 0.1066, 0.1135
Δ²y: 0.0042, 0.0069
Δ³y: 0.0027 - Step 2 — Pick out the top diagonal
Newton forward only uses the first entry of each column.
y₀ = 0.1003 Δy₀ = 0.1024 Δ²y₀ = 0.0042 Δ³y₀ = 0.0027 - Step 3 — Find p
h = 0.1 (gap between x values), x₀ = 0.1 (first x), x = 0.36 (what we want).
p = (0.36 − 0.1) / 0.1 = 0.26 / 0.1 = 2.6 - Step 4 — Substitute into the formula
Work out each bracket separately before multiplying. Do not rush this line.
f = 0.1003 + 2.6(0.1024)
+ [2.6(1.6)/2](0.0042)
+ [2.6(1.6)(0.6)/6](0.0027) - Step 5 — Compute each term
Four numbers to add.
= 0.1003 + 0.26624 + 0.008736 + 0.0011232 - Step 6 — Add them up
That is your answer.
f(0.36) ≈ 0.37640
- Using xₙ instead of x₀ when finding p. Forward always uses x₀.
- Writing p(p+1) instead of p(p−1). Forward has MINUS.
- Reading differences off the wrong diagonal — forward uses the top only.
- Forgetting the factorials 2!, 3!, 4! underneath.
Forward → First point (x₀) → the Falling factor (p−1). Three F's, one rule.
Facts that appear as MCQs
- If you keep r+1 terms, you get a polynomial of degree r.
- Requires equally spaced x-values.
- Setting p = 0 gives back y₀ exactly.
- Can be used to find a missing value: set the highest available difference to zero and solve.
Missing-value technique
With 5 points and one missing, assume degree 3, so set Δ⁴y₀ = 0:
Then solve for the unknown y.
Practice questions
In Newton's forward difference formula, p is defined as:
If we retain r+1 terms in Newton's forward difference formula, we obtain a polynomial of degree:
In Newton's forward formula, the coefficient of Δ²y₀ is:
Newton's forward difference formula is best used to interpolate near the ___ of the table.
Likely subjective questions
“Find p if f(18) is calculated from a table with x = 1,2,3,4,5.”
h = 1, x₀ = 1 → p = (18 − 1)/1 = 17. Add for the extra mark: such a large p means this is extrapolation, not interpolation.
“Find p if f(0.39) is calculated from x = 0.4, 0.5, 0.6, 0.7, 0.8.”
h = 0.1, x₀ = 0.4 → p = (0.39 − 0.4)/0.1 = −0.1
“Find the interpolating polynomial” given y₀=1, Δy₀=2, Δ²y₀=3, Δ³y₀=0, h=1, x₀=0:
p = x, so f(x) = 1 + 2x + [x(x−1)/2](3) = 1 + 2x + 1.5x² − 1.5x = 1.5x² + 0.5x + 1
04 Newton's backward difference formula ▶
Exactly the same idea as the forward formula — but you start from the last value in the table and work backwards.
Use it when the value you want is near the bottom of the table.
- xₙ
- the last x-value in the table
- ∇ᵏyₙ
- the kth backward difference, taken from the bottom row
- p
- usually negative, because x is smaller than xₙ
⭐ Forward vs Backward — the comparison that gets tested
| Feature | Forward | Backward |
|---|---|---|
| Base point | x₀ (first) | xₙ (last) |
| p = | (x − x₀)/h | (x − xₙ)/h |
| Sign of p | usually positive | usually negative |
| Factor pattern | p(p−1)(p−2)… | p(p+1)(p+2)… |
| Differences | Δ, top diagonal | ∇, bottom row |
| Best for | start of table | end of table |
← swipe the table sideways →
Backward → Bottom → Bigger factors (p+1). Forward → First → (p−1).
Worked example — compute f(11)
| x | 2 | 4 | 6 | 8 | 10 | 12 | 14 |
|---|---|---|---|---|---|---|---|
| F(x) | 23 | 93 | 259 | 559 | 1031 | 1713 | 2643 |
← swipe the table sideways →
Use xₙ = 10 and p = 0.5 (so the table is truncated at x = 10, h = 2).
- Step 1 — Build differences up to x = 10
Subtract downwards as usual.
Δ: 70, 166, 300, 472
Δ²: 96, 134, 172
Δ³: 38, 38
Δ⁴: 0 - Step 2 — Read the BOTTOM row (these are the ∇ values at xₙ = 10)
The last entry of each column is what backward needs.
∇y = 472 ∇²y = 172 ∇³y = 38 ∇⁴y = 0 - Step 3 — Confirm p
h = 2, xₙ = 10, x = 11.
p = (11 − 10)/2 = 0.5 - Step 4 — Substitute, using the PLUS pattern
Backward brackets are (p+1), (p+2) — note the difference from forward.
f = 1031 + 0.5(472)
+ [0.5(1.5)/2](172)
+ [0.5(1.5)(2.5)/6](38) - Step 5 — Compute each term
The Δ⁴ term is zero, so it drops out.
= 1031 + 236 + 64.5 + 11.875 - Step 6 — Add up
Final answer.
f(11) = 1343.375
- Using Δ values with the backward formula. Use ∇ from the bottom row.
- Writing (p−1) instead of (p+1).
- Forgetting that p is negative when x < xₙ, then losing the sign mid-calculation.
- Using x₀ as the base. Backward always uses xₙ.
Practice questions
In Newton's backward difference formula, p =
In Newton's backward formula, the coefficient of ∇²yₙ is:
In Newton's backward formula, the differences are taken along the ___ diagonal.
Likely subjective questions
“Construct a backward difference table for x = 1, 3, 5 and y = 2, 4, 7.”
∇y: 4−2 = 2, then 7−4 = 3. ∇²y: 3−2 = 1. So at x = 5: ∇y₅ = 3, ∇²y₅ = 1. (h = 2, equally spaced ✔)
“Construct a backward difference table for x = 1, 5, 10 and y = 3, 7, 14.”
∇y: 4, 7. ∇²y: 3.
⚠ Worth an extra mark here: the spacing is not equal (4 then 5). You can build the table mechanically, but you should note that difference formulas strictly require equal spacing — Newton's divided difference is the correct method for this data. Examiners like that observation.
05 Lagrange's interpolation formula ▶
Newton's formulas break if your x-values are not evenly spaced. Lagrange's formula does not care about spacing at all — that is its whole reason for existing.
The idea: build one "weight" for each data point. Each weight equals 1 at its own point and 0 at every other point. Multiply each weight by its y-value, add them all up, and you have a curve that passes through every point exactly.
Written out for 3 points — the case VU usually sets
Key properties
- No equal spacing required — the whole selling point.
- Σ Lₖ(x) = 1 for every x. Excellent self-check: add your coefficients, they must total 1.
- Lₖ(xₖ) = 1 and Lₖ(xᵢ) = 0 for i ≠ k.
- With n+1 points it gives a polynomial of degree ≤ n.
- Main disadvantage: add one new data point and you must recompute everything from scratch. Newton's divided difference does not have this problem.
Worked example — evaluate f(2.5)
Given x = 1, 3, 7 and y = 4, 18, 21.
- Step 1 — Label everything
Write down which value is which before you start. This prevents most errors.
x₀=1, x₁=3, x₂=7
y₀=4, y₁=18, y₂=21
x = 2.5 - Step 2 — First coefficient L₀
Numerator skips (x−x₀). Denominator is the same brackets with x₀ = 1 put in.
L₀ = (2.5−3)(2.5−7) / [(1−3)(1−7)]
= (−0.5)(−4.5) / [(−2)(−6)]
= 2.25 / 12 = 0.1875 - Step 3 — Second coefficient L₁
Now skip (x−x₁), and use x₁ = 3 in the denominator.
L₁ = (2.5−1)(2.5−7) / [(3−1)(3−7)]
= (1.5)(−4.5) / [(2)(−4)]
= −6.75 / −8 = 0.84375 - Step 4 — Third coefficient L₂
Skip (x−x₂), use x₂ = 7 below.
L₂ = (2.5−1)(2.5−3) / [(7−1)(7−3)]
= (1.5)(−0.5) / [(6)(4)]
= −0.75 / 24 = −0.03125 - Step 5 — CHECK before finishing
The coefficients must add to exactly 1. If they do not, you have made an arithmetic error — go back now.
0.1875 + 0.84375 − 0.03125 = 1.0 ✔ - Step 6 — Multiply each by its y and add
Final answer.
f(2.5) = 0.1875(4) + 0.84375(18) − 0.03125(21)
= 0.75 + 15.1875 − 0.65625
= 15.28125
Inverse interpolation
To find x for a given y, just swap the roles of x and y in the formula. This works precisely because Lagrange doesn't need equal spacing — and y-values are almost never equally spaced.
- Including the factor (x − xₖ) in the kth numerator. It must be left out.
- Sign errors in denominators — write every bracket out, never do it mentally.
- Forgetting to multiply each Lₖ by its own yₖ.
- Not checking Σ Lₖ = 1. That check is free error detection — use it.
Practice questions
The sum of the Lagrange coefficients Lₖ(x) is:
Lagrange's formula with n+1 points produces a polynomial of degree at most:
The main disadvantage of Lagrange's interpolation is that:
Likely subjective question
“Compare Lagrange's interpolation with Newton's divided difference interpolation.”
Model answer: Both handle unequally spaced data and produce the same polynomial. Lagrange is easier to write down directly but must be redone entirely if a data point is added. Newton's divided difference builds up term by term, so a new point only adds one extra term, and it is computationally cheaper — which is why the handouts prefer it.
“Find the interpolating polynomial for (0,7), (2,6), (3,9), (6,21).”
Four points → a cubic. Divided differences are far quicker here:
1st: −1/2, 3, 4 → 2nd: 7/6, 1/4 → 3rd: −11/72
P(x) = 7 + x(−½) + x(x−2)(7/6) + x(x−2)(x−3)(−11/72)
Expanded: P(x) = −(11/72)x³ + (139/72)x² − (15/4)x + 7
Check: P(0)=7 ✔ P(2)=6 ✔ P(3)=9 ✔ P(6)=21 ✔
06 Divided differences & Newton's divided difference formula ▶
A divided difference is a difference divided by the gap. That is literally the name and the definition.
Ordinary differences (Δ) assume every gap is the same size, so they don't bother dividing. When gaps are uneven, you must divide by the actual gap — and that gives you divided differences.
The pattern: take the two divided differences one order below, subtract them, and divide by (outermost x − outermost x). Note the denominator spans all the intervals involved, not just the adjacent gap.
Only the top diagonal of the table is used — exactly like Newton's forward formula.
Key properties
- Works for unequally spaced data — certain MCQ.
- Divided differences are symmetric: y[x₀,x₁] = y[x₁,x₀]. Order does not matter.
- The nth divided difference of a degree-n polynomial is constant; the (n+1)th is zero.
- Link to forward differences (equal spacing only):
y[x₀,…,xₙ] = Δⁿy₀ / (n! hⁿ). This is why the divided difference formula reduces to Newton's forward formula when spacing is equal. - Adding a new data point adds only one new term.
Worked example — small table
x = 0, 1, 3 and y = 2, 4, 20.
- Step 1 — First order divided differences
Difference in y, divided by difference in x.
y[0,1] = (4 − 2)/(1 − 0) = 2
y[1,3] = (20 − 4)/(3 − 1) = 16/2 = 8 - Step 2 — Second order
Difference of the two first-order values, divided by the full span (3 − 0), not (3 − 1).
y[0,1,3] = (8 − 2)/(3 − 0) = 6/3 = 2 - Step 3 — Write the formula with these values
Top diagonal: y[x₀] = 2, y[x₀,x₁] = 2, y[x₀,x₁,x₂] = 2.
f(x) = 2 + (x−0)(2) + (x−0)(x−1)(2) - Step 4 — Expand
Multiply out and collect like terms.
= 2 + 2x + 2x² − 2x = 2x² + 2 - Step 5 — Check against every data point
Always do this. It takes ten seconds and catches everything.
f(0) = 2 ✔ f(1) = 4 ✔ f(3) = 18+2 = 20 ✔
Worked example — larger table
x = 0, 1, 2, 4, 5 and y = 1, 9, 12, 5, 6.
| x | y | 1st | 2nd | 3rd | 4th |
|---|---|---|---|---|---|
| 0 | 1 | ||||
| 8 | |||||
| 1 | 9 | −2.5 | |||
| 3 | 0.08333 | ||||
| 2 | 12 | −2.1667 | 0.16667 | ||
| −3.5 | 0.91667 | ||||
| 4 | 5 | 1.5 | |||
| 1 | |||||
| 5 | 6 |
← swipe the table sideways →
- Dividing by the wrong x-gap. For a kth-order divided difference the denominator spans k intervals: (x₍ᵢ₊ₖ₎ − xᵢ), not the adjacent gap.
- Using values from the wrong diagonal — always the top.
- Miscounting brackets. Term number k carries (k−1) brackets: term 1 has none, term 2 has one, term 3 has two, and so on.
Practice questions
Newton's divided difference interpolation formula is used when the values of the independent variable are:
Divided differences are ___ with respect to their arguments.
For equally spaced data, y[x₀,x₁,…,xₙ] equals:
The nth divided difference of a polynomial of degree n is:
07 Error term in interpolation & 2-D interpolation ▶
When you fit a polynomial through your data points, the polynomial is exactly right at those points — but slightly wrong between them. The error formula tells you how wrong.
You will almost never calculate it. You just need to recognise and write it.
- ξ
- some unknown point inside the interval. You never compute it.
- y⁽ⁿ⁺¹⁾
- the (n+1)th derivative — one order higher than the polynomial degree
- Π(x)
- shorthand for the whole product (x−x₀)(x−x₁)…(x−xₙ)
Facts examiners ask
- Error is defined as E(x) = y(x) − Pₙ(x) — real value minus approximation. Sign-flipped distractors are common.
- Π(x) vanishes n+1 times in the interval — once at each node.
- If y(x) is itself a polynomial of degree ≤ n, then y⁽ⁿ⁺¹⁾ = 0, so the error is exactly zero. The handouts make a point of this: "no error in interpolation!!"
Newton's divided difference formula with error term
The error can be written as one extra divided difference that includes x itself:
+ (x−x₀)(x−x₁)…(x−xₙ)·y[x, x₀, x₁, …, xₙ]
Interpolation in two dimensions
You have a grid of values — f depends on both x and y. To find a value in the middle of the grid, interpolate twice:
- Go along each row, interpolating in the x direction. You now have a single column of values at your required x.
- Go down that column, interpolating in the y direction. That is your answer.
Doing y first then x gives the same answer.
p = (x − x₀)/h q = (y − y₀)/k
Practice questions
If y(x) is approximated by a polynomial Pₙ(x) of degree n, the error is given by:
Let I be the closed interval spanned by x₀, x₁, …, xₙ, x. Then Π(x) vanishes ___ times in the interval.
The error term in interpolation involves the derivative of order:
If the tabulated function is itself a polynomial of degree ≤ n, the interpolation error is:
08 Numerical differentiation ▶
You want the steepness of a curve, but all you have is a table of numbers — no equation to differentiate.
Solution: estimate the steepness from nearby values. If y jumps from 5 to 9 while x moves by 0.1, the slope is about (9−5)/0.1 = 40. Every formula below is a refinement of that one idea.
Where the series formulas come from
We know hD = log(1 + Δ) and hD = −log(1 − ∇). Expanding those logarithms as series gives derivative formulas made purely of differences.
The forward series alternates in sign (it comes from log(1+Δ)). The backward series is all plus (it comes from −log(1−∇)). The second-derivative coefficients 1, 1, 11/12, 5/6 are worth learning by heart.
The printed backward series in the handout PDF shows a stray minus sign on the ∇²/2 term. The correct expansion of −log(1−∇) has all plus signs, as written above. If a paper reproduces the handout's typo, either version will be accepted — but the all-plus form is the correct one.
⭐ Two-point and three-point formulas — heavily examined
These come from keeping only the first term of the series above.
First derivative — two-point formulas
| Type | Formula | Accuracy |
|---|---|---|
| Forward | y′ᵢ = [y(xᵢ+h) − y(xᵢ)] / h | O(h) |
| Backward | y′ᵢ = [y(xᵢ) − y(xᵢ−h)] / h | O(h) |
| Central ⭐ | y′ᵢ = [y(xᵢ+h) − y(xᵢ−h)] / 2h | O(h²) — best |
← swipe the table sideways →
When a VU question says “using the two-point equation”, it means the central version — unless the point sits at the edge of the table, where there is no neighbour on one side and you must use forward or backward instead.
Second derivative — three-point formulas
| Type | Formula |
|---|---|
| Forward | y″ᵢ = [y(xᵢ+2h) − 2y(xᵢ+h) + y(xᵢ)] / h² |
| Backward | y″ᵢ = [y(xᵢ) − 2y(xᵢ−h) + y(xᵢ−2h)] / h² |
| Central ⭐ | y″ᵢ = [y(xᵢ+h) − 2y(xᵢ) + y(xᵢ−h)] / h² |
← swipe the table sideways →
Worked example — first derivative by two-point formula
Find f′ at x = 0.1 for this table:
| x | 0.1 | 0.2 | 0.3 | 0.4 | 0.5 | 0.6 |
|---|---|---|---|---|---|---|
| f(x) | 0.105 | 0.342 | 0.652 | 0.756 | 0.812 | 0.924 |
← swipe the table sideways →
- Step 1 — Check which formula is possible
x = 0.1 is the first point. There is nothing to its left, so the central formula cannot be used.
No point at x = 0.0 → use the forward two-point formula - Step 2 — Identify the values
Read straight off the table.
h = 0.1 f(0.1) = 0.105 f(0.2) = 0.342 - Step 3 — Substitute
Next value minus current, divided by h.
f′(0.1) = [f(0.2) − f(0.1)] / h = (0.342 − 0.105) / 0.1 - Step 4 — Compute
Final answer.
= 0.237 / 0.1 = 2.37
Worked example — second derivative by three-point formula
Find f″ at x = 0.2 for this table:
| x | 0.1 | 0.2 | 0.3 | 0.4 | 0.5 | 0.6 |
|---|---|---|---|---|---|---|
| f(x) | 0.125 | 0.302 | 0.602 | 0.706 | 0.812 | 0.924 |
← swipe the table sideways →
- Step 1 — Check which formula is possible
x = 0.2 has a neighbour on both sides (0.1 and 0.3), so the central formula works — and it is the most accurate.
Use central: y″ = [y(x+h) − 2y(x) + y(x−h)] / h² - Step 2 — Identify the three values
One before, the point itself, one after.
h = 0.1
f(0.1) = 0.125 f(0.2) = 0.302 f(0.3) = 0.602 - Step 3 — Substitute
Note the 2 multiplying the middle value, and h is squared.
f″(0.2) = [0.602 − 2(0.302) + 0.125] / (0.1)² - Step 4 — Simplify the top
Careful with the subtraction.
= [0.602 − 0.604 + 0.125] / 0.01 = 0.123 / 0.01 - Step 5 — Divide
Final answer.
= 12.3
Differentiation using interpolation (unequal spacing)
If the data is not equally spaced you cannot use Δ or ∇ series. Instead: fit an interpolating polynomial (the handouts prefer Newton's divided difference "in view of economy of computation"), then differentiate that polynomial.
Even when Pₙ(x) and y(x) agree exactly at every tabular point, their derivatives can differ substantially at those same points. The handouts show a figure of two curves crossing at the nodes with visibly different slopes. For higher derivatives it is worse.
Takeaway: numerical differentiation amplifies error; numerical integration smooths it. So integration is more accurate than differentiation.
Three-point and five-point formulas (Lecture 42)
These are the interpolation-based derivative formulas — distinct from the difference-operator series above.
Coefficients: 1, −8, 0, 8, −1 over 12h
Which is more accurate? The mid-point formula has error coefficient h²/6 versus h²/3 for the end-point version — so mid-point is about twice as accurate. And the five-point formula beats both.
The handouts demonstrate this with f(x) = x·eˣ, approximating f′(2.0) where the true value is 22.167168:
| Formula | h | Result | Error |
|---|---|---|---|
| Three-point end-point | 0.1 | 22.032310 | 1.35×10⁻¹ |
| Three-point end-point | −0.1 | 22.005452 | 1.13×10⁻¹ |
| Three-point mid-point | 0.1 | 22.228790 | −6.16×10⁻² |
| Three-point mid-point | 0.2 | 22.414163 | −2.47×10⁻¹ |
| Five-point | 0.1 | 22.166999 | 1.69×10⁻⁴ |
← swipe the table sideways →
The handouts state it explicitly: “the five-point formula gives the superior result.” Expect this as a conceptual MCQ.
- Forgetting to square h in second-derivative formulas.
- Using the central formula at an endpoint where a neighbour does not exist.
- Mixing forward differences into the backward series, or vice versa.
- Losing the alternating signs in the forward series.
Practice questions
The two-point central formula for the first derivative is:
The three-point central formula for the second derivative is:
hD in terms of the forward difference operator is:
The coefficient of Δ⁴y₀ in the second-derivative forward series is:
The five-point formula for f′(x₀) has coefficients:
Compared to numerical integration, numerical differentiation is generally:
09 Richardson's extrapolation ▶
Do the calculation twice — once with step h, once with step h/2 — then combine the two answers in a clever way that cancels most of the error.
You get a far better answer without needing any extra data. It feels like getting something for nothing, and mathematically it nearly is.
The derivation in three lines
Using the two-point central formula, y′(x) = F(h) + E, where the error E = c₁h² + c₂h⁴ + …
- Step 1 — Write it at step size h
The leading error term is c₁h².
y′(x) = F(h) + c₁h² + c₂h⁴ + … - Step 2 — Write it again at step size h/2
Halving h divides h² by 4 and h⁴ by 16.
y′(x) = F(h/2) + c₁h²/4 + c₂h⁴/16 + … - Step 3 — Multiply the second by 4
Now the c₁ terms match in size.
4y′(x) = 4F(h/2) + c₁h² + c₂h⁴/4 + … - Step 4 — Subtract line 1 from line 3
The c₁h² terms cancel — that was the whole point.
3y′(x) = 4F(h/2) − F(h) + … - Step 5 — Divide by 3
A fourth-order accurate formula built from two second-order results.
F₁(h/2) = [4·F(h/2) − F(h)] / 3
| m | Denominator (4ᵐ − 1) | Resulting accuracy |
|---|---|---|
| 1 | 3 | O(h⁴) |
| 2 | 15 | O(h⁶) |
| 3 | 63 | O(h⁸) |
Repeating this refinement is called “extrapolation to the limit”.
Worked example
- Step 1 — Pick the right formula
First stage, so m = 1 and the denominator is 4¹ − 1 = 3.
F₁ = [4·F(h/2) − F(h)] / 3 - Step 2 — Substitute
The finer value (h/2) is the one that gets multiplied by 4.
F₁ = [4(0.51) − 0.5] / 3 - Step 3 — Compute
Final answer.
= (2.04 − 0.5)/3 = 1.54/3 = 0.5133
- Using denominator 4 instead of 3 for m = 1. It is 4ᵐ − 1, not 4ᵐ.
- Swapping F(h) and F(h/2). The finer value (h/2) always gets the 4ᵐ multiplier.
Practice questions
The first stage of Richardson's extrapolation gives:
The general Richardson formula has denominator:
Repeating Richardson's process until the required accuracy is achieved is called:
Combining two second-order accurate results by Richardson's method yields a formula of order:
10 Numerical integration — Trapezoidal & Simpson ▶
Guaranteed exam content. You will get at least one numerical from this section, plus an almost-certain “how many intervals?” MCQ.
You want the area under a curve, but you cannot integrate the function by hand — or you only have a table of numbers.
So you chop the area into vertical strips and add the strips up.
- Trapezoidal rule — each strip has a straight slanted top (a trapezium). Quick, least accurate.
- Simpson's 1/3 — takes strips two at a time and puts a curve (parabola) over the pair. More accurate.
- Simpson's 3/8 — takes strips three at a time with a cubic over them.
That is the entire chapter. The formulas are just bookkeeping for adding the strips.
Why numerical integration is needed
All three of these are valid reasons, and the MCQ answer is usually “all of the above”:
- The antiderivative does not exist in closed form (e.g. ∫e^(−x²)dx).
- Analytical evaluation is possible but extremely complicated.
- The integrand is given only in tabular form — there is no formula at all.
Newton–Cotes background
Approximate f(x) by an interpolating polynomial at equally spaced nodes, then integrate that polynomial. The weights cₖ = ∫Lₖ(x)dx are called Cotes numbers.
| Strips per panel | Polynomial fitted | Resulting rule |
|---|---|---|
| 1 | straight line (2 points) | Trapezoidal |
| 2 | parabola (3 points) | Simpson's 1/3 |
| 3 | cubic (4 points) | Simpson's 3/8 |
Line = Trapezoid. Parabola = 1/3. Cubic = 3/8.
And the rule name tells you how many strips it eats at a time: the 1/3 rule eats 2 strips, the 3/8 rule eats 3 strips.
⭐ The three formulas
For Simpson's 1/3: “Odd → fOur, Even → twO.” Odd-numbered subscripts get 4, even-numbered get 2. The first and last always get 1.
🔴 The interval rule — the most repeated MCQ family in the paper
| Rule | Valid number of intervals |
|---|---|
| Trapezoidal | any — 1, 2, 3, 4, 5, 7, 11 all fine |
| Simpson's 1/3 | EVEN — 2, 4, 6, 8, 10, 12 … |
| Simpson's 3/8 | MULTIPLE OF 3 — 3, 6, 9, 12, 15 … |
Past-paper items you will almost certainly meet:
- “To apply Simpson's 1/3 rule, valid number of intervals are: 7 / 8 / 5 / 3”
- “For Simpson's 1/3 rule no. of intervals must be: 2 (even) / 3 / 5 / 7”
- “To apply Simpson's 3/8 rule, the number of intervals must be: 10 / 11 / 12 / 13”
- “To apply Simpson's 3/8 rule, the number of intervals can be: 4 / 7 / 9 / 8”
⚠ Note the trap: 12 is both even and a multiple of 3, so it is valid for both Simpson rules. Read whether the question asks 1/3 or 3/8.
Worked example — Trapezoidal rule
Evaluate ∫₀⁴ x² dx with h = 1.
- Step 1 — Build the table of values
Just square each x.
x: 0 1 2 3 4
f(x): 0 1 4 9 16 - Step 2 — Count the intervals
Five points means four gaps. Trapezoidal accepts any n, so we are fine.
n = 5 points − 1 = 4 intervals h = 1 - Step 3 — Apply the 1, 2, 2, 2, 1 pattern
First and last get 1. Everything in between gets 2.
I = (h/2)[y₀ + 2(y₁ + y₂ + y₃) + y₄] - Step 4 — Substitute the numbers
Middle values are 1, 4 and 9.
I = (1/2)[0 + 2(1 + 4 + 9) + 16] - Step 5 — Compute
Inside the bracket first, then multiply by h/2.
= (1/2)[0 + 28 + 16] = (1/2)(44) = 22 - Step 6 — Sanity check
The exact answer is 64/3 ≈ 21.33. Trapezoidal over-estimates here because the curve bends upward and the straight tops sit above it. Nice conceptual point to know.
Trapezoidal = 22 vs exact ≈ 21.33
Worked example — Simpson's 1/3 rule
Evaluate ∫₀⁴ f(x)dx with h = 1, where f = −1, 0, 1, 2, 3 at x = 0,1,2,3,4.
- Step 1 — Check the interval rule FIRST
Always do this before anything else. 4 is even, so Simpson's 1/3 is allowed.
n = 5 points − 1 = 4 → even ✔ - Step 2 — Label which subscripts are odd and even
y₁ and y₃ are odd. y₂ is even. y₀ and y₄ are the ends.
odd: y₁ = 0, y₃ = 2
even: y₂ = 1
ends: y₀ = −1, y₄ = 3 - Step 3 — Apply the 1, 4, 2, 4, 1 pattern
Odd gets 4, even gets 2, ends get 1.
I = (h/3)[y₀ + 4(y₁ + y₃) + 2(y₂) + y₄] - Step 4 — Substitute
Put the numbers in carefully.
I = (1/3)[−1 + 4(0 + 2) + 2(1) + 3] - Step 5 — Compute
Bracket first.
= (1/3)[−1 + 8 + 2 + 3] = (1/3)(12) = 4 - Step 6 — Sanity check
This data is the straight line y = x − 1, whose exact integral over [0,4] is 4. Simpson is exact here — as expected, since its degree of precision is 3.
Simpson = 4 = exact ✔
Worked example — Simpson's 3/8 rule
- Step 1 — Check the interval rule
Four points → 3 intervals. 3 is a multiple of 3, so the 3/8 rule is allowed.
n = 3 → multiple of 3 ✔ - Step 2 — Apply the 1, 3, 3, 1 pattern
With only one panel there is no “2” yet — that appears from the 4th interval onwards.
I = (3h/8)[y₀ + 3(y₁ + y₂) + y₃] - Step 3 — Substitute and compute
Note the multiplier is 3h/8, not h/3.
= (3/8)[1 + 3(2 + 5) + 10] = (3/8)[1 + 21 + 10]
= (3/8)(32) = 12
Comparison table — learn this, it is pure MCQ marks
| Trapezoidal | Simpson 1/3 | Simpson 3/8 | |
|---|---|---|---|
| Curve fitted | Line | Parabola | Cubic |
| Points per panel | 2 | 3 | 4 |
| Strips per panel | 1 | 2 | 3 |
| Interval condition | any n | n even | n ×3 |
| Multiplier | h/2 | h/3 | 3h/8 |
| Pattern | 1,2,…,2,1 | 1,4,2,…,4,1 | 1,3,3,2,…,3,3,1 |
| Accuracy | O(h²) | O(h⁴) | O(h⁴) |
| Degree of precision | 1 | 3 | 3 |
| Error denominator | 12 | 180 | 80 |
← swipe the table sideways →
- Using Simpson's 1/3 with an odd number of intervals. Instant zero.
- Giving 4 to even subscripts and 2 to odd. It is the reverse.
- Using h/3 for the 3/8 rule, or 3h/8 for the 1/3 rule.
- Miscounting n. Five tabulated values means four intervals. n = points − 1.
- Forgetting to compute h = (b−a)/n when it is not given.
Practice questions
In Simpson's 1/3 rule, f(x) is of the form:
To apply Simpson's 1/3 rule, valid number of intervals are:
To apply Simpson's 3/8 rule, the number of intervals must be:
The Trapezoidal rule is exact for polynomials of degree:
Simpson's 1/3 rule, although based on fitting a quadratic, is exact for polynomials up to degree:
While integrating ∫ from a to b, the width of the interval h is found by:
The weighting coefficients in Newton–Cotes formulae are called:
Which of the following lead(s) towards numerical integration methods?
11 Romberg's integration & double integration ▶
Romberg = Richardson applied to the Trapezoidal rule. That is genuinely all it is.
Trapezoidal is cheap but rough. So compute it at h, then h/2, then h/4 — and combine those rough answers using the same 4ᵐ trick from Richardson's extrapolation. The combined answer is excellent.
This is the identical formula to Richardson's extrapolation — only the thing being refined changes (a derivative there, an integral here). Learn one formula, get two topics.
The Romberg table
| Step | m = 0 (Trapezoidal) | m = 1 | m = 2 | m = 3 |
|---|---|---|---|---|
| h | I(h) | |||
| h/2 | I(h/2) | I_T,1 | ||
| h/4 | I(h/4) | I_T,1 | I_T,2 | |
| h/8 | I(h/8) | I_T,1 | I_T,2 | I_T,3 |
← swipe the table sideways →
Fill the first column with Trapezoidal values at successively halved h, then work rightwards. The best answer is the bottom-right entry.
- The m = 1 column of a Romberg table is exactly Simpson's 1/3 rule. Neat conceptual MCQ.
- Denominators by column: 3, 15, 63, 255 (= 4ᵐ − 1).
Double integration
Now the area becomes a volume — you are integrating over a rectangle in x and y, not a line.
The method is exactly what you would guess: integrate in one direction, then integrate the results in the other direction. You apply the same Trapezoidal or Simpson rule twice.
The procedure (worth full marks even without finishing the arithmetic)
- Build a grid table of f(xᵢ, yⱼ).
- Integrate along x for each fixed y → one value per row.
- Integrate those results along y → final answer.
Worked example — structure only
Evaluate I = ∫₁² ∫₁² dx dy/(x+y) by the Trapezoidal rule with h = k = 0.25.
- Step 1 — Set up both grids
Both variables run over the same values here.
x and y = 1.00, 1.25, 1.50, 1.75, 2.00
5×5 grid, so N = M = 4 - Step 2 — Fill the grid with f(x,y) = 1/(x+y)
Just add the two values and take the reciprocal.
f(1,1) = 0.5 f(1,1.25) = 0.4444
f(1,1.5) = 0.4 f(1,1.75) = 0.3636
f(1,2) = 0.3333 … and so on - Step 3 — Integrate across each ROW (in x)
Apply the ordinary h/2 trapezoidal formula to each row of five numbers.
Gives one value per row → a column of 5 numbers - Step 4 — Integrate DOWN that column (in y)
Apply the k/2 trapezoidal formula to those five results.
That single number is the answer
VU almost never asks you to finish this by hand under time pressure — but the method statement is worth full marks. Learn the four steps and the corner/edge/interior weighting.
Practice questions
Romberg's integration is obtained by applying ___ to the Trapezoidal rule.
In Romberg's formula the denominator for the first extrapolation (m=1) is:
In double integration, the step sizes are given by:
Double integrals are evaluated by applying a one-dimensional rule:
12 Taylor's series method for ODEs ▶
From here on, every section solves the same kind of problem: you know the steepness rule and one starting point, and you must find where the curve goes.
Taylor's method says: build the answer from the starting value plus corrections using the 1st, 2nd, 3rd… derivatives. Very accurate — but you have to work out all those derivatives yourself, which is the painful part.
Getting the derivatives
You don't know y(t), so you get each derivative by differentiating the ODE itself:
Worked example
Solve dy/dt = t + y, y(1) = 0. Find y(1.2) with h = 0.1.
- Step 1 — Write down the first derivative and evaluate it
The ODE gives y′ directly. Put in t = 1, y = 0.
y′ = t + y → y′(1) = 1 + 0 = 1 - Step 2 — Differentiate the ODE to get y″
d/dt of (t + y) = 1 + y′. We already know y′(1) = 1.
y″ = 1 + y′ → y″(1) = 1 + 1 = 2 - Step 3 — Keep going for y‴ and y⁗
d/dt of (1 + y′) = y″, and it repeats from here.
y‴ = y″ = 2 y⁗ = y‴ = 2 - Step 4 — Work out (t − t₀)
We want t = 1.2, and we started at t₀ = 1.
t − t₀ = 1.2 − 1 = 0.2 - Step 5 — Substitute everything
Remember 2! = 2, 3! = 6, 4! = 24.
y(1.2) = 0 + (0.2)(1) + (0.2²/2)(2)
+ (0.2³/6)(2) + (0.2⁴/24)(2) - Step 6 — Compute each term
0.2² = 0.04, 0.2³ = 0.008, 0.2⁴ = 0.0016.
= 0.2 + 0.04 + 0.002667 + 0.000133 - Step 7 — Add up and check
The exact solution is y = 2e^(t−1) − t − 1, giving 0.24281. Excellent agreement.
y(1.2) ≈ 0.2428 (exact 0.24281 ✔)
Advantages and disadvantages
| Advantage | Disadvantage |
|---|---|
| Very accurate if enough terms are kept | Requires higher-order derivatives — laborious |
| Single-step, no starting values needed | Impractical when f(t,y) is complicated |
| Gives a function, not just points | New derivatives needed at every new point |
This is exactly why Runge–Kutta was invented: RK achieves the same accuracy as Taylor without ever computing a derivative of f. That sentence is a repeated True/False and short question.
Practice questions
The main drawback of Taylor's series method is that it requires:
Taylor's series method is a:
In Runge–Kutta methods we do not need to calculate higher-order derivatives and we find greater accuracy.
13 Euler's method & modified Euler ▶
Euler's method: you are standing on the curve and you know its steepness right now. So you just walk in a straight line in that direction for a small step h, and pretend you are still on the curve. Repeat.
It is the simplest possible idea, and also the least accurate — the curve bends away from your straight line.
Modified Euler fixes most of that: instead of using only the slope where you are, it averages the slope at the start with the slope where you land. Two slopes instead of one.
- yₙ
- the y value you already know
- yₙ₊₁
- the next y value — what you are finding
- h
- the step size
- f(tₙ, yₙ)
- the steepness at your current point — put your current t and y into the ODE
Stop iterating when two successive values agree to the accuracy you need.
The whole ODE chapter in one line: Euler uses one slope. Modified Euler uses the average of two. Runge–Kutta 4 uses a weighted average of four. That progression is all you need to remember about how they relate.
Worked example — Euler's method
Solve dy/dx = x + y, y(0) = 1. Find y(0.2) with h = 0.1.
- Step 1 — Find the slope at the starting point
Put x = 0 and y = 1 into the ODE.
f(0, 1) = 0 + 1 = 1 - Step 2 — Take one Euler step
Current y, plus h times the slope.
y₁ = 1 + (0.1)(1) = 1.1 (this is y at x = 0.1) - Step 3 — Find the slope at the NEW point
Now x = 0.1 and y = 1.1. Update both.
f(0.1, 1.1) = 0.1 + 1.1 = 1.2 - Step 4 — Take the second step
Same formula again, starting from 1.1.
y₂ = 1.1 + (0.1)(1.2) = 1.22 (y at x = 0.2)
Worked example — Modified Euler
Same problem, one step to x = 0.1.
- Step 1 — Predictor: do a plain Euler step first
This is only a rough guess. It gets improved next.
y⁽⁰⁾₁ = 1 + (0.1)(1) = 1.1 - Step 2 — Find the slope at that predicted point
Put x = 0.1 and the predicted y = 1.1 into the ODE.
f(0.1, 1.1) = 0.1 + 1.1 = 1.2 - Step 3 — Corrector: average the two slopes
Slope at the start was 1, slope at the end is 1.2. Average them, and note the h/2.
y⁽¹⁾₁ = 1 + (0.1/2)[1 + 1.2]
= 1 + 0.05(2.2) = 1.11 - Step 4 — Iterate once more using the improved y
Feed 1.11 back in and repeat the corrector.
f(0.1, 1.11) = 1.21
y⁽²⁾₁ = 1 + 0.05(1 + 1.21) = 1.1105 - Step 5 — Iterate again to confirm it has settled
The value has stopped changing, so we stop.
f(0.1, 1.1105) = 1.2105
y⁽³⁾₁ = 1 + 0.05(2.2105) = 1.11053 — converged - Step 6 — Compare with the exact answer
Exact: y = 2eˣ − x − 1 gives 1.11034. Modified Euler (1.1105) is far closer than plain Euler (1.1).
Euler 1.1 | Modified Euler 1.1105 | Exact 1.11034
- Using f(tₙ₊₁, yₙ) instead of f(tₙ₊₁, y⁽⁰⁾ₙ₊₁) in the corrector. You must use the predicted y.
- Forgetting the h/2 and writing h instead.
- Forgetting to update both t and y at each step — students often advance y but leave t behind.
- Working in degrees when the ODE contains trig functions. Always radians.
Practice questions
Euler's method is based on approximating the solution curve by its:
Euler's method has order of accuracy:
In the modified Euler method, the slope used is:
Solving y′ = x + y, y(0) = 1 by Euler's method with h = 0.1, y(0.1) is:
The solution of dy/dt = f(t,y) is represented as a curve in the:
14 Runge–Kutta methods ▶
Guaranteed exam content. The four k-formulas appear as MCQs almost every semester, and RK4 is the most common ODE numerical.
Euler used one slope and was inaccurate. Modified Euler used two and was better. Runge–Kutta 4 uses four slopes, sampled at cleverly chosen points inside the step, and takes a weighted average.
Think of it as: peek at the slope at the start, peek twice in the middle, peek at the end, then take a smart average of all four peeks. The middle ones count double because they represent the step best.
Its big advantage: it is as accurate as Taylor's method but you never have to differentiate anything.
Second-order Runge–Kutta
This is algebraically identical to the Modified Euler method.
Alternative RK2 form used in the handouts: yₙ₊₁ = yₙ + (1/3)(2k₁ + k₂), with k₁ = h·f(xₙ,yₙ) and k₂ = h·f(xₙ + (3/2)h, yₙ + (3/2)k₁).
🔴 Fourth-order Runge–Kutta (RK4) — memorise this exactly
| Item | Value |
|---|---|
| Function evaluations per step | 4 |
| Accuracy | O(h⁴) |
| Type | Single-step, self-starting |
| Weights | 1 : 2 : 2 : 1, divided by 6 |
Memory chant: “k₁ at the start, k₂ and k₃ at the middle, k₄ at the end — one, two, two, one, over six.”
The half-step pattern: k₂ and k₃ have halves in both slots (h/2 and k/2). k₄ has no halves at all — full h, full k₃.
Worked example — RK4
Solve dy/dx = x + y, y(0) = 1. Find y(0.1) with h = 0.1.
- Step 1 — Write down what you have
Never skip this. Half of RK4 errors come from losing track of the numbers.
f(x,y) = x + y h = 0.1 x₀ = 0 y₀ = 1 - Step 2 — Compute k₁ (slope at the start)
Straight substitution: x = 0, y = 1.
k₁ = 0.1·f(0, 1) = 0.1(0 + 1) = 0.1 - Step 3 — Compute k₂ (first middle slope)
Move x forward by h/2 = 0.05, and y forward by k₁/2 = 0.05.
k₂ = 0.1·f(0.05, 1.05)
= 0.1(0.05 + 1.05) = 0.1(1.1) = 0.11 - Step 4 — Compute k₃ (second middle slope)
Same x as k₂, but now y moves by k₂/2 = 0.055. This is where students mistakenly reuse k₁.
k₃ = 0.1·f(0.05, 1.055)
= 0.1(0.05 + 1.055) = 0.1(1.105) = 0.1105 - Step 5 — Compute k₄ (slope at the end)
Full step now — no halves. x + h, and y + k₃.
k₄ = 0.1·f(0.1, 1.1105)
= 0.1(0.1 + 1.1105) = 0.1(1.2105) = 0.12105 - Step 6 — Combine with weights 1, 2, 2, 1
Middle two are doubled. Then divide the whole bracket by 6.
y₁ = 1 + (1/6)[0.1 + 2(0.11) + 2(0.1105) + 0.12105]
= 1 + (1/6)[0.1 + 0.22 + 0.221 + 0.12105] - Step 7 — Finish
Add the bracket, divide by 6, add to y₀.
= 1 + (1/6)(0.66205) = 1 + 0.110342 = 1.110342 - Step 8 — Check against exact
Exact answer is 1.1103418. RK4 matched it to six decimal places in a single step. That is why it is the standard method.
RK4 = 1.110342 Exact = 1.1103418 ✔
RK for a system of equations
For dy/dt = f(t,y,z) and dz/dt = g(t,y,z), compute k's for y and l's for z simultaneously at each stage:
You must advance both variables together at every stage. Never finish all of y first and then go back and do z — the values feed into each other.
- Forgetting the h in kᵢ = h·f(…). The k values already include h — then dividing by h again is a classic double error.
- Using k₁ instead of k₂ inside k₃.
- Writing k₄ = h·f(tₙ + h/2, yₙ + k₃/2). k₄ uses full steps, no halves.
- Dividing by 4 instead of 6. (The weights sum to 1+2+2+1 = 6.)
Practice questions
In fourth-order Runge–Kutta method, k₁ is given by:
In fourth-order Runge–Kutta method, k₂ is given by:
In fourth-order Runge–Kutta method, k₄ is given by:
The fourth-order RK method requires how many function evaluations per step?
In RK4, the final combination is:
Runge–Kutta methods are ___ methods.
15 Predictor–corrector methods (Milne & Adams–Moulton) ▶
All the methods so far only looked at where you are now. Predictor–corrector methods look at the last four points as well — more information, better answer.
They work in two moves:
- Predict — make a first guess at the next value using the past points.
- Correct — feed that guess back in to get a much better value.
The catch: you need four points before you can even start. So you generate the first four with RK4, then switch over.
Single-step vs multi-step
| Single-step | Multi-step (P–C) | |
|---|---|---|
| To find yₙ₊₁ you need | only yₙ | yₙ, yₙ₋₁, yₙ₋₂, yₙ₋₃ |
| Self-starting? | Yes | No — needs 4 starting values |
| Examples | Taylor, Euler, Modified Euler, RK | Milne, Adams–Moulton |
← swipe the table sideways →
Repeated short question: “Why is a Runge–Kutta method used to start a predictor–corrector method?”
Answer: because Milne and Adams–Moulton need four past values, and RK4 is self-starting, so it can generate those four from the single initial condition.
The simplest P–C pair (Euler + Modified Euler)
⭐ Milne's predictor–corrector method
| Item | Detail |
|---|---|
| Derived from | Newton's FORWARD difference formula |
| Predictor obtained by | integrating from t₀ to t₄ (4 intervals) |
| Corrector obtained by | integrating from t₀ to t₂ (2 intervals) |
| Predictor truncation error | (28/90)·h·Δ⁴y′₀ |
| Corrector truncation error | −(1/90)·h·Δ⁴y′₀ |
| Starting values needed | 4 |
Recognise the corrector: (h/3)(y′ + 4y′ + y′) is exactly Simpson's 1/3 rule applied to the derivative. That is not a coincidence — it is Simpson's rule.
Coefficients: predictor 2, −1, 2 with 4h/3. Corrector 1, 4, 1 with h/3.
⭐ Adams–Moulton predictor–corrector method
| Item | Detail |
|---|---|
| Derived from | Newton's BACKWARD difference formula |
| Predictor coefficients | 55, −59, 37, −9 over 24 |
| Corrector coefficients | 9, 19, −5, 1 over 24 |
| Corrector truncation error | −(19/720)·h·∇⁴y′ₙ₊₁ |
| Predictor vs corrector error | predictor ≈ 13× larger, opposite sign |
| Starting values needed | 4 |
A checksum that saves you in the exam: the coefficients must add up to the denominator.
Predictor: 55 − 59 + 37 − 9 = 24 ✔
Corrector: 9 + 19 − 5 + 1 = 24 ✔
If your recalled numbers do not sum to 24, you have mis-remembered one. Use this before you write the final answer.
The single most repeated MCQ in this section:
“Adam–Moulton P–C method is derived by employing: Newton's backward difference interpolation formula” ✅
And by contrast, Milne's method uses Newton's FORWARD difference formula.
Learn them as a contrasting pair — VU tests exactly this confusion. Milne = forward, Adams = backward. (Or: they are alphabetically opposite to what you would guess, which is itself a memory hook.)
Worked example — Milne's predictor
Given the table, predict y at x = 0.4 with h = 0.1:
| x | 0 | 0.1 | 0.2 | 0.3 | 0.4 |
|---|---|---|---|---|---|
| y | 1 | 1.1103 | 1.2428 | 1.3997 | 1.5836 |
← swipe the table sideways →
- Step 1 — Identify which point is which
We want y₄, so n+1 = 4, meaning n = 3.
yₙ₋₃ = y₀ = 1
The derivatives we need are at x = 0.1, 0.2, 0.3 - Step 2 — Compute the three derivative values
Put each (x, y) pair into the ODE you were given. These are y′₁, y′₂, y′₃.
y′₁ = f(0.1, 1.1103)
y′₂ = f(0.2, 1.2428)
y′₃ = f(0.3, 1.3997) - Step 3 — Write the predictor formula BEFORE substituting
Marks are given for the formula itself, so always write it out.
y₄ = y₀ + (4h/3)·( 2y′₁ − y′₂ + 2y′₃ ) - Step 4 — Substitute and compute
4h/3 = 4(0.1)/3 = 0.13333. Note the middle term is subtracted.
y₄ = 1 + 0.13333·( 2y′₁ − y′₂ + 2y′₃ )
Method note: the exam almost always gives you the ODE alongside the table. Compute the three derivative values first and list them clearly, then substitute. Marks are for the substitution, so show the formula before plugging numbers in.
- Using yₙ₋₁ instead of yₙ₋₃ in Milne's predictor.
- Mixing up the base points: Milne's predictor starts from yₙ₋₃, its corrector from yₙ₋₁. Adams–Moulton starts from yₙ for BOTH.
- Using 4h/3 for the corrector — it is h/3.
- Forgetting that y′ means f(x, y) evaluated at that point, not a difference.
- Attempting Milne or Adams with fewer than 4 starting values.
Practice questions
Adam–Moulton P–C method is derived by employing:
Milne's method is derived by employing:
Milne's predictor formula is:
In the Adams–Moulton predictor, the coefficients are:
Predictor–corrector methods are:
The truncation error in Adams' predictor compared with the corrector is approximately:
Milne's corrector formula is essentially:
16 Carry-over topics: root-finding, matrices, MAPLE ▶
These topics come from the first half of the course, but Lectures 43–45 revisit them — and they leak heavily into the final paper's True/False block. Spend 10 minutes here; it is cheap marks.
Root-finding methods (Lectures 44–45)
| Method | Formula | Starting points | Type |
|---|---|---|---|
| Bisection | x₂ = (x₀+x₁)/2 | 2 (with f(x₀)·f(x₁) < 0) | Bracketing |
| Regula-Falsi | xₙ₊₁ = xₙ − f(xₙ)(xₙ−xₙ₋₁)/(f(xₙ)−f(xₙ₋₁)) | 2 (opposite signs) | Bracketing |
| Iteration | xₙ₊₁ = φ(xₙ) | 1 | Open |
| Newton–Raphson | xₙ₊₁ = xₙ − f(xₙ)/f′(xₙ) | 1 | Open |
| Secant | xₙ₊₁ = [xₙ₋₁f(xₙ) − xₙf(xₙ₋₁)]/[f(xₙ)−f(xₙ₋₁)] | 2 | Open |
| Muller's | fits a quadratic through 3 points | 3 🔴 | Open |
| Graeffe's | repeated squaring of the polynomial | 0 | Direct |
← swipe the table sideways →
High-yield one-liners
- Muller's method requires 3 starting points. Appears verbatim in past finals.
- Newton–Raphson converges quadratically (order 2). Condition: |f(x)·f″(x)| < |f′(x)|².
- Secant order of convergence = (1+√5)/2 ≈ 1.618 — faster than linear, slower than Newton.
- Newton–Raphson is also called the method of tangents.
- Direct methods (no initial guess needed): Graeffe's root squaring, Gauss elimination, Gauss–Jordan.
- Iterative methods (need an initial guess): Bisection, Newton–Raphson, Secant, Jacobi, Gauss–Seidel.
- Newton–Raphson fails when f′(x) = 0 near the root — use Regula-Falsi instead.
- All trigonometric calculations in RADIANS.
Matrix & eigenvalue True/False bank
These dominate the objective section. Every one below is TRUE as stated:
| Statement | Answer |
|---|---|
| The determinant of a diagonal matrix is the product of the diagonal elements | TRUE |
| Power method is applicable if the eigenvectors are linearly independent | TRUE |
| A 3×3 identity matrix has three and same eigenvalues | TRUE (all = 1) |
| If A and B are similar, they have the same eigenvalues | TRUE |
| Jacobi's method works on a matrix with no zeros along its main diagonal | TRUE |
| An eigenvector is normalized if its largest-magnitude coordinate = unity | TRUE |
| Eigenvalues of a symmetric matrix are all real | TRUE |
| Jacobi iteration converges if A is strictly diagonally dominant | TRUE |
| Gauss–Seidel applies to diagonally dominant or symmetric positive definite A | TRUE |
| The Power method finds only the eigenvalue largest in absolute value | TRUE |
| Iterative algorithms can be more rapid than direct methods | TRUE |
| If λ is an eigenvalue of A with eigenvector V, then λ − a is an eigenvalue of A − aI | TRUE |
| To find the smallest eigenvalue, apply the Power method to A⁻¹ | TRUE |
| A matrix is orthogonal if AᵀA = I, i.e. Aᵀ = A⁻¹ | TRUE |
| Octal number system has base 8 | TRUE |
← swipe the table sideways →
Watch the sign-flip trap: VU also prints the false version of some of these — e.g. “If A and B are similar, they have the different eigenvalues” → that one is FALSE. Read the statement carefully; do not pattern-match on the topic alone.
MAPLE (Lecture 43)
Barely tested. Worth knowing only:
- MAPLE is a computer algebra system that does both symbolic and numerical mathematics.
- Commands end with a semicolon (;) to display output, or a colon (:) to suppress it.
- It can solve equations, differentiate, integrate and plot.
Practice questions
Muller's method requires how many starting points?
Eigenvalues of a symmetric matrix are all:
The Power method can be used to find which eigenvalue?
Newton–Raphson method is also known as the:
17 Master formula sheet ▶
Every formula in the syllabus, on one page. If you only have twenty minutes left, read this and Section 18.
Operators
Interpolation
Differentiation
Integration
ODEs
Conditions & restrictions
| Rule | Condition |
|---|---|
| Δ, ∇, δ formulas | Equally spaced data required |
| Lagrange & Divided Differences | Work for unequally spaced data |
| Simpson's 1/3 | n even |
| Simpson's 3/8 | n multiple of 3 |
| Trapezoidal | any n |
| Milne / Adams–Moulton | 4 starting values (use RK4) |
| Jacobi / Gauss–Seidel | converge if strictly diagonally dominant |
| Power method | eigenvectors must be linearly independent |
| All trigonometry | Radians |
← swipe the table sideways →
Method summaries in one line each
| Method | One-line description |
|---|---|
| Newton Forward | Interpolate near start; Δ and top diagonal |
| Newton Backward | Interpolate near end; ∇ and bottom row |
| Lagrange | Direct interpolation, any spacing, no table needed |
| Divided Difference | Newton-style interpolation for unequal spacing |
| Richardson | Combine h and h/2 results to cancel leading error |
| Trapezoidal | Integrate by fitting straight lines |
| Simpson 1/3 | Integrate by fitting parabolas (even n) |
| Simpson 3/8 | Integrate by fitting cubics (n divisible by 3) |
| Romberg | Richardson applied to Trapezoidal |
| Taylor Series | Expand solution; needs higher derivatives |
| Euler | One tangent step; least accurate |
| Modified Euler | Average of two slopes; predictor–corrector |
| RK4 | Weighted average of four slopes; most used |
| Milne | Multi-step P–C from Newton forward |
| Adams–Moulton | Multi-step P–C from Newton backward |
← swipe the table sideways →
Diagrams you should be able to recognise
| Diagram | What it shows | How it's tested |
|---|---|---|
| Difference table | Triangular array of differences | “Construct the table” — very common |
| Euler tangent line | Curve approximated by tangent at (t₀,y₀) | “Euler is based on the ___” → tangent |
| Modified Euler lines | L₁ slope at t₀, L₂ slope at t₁, L average | “Uses the average of two slopes” |
| Trapezoid vs rectangle | Less “missing” area under the curve | “…because we used ___” → trapezoids |
| Simpson's parabola | Quadratic fitted over a pair of strips | “f(x) is of the form ax²+bx+c” |
| Pₙ(x) vs y(x) slopes | Curves meet at nodes, slopes differ | “Derivatives may differ” → TRUE |
| Romberg table | Triangular array of refined integrals | “Best value is bottom-right” |
← swipe the table sideways →
18 Last-minute sheet — read this in the final 15 minutes ▶
Read only this in the last 15 minutes before you walk in. Nothing else.
Exam tactics
- Do the MCQs first. They are the bulk of the marks and the fastest.
- On any integration question, count the intervals before choosing a rule. n = points − 1.
- On any interpolation question, check spacing first: equal → Newton; unequal → Lagrange or divided difference.
- Write the formula before substituting. Method marks are awarded even if the arithmetic slips.
- In RK4, compute k₁ → k₄ in strict order and list them. Never try to do it in one line.
- If you cannot finish a numerical, still write the formula and the substitution — that is most of the marks.
- Keep 4 decimal places throughout; round only at the very end.
You have covered the whole syllabus. Walk in knowing the operator identities, the interval rules, and the four k's — those three things alone carry a large share of the paper. Good luck.