Geometric Bites

A Gentle Introduction to Function Notation Understanding f : A → B — the language of modern mathematics. One of the most powerful ideas in mathematics is the concept of a function . We usually meet it in the form f(x) = 2x + 1 , but the structure behind this idea is far richer. The notation f : A → B captures the entire architecture of a function in a single line. In this article, we unpack this notation and explain exactly what it means, why it matters, and how it connects to the familiar expression f(x) = y . 1. What does f : A → B mean? When we write f : A → B we are saying: f is a function, A is the domain — the set of inputs the function accepts, B is the codomain — the set in which all outputs must lie. In words: A function assigns to every element of the domain A exactly one output in the codomain B . Two rules always hold for genuine functions: Every input must have an output. No input may have more than one output. D...

Arithmetic and Geometric Sequences Arithmetic and geometric sequences are two fundamental types of numerical progressions. They describe how quantities grow or shrink by addition or by multiplication, and they form the foundation for topics such as series, summation formulas, and exponential growth. 1. Arithmetic Sequence An arithmetic sequence is a list of numbers in which each term differs from the previous one by a fixed amount called the common difference d . a, a + d, a + 2d, a + 3d, … , a + (n − 1)d a – first term d – common difference The n th term, denoted T n , is given by: T n = a + (n − 1)d Each new term is obtained by adding d to the previous term. The difference between consecutive terms remains constant: T k+1 − T k = d Example: If a = 4 and d = 3, the sequence is 4, 7, 10, 13, 16, … The 20th term is T 20 = 4 + (20 − 1)×3 = 61. 2. Geometric Sequence A geometric sequence is a list of numbers where each term is fo...

What Is an Isomorphism? In mathematics, an isomorphism is a function that shows two mathematical objects have the same structure. Although the objects may look different, an isomorphism demonstrates that they behave in exactly the same way with respect to the operations that define them. If such a map exists, the objects are called isomorphic . An isomorphism tells us that two systems are essentially the same, differing only by a relabelling of their elements. The Basic Idea An isomorphism is a function: f : A → B that must be: Injective — different elements of A map to different elements of B. Surjective — every element of B comes from some element of A. Together these mean f is bijective , and so it has an inverse: f⁻¹ : B → A No information is lost moving from A to B or back. Preserving Structure Bijectivity alone is not enough. An isomorphism must also preserve structure. For groups, this means: f(a ⋆ b) = f(a) ∘ f(b) for all a, b ∈ A ,...

Proof by Induction: The Sum of Squares Formula Theorem. For any integer n ≥ 1, ∑ i=1 n i² = (1/6)·n·(n+1)·(2n+1) Proof (by mathematical induction) 1. Basis Step For n = 1: LHS = 1² = 1 RHS = (1/6)·1·(1+1)·(2·1+1) = (1/6)·1·2·3 = 1 Therefore, the formula holds for n = 1. 2. Inductive Hypothesis Assume the statement is true for some integer k ≥ 1: ∑(i=1→k) i² = (1/6)·k·(k+1)·(2k+1) 3. Inductive Step We must show it is true for n = k + 1: ∑(i=1→k+1) i² = [∑(i=1→k) i²] + (k+1)² = (1/6)·k·(k+1)·(2k+1) + (k+1)² = (1/6)(k+1)[k(2k+1) + 6(k+1)] = (1/6)(k+1)[2k² + 7k + 6] = (1/6)(k+1)(k+2)(2k+3) This matches the same form with k replaced by k+1: (1/6)·(k+1)·((k+1)+1)·(2(k+1)+1) Hence, the formula holds for n = k + 1. 4. Conclusion Since the statement is true for n = 1 (basis step), and true for n = k + 1 whenever it is true for n = k (inductive step), it follows by the principle of mathematical induction that ∑(i=1→n) i² = (1/6)·n·(n+1)·(2n+1) for al...

The Trapezoidal Rule — A Visual, First-Principles Introduction 11 November 2025 · @mathematics.proofs To understand integration deeply, it helps to think geometrically. Instead of memorising formulas, we begin by observing shapes. The goal is simple: break the interval into small pieces, estimate the area on each piece, and add everything together. Rectangles give a basic approximation. But functions rarely behave perfectly flat on every interval. A better idea is to allow the top edge to tilt. This leads us naturally to trapezoids . 1) Partitioning the Interval Consider an interval from a to b . We divide it into n equal parts. Δx = (b − a) / n x i = a + i·Δx for i = 0, 1, 2, …, n At each x i , we record the height of the function f(x i ). These sample values will guide our area estimates. 2) One Slice: Rectangle + Triangle Focus on a single subinterval [x i , x i+1 ]. If we draw a vertical line at x i and take a height of f(x i ), we obtain a rectangle o...

The Method of Differences — A Clean Proof of the Sum of Cubes The method of differences is a remarkably elegant tool for evaluating finite sums. When each term of a series can be written in the form f(r+1) − f(r) , the sum “collapses” — all interior terms cancel, leaving only a boundary expression. This behaviour is called a telescoping sum . 1) Telescoping Sums Assume the general term u r can be written as: u r = f(r+1) − f(r). Then the finite sum from r = 1 to r = n becomes: Σ u r = Σ ( f(r+1) − f(r) ). To see what happens, write out a few terms: u₁ = f(2) − f(1) u₂ = f(3) − f(2) u₃ = f(4) − f(3) ⋮ uₙ = f(n+1) − f(n) When these are added, everything cancels except the first and last pieces: Σ u r = f(n+1) − f(1). This is the essence of the method: interior structure disappears, leaving just the difference between the final and initial states. 2) A Classic Application — The Sum of Cubes We will use this technique to prove the well-known ...

The Maclaurin Series — A Clean Derivation Many smooth functions can be written as an infinite polynomial. When this expansion is centred at x = 0 , we obtain the Maclaurin series . This article derives the Maclaurin formula directly from repeated differentiation, showing precisely why the coefficients involve derivatives and factorials. 1) Begin with a General Power Series Suppose a function f(x) can be expressed as f(x) = a₀ + a₁x + a₂x² + a₃x³ + … + aᵣxʳ + … The constants aᵣ are real coefficients whose values we wish to determine. 2) Evaluate at x = 0 f(0) = a₀ so a₀ = f(0). 3) Differentiate Once f′(x) = a₁ + 2a₂x + 3a₃x² + … + r·aᵣxʳ⁻¹ + … Setting x = 0 eliminates all higher powers: f′(0) = a₁. Thus, a₁ = f′(0). 4) Differentiate Again f″(x) = 2·1·a₂ + 3·2·a₃x + … + r(r−1)aᵣxʳ⁻² + … Evaluate at x = 0 : f″(0) = 2! · a₂ Hence a₂ = f″(0) / 2!. 5) The General Pattern Differentiate repeatedly. After r differentiations, a...