Geometric Bites

Two of the most important results in differential calculus are d/dx(ln x) = 1/x d/dx(aˣ) = aˣ ln a. These formulas are closely connected. One describes the derivative of the natural logarithm, while the other gives the derivative of an exponential function with positive base. Together, they reveal the deep relationship between logarithms, exponentials, inverse functions, and differentiation. 1. Derivative of the natural logarithm Begin with ln x = y. This is equivalent to eʸ = x. Differentiate both sides with respect to y: dx/dy = eʸ. Now invert this result: dy/dx = 1/eʸ. Since eʸ = x, substitute back: dy/dx = 1/x. Therefore, d/dx(ln x) = 1/x, for x > 0. So the derivative of the natural logarithm is the reciprocal of x. 2. Derivative of the exponential function aˣ Now let y = aˣ, where a > 0 and a ≠ 1. Take logarithms to base a: logₐ y = x. Now apply the change-of-...

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 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...