Geometric Bites

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...

Essential Elements of Algebra Problem Solving Solving equations in algebra is about preserving truth while working step by step toward the unknown. Although equations can look complicated, most of the time we are simply applying a small set of rules with care. If these rules are followed, the solution you reach will be valid; if they are broken, the result becomes unreliable. This post introduces three essential principles that underpin almost all algebraic manipulation. Follow them consistently, and you will have a strong foundation for solving equations with confidence. 1) Start with a True Statement Everything in algebra begins with a statement that is already true. If the starting point is false, no amount of manipulation can produce a trustworthy conclusion. When you write an equation such as: 2x + 3 = 13 you are asserting that doubling a number and then adding three gives thirteen. This is your initial truth. All further steps must preserve this truth. If y...

Continuity and Differentiability (Clear, Compact Guide) We study functions of a single real parameter u . A scalar function is written as φ(u) , and a vector function as R(u) = (x(u), y(u), z(u)) . All the ideas below are based on ordinary one-variable limits. 1) What it means for a function to be continuous at u (scalar case) A function φ is said to be continuous at u if its value changes smoothly as u changes slightly. Formally, for every ε > 0 , there exists a δ > 0 such that |φ(u + Δu) − φ(u)| < ε whenever |Δu| < δ . 2) Continuity of a vector function Let R(u) = (x(u), y(u), z(u)) . The function R is continuous at u if each component x(u) , y(u) , and z(u) is continuous at that same point. Equivalently, using any fixed norm |·| on ℝ³ : for every ε > 0 , there exists a δ > 0 such that |R(u + Δu) − R(u)| < ε whenever |Δu| < δ . 3) Differentiability (first order) A scalar or vector function is differentiable at u if the limit (F(u + Δ...

How to Derive the Derivative of a Vector Function 🧮 Let’s start with a vector function of a single variable: R(u) = x(u)î + y(u)ĵ + z(u)k̂ Here, x , y , and z are differentiable scalar functions of a real parameter u . This means that as u changes, the point R(u) moves through space — tracing a smooth curve. ✨ The Goal We want to find dR/du — the rate at which the vector R(u) changes with respect to u . Proof By definition of the derivative: dR/du = lim Δu→0 [R(u+Δu) − R(u)] / Δu = lim Δu→0 [x(u+Δu)î + y(u+Δu)ĵ + z(u+Δu)k̂ − (x(u)î + y(u)ĵ + z(u)k̂)] / Δu = lim Δu→0 [(x(u+Δu)−x(u))/Δu]î + [(y(u+Δu)−y(u))/Δu]ĵ + [(z(u+Δu)−z(u))/Δu]k̂ = (dx/du)î + (dy/du)ĵ + (dz/du)k̂ Interpretation The derivative dR/du is itself a vector — one that points in the direction of motion of R(u) and whose magnitude gives the speed of change. Each component ( x , y , z ) behaves just like an ordinary function, so we can differentiate them individually and recombine t...