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Approximate calculations and modelling

Lessons 10–11 · 2 lessons · M. A. Mirzaahmedov, Sh. N. Ismailov, A. Q. Amanov (algebra and analysis), B. Q. Haydarov (geometry). Mathematics Grade 11, Parts 1 and 2, 1st edition. ZAMIN NASHR, Tashkent, 2018
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Approximate calculations

Textbook, Part 1: pp. 56–61
GoalUse the small-increment formula f(x₀ + Δx) ≈ f(x₀) + f′(x₀) · Δx to compute function values, roots, powers and trigonometric expressions approximately.
New words
small-increment formula · kichik orttirmalar formulasiapproximate value · taqribiy qiymatlinear approximation · chiziqlilashtirisherror · xatolik
Explanation

Near x₀ the graph can be replaced by its tangent, so for small Δx we have f(x₀ + Δx) ≈ f(x₀) + f′(x₀) · Δx. To use it, choose x₀ so that f(x₀) and f′(x₀) are easy (perfect squares, cubes, 0, 1, 30°, 45°, …) and Δx small: the smaller Δx is, the better the approximation. At x₀ = 0: sin x ≈ x, tan x ≈ x (x in radians), eˣ ≈ 1 + x, ln(1 + x) ≈ x, (1 + x)ᵐ ≈ 1 + mx, √(1 + x) ≈ 1 + x/2. Do not forget to convert angles to radians: 1° = π/180 ≈ 0.01745 rad. Write results with ≈ to show they are approximate; the method gives a value to several decimal places even without a calculator.

Worked examples
Compute 2.97³: f(x) = x³, x₀ = 3, Δx = −0.03, f(3) = 27, f′(3) = 27. f(2.97) ≈ 27 + 27 · (−0.03) = 26.19. (The exact value is 26.198…, the difference is below 0.01.) √16.08: x₀ = 16, Δx = 0.08, f′(16) = 1/8; √16.08 ≈ 4 + 0.08/8 = 4.01.
1.004²⁵ ≈ 1 + 25 · 0.004 = 1.1. tan 46°: x₀ = 45° = π/4, f(π/4) = 1, f′ = 1/cos²x = 2, Δx = π/180 ≈ 0.01745; tan 46° ≈ 1 + 2 · 0.01745 ≈ 1.035. Also ln 1.05 ≈ 0.05 (exact 0.0488…).
Class activity

“No calculator”: the class splits into two groups. One computes 5 expressions given by the teacher (say √9.06, 0.98⁴⁰, sin 0.05) with the small-increment formula, the other with a calculator. Compare the results and discuss when the error is larger.

Practice
1
Approximate √25.2 with the small-increment formula (x₀ = 25).
2
Approximate 1.003⁴⁰ with (1 + x)ᵐ ≈ 1 + mx.
3
Approximate the value of f(x) = x² + 3x at x = 4.02.
4
Why is x₀ = 8 convenient for computing ∛8.3 while x₀ = 1 is not?