Taylor Approximation — Curves as Polynomials
“Taylor approximation swaps a curve out for a polynomial.”
The formula
f(x) ≈ f(a) + f'(a)(x−a) + ½f''(a)(x−a)²How to read it: mimic a curve near one point with a polynomial by matching its value, slope, and bend in turn
- f(a)
- — the value at that point — 0th order: match the height
- f'(a)(x−a)
- — match the slope — 1st order: the tangent line
- ½f''(a)(x−a)²
- — match the bend — 2nd order: the curvature
The hook
How does a computer compute sin or eˣ? The answer is surprising — it swaps the curve for a polynomial.
In plain words
A complicated function approximated, near one point, by a simple polynomial tuned to share the same value, slope, and bend there.
The intuition
Draw a tangent at a point (1st order) and it roughly mimics the neighborhood. Add the curve's 'bend' (2nd order) and it hugs tighter. Each extra term wraps the polynomial more snugly around the real curve, over a wider stretch.
How it's built
The 0th order is the height f(a); the 1st-order term f'(a)(x−a) builds the tangent; the 2nd-order term ½f''(a)(x−a)² matches the concavity. Each term makes one more derivative agree at that point.
Example
For eˣ near a=0: f(0)=1, f'(0)=1, f''(0)=1 → 1+x+x²/2. At x=0.1 that's 1+0.1+0.005 = 1.105, essentially the true e^0.1 ≈ 1.10517.
Common mistake
The approximation is only good near the base point a. It drifts off as you move away, and adding terms raises accuracy but doesn't make it valid everywhere (radius of convergence).
Where it's used
How calculators and computers evaluate sin, eˣ, ln; the small-angle approximation sinθ≈θ in physics; linearization in engineering — practical computation everywhere.
Where it came from
Brook Taylor formalized it in the 1700s; the special case based at a=0 is called the Maclaurin series.
Prerequisites
Quick check
Up to its 1st-order (linear) term, a Taylor approximation is the same as what?
- the tangent line✓
- the area
- the limit
- the second derivative
Practice
Using the 1st-order (tangent) approximation 1+x for eˣ, what is the estimate at x=0.2?
Answer: 1.2
- Approximation: 1+x
- Plug in x=0.2 → 1.2 (true e^0.2 ≈ 1.221)
Takeaway: The 1st-order approximation is the tangent — quite accurate up close.
Using eˣ ≈ 1+x+x²/2 (based at a=0), what is the estimate at x=1?
Answer: 2.5
- 1 + 1 + 1²/2 = 1 + 1 + 0.5
- = 2.5 (true e ≈ 2.718 — the error is large this far out)
Takeaway: The farther from the base point, the more terms you need.
Explain the small-angle approximation sinθ≈θ using Taylor. What is the 1st-order approximation of sin at θ=0?
Answer: undefined
- sin 0 = 0, (sin)'=cos, cos 0 = 1
- 1st-order: 0 + 1·(θ−0) = θ
- so for small θ, sinθ ≈ θ
Takeaway: Physics' small-angle approximation is really just the 1st Taylor term.