Taylor series approximations

The dashed curve is $f(x)=x^3$. The solid line is its tangent at $x_0$ — drag the slider and watch the line pivot, matching the curve closely right at $x_0$ but drifting away fast on either side.

A Taylor series keeps adding derivative terms at $x_0$ to build a better local approximation:

• $x_0$ — the point the approximation is built around (the point of expansion).
• $f(x_0)$ — the function's actual value at $x_0$; the zeroth-order term.
• $f'(x_0)(x-x_0)$ — the first-order (linear) term: the tangent line's contribution.
• $\dfrac{f''(x_0)}{2}(x-x_0)^2$ — the second-order (quadratic) term: adds curvature.

1. Truncating after one term is the tangent line

   Stopping after the linear term recovers exactly the tangent-line approximation from Chapter 1 — same formula, new name.

2. Adding the quadratic term adds curvature

   The quadratic term is the first one that can bend — it uses $f''(x_0)$ to curve the approximation the same way the true function curves near $x_0$.

3. Every term after that keeps closing the gap

   Each additional term corrects more of what the previous ones missed; for a well-behaved function, the approximation converges to $f(x)$ itself as more terms are added.

4. It only works locally

   All of this accuracy is local — no matter how many terms you add, expect the approximation to get worse the farther $x$ strays from $x_0$.

Move $x_0$ and the evaluation point independently. Near $x_0$, both approximations track the curve almost perfectly; the quadratic one keeps tracking noticeably farther out, because it bends with the curve instead of running straight through it.

Expand $f(x)=x^3$ around $x_0=1$ and evaluate at $x=1.5$:

1. Compute the ingredients at x0=1

   $$f(1)=1,\qquad f'(1)=3(1)^2=3,\qquad f''(1)=6(1)=6$$
2. Build the linear approximation

   $$L(1.5) = 1 + 3(0.5) = 2.5$$
3. Build the quadratic approximation

   $$Q(1.5) = 1 + 3(0.5) + \frac{6}{2}(0.5)^2 = 1 + 1.5 + 0.75 = 3.25$$
4. Compare against the true value

   $f(1.5)=3.375$.

   • The linear approximation misses by $0.875$.
   • The quadratic one misses by only $0.125$.

   Seven times closer, from one extra term.