Recipe

Build a Taylor polynomial centered at a

Use when: a degree-n polynomial approximation to f(x) is needed, matching f's value and its first n derivatives at one center point x=a

Goal: the degree-n Taylor polynomial T_n(x), built entirely from derivatives evaluated at the center

Taught in Taylor polynomials · a later section extends this recipe with more cases

1. 1

   Differentiate f repeatedly, up to the degree n wanted, evaluating each derivative at the center x=a as you go.

   Example

   f(x)=\sqrt x, centered at a=4, degree n=2: f(4)=2; f'(x)=\tfrac12x^{-1/2}\Rightarrow f'(4)=\tfrac14; f''(x)=-\tfrac14x^{-3/2}\Rightarrow f''(4)=-\tfrac1{32}.

   Why: every coefficient T_n will use comes from information available at the single point a; nothing else about f enters the formula.

2. 2

   Form the coefficient of (x-a)^k as \dfrac{f^{(k)}(a)}{k!}, for each k from 0 up to n.

   Example

   Coefficient of (x-4)^0: \dfrac{f(4)}{0!}=2. Coefficient of (x-4)^1: \dfrac{f'(4)}{1!}=\dfrac14. Coefficient of (x-4)^2: \dfrac{f''(4)}{2!}=\dfrac{-1/32}{2}=-\dfrac1{64}.

   Why: dividing by k! is exactly what makes T_n^{(k)}(a)=f^{(k)}(a) hold for every derivative order up to n, the defining property of the Taylor polynomial.

3. 3

   Write T_n(x)=\displaystyle\sum_{k=0}^n\frac{f^{(k)}(a)}{k!}(x-a)^k, using the coefficients from Step 2, left in terms of (x-a) rather than expanded out in x.

   Example

   T_2(x) = 2+\frac14(x-4)-\frac1{64}(x-4)^2

   Why: leaving the polynomial centered at a, instead of expanding it into powers of x, keeps every coefficient's origin visible and avoids needless algebra.

4. 4

   Confirm T_n(a)=f(a), and that differentiating T_n the right number of times at a reproduces each f^{(k)}(a) used to build it.

   Example

   T_2(4)=2+0-0=2=f(4). T_2'(x)=\tfrac14-\tfrac{2}{64}(x-4), so T_2'(4)=\tfrac14=f'(4).

   ⚡ Shortcut

   Some functions have derivatives that cycle at the center, like \sin x's four-term pattern \sin,\cos,-\sin,-\cos. Write that pattern once and read off every coefficient from it, instead of differentiating from scratch at every order.

   Why: this confirms the construction actually satisfies its own defining property, catching an arithmetic slip in a derivative or a factorial before it's reported as the final answer.