Taylor polynomials: matching derivatives at a center

Understand the idea

A Taylor polynomial approximates a function near a chosen center by matching its derivatives there. The coefficients include factorial denominators, which ensure that differentiating the polynomial the required number of times reproduces the corresponding derivative value.

Evaluate the function and its derivatives at the center before assembling the polynomial. Powers are built from $\displaystyle x-a$ , not necessarily from $\displaystyle x$ . A Maclaurin polynomial is the special case with center zero. The degree refers to the highest power retained, including any zero coefficients.

A polynomial approximation is usually most useful near its center, but the formula alone does not quantify accuracy. A remainder estimate or another error argument is needed to justify a requested tolerance. Having derivatives at the center does not automatically establish that an infinite Taylor series equals the function everywhere.

T_n(x)=\sum_{k=0}^{n}\frac{f^{(k)}(a)}{k!}(x-a)^k

A method you can apply

Identify the center and the requested degree.
Compute derivative values through that degree and divide by the appropriate factorials.
Write powers of $\displaystyle x-a$ , evaluate if requested, and keep the approximation distinct from equality.

Worked example

Build the degree-three Maclaurin polynomial for $\displaystyle e^x$ and use it to approximate $\displaystyle e^{0.2}$ .

Every derivative of $\displaystyle e^x$ equals $\displaystyle e^x$ , so all derivative values at zero are $\displaystyle 1$ .
The polynomial is $\displaystyle T_3(x)=1+x+x^2/2+x^3/6$ .
At $\displaystyle x=0.2$ , this gives $\displaystyle 1+0.2+0.02+0.008/6$ .

The approximation is

\displaystyle e^{0.2}\approx1.221333333

. An error estimate is a separate step from constructing the polynomial.

Try it yourself

Find the degree-two Taylor polynomial for $\displaystyle \ln x$ centered at $\displaystyle 1$ , then estimate $\displaystyle \ln(1.1)$ .

Show solution and reasoning

The derivative values are $\displaystyle f(1)=0$ , $\displaystyle f'(1)=1$ , and $\displaystyle f''(1)=-1$ . Thus $\displaystyle T_2(x)=(x-1)-(x-1)^2/2$ , giving $\displaystyle \ln(1.1)\approx0.095$ .