Taylor Series & Taylor Polynomials 

What Is a Taylor Polynomial?

A Taylor polynomial is a polynomial that approximates a function near a point.
It is constructed using the function’s derivatives.

For a function f(x) that is infinitely differentiable at x=a:

Taylor Polynomial of degree n:

$$P_{n}(x)=f(a)+f^\prime(a)(x-a)+\frac{f^{\prime\prime}(a)}{2!}(x-a)^{2}+\ldots+\frac{f^{(n)}(a)}{n!}(x-a)^{n}$$

This is called the Taylor polynomial centered at a.

Maclaurin Polynomial (Special Case)

If the center a=0, then the Taylor polynomial becomes a Maclaurin polynomial:

$$P_{n}(x)=f(0)+f^\prime(0)(x)+\frac{f^{\prime\prime}(0)}{2!}(x)^{2}+\ldots+\frac{f^{(n)}(0)}{n!}(x)^{n}$$

Why Are Taylor Polynomials Useful?

Taylor polynomials:

• Approximate complicated functions

• Help compute values (e.g., $$e^{0.1} , \sin(0.2)$$)

• Lead to Taylor series

• Are used in physics, engineering, numerical methods

They are the “polynomial version” of the function near a point.

Taylor Series (Infinite Taylor Polynomial)

A Taylor series is an infinite Taylor polynomial:

$$\sum^{\infty}_{n=0}\frac{f^{(n)}(a)}{n!}(x-a)^{n}$$

When centered at 0:

$$\sum^{\infty}_{n=0}\frac{f^{(n)}(0)}{n!}(x)^{n}$$

This is the Maclaurin series.

Famous Maclaurin Series You Must Memorize

These appear on many AP BC exam questions.

Geometric Series

$$\frac{1}{1-x} = \sum^{\infty}_{n=0}x^{n} , |x| < 1$$

Natural Log

$$\ln(1+x) = \sum^{\infty}_{n=1)(-1)^{n-1}\frac{x^{n}}{n} , |x| < 1 , x> -1$$

Exponential Function

$$e^{x} = \sum^{\infty}_{n=0}\frac{x^{n}}{n!} , all x$$

Sine

$$\sin(x) = \sum^{\infty}_{n=0}(-1)^{n}\frac{x^{2n+1}}{(2n+1)!}$$​

Cosine

$$\cos(x) = \sum^{\infty}_{n=0}(-1)^{n}(-1)^{n}\frac{x^{2n}}{(2n)!}$$​​

Arctan

$$\tan^{-1}(x) = \sum^{\infty}_{n=0}(-1)^{n}\frac{x^{2n+1}}{2n+1} , |x| \leq 1$$​

How to Build a Taylor Polynomial Step-by-Step

Example: Find the 4th-degree Taylor polynomial for $$\cos(x)$$ at a=0.

1. Compute derivatives:

$$f(x)=\cos(x)$$​

$$f^\prime(x)=-\sin(x)$$

$$f^{\prime\prime}(x)=-\cos(x)$$

$$f^{(3)}(x)=\sin(x)$$

$$f^{(4)}(x)=\cos(x)$$

2. Evaluate at x=0:

$$\cos(0)=1 , -\sin(0)=0 , -\cos(0)=-1 , \sin(0)=0 , \cos(0)=1$$

3. Insert into formula:

$$P_{4}(x)=1+0\codt{x}-\frac{x^{2}}{2!}+0\cdot{x^{3}}+\frac{x^{4}}{4!}$$​

Error Bound (Lagrange Error)

On AP BC, you sometimes must justify accuracy.

The error after using a Taylor polynomial is:

$$|R_{n}(x)| = \left|\frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}\right|$$​

for some number $c$ between x and a.

Simplified use: find a max bound on a derivative.

Interval of Convergence

Taylor series (infinite) must be tested for convergence using:

• Ratio test

• Endpoint convergence checks

Taylor polynomials (finite) always converge (they are just polynomials).

Example: Build a Taylor Series

Find the Maclaurin series for e2xe^{2x}e2x.

We know:

$$e^{x}=\sum^{\infty}_{n=0}\frac{x^{(n)}}{n!}$$

then:

$$e^{2x}=\sum^{\infty}_{n=0}\frac{(2x)^{(n)}}{n!}$$

Simplify:

$$\sum^{\infty}_{n=0}frac{2^(n)x^(n)}{n!}$​

Example: Using Taylor Polynomials to Approximate

Approximate $$e^{0.1}$$ using 2nd-degree Maclaurin polynomial:

$$P_{2}(x)=1+x+\frac{x^{2}}{2}$$​

Plug in:

$$e^{0.1} \approx 1+0.1+\frac{(0.1)^{2}}{2}=1.105$$

Actual value: ≈1.10517.