This is a special case called a Maclaurin series. The formula simplifies by replacing $(x-a)^n$ with $x^n$.

What if the function is not infinitely differentiable at $x=a$?

A standard Taylor series cannot be formed at that center point. You must choose a different center point where the function is smooth.

How do I find the Taylor series of a product of functions?

Multiply their known Taylor series and collect terms of the same degree, rather than computing derivatives of the product directly via the product rule.

How to find a Taylor series

A Taylor series represents a function as an infinite sum of terms calculated from the values of its derivatives at a single point. It applies whenever a function is infinitely differentiable in a neighborhood around the chosen center point $x=a$ .

The setup

Identify the function $f(x)$ and the center point $x=a$ . You must be able to differentiate $f(x)$ repeatedly. Write down the general Taylor series definition: $f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x-a)^n$ .

The steps

Evaluate the base function at the center: find $f(a)$ . 2. Compute the first few derivatives $f'(x), f''(x), f'''(x)$ and evaluate each at $x=a$ . 3. Substitute these scalar values into the Taylor series formula. 4. Identify the mathematical pattern in the evaluated derivatives to write the general $n$ -th term for the sum.

Checking the result

Apply the Ratio Test to the general term to find the interval of convergence. Verify that substituting $x=a$ into your series collapses all terms except the first, yielding $f(a)$ .

Common errors

Omitting the $n!$ denominator in the series terms. Forgetting the $(x-a)^n$ factor or writing $x^n$ instead of $(x-a)^n$ when $a eq 0$ . Making sign errors when differentiating trigonometric functions repeatedly.

Worked example

Find the Taylor series for $f(x) = \sin(x)$ centered at $a = 0$ .

$f(0) = \sin(0) = 0$ . 2. $f'(x) = \cos(x) \implies f'(0) = 1$ . $f''(x) = -\sin(x) \implies f''(0) = 0$ . $f'''(x) = -\cos(x) \implies f'''(0) = -1$ . $f^{(4)}(x) = \sin(x) \implies f^{(4)}(0) = 0$ . 3. Substitute into formula: $f(x) = 0 + \frac{1}{1!}x - \frac{0}{2!}x^2 - \frac{1}{3!}x^3 + 0 + \frac{1}{5!}x^5 - ...$ 4. The even terms are zero and odd terms alternate signs. The general term is $\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} x^{2n+1}$ .

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References: Calculus: Early Transcendentals by James Stewart · OpenStax Calculus Volume 2