Free Taylor Series and Maclaurin Expansion Calculator

Taylor and Maclaurin Series Calculator

Enter the function, the point $a$ for the expansion, and the desired order $n$.

Function (f(x)):

Expansion Point (a):

Order of Expansion (n):

The Taylor Series Calculator is your advanced tool for approximating complex functions with simple polynomials. The Taylor series is the bedrock of numerical analysis, allowing functions to be accurately modeled and simplified around a specific point. Our tool is essential for fields like physics, engineering, and advanced calculus, instantly providing the polynomial expansion up to your chosen order.

Mastering the Foundation: What is the Taylor Series?

The Taylor Series is a powerful tool in mathematical analysis that represents a function as an infinite sum of terms. Each term is derived from the values of the function’s derivatives at a single expansion point. This polynomial approximation is incredibly useful because it allows complex, non-polynomial functions (like $\sin(x)$ or $\ln(x)$ ) to be easily integrated, differentiated, and evaluated.

The general formula for the Taylor series for a function $f(x)$ centered at the point $a$ is given by:

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

Where $f^{(n)}(a)$ represents the $n$ -th derivative of the function evaluated at the point $a$ . Our calculator automates the tedious, multi-step process of calculating these derivatives and factorials, giving you the result instantly.

The Maclaurin Series: The Special Case

The Maclaurin series is simply the Taylor series when the expansion point $a$ is set to zero ( $a=0$ ). This series is defined by:

f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \dots

Since the Maclaurin series is frequently used, our tool defaults the expansion point to $a=0$ . To find the Maclaurin expansion, simply set the expansion point input to 0 and click calculate.

The Calculus Hub: Linking Series to Derivatives and Limits

The entire concept of the Taylor series proves why a comprehensive Calculus Hub is necessary. The series explicitly relies on all three core concepts:

Derivatives Define the Terms: Every single term in the Taylor series requires calculating the successive Derivatives of the function. Our Derivative Calculator is the perfect companion for understanding the individual components of the series formula.
Limits Define Convergence: The accuracy of the Taylor series hinges on its Limit of convergence. For the series to perfectly represent the function, the remainder term must approach zero as the order approaches infinity. This concept is formalized through the Radius of Convergence—a value defined by a limit test. Using our Limit Calculator, you can explore the formal proof of convergence for various functions.
Integrals Utilize the Series: Many non-elementary integrals (integrals that cannot be solved by standard rules) can be solved by first converting the function into its Taylor series expansion. The power series is then easy to integrate term-by-term, showcasing the immense practical utility of this tool. For the final calculation, refer to our Integral Calculator.

Practical Applications and Worked Examples

The Taylor series is not just theoretical; it powers the computational world. Every scientific calculator relies on series approximations to calculate values for $\sin(x)$ , $\cos(x)$ , and $e^x$ .

Example 1: The Maclaurin Expansion of $\cos(x)$

The cosine function is one of the most common applications of the Maclaurin series. When expanded around $a=0$ to the fifth order ( $n=5$ ), the resulting polynomial is:

P_5(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!}

Notice the terms for the odd powers ( $x^1, x^3, x^5$ ) are all zero because the odd-order derivatives of $\cos(x)$ are zero at $x=0$ . You can verify this result instantly using our calculator by entering cos(x) with an expansion point of 0 and an order of 5.

Example 2: Taylor Expansion of $\ln(x)$ Around $a=1$

For the function $f(x) = \ln(x)$ , the Maclaurin series ( $a=0$ ) does not exist because $\ln(0)$ is undefined. Therefore, we must choose another point, typically $a=1$ . The expansion is calculated as:

\ln(x) = (x-1) - \frac{(x-1)^2}{2} + \frac{(x-1)^3}{3} - \frac{(x-1)^4}{4} + \dots

This example proves the necessity of the Taylor series approach and highlights why having the flexibility to choose the expansion point is key to advanced problem-solving.

Understanding Accuracy and the Remainder

The higher the order ( $n$ ) you choose, the more terms are included in the polynomial, and the closer the polynomial’s graph will be to the original function’s graph. However, the approximation is only truly accurate near the expansion point $a$ .

The error in the approximation is known as the remainder ( $R_n(x)$ ). Finding the remainder often involves an advanced application of the Mean Value Theorem (Lagrange form) to determine the maximum possible error, ensuring the reliability of your numerical calculations.