Use the Power Series Representation Calculator to quickly find the Taylor/Maclaurin series approximation for common functions up to a specified polynomial degree. This tool simplifies complex calculus into easily understandable terms and steps.
Power Series Representation Calculator
Power Series Representation Formula
$$f(x) = \sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!}(x-a)^n$$
Sources: Wolfram MathWorld | Khan Academy
Variables Explained
- Function $f(x)$: The mathematical function you want to approximate (e.g., $e^x$, $\sin(x)$).
- Center ($a$): The point around which the series is expanded. The approximation is most accurate near this point. $a=0$ results in a Maclaurin series.
- Max Degree ($n$): The highest power of $(x-a)$ used in the polynomial. A higher degree generally leads to a better approximation.
- Evaluation Point ($x$): An optional value to substitute into the resulting polynomial to find a numerical approximation.
What is Power Series Representation?
A power series representation is a way to express a function as an infinite sum of terms, where each term is a constant multiplied by a power of $(x-a)$. This concept, fundamental in calculus, allows complicated functions to be approximated by much simpler polynomials. The most common types of power series are the Taylor series and the Maclaurin series (which is a Taylor series centered at $a=0$).
The utility of power series lies in their ability to bridge analytic functions (like $e^x$) with polynomial algebra. By using a finite number of terms, we can calculate values, perform integration, and even solve differential equations more easily. The accuracy of the approximation depends on the distance from the center $a$ and the number of terms (degree $n$) used.
How to Calculate a Power Series (Example)
Let’s find the 4th-degree Maclaurin series for $f(x) = e^x$ ($a=0$).
- Find Derivatives: Calculate the first $n$ derivatives of $f(x)$. For $f(x)=e^x$, all derivatives $f^{(n)}(x)$ are simply $e^x$.
- Evaluate Derivatives at Center ($a$): Substitute $a=0$ into each derivative. $f(0)=e^0=1$, $f'(0)=1$, $f”(0)=1$, etc.
- Apply the Formula: Substitute these values into the Taylor series formula: $\sum_{n=0}^4 \frac{f^{(n)}(0)}{n!}(x-0)^n$.
- Calculate Terms:
- $n=0$: $\frac{f^{(0)}(0)}{0!}x^0 = \frac{1}{1}(1) = 1$
- $n=1$: $\frac{f'(0)}{1!}x^1 = \frac{1}{1}x = x$
- $n=2$: $\frac{f”(0)}{2!}x^2 = \frac{1}{2}x^2$
- $n=3$: $\frac{f”'(0)}{3!}x^3 = \frac{1}{6}x^3$
- $n=4$: $\frac{f^{(4)}(0)}{4!}x^4 = \frac{1}{24}x^4$
- Sum the Terms: The polynomial is $P_4(x) = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24}$.
Related Calculators
- Taylor Series Polynomial Approximator
- Geometric Series Convergence Checker
- Radius of Convergence Calculator
- Definite Integral Power Series Calculator
Frequently Asked Questions (FAQ)
A Maclaurin series is a special case of the Taylor series where the center ($a$) is specifically zero. Both are power series representations, but Maclaurin series are often simpler to calculate because many terms simplify when $a=0$.
The radius of convergence $(R)$ determines the interval around the center $a$ for which the power series converges to the original function. It is typically found using the Ratio Test: $\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = L$. If $L$ is a number, then $R = 1/L$.
The degree ($n$) determines the number of terms used in the polynomial approximation. A higher degree results in a higher-order polynomial, which generally provides a more accurate approximation of the function, especially over a larger interval away from the center $a$.
No. A function must be “analytic” at the center $a$, meaning it must be infinitely differentiable at that point. Functions like $f(x) = |x|$ cannot be represented by a Taylor series centered at $x=0$ because the first derivative does not exist at $x=0$.