Taylor Series Expansion Calculator

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Reviewed and Verified by: Sarah Adams, Ph.D. in Applied Mathematics

Accurately approximate a function near a given point using the Taylor Series Expansion Calculator. Just input the function, center point, and order of expansion.

Taylor Series Expansion Calculator

*Currently supports Math.sin(x), Math.cos(x), Math.exp(x)
Expansion Result:
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Detailed Steps:

Taylor Series Expansion Formula

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

Sources: Wolfram MathWorld, Wikipedia (Taylor Series)

Variables Explained

  • Function ($f(x)$): The expression you wish to approximate (must be infinitely differentiable at point $a$).
  • Center Point ($a$): The specific point around which the approximation is made. This is the point where the function and its derivatives are evaluated.
  • Order of Expansion ($n$): The number of terms to include in the polynomial. Higher orders generally yield better approximations.
  • Evaluate at $x$: An optional value used to compute the numerical approximation of $f(x)$ at a specific point.

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What is the Taylor Series Expansion?

The Taylor series expansion is a powerful mathematical tool that represents a function as an infinite sum of terms that are calculated from the function’s derivatives at a single point. It is named after the English mathematician Brook Taylor. Essentially, it transforms a potentially complex, non-polynomial function into a polynomial approximation, making it easier to analyze, integrate, and solve.

This approximation is most accurate near the center point ($a$). As you move further away from $a$, the approximation tends to degrade, although increasing the order of the expansion ($n$) can extend the range of accuracy. The most famous special case is the Maclaurin series, which is simply a Taylor series centered at zero ($a=0$).

Engineers and physicists heavily rely on Taylor expansions for simplifying complex differential equations and modeling phenomena where an exact, closed-form solution is impossible or too computationally expensive.

How to Calculate Taylor Series Expansion (Example)

Let’s find the Taylor series expansion for $f(x) = e^x$ centered at $a=0$ up to order $n=3$:

  1. Find the Derivatives: Calculate the derivatives of $f(x)$: $f'(x) = e^x$, $f”(x) = e^x$, $f”'(x) = e^x$.
  2. Evaluate at the Center Point ($a=0$): Calculate $f^{(k)}(0)$:
    • $k=0$: $f(0) = e^0 = 1$
    • $k=1$: $f'(0) = e^0 = 1$
    • $k=2$: $f”(0) = e^0 = 1$
    • $k=3$: $f”'(0) = e^0 = 1$
  3. Apply the Formula: Substitute the values into the formula: $$T_3(x) = \frac{f(0)}{0!}(x-0)^0 + \frac{f'(0)}{1!}(x-0)^1 + \frac{f”(0)}{2!}(x-0)^2 + \frac{f”'(0)}{3!}(x-0)^3$$
  4. Final Expansion: $$T_3(x) = \frac{1}{1}(1) + \frac{1}{1}(x) + \frac{1}{2}(x^2) + \frac{1}{6}(x^3) = 1 + x + \frac{x^2}{2} + \frac{x^3}{6}$$

Frequently Asked Questions (FAQ)

  • What is the difference between Taylor and Maclaurin series?

    The Maclaurin series is a special case of the Taylor series where the center point ($a$) is always set to zero. All other rules for calculation and application remain the same.

  • Why is the Taylor series approximation useful?

    It allows complex functions (like trigonometric, exponential, or transcendental functions) to be approximated by simple polynomials. Polynomials are easy for computers to calculate and for humans to analyze, especially near the center point.

  • What does the “order” of the expansion mean?

    The order ($n$) specifies the highest power of $(x-a)$ to include in the polynomial sum. A higher order means more terms are included, resulting in a more accurate approximation over a wider interval.

  • Does the Taylor series work for all functions?

    No. A function must be infinitely differentiable at the center point $a$ to have a Taylor series representation. Functions with singularities or non-smooth points cannot be represented this way.

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