Taylor Series Explained: Approximating Functions with Polynomials

A Taylor series approximates a function near a point $a$ by a polynomial whose derivatives at $a$ match the function's. Each coefficient is $\frac{f^{(k)}(a)}{k!}$. When $a = 0$ it is called a Maclaurin series. Near the centre, a few terms give an excellent approximation.

The formula

Worked example 1: $e^x$ about 0

Every derivative of $e^x$ is $e^x$, and $e^0 = 1$, so every coefficient is $\frac{1}{k!}$:

At $x = 0.5$ this gives $1.6484$, while $e^{0.5} = 1.6487$.

Worked example 2: $\sin x$ about 0

The derivatives cycle $\sin, \cos, -\sin, -\cos$, which at 0 give $0, 1, 0, -1, 0, 1, \ldots$:

This is why $\sin x \approx x$ for small angles in radians.

Why does the approximation drift away?

The polynomial is designed to match the function at the centre. Further away, the ignored terms matter more. Adding more terms keeps the polynomial close over a wider range.

Where Taylor series are used

• Calculators evaluate $\sin$, $e^x$ and $\ln$ with polynomial approximations.
• Physics uses $\sin\theta \approx \theta$ for small angles (pendulums).
• Limits like $\frac{\sin x}{x} \to 1$ become obvious from the series.

Common mistakes

• Forgetting the factorial in the denominator.
• Using degrees instead of radians for trig series.
• Centring at 0 when the function (like $\ln x$) is undefined there.

Practice questions

1. Write the first three terms of the Maclaurin series of $\cos x$.
2. Use $e^x \approx 1 + x + \frac{x^2}{2}$ to estimate $e^{0.1}$.
3. What is the coefficient of $x^3$ in the Maclaurin series of $e^x$?

Answers: 1) $1 - \frac{x^2}{2} + \frac{x^4}{24}$ 2) $1.105$ 3) $\frac{1}{6}$

The Taylor series calculator builds the polynomial for any function and draws it against the real curve. Derivatives come from the derivative calculator; revise them in derivatives for beginners.

Frequently asked questions

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