Taylor and Maclaurin Series Explained

How does a calculator find \( \sin(0.3) \) or \( e^{0.1} \)? It cannot draw a triangle or a graph. Instead, it uses polynomials that imitate the function very closely, and those polynomials come from Taylor and Maclaurin series. This guide explains the idea, the formula, and how to build and use these series step by step.

The Big Idea

Polynomials are easy to work with: you only need addition and multiplication. A Taylor series builds a polynomial that matches a function's value, slope, curvature and higher derivatives at one chosen point. The more terms you include, the better the approximation near that point.

The Taylor Series Formula

The Taylor series of \( f(x) \) about \( x = a \) is:

$$ f(x) = f(a) + f'(a)(x - a) + \frac{f''(a)}{2!}(x - a)^2 + \frac{f'''(a)}{3!}(x - a)^3 + \cdots $$

In compact form:

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

Here \( f^{(n)}(a) \) means the \( n \)th derivative evaluated at \( a \), and \( n! = 1 \times 2 \times \cdots \times n \).

The Maclaurin Series

A Maclaurin series is simply a Taylor series centred at \( a = 0 \):

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

Most A Level and FSc questions use Maclaurin series, while university courses use both.

How to Build a Series: The Method

1. Find several derivatives of \( f(x) \).
2. Evaluate each derivative at the centre \( a \).
3. Substitute into the formula, dividing the \( n \)th term by \( n! \).
4. Look for a pattern to write the general term.

Worked Example 1: Maclaurin Series for \( e^x \)

Step 1: Derivatives. Every derivative of \( e^x \) is \( e^x \).

Step 2: Evaluate at 0. \( f(0) = f'(0) = f''(0) = \cdots = e^0 = 1 \).

Step 3: Substitute.

$$ e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots = \sum_{n=0}^{\infty}\frac{x^n}{n!} $$

Using it: estimate \( e^{0.1} \) with four terms.

$$ e^{0.1} \approx 1 + 0.1 + \frac{0.01}{2} + \frac{0.001}{6} = 1 + 0.1 + 0.005 + 0.000167 = 1.105167 $$

The true value is \( 1.105171 \), so four terms already give five correct decimal places.

Worked Example 2: Maclaurin Series for \( \sin x \)

Step 1 and 2: Derivatives and values at 0.

• \( f(x) = \sin x \), so \( f(0) = 0 \)
• \( f'(x) = \cos x \), so \( f'(0) = 1 \)
• \( f''(x) = -\sin x \), so \( f''(0) = 0 \)
• \( f'''(x) = -\cos x \), so \( f'''(0) = -1 \)
• \( f^{(4)}(x) = \sin x \), and the pattern repeats

Step 3: Substitute. The even-power terms vanish:

$$ \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots $$

Step 4: General term.

$$ \sin x = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!} $$

This explains why \( \sin x \approx x \) for small angles in radians, a result used constantly in physics, for example with the simple pendulum.

Worked Example 3: Taylor Series for \( \ln x \) About \( a = 1 \)

We cannot use a Maclaurin series here because \( \ln 0 \) is undefined, so we centre at \( a = 1 \).

Step 1 and 2:

• \( f(x) = \ln x \), so \( f(1) = 0 \)
• \( f'(x) = \frac{1}{x} \), so \( f'(1) = 1 \)
• \( f''(x) = -\frac{1}{x^2} \), so \( f''(1) = -1 \)
• \( f'''(x) = \frac{2}{x^3} \), so \( f'''(1) = 2 \)

Step 3: Substitute.

$$ \ln x \approx 0 + (x - 1) - \frac{1}{2!}(x - 1)^2 + \frac{2}{3!}(x - 1)^3 $$

$$ \ln x \approx (x - 1) - \frac{(x - 1)^2}{2} + \frac{(x - 1)^3}{3} $$

Using it: estimate \( \ln 1.1 \), where \( x - 1 = 0.1 \).

$$ \ln 1.1 \approx 0.1 - 0.005 + 0.000333 = 0.095333 $$

The true value is \( 0.095310 \), very close.

Worked Example 4: Approximating \( \sqrt{4.1} \)

Use a Taylor polynomial of degree 2 for \( f(x) = \sqrt{x} \) about \( a = 4 \).

Step 1 and 2:

• \( f(x) = x^{1/2} \), so \( f(4) = 2 \)
• \( f'(x) = \frac{1}{2}x^{-1/2} \), so \( f'(4) = \frac{1}{4} \)
• \( f''(x) = -\frac{1}{4}x^{-3/2} \), so \( f''(4) = -\frac{1}{4} \cdot \frac{1}{8} = -\frac{1}{32} \)

Step 3: Substitute.

$$ \sqrt{x} \approx 2 + \frac{1}{4}(x - 4) - \frac{1}{32 \cdot 2!}(x - 4)^2 = 2 + \frac{x - 4}{4} - \frac{(x - 4)^2}{64} $$

Step 4: Put \( x = 4.1 \).

$$ \sqrt{4.1} \approx 2 + 0.025 - \frac{0.01}{64} = 2.025 - 0.00015625 = 2.02484375 $$

The true value is \( 2.0248457\ldots \), accurate to six decimal places.

Standard Maclaurin Series to Remember

• \( e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \) for all \( x \)
• \( \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots \) for all \( x \)
• \( \cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots \) for all \( x \)
• \( \ln(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots \) for \( -1 < x \le 1 \)
• \( \frac{1}{1 - x} = 1 + x + x^2 + x^3 + \cdots \) for \( |x| < 1 \)

Note that some series only work within an interval of convergence. Outside it, adding more terms makes things worse, not better.

Common Mistakes

• Forgetting the factorials in the denominators.
• Using degrees instead of radians in trig series. The series for \( \sin x \) only works in radians.
• Mixing up \( x \) and \( (x - a) \) in a Taylor series about \( a \neq 0 \).
• Ignoring the interval of convergence, for example using the \( \ln(1 + x) \) series at \( x = 3 \).
• Arithmetic slips when evaluating derivatives at the centre. Write each value clearly before substituting.

Key Takeaways

• A Taylor series approximates a function by a polynomial built from its derivatives at one point.
• A Maclaurin series is the special case centred at \( x = 0 \).
• Each term is \( \frac{f^{(n)}(a)}{n!}(x - a)^n \); more terms give better accuracy near the centre.
• Learn the standard series for \( e^x \), \( \sin x \), \( \cos x \), \( \ln(1 + x) \) and \( \frac{1}{1 - x} \), along with their intervals of convergence.

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