A Maclaurin series represents a function as an infinite sum of powers of x (expanded about x = 0)

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

\[f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n\]

\[e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots\]

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

\[\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots\]

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

\[(1 + x)^p = 1 + px + \frac{p(p-1)}{2!}x^2 + \frac{p(p-1)(p-2)}{3!}x^3 + \cdots\]

All five standard series above are given in the IB formula booklet. Use them directly unless specifically asked to derive!

Replace x in a standard series with another expression (e.g., x², -x, 2x, etc.)

Start with: \(e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots\)
Substitute x² for x:

\[e^{x^2} = 1 + x^2 + \frac{x^4}{2!} + \frac{x^6}{3!} + \cdots\]

Start with: \(\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots\)
Substitute 2x for x:

\[\sin(2x) = 2x - \frac{(2x)^3}{3!} + \frac{(2x)^5}{5!} - \cdots = 2x - \frac{8x^3}{6} + \frac{32x^5}{120} - \cdots\]

Start with: \(\ln(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots\)
Substitute -x for x:

\[\ln(1 - x) = -x - \frac{x^2}{2} - \frac{x^3}{3} - \cdots\]

Multiply two known Maclaurin series term-by-term, collecting like powers of x

\(e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \cdots\)
\(\sin x = x - \frac{x^3}{6} + \cdots\)

Multiply term-by-term:
• \(1 \times x = x\)
• \(1 \times (-\frac{x^3}{6}) + x \times x = x^2 - \frac{x^3}{6}\)
• \(x \times (-\frac{x^3}{6}) + \frac{x^2}{2} \times x = -\frac{x^4}{6} + \frac{x^3}{2}\)

\[e^x \sin x = x + x^2 + \frac{x^3}{3} + \cdots\]

Only multiply terms whose powers add up to the required maximum power or less

Differentiate a known Maclaurin series term-by-term to find the series for its derivative

\[\frac{d}{dx}\left[\sum_{n=0}^{\infty} a_n x^n\right] = \sum_{n=1}^{\infty} na_n x^{n-1}\]

Integrate a known Maclaurin series term-by-term to find the series for its integral

\[\int\left[\sum_{n=0}^{\infty} a_n x^n\right]dx = C + \sum_{n=0}^{\infty} \frac{a_n}{n+1} x^{n+1}\]

Since \(\frac{d}{dx}(\sin x) = \cos x\):
\(\frac{d}{dx}\left[x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots\right] = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots = \cos x\)

• \(e^x\), \(\sin x\), \(\cos x\): Converge for all x ∈ ℝ
• \(\ln(1+x)\): Converges for -1 < x ≤ 1
• \((1+x)^p\): Converges for |x| < 1

• Complete series (infinite terms): Exactly equals the function
• Truncated series (finite terms): Approximation of the function
• More terms → better approximation
• Approximation best near x = 0, less accurate as |x| increases

Truncated series are exactly equal to the function at x = 0

Use truncated series to estimate values like e, sin(0.1), √(1.05), etc.

Replace functions with their series to evaluate indeterminate limits (e.g., \(\lim_{x \to 0} \frac{\sin x}{x}\))

Replace function with series, then integrate term-by-term

Assume solution as power series, substitute into differential equation, match coefficients

1. Check if function matches a standard form in formula booklet
2. If yes, use directly; if no, proceed with derivation
3. For derivation: find f(0), f'(0), f''(0), etc. (use GDC if allowed)
4. Substitute into general Maclaurin formula
5. For composites: use substitution method
6. For products: multiply series term-by-term
7. Simplify and state validity/convergence if asked

• "Up to and including x³" → Find first 4 terms (x⁰ through x³)
• "The first three non-zero terms" → Count only non-zero terms
• "Hence or otherwise" → Use previous result to simplify work
• "Valid for" → State convergence interval

• Forgetting factorials in denominators
• Sign errors in alternating series
• Not simplifying fractions properly
• Confusing Taylor series (about x = a) with Maclaurin (about x = 0)
• Forgetting to state convergence conditions when required