Revise Maclaurin series for IB Mathematics: Analysis and Approaches (5.10)

A function $f(x)$ that is infinitely differentiable at $x=0$ can be represented by its Maclaurin series. The idea is to force a polynomial to agree with the function AND with all of its derivatives at the single point $x=0$. Matching the value fixes the constant term, matching the first derivative fixes the coefficient of $x$, matching the second derivative fixes the coefficient of $x^2$, and so on. This locks the polynomial to the function at $x=0$ and makes it an excellent approximation nearby.

$f(x) = f(0) + f'(0)\,x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \dots = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n$

Five standard series appear again and again, and each can be derived by the same routine: differentiate, evaluate at $0$, substitute. Take $e^x$. Every derivative of $e^x$ is $e^x$, and $e^0=1$, so every coefficient $\frac{f^{(n)}(0)}{n!}=\frac{1}{n!}$, giving $e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots$. For $\sin x$, the derivatives cycle $\cos x,-\sin x,-\cos x,\sin x,\dots$; at $x=0$ these are $1,0,-1,0,\dots$, so only the odd powers survive and the signs alternate. The same cycle gives $\cos x$ with only even powers. For $\ln(1+x)$ the derivatives are $\frac{1}{1+x},-\frac{1}{(1+x)^2},\frac{2}{(1+x)^3},\dots$, which at $0$ give $1,-1,2,-6,\dots=(-1)^{n-1}(n-1)!$; dividing by $n!$ leaves $\frac{(-1)^{n-1}}{n}$ and hence no factorials in the final answer. And $(1+x)^n$ generates the binomial series.

The real power of Maclaurin series is that you rarely differentiate from scratch. Once you know the standard five, you can build a huge range of new series by (i) substitution — replace $x$ by an expression such as $2x$, $-x$ or $x^2$; (ii) multiplication — multiply two series and collect terms up to the required power; and (iii) term-by-term differentiation or integration within the interval of convergence. These techniques are faster and far less error-prone than repeated differentiation, and examiners expect you to use them.

Truncating a Maclaurin series after a few terms gives a Maclaurin polynomial, and substituting a small value of $x$ produces a numerical approximation. The closer $x$ is to $0$, and the more terms you keep, the better the estimate. A short table of the powers of $x$ makes the substitution tidy and reduces arithmetic slips.

In Paper 1 the marks are analytic and follow the IB Mathematics conventions: an M mark rewards a valid METHOD, an A mark rewards ACCURACY and is usually dependent on the method mark it follows, and correct equivalent forms are always accepted. That structure only protects you if the method is written down. Study how each mark below is tied to a specific line of working.

Maclaurin series turn hard functions into polynomials, and that unlocks the rest of HL calculus. Truncated series evaluate limits that would otherwise need l'Hopital's rule, they approximate definite integrals with no elementary antiderivative, and separating the series for $e^{ix}$ into real and imaginary parts delivers Euler's formula $e^{ix}=\cos x + i\sin x$, tying calculus to complex numbers. Master the habit — differentiate, evaluate at $0$, substitute the factorials, or better still reach for a known series and manipulate it — and these later applications become variations on a method you already own.