What a Maclaurin series is doing

A Maclaurin series is a power series centred at zero. It represents a function through the value of the function and the values of its derivatives at zero. It is the special case of a Taylor series with centre $0$.

For a sufficiently well-behaved function,

A partial sum is the polynomial produced when an infinite series is stopped after finitely many terms. Keeping terms up to degree $N$ gives

This creates the trade-off in the TOK question. Fewer terms make the expression simpler, but generally less accurate, particularly when $x$ moves away from zero.

A convergent series is an infinite series whose partial sums approach a finite limiting value. A Maclaurin expression represents the function only where the series converges to that function. Outside that interval, it may still show a formal pattern, but it cannot validly replace the function.

The standard series you must be able to obtain and use

Here $e$ is Euler's number. The inputs of $\sin x$, $\cos x$ and $\arctan x$ are measured in radians, making the input $x$ dimensionless.

This converges for all real $x$.

This converges for all real $x$.

This converges for all real $x$.

This converges for $|x|\le 1$, with conditional convergence at the endpoints.

This converges for $-1<x\le 1$. Watch the centre: this is a series for $\ln(1+x)$, not directly for $\ln x$.

For the binomial expansion,

In general, the series converges for $|x|<1$. If $p$ is a non-negative integer, it terminates and becomes the ordinary binomial expansion.

How the formula creates the pattern

Every derivative of $e^x$ is $e^x$, so each derivative has value $1$ at zero. The coefficients are therefore $1/n!$.

The derivatives of $\sin x$ cycle through $\sin x$, $\cos x$, $-\sin x$, $-\cos x$, and then repeat. At $x=0$, the values are $0,1,0,-1,\ldots$. Only the odd powers survive.

For $\cos x$, the derivative cycle produces $1,0,-1,0,\ldots$ at zero, leaving only even powers. This also gives a quick check. Since $\sin x$ is an odd function, its Maclaurin series should have only odd powers. Since $\cos x$ is even, its series should have only even powers.

The historical comparison between the Bourbaki style and the Kerala School shows how different mathematical cultures can view series. One style may stress formal structure and abstraction; another may focus on computational procedures and approximation. Both viewpoints matter here.

Starting from known series

Once you know a few standard Maclaurin series, there’s usually no need to return to the derivative formula each time. A term-by-term operation is an algebraic, differentiation or integration operation applied separately to each term of a power series within its interval of convergence.

This syllabus uses four such operations: simple substitution, products, integration and differentiation. In practice, first decide the highest power of $x$ required. Then keep enough terms to reach it and collect like powers carefully.

Simple substitution

In a substitution into a series, the input of a known series is replaced by a new expression. Replacing $x$ by $x^2$ in the exponential series, for example, gives

This is exactly the type of substitution named in the guide. The series for $e^x$ converges for all real inputs, so the substituted series for $e^{x^2}$ also converges for all real $x$.

Substitution helps when an expression isn’t quite in standard form. For instance,

provided $-1<3x\le 1$. Here, the interval changes because the input to the known series is $3x$ rather than $x$.

Products of series

A product of series is a series obtained by multiplying two power series and collecting equal powers of the variable. Work as you would with polynomial multiplication, but keep enough terms to achieve the requested accuracy.

For example, use the known series for $e^x$ and $\sin x$:

Watch the missing terms. To find terms up to $x^3$, you don’t need every term from both series. You do, however, need every product that can contribute to $x$, $x^2$ or $x^3$.

Integration of series

To integrate a Maclaurin series, integrate each power of $x$ separately. If necessary, use the function value at $x=0$ to determine the constant of integration.

A neat way to obtain the logarithm series begins with the geometric series

Integrating term by term,

so

Since $\ln(1+0)=0$, the constant $C$ is $0$. Don’t skip this line when a series has been found by integration; marks are often lost here.

The same method produces the arctangent series from

Integrate term by term and use $\arctan 0=0$ to get

Differentiation of series

To differentiate a series, differentiate each term and check the interval on which the resulting series remains valid. For example, differentiating

gives

Differentiation can produce related binomial series too. Starting from

differentiating term by term gives the series for $p(1+x)^{p-1}$. This is often quicker than returning to the Maclaurin formula.

Using the differential equation to generate derivatives

A differential equation links an unknown function with one or more of its derivatives. When it provides enough information at $x=0$, we can generate the derivative values for the Maclaurin formula without finding a closed-form solution first.

Let $y$ be the unknown function of the dimensionless input $x$. Suppose it satisfies

The solution is $y=e^x\sin x$. For the Maclaurin series, though, we can work straight from the differential equation.

First, rearrange it:

At $x=0$,

Now differentiate the differential equation to move one derivative higher:

Therefore,

Differentiate once more:

which gives

Substitute these values into the Maclaurin formula:

So the expansion begins

Derivative values and Maclaurin terms from the DE example.

You can obtain the same early expansion by multiplying the standard series for $e^x$ and $\sin x$. The differential-equation method can be cleaner, especially when direct multiplication becomes messy, because the equation gives a recurrence for the derivatives.

Coefficient comparison as the same idea

The same method can be written another way. Assume

Substitute the series and its derivatives into the differential equation. Then compare coefficients of equal powers of $x$ to produce equations for the coefficients. This isn’t a different theory. It uses the same Maclaurin information, but starts with coefficients rather than derivatives.

One warning: initial conditions matter. Without values such as $y(0)$ and $y'(0)$, the differential equation may describe a whole family of functions, leaving unknown constants in the Maclaurin series.