Maclaurin Series
A Maclaurin series is an infinite polynomial representation of a function $f(x)$. The reason why one would want to represent a function as a polynomial rather is because polynomials are easy to do calculus with, and mathematicians are lazy to do extra work.
The key idea behind a Maclaurin series is that we want to create a polynomial such that at the point $x=0$ the value of the polynomial $P(x)$ is $P(0) = f(0)$, the value of the first derivative is $P'(0) = f'(0)$, the value of the second derivative is $P''(0) = f''(0)$, and the same must hold true for the third derivatives, fourth derivatives, etc etc.
A polynomial can be defined using its coefficients, so we want to find a polynomial $P(x) = a_0 + a_1 x + a_2x^2 \cdots$ that satisfies the given condition.
To find the coefficients $a_n$, consider the following process:
Substitute $x=0$ into $P(x)$ to get:
$$P(0) = f(0) = a_0$$
Substitute $x=0$ into $P'(x)$ to get:
$$P'(x) = a_1 + 2a_2x + 3a_3x^2\cdots$$
$$P'(0) = f'(0)= a_1 $$
Substitute $x=0$ into $P''(x)$ to get:
$$P''(x) = 2a_2 + 6a_3x+ 12a_4x^2\cdots$$
$$P''(0) = f''(0) = 2a_2 $$
$$a_2=\frac{f''(0)}{2} $$
Substitute $x=0$ into $P'''(x)$ to get:
$$P'''(x) = + 6a_3+ 24a_4x + 60a_5 x^2\cdots$$
$$P'''(0) =f'''(0)= 6a_3$$
$$a_3=\frac{f'''(0)}{6} $$
Repeating this process, one can see that
$$a_n=\frac{f^{(n)}(0)}{n!}$$
and so
$$P(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots$$
or
$$P(x) =\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n$$
Definition
The Maclaurin series for a function f(x) is given by:
$$ f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n $$
where $f^{(n)}(0)$ represents the $n$th derivative of f evaluated at x = 0.
Standard Maclaurin Series Expansions
These expansions are given in the data booklet, and should not be memorised.
Exponential Function
The Maclaurin series for $e^x$ is:
$$ e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots = \sum_{n=0}^{\infty} \frac{x^n}{n!} $$
To approximate $e^{0.1}$, we can use the first few terms:
$e^{0.1} \approx 1 + 0.1 + \frac{0.1^2}{2!} + \frac{0.1^3}{3!} = 1.1051667$
The actual value is approximately 1.1051709, showing the accuracy of even a few terms.
Trigonometric Functions
For sin x:
$$ \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{(2n+1)!} $$
For cos x:
$$ \cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n}}{(2n)!} $$
Notice the alternating signs in these series. This pattern is crucial for understanding the behavior of these functions.
Natural Logarithm
The Maclaurin series for ln(1+x) is:
$$ \ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots = \sum_{n=1}^{\infty} (-1)^{n+1} \frac{x^n}{n} $$
Students often forget that this series is only valid for $-1 < x \leq 1$. Using it outside this range can lead to incorrect results.
Binomial Expansion
For $(1+x)^p$ where p is rational:
$$ (1+x)^p = 1 + px + \frac{p(p-1)}{2!}x^2 + \frac{p(p-1)(p-2)}{3!}x^3 + \cdots $$
Because you're using infinite terms , there are certain values where the Maclaurin series does not converge. (For example, the value of e^3 according to the Maclaurin series is:
$$1 + 3 + \frac{3^2}{2!} + \frac{3^3}{3!} + \frac{3^4}{4!} \cdots$$
which only gets bigger as you add more terms, so it does not converge.
To determine the values of $x$ such that it can converge, you look at the infinite summation and check what values of will cause make each term bigger and bigger endlessly.
In this case, notice that the values will converge provided that $|x|<1$.
Generally speaking, if you're taking a Maclaurin series function where the coefficient of $x$ is $a$, then the range of values for which it is valid is given by
$$|x|<\frac{1}{a}$$
Although this is not always the case for more complicated functions.
