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