Started elsewhere:

That is entirely possible considering the only line of working I have been given is: e^x is approximately 1 + x where any term of order 2+ is ignored because it’d be ridiculously small and thus negligible. (X is meant to be tiny e.g. 10^- 8, hence why it’d be negligible. At least in the context where I got these from - a genetics book. See below.)

I got that answer because my notes are old and written. I must've been haphazardly copying them down quickly during the lectures. Must've missed the factorials.

Still not quite so sure how we would get the one your wrote for e^x above though, but maybe it's because I've forgotten large chunks of content.

Optionally you may also memorise the one for \(\ln(1-x)\), but that can be obtained by integrating \( \frac1{1-x}\).

Remark on first order approximations

This means, that for small \(x\), the following are reasonable:

\begin{align*}e^x&\approx 1+x\\ \sin x &\approx x\\ \cos x&\approx 1\end{align*}

These four are generally regarded as the most useful. The first one really isn't anything fancy as it's literally just the geometric series, but exponentials and sinusoids appear quite common naturally. All the other stuff either stem out of these (e.g. \( \sinh\)), are inverses of these (e.g. \(\tan^{-1}\)) or are just random shit mathematicians use for convenience.

We omit justification as to why that is the case for the sake of mere computations.

So all we really need to compute is \( f(0), f^\prime(0), f^{\prime\prime}(0), f^{\prime\prime\prime}(0), f^{(4)}(0) \) and pretty much, all of the derivatives of \(f\), evaluated at 0.

The actual computation starts here.Expanding this sum out gives \(1 + \frac{x}{1!}+\frac{x^2}{2!}+\dots\) as required.

Small note - The factorials basically appear because they're actually a part of the Taylor series formula.