Professional high school math teacher/ amateur mathematician here looking for feedback on my attempt to to derive the general case of a Maclaurin polynomial using finite difference methods. Here's my overall approach (not a proof). Consider an nth-differentiable function $f(x)$ on a closed interval $[0,t]$ and allow $t$ to approach zero. The first key observation is that the difference quotient calculations for any given $f^{(n)}(0)$ require $n+1$ points from $f(x)$ in $[0,t]$, which gives $n$ partitions of the closed interval if we use the endpoints. Applying the difference quotient $n$ times yields $n$ points in $f'(x)$, with the 2nd derivative having $n-1$ points, the 3rd having $n-2$ points, ... , $(n-1)th$ derivative having two points until we obtain the point $(0,f^{(n)}(0))$ . If we allow the partitions to be regular, then the step size required to calculate $(0,f^{(n)}(0))$ is $Δx=\frac{t}{n}$.
Now let's work backwards using successive iterations of Euler's method and known derivatives at $t=0$ until we obtain a polynomial that approximates $f(x)$. I will use $n=4$ to illustrate how to get the approximating polynomial. Starting with the 4th derivative at $t=0$, Euler's method gives
$f^{(3)}(\frac{t}{4})\approx f^{(3)}(0)+ \frac{t}{4}f^{(4)}(0) \hspace{4 cm} (1)$
The 2nd key observation that is required is now given: $lim_{t\rightarrow0}f^{(3)}(\frac{t}{4})=f^{(3)}(0)$. So, the next move is to replace $f^{(3)}(0)$ with (1). We can now apply another iteration of Euler's method, except this time we will use the replaced (approximated) value instead of the actual $f^{(3)}(0)$. On this iteration however, $n$ is reset to $n=3$ which also resets the step-size to $Δx=\frac{t}{3}$. The result is
$f^{(2)}(\frac{t}{3})\approx f^{(2)}(0)+ \frac{t}{3}\left(f^{(3)}(0)+\frac{t}{4}f^{(4)}(0) \right) \hspace{3 cm} (2)$
As before, we can replace $f^{(2)}(0)$ with (2) since the limit still applies, and iterate Euler's method again but with the step-size $Δx=\frac{t}{2}$ (b/c we reset to $n=2$). This gives
$f^{(1)}(\frac{t}{2})\approx f^{(1)}(0)+ \frac{t}{2}\left(f^{(2)}(0)+\frac{t}{3}\left(f^{(3)}(0)+\frac{t}{4}f^{(4)}(0) \right)\right) \hspace{3 cm} (3)$
The final Euler iteration that replaces $f^{(1)}(0)$ with (3) gives the desired approximation in Horner's form in terms of t.
$f(t)\approx f^{(0)}(0)+ t\left(f^{(1)}(0)+ \frac{t}{2}\left(f^{(2)}(0)+\frac{t}{3}\left(f^{(3)}(0)+\frac{t}{4}f^{(4)}(0) \right)\right)\right) \hspace{3 cm} (4)$
Expanding gives the familiar Maclaurin polynomial approximation,
$f(t) \approx f(0)+f'(0)t+\frac{f''(0)t^2}{2}+\frac{f'''(0)t^3}{6}+\frac{f''''(0)t^4}{24} \hspace{3 cm} (5)$
The general Taylor form is found by horizontally translating the polynomial a distance c, that is by substituting (t-c) for t in (5).
