Why does a power series with binomial coefficients reduce to a Taylor series? I got this surprising result while numerically modeling a differentiable function $f(x)$ with a power series on the interval $[0,t]$. I couldn't find any references to a connection between power series with binomial coefficients and Taylor/Maclaurin series.
$f(t)= \displaystyle \lim_{n\rightarrow \infty} \sum_{k=0}^{n}{n \choose k} \left( \frac{t}{n}\right)^kf^{(k)}(0)= \sum_{k=0}^{\infty}\frac{f^{(k)}(0)t^k}{k!}\lim_{n\rightarrow \infty}\frac{\frac{n!}{(n-k)!}}{n^k}=\sum_{k=0}^{\infty}\frac{f^{(k)}(0)t^k}{k!}$
I obtained the expression on the left by first approximating an $(n-1)th$ derivative as a linear sequence using Euler's method, $\displaystyle f^{(n-1)}\left( \frac{tk}{n} \right)=f^{(n-1)}(0)+\frac{tk}{n}f^{(n)}(0)$ for natural numbers k such that $0\leq k \leq n$. I then numerically integrated to obtain each of the antiderivatives until I got $f(t)$, which revealed binomial coefficients.  I'm betting this connection is well-known, so please share any insights you may have about the relationship here between Taylor series and binomial coefficients. Thank you.
 A: I derived the same result you did while attempting to derive the Taylor series. I think this is basically the same thing you did, but here's how I did it if it might help someone else. It actually involves the binomial series in more than one way.
If we know $f(a)$ and it's derivatives at $a$ and we want to find $f$ just a little ways away at $x$, i.e. $f(x)$, where $x = a + w$ and $w$ is the width of the interval between $a$ and $x$, from the definition of derivative we have our basic
$$f(x) \approx f(a) + wf^{(1)}(a)$$
or to make it clear we're starting at or centered at $a$
$$f(a+w) \approx f(a) + wf^{(1)}(a) \tag{1}\label{eq1}$$
The geometric interpretation is straightforward, we take $f$ at $a$ and then add on one output segment.
To make this more accurate, let's divide the distance $x-a$ into two intervals, so now the width, $w$, of each interval is $w = \dfrac{(x-a)}{2}$, we have $$f(x) \approx f(a) + wf^{(1)}(a) + wf^{(1)}(a+w)$$
We just start at $f(a)$ and add on two segments. We're basically starting at $a$ and integrating over to x.
We can make this as accurate as we like by dividing $(x-a)$ into more and more intervals. So generally, where $w=\dfrac{(x-a)}{n}$ and $n$ is the number of intervals
\begin{align}
f(x) \approx f(a) &+ wf^{(1)}(a) + wf^{(1)}(a+w) + wf^{(1)}(a+2w)  \\&\quad+\ldots+wf^{(1)}(a+(n-1)w) \tag{2}\label{eq2}\end{align}
$f(a)$ is the starting point and $wf^{(1)}(a)$ is the first segment, $wf^{(1)}(a+w)$ is the second segment, etc.
But now the problem is we're not just taking the derivative at $a$, we're also taking the derivative at every interval point between $a$ and $x$ (at $a$, $a+w$, $a+2w$, etc). So we have to perform reduction steps on each segment until they only rely on derivatives of $f$ at $a$. We use the following rule:
$$f^{(k)}(a+mw) = f^{(k)}(a+(m-1)w)+wf^{(k+1)}(a+(m-1)n) \tag{3}\label{eq3}$$
Which looks messy but it's really the same as our original equation \eqref{eq1}.
E.g to reduce the 3rd segment, $wf^{(1)}(a+2w)$, we have:
\begin{align}
wf^{(1)}(a+2w)
&= w[f^{(1)}(a+w)+wf^{(2)}(a+w)]\\
&= wf^{(1)}(a+w)+w^2f^{(2)}(a+w)
\end{align}
We've reduced it from $a+2w$ to $a+w$, so we have to apply the reduction a 2nd time and then collect like terms.
\begin{align}
wf^{(1)}(a+2w) 
&= w[f^{(1)}(a) + wf^{(2)}(a)]+w^2[f^{(2)}(a) + wf^{(3)}(a)]\\
&= wf^{(1)}(a) + w^2f^{(2)}(a)+w^2f^{(2)}(a) + w^3f^{(3)}(a)\\
&= wf^{(1)}(a) + 2w^2f^{(2)}(a) + w^3f^{(3)}(a)
\end{align}
So we have the coefficients $(1, 2, 1)$ which are the binomial coefficients! We can write this as a summation and generalize it to any n, giving us the output segment for any interval, $i$:
$$\sum_{k=1}^{i}{i-1 \choose k-1}w^kf^{(k)}(a) \tag{4}\label{eq4}$$
Where $n$ is the number of intervals and $w=\frac{(x-a)}{n}$ and $i$ is the $i^{th}$ interval (e.g, an $i$ of $3$ is the third interval, which starts at $a+2w$)
Substituting \eqref{eq4} into \eqref{eq2}, we arrive at:
$$f(x) \approx f(a) + wf^{(1)}(a)  + \sum_{k=1}^{2}{2-1 \choose k-1}w^kf^{(k)}(a)  + ... +  \sum_{k=1}^{n}{n-1 \choose k-1}w^kf^{(k)}(a)\tag{5}\label{eq5} $$
So each segment we're adding on is using binomial coefficients, but the really interesting part (for me at least) is that when we add up all the segments and collect like terms, our final result is also binomial coefficients! I won't prove it but if you solve the problem using 3 or 4 segments you can easily convince yourself that is indeed the case (it'd just be a proof by induction showing that this mechanical procedure we're doing here holds at each step of the way). Here's an example with 3 intervals. I've arranged it in a grid to more easily illustrate this where each row is one of the segments we're adding on. Each row is the summation for each segment from \eqref{eq4} and then we sum the columns to produce the final result from \eqref{eq5}. Summing the columns is simply collecting like terms.
\begin{array}{rrrrrr|r}
              & f^{(0)} & wf^{(1)} & w^2f^{(2)}& w^3f^{(3)} & w^4f^{(4)} &    \\ \hline
f(a)          & 1 &   &   &   &  &  f(a) \\
wf^{(1)}(a)   &   & 1 &   &   &  &  wf^{(1)}(a) \\
wf^{(1)}(a+w) &   & 1 & 1 &   &  &  wf^{(1)}(a) + w^2f^{(2)}(a) \\ 
wf^{(1)}(a+2) &   & 1 & 2 & 1 &  &  wf^{(1)}(a) + 2w^2f^{(2)}(a) + 1w^3f^{(3)}(a) \\ \hline
              & 1 & 3 & 3 & 1 &  &  f(a) + 3wf^{(1)}(a) + 3w^2f^{(2)}(a) + w^3f^{(3)}(a)
\end{array}
Each segment has binomial coefficients and they sum together (along with the base $f(a)$) to more binomial coefficients! The above result ends up being a consequence of the hockey stick identity. So the final approximation of $f(x)$ when we divide $a$ to $x$ into 3 interval is
$$f(x) \approx f(a) + 3wf^{(1)}(a) + 3w^2f^{(2)}(a) + w^3f^{(3)}(a)$$
Which of course we can rewrite as a summation and for the general case is:
$$
f(x) \approx \sum_{k=1}^{n}{n \choose k}w^kf^{(k)}(a)
$$
As always, $n$ is the number of intervals and $w=\frac{(x-a)}{n}$.
This approximation becomes exact as the number of intervals tends to infinity
$$
f(x) = \lim_{n\rightarrow \infty} \sum_{k=1}^{n}{n \choose k}w^kf^{(k)}(a)
$$
Substituting $w=\frac{(x-a)}{n}$:
$$
f(x) = \lim_{n\rightarrow \infty} \sum_{k=1}^{n}{n \choose k}\frac{1}{n^k}(x-a)^kf^{(k)}(a)  \tag{6}\label{eq6}
$$
Which I think is the Newton series for finite differences or at least related. This is as far as I got, so I failed to derive the Taylor series, but it's basically the same form as the Taylor series, which is:
$$
f(x) = \sum_{n}^{\infty}\frac{f^{(n)}(a)}{n!}(x-a)^n \tag{Taylor Series}\label{Taylor series}$$
Since $f(x) = f(x)$ and we have 2 infinite sums in $(x-a)$, the coefficients have to be equal to each other, so I inadvertently showed:
$$
\lim_{n\rightarrow \infty} {n \choose k}\frac{1}{n^k} = \frac{1}{k!} \tag{7}\label{eq7}
$$
I do not understand Qiaochu Yuan's remark that \eqref{eq6} is a generalization of
$e^t = \lim_{n \to \infty} \left( 1 + \frac{t}{n} \right)^n = \sum_{n \ge 0} \frac{t^n}{n!}$ but it sounds interesting and is probably worth a look. So if someone wants to expound on that, that'd be cool!
