The Euler's constant \(e\) is the value of the exponential function at the argument \(1\):
\[e:=\exp(1)=\sum_{n=0}^\infty \frac{1}{n!}=2.718~281~828~459...\]
Practically, it can be calculated using the first \(N\) terms of the exponential series \[\sum_{n=0}^N \frac{x}{n!}=\frac{x^N}{N!}+\frac{x^{N-1}}{(N-1)!}+\ldots+ x + 1\] and considering them as a real polynomial of degree \(N\). We can then significantly reduce the number of multiplications required to calculate it by regrouping this polynomial using the Horner scheme and getting
\[\sum_{n=0}^N \frac{x}{n!}=\left(\left(\ldots\left(\left(\frac xN + 1\right)\frac{x}{N-1}+1\right)\frac{x}{N-2}+\ldots\right)\frac x2 + 1 \right)+ x + 1.\]
For \(x=1\) we get the expression
\[e=\sum_{n=0}^\infty \frac{1}{n!}=\left(\left(\ldots\left(\left(\frac 1N + 1\right)\frac{1}{N-1}+1\right)\frac{1}{N-2}+\ldots\right)\frac 12 + 1 \right)+ 1 + 1 + r_{N+1}\]
with some remainder term \( r_{N+1}\), which can be estimated by \[r_{N+1}\le \frac 2{(N+1)!}.\]
For \(N=15\) we have \(r_{16}\le \frac 2{16!} < 10^{-13}\) the estimation
\[e=2.718~281~828~459\pm 2\cdot 10^{-13}.\]
