m (→III. Computer Programming) |
m (→III. Computer Programming) |
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A computer would make a great job of checking this, in particular when convergence is slow. It will also allow us to deepen this important business of convergence. | A computer would make a great job of checking this, in particular when convergence is slow. It will also allow us to deepen this important business of convergence. | ||
| − | Let us compute \ref{eq:e1}: for $n=1$, this is $(1+(1/2))= | + | Let us compute \ref{eq:e1}: for $n=1$, this is $(1+(1/1))=2$. For $n=2$, this is $(1+(1/2))^2=9/4=2.25$. For $n=3$, this is $(1+(1/3))^3=(4/3)^3=64/27=2.\bar{370}$ |
{{WLP6}} | {{WLP6}} | ||
Computers are good at computing! We have seen various definitions of Euler's number $e$, of which, in particular:
$$\tag{1}e=\lim_{n\rightarrow\infty}\left(1+{1\over n}\right)^n$$
$$\tag{2}e=\sum_{k=0}^\infty{1\over k !}$$
A computer would make a great job of checking this, in particular when convergence is slow. It will also allow us to deepen this important business of convergence.
Let us compute (1): for $n=1$, this is $(1+(1/1))=2$. For $n=2$, this is $(1+(1/2))^2=9/4=2.25$. For $n=3$, this is $(1+(1/3))^3=(4/3)^3=64/27=2.\bar{370}$