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 $(1+(1/1))=2$. 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.\ | + | 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 $(1+(1/3))^3=(4/3)^3=64/27=2.\overbar{370}$. Unlike computers, we get pretty tired. So we can ask the computer to do it. This is how one popular computer program, [[Wolfram]]'s [[Mathematica]], would do it: |
| + | |||
| + | <pre> | ||
| + | Table[N[(1 + 1/n)^n], {n, 1, 10}] | ||
| + | {2., 2.25, 2.37037, 2.44141, 2.48832, 2.52163, 2.5465, 2.56578, \ | ||
| + | 2.58117, 2.59374} | ||
| + | </pre> | ||
{{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 $(1+(1/3))^3=(4/3)^3=64/27=2.\overbar{370}$. Unlike computers, we get pretty tired. So we can ask the computer to do it. This is how one popular computer program, Wolfram's Mathematica, would do it:
Table[N[(1 + 1/n)^n], {n, 1, 10}]
{2., 2.25, 2.37037, 2.44141, 2.48832, 2.52163, 2.5465, 2.56578, \
2.58117, 2.59374}