WOLFRAM|DEMONSTRATIONS PROJECT

Euler's Estimate of Pi

​
choose formula for π/4
-1
tan
1
2
+
-1
tan
1
3
2
-1
tan
1
3
+
-1
tan
1
7
3
-1
tan
1
7
+ 2
-1
tan
2
11
5
-1
tan
1
7
+ 2
-1
tan
3
79
decimals
27
round off
number of terms of part 1
8
number of terms of part 2
8
show Euler's calculation
term
1
2400000000000000000000000000
––––
2
0160000000000000000000000000
––––
3
0012800000000000000000000000
––––
4
0001097142857142857142857142
––––
5
0000097523809523809523809523
––––
6
0000008865800865800865800865
––––
7
0000000818381618381618381618
––––
8
0000000076382284382284382284
Part I.
2574004427231435231435231432
term
1
0560000000000000000000000000
––––
2
0007466666666666666666666666
––––
3
0000119466666666666666666666
––––
4
0000002048000000000000000000
––––
5
0000000036408888888888888888
––––
6
0000000000661979797979797979
––––
7
0000000000012221165501165501
––––
8
0000000000000228128422688422
Part II.
0567588218416651314125874122
I. + II.
3141592645648086545561105554
π
3141592653589793238462643383
In [1] Euler derived the formula
arctan(t)=
t
1+
2
t
∞
∑
i=0
(2i)!!
(2i+1)!!
i
2
t
1+
2
t
. He claimed that his formula was better for calculation than the Leibniz–Gregory formula
arctan(t)=
∞
∑
i=0
2i-1
t
2i-1
, since for
t=
1
3
,
1
7
,
3
79
, the factor
2
t
1+
2
t
in the series has values
1
10
,
2
100
,
144
100000
, which are simpler to calculate with. He illustrated this with the formula
π=8arctan
1
3
+4arctan
1
7
. He calculated eight terms of the sum for each of the arc tangents on the right to 27 decimal places each and concluded that
π≈3.14159265
. On the next page he calculated terms 9–16 of the first part and terms 9–10 of the second part and concluded that
π≈3.14159265358979315
. To eighteen places,
π≈3.14159265358979324
. To 30 places,
π≈3.14159265358979323846264338328
.