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WOLFRAM|DEMONSTRATIONS PROJECT

Newton's Pi Approximation

​
calculation
semicircle
decimal places
16
number of terms in series
S
1
20
1
160
62500000000000
1
3584
2790178571428
1
36864
271267361111
5
1441792
34679066051
7
13631488
5135169396
7
83886080
834465026
33
2281701376
144628917
429
163208757248
26285354
715
1443109011456
4954580
2431
25288767438848
961296
4199
219902325555200
190948
29393
7599824371187712
38675
52003
65302194596872192
7963
185725
1116892707587883008
1662
111435
3170534137668829184
351
1938969
258254417031933722624
75
17678835
10920472491636054556672
16
21607465
61390764277305387778048
3
119409675
1548936206381243630092288
0
883631595
51983810243429054512365568
0
S
1
65602271702852
24
S
1
1574454520868448
S
2
= 2+
3
3
4
32990381056766579
S
2
- 24
S
1
31415926535898131
π
31415926535897932
This Demonstration gives Newton's approximation of
π
based on calculating the area of a semicircle using an integral.
The area
α
under the arc
AD
(light blue) is equal to the area of the sector
ADC
minus the area of triangle
BCD
(light green), that is,
π
24
-
3
32
.
On the other hand, the semicircle has equation
y=
x-
2
x
=
x
1-x
, so
α=
1/4
∫
0
x
1-x
dx=
1
12
-
1
160
-
1
3584
-
1
36864
-…
. Set
S
1
=
1
160
+
1
3584
+
1
36864
+...
and
S
2
=
3
3
4
+2
.
Therefore,
π=24
3
32
+
1
12
-
S
1
=
3
3
4
+2-24
S
1
=
S
2
-24
S
1
.
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