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

4

+2

.

Therefore,

π=24

3

32

+

1

12

-

S

1

=

3

4

+2-24

S

1

=

S

2

-24

S

1

.