Making Pi. Then we automatically get the formula C = 2πr, which enables us to evaluate C whenever we know the value of r.

Transcription

1 Maths 1 Extension Notes #3.b Not Examinable Making Pi 1 Intoduction 1.1 Definition of Pi Conside any cicle with adius and cicumfeence C. In the following sections, we show that the atio C 2 is just a numbe which does not change with diffeent cicles. This is a vey useful fact, since once we know what the numbe is, we can use it to find C wheneve we know the value. We define π to be this special atio: π = C 2. Then we automatically get the fomula C = 2π, which enables us to evaluate C wheneve we know the value of. 1.2 Evaluation of Pi It can be shown that the numbe π is iational (in fact tanscendental); this means that the decimal expansion of π goes on foeve and does not epeat itself, and so we can neve evaluate π exactly! What we do hee is we find a sequence of bette and bette appoximations fo π. To do this, we use the same method used by Achimedes (about 200 B.C.). 1

2 2 Achimedes Method fo finding π We fist find appoximations fo the cicumfeence C of a cicle. We can appoximate the value of C with the peimete of a polygon inscibed inside the cicle. In the following diagam, we have a hexagon inscibed inside a cicle of adius. The hexagon is made up of six equilateal tiangles, each of which has side length. Now C 6 = 6, and so π 6 2 = 3. To impove this appoximation fo π, we continually double the numbe of sides of the polygon inscibed inside the cicle. Then we get an inceasing sequence C 0 = l 0 6, C 1 = l 1 6 2, C 2 = l ,..., C n = l n 6 2 n,... of appoximations fo C. In fact, we actually define the cicumfeence C by C = lim n C n. Achimedes did not have a compute, calculato o even tables of the tigonometic functions (sine, cosine and tangent) at his disposal to calculate l n ; so how did he calculate l n without using tigonomety? 2

3 We stat with l 0 = (in the six-sided polygon). In ode to find l 1 we will use a 30 ight-angled tiangle. This can be found as follows. In the following diagam, the angle CP 0 D is a ight-angle (by the Cicle Lemma (a)). Also, by the Cicle Lemma (b), the angle CDP 0 = = 30. Thus, we can constuct a 30 ight-angled tiangle, with side lengths, and 3, as shown in the following diagam: 2 2 We use the 30 ight-angled tiangle to find l 1 : Note that we can calculate l 1 by using Pythagoas Theoem: ( ( 2 l 2 1 = + 2) ) = = (2 3) 2. Theefoe, without using tigonomety, we have calculated that l 1 =

4 As a bonus, we get even moe infomation: we can now constuct a 1 ight-angled tiangle. In the following diagam, the angle CP 1 D is a ight-angle (by the Cicle Lemma (a)). Also, by the Cicle Lemma (b), the angle CDP 1 = = 1. Thus, we can constuct a 1 l ight-angled tiangle, with side lengths, 12 and l 2 1, as shown in the following diagam: We can use this 1 ight-angled tiangle to calculate l 2. Now l 2 2 ( ) 2 ( l1 = + 1 ) l 1 = l l (42 l 2 1 ) ( ) 2 = l1 4 ) 2 = ( l1 4

5 and so l 2 = 2 4 ( l 1 ) 2. If we epeat this pocess by constucting a 7. ight-angled tiangle, we get l 3 2 = and so l 3 = ( ) 2 ( l ) l 2 = l 2 2 = ( l 2 ) 2. In geneal, fo any natual numbe i, we have We now have a method fo calculating l i : ( ) l i+1 = 2 li 2 4. l 0 = ( l 1 = l0 2 4 ( ) l 2 = 2 l1 2 4 = ( ) l 3 = 2 l2 2 4 = ( ) l 4 = 2 l3 2 4 = ) 2 = = ( l2 ) 2 Now, since C n = l n 6 2 n (see page 2), we have the following appoximations fo C: C 0 = l 0 6 = 6 C 1 = l = C 2 = l = C 3 = l = C 4 = l = Recall that, on page 2, we defined C by C = lim n C n. This means that π = C 2 = 1 2 lim C C n n = lim n n 2.

6 Theefoe, if we let π n = Cn 2, then the sequence π 0, π 1, π 2,... appoaches π as n. Ou appoximations fo π ae now π 0 = C 0 2 = 6 2 = 3 π 1 = C 1 2 = π 2 = C 2 2 = 12 2 = = = π 3 = C 3 2 = π 4 = C 4 2 = = = Since each π i (i = 0, 1, 2, 3,...) is just a eal numbe (independent of and C), we have that π is also just a eal numbe, as stated in Section Achimedes Calculation When Achimedes found appoximations fo π, he fist needed to find appoximations fo the squae oots of numbes (because he didn t have a calculato!). At the time, vaious methods fo finding squae oots wee known (see the Maths 1 Extension topic How does a calculato calculate sin x? ). He actually found the appoximation fo π Futhemoe, he found appoximations fo the cicumfeence C of a cicle by cicumscibing the cicle with polygons, i.e., by putting the polygons on the outside of the cicle. These othe appoximations ae always geate than C, wheeas the appoximations that we have found ae always smalle than C. By compaing the two diffeent appoximations, we can get an uppe bound and a lowe bound fo π. When n = 4, Achimedes found that < π < It is inteesting to note that = 22 is a vey common appoximation fo π that is still 7 7 used today! Compae the following values: = π 4 = π = = Rounding Eos Note that C 4 is the peimete of a polygon with 96 sides. If we calculated C, it would be the peimete of a polygon with 96 2 = 192 sides, and so C is a bette appoximation that C 4. 6

7 In geneal C n+1 is a bette appoximation that C n, and so π n+1 is a bette appoximation that π n. In theoy, this means that we can compute π to any degee of accuacy we like, by choosing n lage enough. Unfotunately, this does not wok in pactice. Conside the fomula C n = l n 6 2 n. When we calculate l n with a calculato o compute, we can only do so fo a fixed numbe of decimal places; this causes ounding eo. Fo example, suppose that we calculate l 20 with an eo of Since C 20 = l , we will have an eo of = fo C 20. As n gets even lage the tem 2 n causes small eos in l n to become lage eos in C n. This means that Achimedes Method is not good fo appoximating π to a lage numbe of decimal places with a compute o calculato. Fo example, using about 14 decimal places we have π 0 = 3 π π π π π π π 31 3 π The tue value of π is π = , and so ou appoximations π i stat to get wose when i Moden methods fo finding π Fo centuies, people have been fascinated by π and a lot of effot has been put into computing π to as many decimal places as possible (its decimal expansion neve epeats). To 39 decimal places of accuacy, it can be shown that π = But how can we find π so accuately? One of the most common methods fo computing π is called Machin s Fomula. 3.1 Machin s Fomula One method of calculating π is to use the facts that (a) π 4 = tan 1 (1) ; and (b) invese tangent can be expanded in a powe seies: tan 1 x = x x3 3 + x x7 7 + x 1. 7

8 It follows that π = 4 ( ). Unfotunately, the above seies conveges quite slowly. Fo example ( ) = With this seies, we need a lot of tems to evaluate π accuately. To get a seies that conveges faste, we note that the seies x x3 3 + x x7 7 + conveges faste when x is close to zeo. Machin ealised that tan 1 (1) = 4 tan 1 tan (see Section 4 fo details) and that the seies fo tan and tan 239 convege quite fast. Thus Machin s Fomula: π 4 = 4 tan 1 tan can be used to appoximate π faily accuately. Example 1. We have tan 1 ( 1 ) 1 ( 1 )3 3 + ( 1 ) = and ( ) 1 tan ( )3 + ( ) = Theefoe an appoximate value fo π is given by 3.2 Othe fomulas n=1 ( ) ( ) π = 16 tan 1 4 tan ( ) 4( ) = Thee ae hundeds of fomulas fo calculating π. Some inteesting ones ae (i) π [ 2 = (2n) 2 ] = (2n 1)(2n + 1) (ii) π2 6 = 1 n = (iii) π 2 = 1 2 n=1 n=0 (n!) 2 2 n+1 (2n + 1)! =

9 4 Execises Machin-like fomulas come fom the identity Let x = tan θ 1 and y = tan θ 2. Then and so tan(θ 1 + θ 2 ) = tan θ 1 + tan θ 2 1 tan θ 1 tan θ 2. tan(θ 1 + θ 2 ) = x + y 1 xy ( ) x + y θ 1 + θ 2 = tan 1. 1 xy Since x = tan θ 1 and y = tan θ 2, we have θ 1 = tan 1 (x) and θ 2 = tan 1 (y). Thus ( ) x + y tan 1 x + tan 1 y = tan 1. (1) 1 xy We can use the above identity to find numbes x and y such that tan 1 x+tan 1 y = tan 1 (1). Ou goal is to make x and y as close to zeo as possible, so that the powe seies fo tan 1 x conveges quickly. (a) (Eule s Fomula.) Use identity (1) and the fact that π 4 = tan 1 (1) to show that tan 1 + tan 1 = π (b) (Hutton s Fomula.) We now find x such that tan ( ) 2 = tan 1 x + tan ( ) 3. By identity (1), the numbe x must satisfy Solve the equation tan 1 ( 1 2 fo x, and then by using (a), show that (c) Show that ) ( ) x + 1 = tan x = x x π 4 = 2 tan 1 + tan α = x + y 1 xy x = α y 1 + αy. (2) 9

10 (d) By using (1) and (2) find x such that tan ( ) 3 = tan 1 x + tan ( ) 7 ; hence by using (b) show that ( ) ( ) π = 3 tan tan Identity (1) implies two othe identities: ( ) x y tan 1 x tan 1 y = tan + xy ( ) 2x 2 tan 1 x = tan x 2 These identities can be used to deive moe Machin-like fomulas: (3) (4) (e) (Hemann s Fomula.) Use (3) and (4) to show that π 4 = 2 tan 1 tan (f) (Machin s Fomula.) Use (4) to show that 4 tan 1 ( 1) = ( ) tan Now use (3) to show that tan 1 tan 1 = tan 1 (1) Hence deduce that π 4 = 4 tan 1 tan Refeence A lage amount of infomation about π can be viewed on the intenet at: (Don t type www befoe mathwold!) 10