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    • cake

      What is the Area of a Circle?

      Tom Körner
      1 June, 2007
      3 comments

      Suppose that we are asked to find the area enclosed by a circle of given radius. A simple way to go about this is to draw such a circle on graph paper and count the number of small squares within it. Then

      area contained ≈ number of small squares within circle × area of a small square.

      A circle drawn on graph paper - the area inside is approximately the number of small squares times the area of each small square

      A circle drawn on graph paper - the area inside is approximately the number of small squares times the area of each small square

      If we doubled all the lengths, involved then the new circle would have 4 times the area contained in the old circle

      If we doubled all the lengths involved then the new circle would have 4 times the area contained in the old circle

      We notice that if we doubled all the lengths involved then the new circle would have twice the radius and each of the small squares would have four times the area. Thus

      area contained in new circle of twice the radius

      ≈ number of small squares × area of a new small square
      = number of small squares × 4 × area of a old small square
      ≈ 4 × area contained in old circle.

      By imagining what would happen if we used finer and finer graph paper, we conclude that doubling the radius of a circle increases the area by a factor of 4.

      The same argument shows that, if we multiply the the radius of a circle by k, its area is multiplied by k2. Thus  area contained within in circle radius r=r2× area contained within in circle radius 1.  If, following tradition, we write  area contained within in circle radius 1 =π, we obtain the memorable formula  area contained within in circle of radius r=πr2.

      We can now play another trick. Consider our circle of radius r as a cake and divide it into a large number of equal slices. By reassembling the slices so that their pointed ends are alternately up and down we obtain something which looks like a rectangle of height r and width half the length of the circle.

      Cake slices rearranged into a rectangle

      The area covered by a cake is unchanged by cutting it up and moving the pieces about.

      But the area of a rectangle is

      width × height

      and the area of of the cake is unchanged by cutting it up and moving the pieces about. So

       half length circle ×r≈ width of rectangle × height of rectangle = area of rectangle ≈ area enclosed by circle =πr2. Dividing by r and multiplying by 2, we get  length circle ≈2πr. By increasing the number of slices, we can make our approximations better and better, obtaining another memorable result  length of circle with radius r=2πr. The formulae we have obtained for the area enclosed and the length of a circle are very nice but we can not actually use them unless we know a good approximation to π.

      Approximating Pi

      One approximation goes back to the ancient Greeks who looked at the length of a regular polygon inscribed in a circle of unit radius. As we increase the number of sides of the polygon the length of the polygon will get closer and closer to the length of the circle that is to 2π.
      Square inscribed in a circle. Equilateral triangle inscribed in a circle

      Can you compute the total length of an inscribed square? Of an inscribed equilateral triangle?

      Francois Vieta

      Francois Vieta

      Many of the ideas in this article go back to Archimedes, but developments in mathematical notation and computation enabled the great 16th century mathematician Vieta to put them in a more accessible form. (Among other things, Vieta broke codes for the King of France. The King of Spain, who believed his codes to be unbreakable, complained to the Pope that black magic was being employed against his country.)


      Octagon inscribed in a circle

      We can approximate the circle with n-sided polygon,
      in this case an hexagon with n=6.

      We use trigonometry to find a general formula for the length of the perimeter of an inscribed n-sided regular polygon. Observe that the polygon is made up of n triangles of exactly the same shape. If we look at one of these n triangles, OBC say, we see that (as we have drawn things)  length perimeter =n× length BC. If X is the mid point of BC then OBX is a right angled triangle with hypotenuse OB of length one and angle ∠BOX=12×∠BOC=3602n. By trigonometry  length BX=sin⁡3602n and so  length BC=2× length BX=2sin⁡3602n. We have shown that  length inscribed polygon =2nsin⁡3602n. Since we are interested in π which is half the length of the perimeter of the circle of radius 1, we consider ln=12× length inscribed polygon =nsin⁡180n. When n is large, we expect that ln will be close to π. Trying out our formula on a hand calculator, we get l100=200×sin⁡1.8=3.141075908.

      If you calculated the length of the perimeter for an inscribed square or triangle, does our general formula for an n-sided polygon agree for n = 3 and 4? If you try to use your calculator to calculate l5 , l6, you'll observe that the results aren't a very good approximation for π.

      There are two problems with our formula for ln . The first is that we need to take n large to get a good approximation to π. The second is that we cheated when we used our calculator to evaluate sin⁡(180/n) since the calculator uses hidden mathematics substantially more complicated than occurs in our discussion.

      Doubling sides

      How can we calculate sin⁡(180/n)? The answer is that we cannot with the tools presented here. However, if instead of trying to calculate sin⁡(180/n) for all n, we concentrate on sin⁡(180/4), sin⁡(180/8), sin⁡(180/16), ... (in other words we double the number of sides each time), we begin to see a way forward. Ideally, we would like to know how to calculate sin⁡x/2 from sin⁡x. We can not quite do that, but we do know the formulae (from the standard trigonometric identities)sin⁡x=2sin⁡x2cos⁡x2 and cos⁡x=cos2⁡x2−sin2⁡x2=2cos2⁡x2−1. Thus cos⁡x2=121+cos⁡x and sin⁡x2=sin⁡x2cos⁡(x/2). We can use these forumulae to find cos⁡(180/2),sin⁡(180/2),cos⁡(180/4),sin⁡(180/4), ... cos⁡(180/16),sin⁡(180/16). Try it out on your caculator (but don't use the trigonometric function keys) and hence find l16 and l32. We can make the pattern of the calculation clear by writing things algebraically. Let us take y=45 and write ar=cos⁡(2−ry), and sr=l2r+2. Then s0=l4=4sin⁡y,s1=l8=8sin⁡(y/2)=4sin⁡ycos⁡(y/2)=s0a1,s2=l16=16sin⁡(y/4)=8sin⁡(y/2)cos⁡(y/4)=s0a1×a2 and, continuing as far as we like, we see that sn=s0a1×a2×a3×…×an. Thus for very large n π≈s0a1×a2×a3×…×an. Since y=45, s0=42 and a0=12, and we can write s0=2a0. Therefore sn=2a0×a1×a2×…×an.

      Vieta's formula and beyond

      Although Vieta lived long before computers, this approach is admirably suited to writing a short computer program. The definitions of an and sn lead to the rules an+1=121+an and sn+1=snan+1 and we can use these to easily compute successive values of sn. If you have a computer or programmable calculator, see if you can compute s1,s2,…,s20. Nowadays we leave computation of square roots to electronic calculators, but, already in Vieta's time, there were methods for calculating square roots by hand which were little more complicated than long division. Vieta used his method to calculate π to 10 decimal places. We have shown that 2π≈a0×a1×a2×…×an, and the larger n is, the better the approximation. It is natural to write 2π=a0×a1×a2×…×an×an+1×an+2×… where, in some sense, we multiply together an infinite number of terms. Using the rule above to write the successive values for an (starting with a0=1/2), we obtain Vieta's formula: 2π=12×12+1212×12+1212+1212×…

      From an elementary point of view this formula is nonsense, but it is beautiful nonsense and the theory of the calculus shows that, from a more advanced standpoint, we can make complete sense of such formulae.

      The Rhind Papyrus

      The Rhind Papyrus, from around 1650 BC,
      is thought to contain the earliest
      approximation of pi
      ; as 3.16.

      The accuracy of this approximation increases fairly steadily, as you will have seen if you used your calculator to compute the successive values of sn. Roughly speaking the number of correct decimal places after the nth step is proportional to n. Other methods were developed using calculus, but it remained true for these methods that the number of correct decimal places after the nth step was proportional to n.

      In the 1970's the mathematical world was stunned by the discovery by Brent and Salamin of method which roughly doubled the number of correct decimal places at each step. To show that it works requires hard work and first year university calculus. However, the method is simple enough to be given here.


      Take A0=1,B0=1/2 and S0=1/2. Let An+1=An+Bn2,Bn+1=An×Bn and Sn+1=Sn−2n+1(An+12−Bn+12) Then Pn=2An2Sn is a good approximation to π. Using this algorithm, it is easy to compute the first few approximations, say P1, P2, P3 and P4, on your calculator.

      Since then even faster methods have been discovered. Although π has now been calculated to a trillion (that is to say 1,000,000,000,000) places, the hunt for better ways to do the computation will continue.


      About the author

      Tom Körner is a lecturer in the Department of Pure Mathematics and Mathematical Statistics at the University of Cambridge, and Director of Mathematical Studies at Trinity Hall, Cambridge. His popular mathematics book, The Pleasures of Counting, was reviewed in Issue 13 of Plus.

      He works in the field of classical analysis and has a particular interest in Fourier analysis. Much of his research deals with the construction of fairly intricate counter-examples to plausible conjectures.

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      Comments

      Anonymous

      8 April 2013

      Permalink

      I'm a 16 years old student from Israel and really liked and enjoyed this article. In my opinion, it prooves us once more how Math is so pure and beautiful.

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      Anonymous

      19 October 2023

      In reply to Lovely =] by Anonymous

      Permalink

      its been 10 years since you left this comment and i hope that you changed your mind about math being "pure" because this article almost gave me a stroke

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      Anonymous

      21 April 2013

      Permalink

      If we are given the circumference of a circle to be 2(PI)r, then by simple calculus the area will be (PI)r^2. This comment is submitted by Peter L. Griffiths.

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