Joseph B. Keller and Ravi Vakil. ( x p + y p ) 1/p = 1. (1) by x, when y is replaced by y, and when x and y are interchanged.

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1 =3 π, the value of π in l Joseh B. Keller and Ravi Vakil The two-dimensional sace l is the set of oints in the lane, with the distance between two oints x, y) and x, y ) defined by x x + y y ) 1/, 1. The distance from x, y) to the origin is then x + y ) 1/. The equation of the unit circle C, i.e., the circle with its center at the origin and radius 1, is x + y ) 1/ = 1. 1) Figure 1 shows C for = 1, 3/2, 2, 3, and. Equation 1) is unchanged when x is relaced by x, when y is relaced by y, and when x and y are interchanged. Therefore C is symmetric about the y-axis, about the x-axis, and about the line x = y. 1 =.8 =2 =3/2.6 =1 y x Figure 1. The unit circle C in the first quadrant, defined by 1), for = 1, 3/2, 2, 3,. 1

2 It is natural to define π as the ratio of the circumference of C in the -metric) to two times its radius also in the -metric), which is its diameter, 2. This definition has been well studied, see for examle [2], [1], and [3]. The circumference is the integral of the element of arclength ds = dx + dy ) 1/ around C. Thus π = 1 dx + dy ) 1/ = C C dy ) 1/ 1 + dx. 2) dx Because of the symmetry of C, its circumference is equal to four times its arclength in the first quadrant, or eight times its arclength in the first quadrant between the lines x = and x = y. When x = y, 1) shows that x = 2 1/, so the integral in 2) is 8 times the integral from to 2 1/. By calculating dy/dx from 1), we can rewrite 2) as π = 4 2 1/ 1 + x 1 1 ) 1/ dx. 3) For = 1, 3) yields π 1 = 42 1/ )2 1/ ) = 4. For =, the integrand is 1 and the uer limit is 1, so π = 4. At = 2, π 2 = π. If geometry had been develoed using the l distance instead of the l 2 distance, π would have relaced π 2, which is just the familiar π. Figure 2 shows a grah of π as a function of, obtained by numerical integration of 3). The grah suggests that as increases from = 1, π decreases monotonically from its maximum value π 1 = 4 to its minimum value π 2 = π, and then increases monotonically to π = 4. In fact this is the case, and was roved by Adler and Tanton in [1]. Thus for each in 1 2, there is a q in 2 q such that π = π q. 4) 2

3 π 3.6 PSfrag relacements π Figure 2. The grah suggests that π = π q for 1/ + 1/q = 1 To find q we recall that the Hölder inequality involves two exonents and q related by q = 1. 5) The numerical results shown in Figure 2 lead one to conjecture that 4) will hold when 5) does. Indeed, Adler and Tanton earlier asked recisely this question as the concluding remark to [1]. We learned of Adler and Tanton s work only after writing this note.) In fact, when 5) holds, the domains bounded by C and C q are olar or dual) to one another. If K is a convex set in R 2, then its olar is defined by {y R 2 : x y 1 x K}.) Then a result of Schäffer [4] see also Thomson [5,. 118, Cor , and. 22, Cor ]) shows that 4) holds. This argument from Minkowski geometry suggested to us that there should be a direct elementary exlanation. We shall now give another roof that 4) holds when 5) does, by showing that then the integral 3) for π equals that for π q. We begin by writing the equation for the arc of C in the first quadrant in terms of a arameter t [, ], setting x = f 1 t) and y = f 2 t). Then 3

4 the length L of that arc can be written as L = f 1 + f 2 ) 1/ dt. 6) We choose the arameter t such that t q/ is the sloe of the line from the origin to the oint f 1 t), f 2 t)) on C, so that t q/ = f 2 t)/f 1 t). From this equation and 1) we find that f 1 t) = t q + 1) 1/, f 2 t) = t q + 1 ) 1/. 7) We arameterize C q in the same way, setting x = g 1 t) and y = g 2 t), with g 1 t) = t + 1) 1/q, g 2 t) = t + 1 ) 1/q. 8) Now we define the function F t) = f 1 g 2 + f 2 g 1. At the ends of the two arcs, t = and t =, we have f 1 = g 1 and f 2 = g 2. Therefore F ) = and F ) =, so F t)dt =. This equation can be rewritten as follows, by differentiating the definition of F t) to get F t) and then transosing: f 1g 2 + f 2g 1 ) dt = g 1f 2 + g 2f 1 ) dt. 9) Of course, this is just integration by arts, but it is enlightening to interret F t) as essentially a cross roduct. We now show that the integrand on the left side of 9) can be rewritten as f 1 g 2 + f 2 g 1 = f 1 + f 2 ) 1/. 1) To rove 1) we first transform the left side as follows: f 1 g 2 + f 2 g 1 = 1 ) t q + 1) +1 qt q 1 t + 1) 1/q 4

5 + 1 ) t q + 1) +1 q)t q+1) t + 1) 1/q = q tq + 1) +1 t + 1) 1/q t q 1+/q + t q+1)/ q+1) ) = q tq + 1) +1 t + 1) 1/ 1 t t + 1) = q tq + 1) +1 t + 1) 1/ t ) Then we transform the right side: f 1 + f 2 ) 1/ = [ ) 1 tq + 1) +1 1 qt q 1 + t q + 1 ) +1 = q tq + 1) +1 t q 1) + t q+1) q+1)) 1/ qt q+1) ) ] 1/ = q tq + 1) +1 t q + t q ) 1/ = q tq + 1) +1 t + 1) 1/ t ) The last forms of 11) and 12) are the same, which roves 1). The integral from to of the right side of 1) is just L, as 6) shows. Therefore the integral of the left side, which is also the left side of 9), is also L. A symmetrical argument shows that the right side of 9) is L q, so 9) shows that L = L q. This roves that 4) is true when and q are related by 5). This argument is geometrically motivated, not just formal maniulation. Suose q >. As t goes from to 1, the oint on C is behind the oint on C q. At t = 1 the -oint asses the q-oint. The cumulative lengths at time t are not the same. The difference is twice the area of the triangle sanned by the origin, the -oint and the q-oint see the cross roduct comment above). This difference vanishes at t =, when the two oints coincide. 5

6 1.9.8 q=3.7.6 =3/2 y x Figure 3. At time t, the q-oint and the -oint are at different angles from the x-axis. The signed difference in cumulative lengths is the signed area of the region shown. ACKNOWLEDGMENTS. Jonathan C. Mattingly and Arnold D. Kim calculated π. Rafe Mazzeo and Zhongmin Shen brought to our attention the work of Schäffer. Gautam Iyer rovided invaluable hel with the figures. We thank them all. We also thank the referees for bringing to our attention many references we were not earlier aware of, and for comments which substantially imroved the exosition of this article. References [1] C. L. Adler and J. Tanton, π is the minimum value of Pi, College Math. J. 31 2) [2] R. Euler and J. Sadek, The πs go full circle, Math. Mag )

7 [3] R. Poodiack, Generalizing π, angle measure, and trigonometry 24), available at htt://www2.norwich.edu/rodiac/i.df. [4] J. J. Schäffer, The self-circumference of olar convex disks, Arch. Math ) [5] A. C. Thomson, Minkowski Geometry, Cambridge University Press, Cambridge, Addresses: Deartment of Mathematics, Stanford University, Stanford CA 9435, 7

Joseph B. Keller and Ravi Vakil. Department of Mathematics. Stanford University, Stanford, CA February 15, 2008

π p, the value of π in l p Joseph B. Keller and Ravi Vakil Department of Mathematics Stanford University, Stanford, CA 9345-15 February 15, 8 The two dimensional space l p is the set of points in the plane,