How Small Is a Unit Ball?Author(s): David J. Smith and Mavina K. VamanamurthySource: Mathematics Magazine, Vol. 62, No. 2 (Apr., 1989), pp. 101-107Published by: Mathematical Association of AmericaStable URL: http://www.jstor.org/stable/2690391 .

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VOL. 62, NO. 2, APRIL 1989 101

REFERENCES

1. Dennis Berkey, Calculus, Saunders, 1984, page 194, square solution. 2. H. Davenport, The Higher Arithmetic, 2nd ed., Hutchison University Library, London, 1964. 3. Leonard E. Dickson, History of the Theory of Numbers, Vol. 2, Chelsea Publishing Company, 1952. 4. , Introduction to the Theory of Numbers, Dover, 1957. 5. John Fraleigh, Calculus with Analytic Geometry, 2nd ed., Addison-Wesley, 1985, page 186, arbitrary

dimensions a and b. 6. Karl Friedrich Gauss, Disquisitiones Arithmeticae, Yale University Press, 1966. 7. A. 0. Gelfond, Solving Equations in Integers, Little Mathematics Library, Mir Publishers, Moscow,

1981. 8. Larson and Hostetler, Calculus, 2nd ed., D. C. Heath and Co., 1982, page 203, square solution. 9. Munem and Foulis, Calculus, 2nd ed., Worth, 1984, page 212, a = 8, b = 15 and page 218, square

solution. 10. Richard Silverman, Calculus with Analytic Geometry, Prentice-Hall, 1985, page 204, square solution

and page 212, a = 8, b = 15. 11. E. W. Swokowski, Calculus with Analytic Geometry, Prindle, Weber, and Schmitt, 1983, page 183,

a = 16, b = 21. 12. G. B. Thomas, Calculus with Analytic Geometry, 2nd ed., Addison-Wesley, 1956, page 96, square

solution and page 103, a = 8, b = 15. 13. G. B. Thomas and R. L. Finney, Calculus and Analytic Geometry, 6th ed., Addison-Wesley, 1984, page

205, square solution and page 212, a = 8, b = 15.

How Small Is a Unit Ball?

DAVID J. SMITH MAVINA K. VAMANAMURTHY

University of Auckland Auckland, New Zealand

The volume of the cube of edge d in Rn is dn so that, as the dimension n increases, this volume increases, stays constant, or decreases to zero according as d > 1, d = 1, or d < 1. The situation for the ball of radius r in R n is quite different.

For n = 0, 1,2,..., and r> 0, let Vn(r) denote the n-dimensional volume of the n-dimensional ball of radius r in Rn. Then VO(r) = 1, V1(r) = 2r, V2(r) = 7r2 V3(r) = (4/3),gr3, and in general

rn nT n/2!

The derivation of this formula for Vn(r) is a useful pedagogical device, and studying the formula reveals some interesting properties. For example, Vn(1) increases for O < n < 5 and decreases for 5 < n < x. Its maximum value is V5(1) = 87 2/15, and

2 limVn -Vn(l) = 0. In fact E 0Vj(r) converges for all r>0, and E??V2n(r)=err. Thus, for fixed r, Vn(r) tends to zero as n tends to infinity.

We derive the volume formula in three ways. The first and second methods use cross-sections and Fubini's theorem, and the third uses a polar-coordinate transforma- tion and some simple properties of determinants.

The asymptotic properties already mentioned and some others are then developed.

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102 MATHEMATICS MAGAZINE

The following basic properties of Gamma (F) and Beta (B) functions are used (cf. [4], p. 235 ff.):

(i) xF(x) = F(x + 1) for all x > 0. (ii) F(n + 1) =n! for n=0,1,2,... (iii) F(1/2) = V7T .

(iv) F(x + 1) - l2x(x/e)x, (x - ). (v) B(p, q) = JfOx,-'(I _ x)q-ldx = F(p)'(q)/F(p + q), for all p, q positive.

1. First Method (cf. [1], p. 412)

Regard R n as R n -2x R2. Then (x1,...,xj) is in the ball Bn(r) if and only if x12 + X2 + +X-2_

2 x1 X> < r2 that is, if and only if

x2 + X22 + ***+Xn2_ < r 2-x 2_1 - x2

Hence,

Vn(r) dxl dx2 *...dxn, Bn(r)

(r)(=( 2)(/r2 x2 xn

dx - ... dXn-2 dxn1 dxn.

By induction

_____n_-_2__2__n - 2)72

Vn(r) f _ x 2 X 2 dx d 22r

Using polar coordinates in R2 this expression becomes

.(n -2)/2 f2 do r(r2 -t2)(n-2)/2 tdt = 2 n/2) n =__ n_2_n

1'(n/2) ~~~~~F(n/2) n =F v( +1

2. Second Method (cf. [1], p. 416)

Regard R n as Rn-1 X R. Then

Vn( ) = L(r)(fBL( x) dxl dx-1) dx n and (by induction),

7TR *n 12 Jr 2 2 )(n - 1)/2

2vn l)2Irr 2)n-l/ d n,(e =r, 2_T(n_1)/2 rf r2 )(n , - -

n_ _ __ _ o 1

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VOL. 62, NO. 2, APRIL 1989 103

_n -_ )/2 (n - 1)/2 n

2

=n IT n( T1/ (2 2 =r T /2 r

= r r(n2 2+1)

3. Third Method LEMMA.

|sn dO f sin2O dO

.. fsiinS dfl = n/

Proof: For n= 1, the left-hand side is

f sinsdn = - cos 0

and the right-hand side is

qT1/2 qT1/2

(3)=1 (1=2.

For n >I2,

f sinnO dO = 2f / sinnO dO, (put sinO8 = F), 0 0

r In/2 =2J 5/iv dt,

- ft(n-1)/2(1 -t)"1/2 dt- B( n?+ 1 1 )

IF( ?1)

Hence by induction,

f sin O dO f sin2o dO sin2O df= 0 ( ..).s in ( n9 n 1 ) ) - 0e 0ntroduce n-dimensional(hericalcoordnat) asfo s(jc.) [] d. 2 1):

Wn te rintroducen-dsimensionlsphrclcodntsa olwcf3,p1)

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104 MATHEMATICS MAGAZINE

xi = t sin Olsin 02 ... sin On- 1

x2 = t cos Olsin 02 * sin On- 1

x3 = t cos O2sin 03 * sin On- 1

Xn_1 = tCosOn-2sinOn-1

x n=tCosOn-1

where O < t < r, O < Oi < 7T for i =2,..., n - 1, andO < 01 < 27T. For brevity, denote sinOi=s2, cos2i=c2, cotOi=ki, i= 1,...,n- 1. Then

x1/t x1k1 x1k2 xlkn-

x2/t -X2/k1 x2k2 * k a(x, ... *,Xn) x3/t 0 x3/k2 ...

a(t, 01X**, .n-1) 0 ... ~~~~~~xn lkn-1

xn/t 0 O ... xn/k

1 1 1

1 -_ 1 1

=t- xx2 x.nk1k2 ... kn- 1 0 -

0

1 0 0 n-1

Using elementary column operations, we can reduce this to

1 0 0 0

1 -(1+? ) 0

-1 t-xi

... xnki

... kn-l|1 -1 _1+ k2)

0

1 -1~~~~~~~~~-

=t-lx ..xnk ... kn 1(-1)n'(1?+k2) (? k2 )

8n-1 C1 1

tHc xl xn( Si ., sc S2S3 .Sn-

Hence,

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VOL. 62, NO. 2, APRIL 1989 105

V(r)=f dxl dx2 ... dxn Bn(r)

ff|dt f d | d2 ..

dln-, d(tf01d--n-1)

n 2,g sin u du f sinn-2u du.

By the above Lemma, this is equal to

r n 7T (n - 2)/2 r nT n/2 r nT n/2

n IF n n-2 + (n/2)r'(n/2) F(n/2 + 1)

4. Asymptotic Properties

(i) 00

E Vn(r) n=O

converges for all r > 0.

This is an immediate consequence of the following result and its proof.

THEOREM. EO on(n+fl)/2Vj(r) converges if and only if 0 < r < 2 .

Proof. Stirling's Formula states that

(2 ) ( 2e)

Hence,

#7, n/2 r q7T n/2 =

F(n/2 + 1) (,gn)1/2(n/2e /

rn(2 re) n) - qT1/2n( n + 1)/2

Thus,

[n(n + 1)/2V ( r) 1 /n r (2 ve )1/

so that

lim [n(n )/2V,( r )] 1n= r V2~w. n - oo

By the Cauchy root test, 00

E n(n+l)/2Vn(r) n=O

converges for r < 1/ 2ve and diverges for r> 1/ 2e. If rI= 1/A2e, then

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106 MATHEMATICS MAGAZINE

lim n(n+ 1)/2V(r) = 1/ Vr # 0, and, hence, the series diverges. (ii)

V,(r) = er I + | e-t dt) n=O 0 /T

Proof. V2n(r)= n! ,and

_2+lr 2r(2Tr 2)

v2n+lt =1. 3 5 .. (2)n+ 1)

Hence, 00

wrr2 V2n(r) =e

n=O

To calculate Y2V2n+i(r), we note that 0

nl n 2 x/4x /2 t2d n=O (2n?l )! Xt Vrxe

(cf. [2], no. 5.21.13). Hence, putting x = 47r 2

00 00 ~~~(27r 2)~

n=O n==O 3 ( )

=2r I

e7 r2 e-t2dt rF o 2rJ7r r2fr t2

= 7- e-r2frIet2 dt.

(iii) For any r > 0 there is an N > 0 such that Vj(r) is monotone decreasing for n > N.

Proof

V 2n ne

n~~~~

_2+lr 2 r(2 Tr 2) v2n+l(t

I = 3 5(.. (2n + 1)

Hence,

V2.(r) 1 3.5 ... (2n + 1) V2n+l(r) 2r 2-4 .. (2n)

and

V2n l(r) 1 2 4*..(2n) V2(r 7T 3

, 5 .. 2n-

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VOL. 62, NO. 2, APRIL 1989 107

V.(r) monotone decreasing means

V2.(r) >Iad V2,-(r) >1 v:n()>1 and V ()>1. V2n + l(r) V2.(r) That is, r < min(a./2, bn/7T), where

35 *.. (2n+ 1) and bn 2*4... (2n)

an 2.4 ... (2n) b 13. 5... (2n- 1)

(Note that in this case, r2 < a bn/2 T = (2n + 1)/2 T, hence n + (1/2) > 7Tr2.) Now an= Ht1(I + 1/2i) and bn = HM1(I + 1/(2i - 1)), hence both are strictly increas- ing and unbounded (cf [4], p. 32), so for each r > 0 there is an N such that Vn(r) decreases for n > N. Direct calculation shows, for example, that Vn(l) increases for n < 5 and decreases for n > 5, and V5(1) = 8v 2/15.

In conclusion, we remark that analogous results are true for the n-area an(r) of the n-dimensional sphere Sn(r), and similar arguments may be used since

anl(r)= !Vn(r).

R E F E R E N C E S

1. T. M. Apostol, Calculus, Volume 2, second edition, Wiley, 1969. 2. E. R. Hansen, A Table of Series and Products, Prentice-Hall, 1975. 3. K. Rogers, Advanced Calculus, Merrill, 1976. 4. E. T. Whittaker and G. N. Watson, A Course of Modern Analysis, Fourth Edition, Cambridge, 1927.

More on Incircles

HUSEYIN DEMIR CEM TEZER

Middle East Technical University Ankara, Turkey

The contents of this note came into being during the authors' search for a "synthetic" proof of the following result by H. Demir (FIGURE 1):

"Consider a triangle ABC and points P, Q on the line segment BC. If the incircles of the subtriangles ABP and AQC are congruent then the incircles of the subtriangles ABQ and APC are congruent."

(Notice that the requirements of the "Five Circle Theorem" ([2]) are partly redundant.)

Singularly enough, this question turned out to be less accessible than a more general result which was conjectured at the very outset of our investigations and later proved by means of the methods which will constitute the body of the present work:

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