Notes and Solutions #5 - Haverford Collegeblogs.haverford.edu/mathproblemsolving/files/2010/05/S4-Sleeping... · Notes and Solutions #5 Meeting of 7 October 2008 Sleeping Beauty Michael

HAVERFORD COLLEGE PROBLEM SOLVING GROUP 2008-9

Oct 16, 2008 W:\Haverford\Problems\2008-9\Solutions 5.doc

Notes and Solutions #5 Meeting of 7 October 2008

Sleeping Beauty

Michael Lavine related this problem at dinner Monday night after his talk. It is originally a philosophical problem about knowledge and consciousness, with possible implications concerning the meaning of probability. A mathematical treatment is possible, so we discussed it at the end of our meeting Tuesday.

In a paper published in 2000, Adam Elga posed a problem that morphed into this more fanciful version. The legendary Sleeping Beauty (SB) agreed to conduct an experiment before her hundred-year sleep. The plan calls for doctors to flip a fair coin after SB falls asleep Sunday evening. If heads is observed, SB will be awakened on Monday and asked for her probability that the coin landed heads. She will then resume her long sleep, awakening only with the prince’s arrival.

If tails is observed the initial protocol is the same: awaken on Monday, ask for the probability of heads. However, the doctors then administer a magical drug that erases SB’s memories of awakening, returning her to the mental state she was in on Sunday evening. She goes back to sleep. The doctors will awaken her again on Tuesday morning and repeat the probability question.

SB understands this protocol (and it is approved by her kingdom’s Institutional Review Board, which must review any experiment governing human subjects). The fairness of the coin is not in question, although of course SB is not privy to the outcome of the flip. Neither does she know what day it is when she is awakened, due to the workings of the drug.

What should SB answer when asked the probability of heads?

Solution

(1) Simple: it’s a fair coin; the probability of heads is ½.

(2) There are three equally likely possibilities: heads and it’s Monday, tails and it’s Monday, or tails and it’s Tuesday. In only one of these three cases does the coin land heads, so the probability is 1

3 .

Introduction

Before we address this apparent paradox, let’s consider some of the things that have been written about it.

“[Y]ou have gone from a situation in which you count your temporal location as irrelevant to the truth of H, to one in which you count your own temporal location as

HAVERFORD COLLEGE PROBLEM SOLVING GROUP SOLUTION SET #5 Fall 2008

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relevant to the truth of H.” [Elga (2000), as quoted by Arntzenius (2002).] In other words, Elga is a “thirder:” he claims that a rational person—SB—would give a probability of ½ before going to sleep, but upon awakening—having no more information than before except that now it’s not Sunday anymore and she has been awakened as expected—somehow she must “update” her probability from ½ to 1

3 .

Many papers have been written arguing that 13 is correct because otherwise SB will

expect to lose money if she bets on heads with any other odds. I won’t go into these arguments, because they only obscure the situation, but consult Vineberg (undated) for an example.

Other papers argue just as strongly that ½ is correct, such as Lewis (2001). Their authors are “halfers.” Given that ½ is obviously the correct answer to the question on Sunday night, many philosophers struggle openly with the problem of how and why SB changes her answer from ½ to 1

3 despite receiving no new information upon being awakened.

Many of these writers seem to see the issue as one of updating a belief based on self-perception of one’s “temporal location” in the world. They have brought many interesting arguments to bear on this, including a quantum-mechanical “many worlds” analogy (Papineau & Dura-Vila 2008) and drawing visible and invisible balls from urns (Franceschi 2005). The urn models typically involve two urns. One is chosen by the coin flip. Thus, the contents of one urn model a Monday-only awakening and the contents of the other urn model the Monday-Tuesday sequence. A challenge for the balls-in-urns modelers is that it’s unnecessary to flip the coin on Sunday: the experimenters can wait as late as Tuesday morning to make the flip. How, then, do you simulate Monday’s awakening by drawing balls from an urn that hasn’t yet been chosen?

Others try to make the problem go away by challenging the meaningfulness of the probability question. Arntzenius (2002) suggests “it seems more plausible to say that her epistemic state upon waking up should not include a definite degree of belief in heads.” Why? Because “the real issue is how one deals with known, unavoidable, cognitive malfunction.” In other words, the drug—or at least the potential to be administered the drug—confuses SB.

What is left for mathematics to contribute? Nothing, I contend, except a clear answer based on axiomatic principles of probability and its application to experimental situations. This solution is most closely approached by a recent paper of Groisman (2007). The following analysis is inspired by Groisman, but whereas he pursues a ball-in-urn analogy, here we address the experiment directly, with the result that our resolution of the paradox is a little different.

Analysis

Probability can mean many things, but it becomes objective and operational in the context of a well-defined experiment. This seems to be such a situation. By definition, an experiment has a specific preparation, which is the set of conditions required for its


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potential repetition, and a definite sample space, which is the set of all possible outcomes (results) of the experiment. No two experiments are exactly like—they cannot, after all, occur simultaneously in the same space—and therefore we have to accept that there will be incidental differences among replications of any experiment. Those differences are, by definition, not part of the preparation. We will exploit this below in a positive way, by introducing an incidental difference that clarifies the nature of the experiment.

Due to the unusual action of the memory erasing drug, we have to be careful in describing the outcomes. It would seem there are two of them: either the coin is heads and SB is awakened on Monday, or the coin is tails and SB is awakened on Monday and on Tuesday. However, the “memory erasure” after the tails/Monday awakening means that Tuesday’s Sleeping Beauty is not the natural continuation of Monday’s Sleeping Beauty. Really, as far as knowledge, consciousness, and memory are concerned, Tuesday’s SB is an independent clone of Sunday’s SB.

Let’s explore this cloning possibility. I claim the experiment is equivalent to the following: prepare two identical copies of SB on Sunday evening. This is done in a way that does not disturb her: for example, put her through the teleporter after she falls asleep on Sunday and reconstruct two copies in the receiver. Whenever you need to awaken SB during the experiment, select a clone who has not yet been awakened for this purpose. The memory erasing drug is no longer necessary. Because this experimental preparation (“setup” in the language of Groisman (2007)) creates exactly the same set of possible results—right down to SB’s mental state and awareness—with exactly the same probabilities, it is valid to use it for our analysis1.

In fact, it helps to consider a slight generalization: prepare n > 1 identical copies of SB. (She knows this will happen; we’re not trying to trick her.) Whenever you need to awaken an SB, choose one of these clones. The choice ought to be random, but if you use any procedure, random or not, which is unknown to SB, from her point of view it will seem random anyway, and that’s all that matters for her evaluation of probability.

What, now, is the sample space? Number the SB clones 1, 2, …, n; let “M” and “T” designate the days of awakening; and let “h” and “t” designate the outcome of the coin flip. The set of experimental outcomes Ω is

hM1, hM2, …, hMn, (tM1, tT2), (tM1, tT3), …, (tM1, tTn), (tM2, tT1), (tM2, tT3), …, (tM2, tTn), … (tMn-1, tT1), (tMn-1, tT2), …, (tMn-1, tTn-2), (tMn-1, tTn) .

There are n Monday outcomes and n(n-1) Monday-Tuesday outcomes (you cannot awaken the same SB clone on both days).

1 If you are interested in the philosophical implications, then this is where you will want to criticize the argument. From this point on, everything is mathematical and if you’re a halfer, you’re sunk.


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Before conducting the experiment, how will SB assess the probabilities? The probability of the Monday-only event, h = hM1, hM2, …, hMn is the probability of heads which, because the coin is fair, must be ½. The probability that any particular SB will be selected, given that the coin is heads, is 1/n, because h comprises n equally likely outcomes. Letting h(i) be the event that the ith clone is awakened and the coin lands heads, we have

h(i) = hMi,

1

[ ]2

=hP , and

1

[ ( ) | ]in

=h hP .

Therefore—this is an axiom of probability—we compute

1 1 1

[ ( )] [ ( ) | ] [ ]2 2

i in n

= ⋅ = =h h h hP P P .

As one of the SB clones, your chance of being awakened on a Monday during a heads-up experiment is 1/(2n).

Similarly, still before the experiment is done, we can compute the chance of being awakened with a tails outcome. The set of tails outcomes is the complement Ω – h = t. The set of tails outcomes where clone i is awakened is

t(i) = (tMi, tTj) | j ≠ i ∪ (tMj, tTi) | j ≠ i.

This clone is awakened on Monday or on Tuesday but not on both days. This event has (n-1) + (n-1) = 2(n-1) outcomes, each of which is equally probable. Thus

1

[ ]2

=tP ,

( )( )

2 1 2[ ( ) | ]

1n

in n n

−= =

−t tP , and therefore

1 2 1

[ ( )] [ ( ) | ] [ ]2

i in n

= ⋅ = =t t t tP P P .

Now any one of the SB clones can compute the chances of heads, given that she is the one awakened, which is the outcome t(i) ∪ h(i):

( )

1[ ( ) ( ) ] [ ( )] 12[ | ( ) ( )]

1 1[ ( ) ( )] [ ( ) ( )] 32

i i i ni ii i i i

n n

∩ ∪ ∪ = = = =

∪ ∪ +

h t h hh t h

t h t hP PP P P .


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Again this is an axiomatic calculation: it is how probability works, period. Moreover, SB can carry it out before she goes to sleep Sunday night before she is cloned (or drugged). Therefore she is consistent in her evaluation of probabilities and nothing changes upon awakening.

Why does this resolve the problem? Let’s let Groisman (2007) sum up:

“The concept of an event is central and crucial in Probability Theory2. The Sleeping Beauty Problem arises due to improper use of the notion of an event. The setup under which the event takes place must always be taken into account. If we do so, then we realize that the original question posed to SB can be interpreted in two different ways. The first interpretation is ‘What is your credence that the coin landed Heads under the setup of coin tossing?’, and the answer should be ½. The second interpretation is ‘What is your credence that this awakening is a Head-awakening under the setup of wakening?’, and the answer should be 1

3 . Thus there is no paradox!”

Conclusion

Our mathematical analysis clarifies Groisman’s point and modifies it subtly. The question “what is the probability of heads” has two interpretations: it can ask for [ ]hP , which is ½ (answer #1), or it can ask for [ | ( ) ( )]i i∪h t hP , which is 1

3 (answer #2). One difference between our analysis and Groisman’s is that both these probabilities are clearly computed within the same experimental setup. They differ because “the coin landed heads” ambiguously refers to two different probabilities: the probability of h or the conditional probability of h given t(i) ∪ h(i). I guess that, following Elga (2000), you could legitimately characterize the latter as “count[ing] your own temporal location as relevant to the truth of h,” but that characterization adds no insight to the problem. It only detracts from the mathematical facts in evidence. There really doesn’t seem to be anything going here that is not adequately accounted for by the axioms of probability and the most basic framework of using probability to understand experiments. Our analysis effectively reduces the paradox to a confusion about conditional probability. It also suggests that the underlying philosophical issue is one of identity: What happens to the clones who are not awakened? What relationships hold among the clones?—but that’s a discussion for a different forum.

If you think this analysis is convincing, or at least illuminating, then look over the references (next page) and ponder the contrast between philosophical approaches to the problem and a mathematical approach.

—Bill Huber, 16 October 2008

2 An event is any set of outcomes to which you can assign a probability. Here, h, t, h(i), t(i), and t(i) ∪ h(i) are all examples of events.


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References

Arntzenius, Frank (2002). Reflections on Sleeping Beauty. Analysis 62.1 pp 53-62.

Elga, Adam (2000). Self-locating belief and the Sleeping Beauty Problem. Analysis 60 pp 143-7.

Franceschi, Paul (2005). Sleeping Beauty and the Problem of World Reduction. Preprint.

Groisman, Berry (2007). The end of Sleeping Beauty’s nightmare.

Lewis, D (2001). Sleeping Beauty: reply to Elga. Analysis 61.3 pp 171-6.

Papineau, David and Victor Dura-Vila (2008). A thirder and an Everettian: a reply to Lewis’s ‘Quantum Sleeping Beauty’.

Pust, Joel (2008). Horgan on Sleeping Beauty. Synthese 160 pp 97-101.

Vineberg, Susan (undated, perhaps 2003). Beauty’s Cautionary Tale.


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Contents of higher dimensional spheres

For n ≥ 0, the (Euclidean) n-sphere of radius r is defined as the locus of points in Rn

of length r and its interior consists of all points less than length r.

(1) Find formulae for surface area and volume for any n > 0.

(2) Use these to determine what happens to the sequences of surface areas and volumes of unit spheres: do they approach a nonzero limit? Increase without limit? Increase and then decrease to zero? Oscillate?

(3) Determine what happens as n gets large to (a) the ratio of the volume of the unit n-sphere to the volume of its circumscribed hypercube; (b) the ratio of the volume of the inscribed hypercube to the volume of the unit n-sphere.

Solution

(1) Volume and surface area formulae.

Let the n-volume of an n-sphere be Vn(r). In our previous meeting we derived the fundamental relationships

Vn(r) = rnVn(1) = rnVn (for all n) and

0

1 sin( ) nn nV V d

πθ θ− −

= ∫ = 1 0sin( )n

nV dπ

θ θ− ∫ (for n ≥ 1); V0 = 1.

The integral can be performed by repeated integration by parts. Writing

0sin( )n

nI dπ

θ θ= ∫ , and taking n > 1, observe that

1 1 2 2

00 0sin( ) cos( ) cos( )sin( ) ( 1) sin( ) cos( )n n n

nI d n dπ ππ

θ θ θ θ θ θ θ− − −= − = + −∫ ∫

( ) ( )2 220

0 ( 1) sin( ) 1 sin( ) ( 1)nn nn d n I I

πθ θ θ−

−= + − − = − −∫ .

Solve for In in terms of In-2 to obtain

2 41 1 3

2n n nn n n

I I In n n− −− − −

= = =−

L

At this point we are done, sort of: with these formulae, together with I0 = π and I1 = 2, we can compute Vn(r) for any natural number n and any radius r, allowing us to address our original questions about the behaviors of sphere volumes and surface areas as n grows large.


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We found it more satisfactory, however, to push a little further. First, it would be nice to have a closed formula for the volumes. Doing this is aided by extending our original table of sphere volumes:

n In = 21

nn

In −−

Vn = In ⋅ Vn-1

0 π 1

1 2 2 ⋅ 1 = 2

2 012 2

Iπ

= 22π

π=

3 12 2

23 3

I = ⋅ 22

3π

4 23 34 2 4

Iπ

=⋅

2 23 2

2 4 3 2π π π

=⋅

5 34 2 4

25 3 5

I⋅

= ⋅⋅

2 3 22 4 2

23 5 2 3 5

π π⋅⋅ =

⋅ ⋅

6 45 3 56 2 4 6

Iπ⋅

=⋅ ⋅

3 2 33 5 2

2 4 6 3 5 2 3π π π⋅

=⋅ ⋅ ⋅ ⋅

7 56 2 4 6

27 3 5 7

I⋅ ⋅

= ⋅⋅ ⋅

3 4 32 4 6 2

23 5 7 2 3 3 5 7

π π⋅ ⋅⋅ =

⋅ ⋅ ⋅ ⋅ ⋅

8 47 3 5 78 2 4 6 8

Iπ⋅ ⋅

=⋅ ⋅ ⋅

4 3 43 5 7 2

2 4 6 8 3 5 7 2 3 4π π π⋅ ⋅

=⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅

9 78 2 4 6 8

29 3 5 7 9

I⋅ ⋅ ⋅

= ⋅⋅ ⋅ ⋅

4 5 42 4 6 8 2

23 5 7 9 2 3 4 3 5 7 9

π π⋅ ⋅ ⋅⋅ =

⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅

Using factorial notation, defined by n! = n(n-1)! and 0! = 1, we can write

/2

( / 2)!

n

nVnπ

= , n even, (1)

but the expression for odd n looks a little more complicated. Here’s one way to write it that looks somewhat like the preceding formula (in terms of the power of π in the numerator):

/2 22 !

!

n nn

nVn

ππ = , n odd.

For example, 7 7

7/2 27

2 !

7!V π

π = =

7/2 72 3!7!

ππ

= 7

3 2 3 2 17 6 5 4 3 2 1

π⋅ ⋅ ⋅

⋅ ⋅ ⋅ ⋅ ⋅ ⋅=

43 2 6 4 2

7 6 5 4 3 2 1π

⋅ ⋅ ⋅⋅ ⋅ ⋅ ⋅ ⋅ ⋅

=

43 2

7 5 3 1π

⋅ ⋅ ⋅, agreeing with the table entry for n = 7. In some sense, then, this formula is


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asking us to interpret the expression (7/2)! as 7 72

7!2 !

π

≈ 11.6317 and similarly for general

odd n.

There is some insight to be gained from a creative approach to evaluating the integral

0sin( )n

nI dπ

θ θ= ∫ . This integrand could be obtained by integrating yn in polar

coordinates. Specifically,

( ) 1 1sin( ) sin( ) sin( )nn n n n n ny dy dx y dxdy r rdrd r drd r dr dθ θ θ θ θ θ+ += = = =∫ ∫ ∫∫ ∫∫ ∫∫ ∫ ∫ .

Unfortunately, the Cartesian double integral on the left separates into two single integrals when the region of integration is a square, while the polar double integral on the right separates only when the region of integration is a circle. Furthermore, we can’t just ignore the difference between the square and circle as some “technical detail,” because the integrand yn gets pretty big near the upper and lower extremes of these regions. So, we use a memorable trick: we would like to decrease the size of the integrand so that we can neglect its contribution to the integral in the region between a circle and a square, but we need to do so in a way that is simple to evaluate both in Cartesian and polar coordinates.

The first requirement suggests we try multiplying the integrand by a decaying exponential function. When we looked at the formulae for converting between Cartesian and polar coordinates, we immediately thought of using exp(–r2) = exp(–x2 – y2). This really does do the job:

( )2 22 2 x yn ny xy e dy e dx y e dxdy− +− − =∫ ∫ ∫∫

( ) 2 21 1sin( ) sin( ) sin( )n n nr n n rr e rdrd r drd r e dr dθ θ θ θ θ θ− + + −≈ = =∫∫ ∫∫ ∫ ∫ .

(Notice how an extra factor of r slipped in there from the rdrdθ expression.) We have to be a little careful about the domain of integration. The Cartesian integral is over a large square, say [-R, R] × [-R, R], and the polar integral is over a circle of comparable size, say of radius R, so that r ranges from 0 to R and θ ranges from 0 to 2π. In the limit as R grows large, the Cartesian integral approaches the polar integral (due to the rapid decrease of the exponential terms in the integrands): now we really can neglect the

difference between a large square and a large circle. Writing Α(n) = 2

0

n xx e dx∞ −∫ (Greek

capital Alpha) and recognizing that 2n yy e dy

∞ −

−∞∫ =2

02 n yy e dy

∞ −∫ , we have just computed

that ( )( )2 ( ) 2 (0)nΑ Α = ( )( 1) 2 nn IΑ + , giving an immediate formula for In:

2 ( ) (0)

( 1)nn

In

Α Α=

Α +.


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This doesn’t seem like much progress, but let’s try to evaluate the integral Α(n):

Α(n) = 2

0

n xx e dx∞ −∫ = ( )21

0

12

2n xx xe dx

∞ − −∫ = ( )2 21 2

00

11

2n x n xx e n x e dx

∞ ∞− − − − − + − ∫

= ( )1

( 2)2

nn

−Α − . (2)

This is integration by parts; it exploits the fact that xn-1 goes to 0 as x goes to 0 (provided n > 1) and exp(-x2) goes rapidly to zero as x goes to infinity. This behavior makes the first term in the integration by parts vanish, leaving behind a multiple of the integral Α(n-2).

A simple trick will make this relationship among the Α(n) even simpler (and will allow us to use a more conventional notation). That, in turn, will create a wonderful cancellation when we compute the product of the In’s needed to evaluate the spherical volume. We can achieve a relationship between something depending on n and something depending on n-1 (rather than n-2) by halving the argument of Α(n). We can also convert the somewhat annoying factor of (n-1)/2 into the better factor n/2 by adding or subtracting one to the argument. This suggests we define a function that is some constant multiple of Α(2n±1). It is conventional to set

Γ(n) = 2Α(2n-1) = 22 1

02 n xx e dx

∞ − −∫ = 0

n x dxx e

x

∞ −∫ . (3)

(The last integral is obtained by using x2 as the variable of integration.) Now we can rewrite the relationship (2) between Α(n) and Α(n-2) in terms of the Γ’s:

Γ(n) = 2Α(2n-1) = 2 [(2n-2)/2 ⋅ Α(2n-3)] = (n-1) [2Α(2n-3)] = (n-1)Γ(n-1).

It is simple to compute that Γ(1) = 0

x dxxe

x

∞ −∫ = 1, whence Γ(2) = 1Γ(1) = 1;

Γ(3) = 2Γ(2) = 2 ⋅ 1; Γ(4) = 3Γ(3) = 3 ⋅ 2 ⋅ 1; and, in general, when n is a Natural number,

Γ(n+1) = (n) ⋅ (n-1) ⋅ … ⋅ 2 ⋅ 1 = n!.

(It was to achieve this simple relationship that Γ(n) is defined as twice Α(2n-1), rather than just Α(2n-1) itself.) Indeed, by virtue of the expression (3), the Gamma function extends the definition of the factorial to all values of n for which the integral can be evaluated. (The exponential factor in the integrand guarantees convergence for all positive values of n. There are ways to make sense of this integral for all complex values of n, understanding that it diverges at the non-positive integers n = 0, -1, -2, … .)

Returning to the calculation of spherical volumes, definition (3) implies


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1 1

( )2 2

nn

+ Α = Γ

, so

2 ( ) (0)

( 1)nn

In

Α Α=

Α + =

1 12 2

1 12 2

n

n

+ Γ Γ

+ Γ +

=

11 2

1 122 2

n

n

+ Γ Γ + Γ +

, and

Vn = In ⋅ In-1 ⋅ … ⋅ I1 ⋅ V0 =

1 3 21 2 2 2 2 1

1 1 1 3 1 2 122 2 2 2 2 2 2 2

nn n

n n

+ Γ Γ Γ Γ Γ ⋅ + Γ + Γ + Γ + Γ +

L .

This product telescopes: the denominator of each term cancels the numerator of the preceding term, as shown by the red strokes, whence

Vn =

21 2

1 122 2

n

n

Γ Γ + Γ +

=

12

12

n

n

Γ

Γ +

.

To compute the numerator, set n = 2 and solve:

π = Area of the unit circle = V2 =

212

21

2

Γ Γ +

= 21

2 Γ

, implying

12

π Γ =

.

Our ultimate formula is

Vn =/2

12

n

nπ

Γ +

= /2

( / 2)!

n

nπ

. (4)

I have extended the factorial notation to half-integers by means of the Gamma function. Now our previous formula (1), which worked only for even n, now holds—without any modification—for all n!


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By the way, there aren’t really any square roots of π in there, despite the presence of the half-integral powers when n is odd. To see this, look at how it works out for some small values of n:

V3 = 3/2

31

2

π Γ +

= 3/2

3 1 12 2 2

π Γ

=3/2

1/2

2 23 1

ππ

⋅ ⋅⋅ ⋅

= 43π

;

V5 = 5/2

51

2

π Γ +

=5/2

5 3 1 12 2 2 2

π Γ

=5/2

1/2

2 2 25 3 1

ππ

⋅ ⋅ ⋅⋅ ⋅ ⋅

= 28

15π

, etc.

When n is even, there is an integral number (n/2) of factors of π in the volume; when n is odd, there is a term Γ(1/2) = π1/2 in the denominator canceling that half-integral power πn/2, so that again there is an integral number (n-1)/2 of factors of π in the volume.

As far as surface area goes, we had argued last time that Sn(r), the surface area of the ball of radius r in n-space, equals the derivative of Vn(r). Therefore

Sn(r) = dVn(r)/dr = Vn d(rn)/dr = nVn rn-1.

(2 and 3) What happens as n gets large.

When n is increased by two, the numerator in our formula (4) for Vn is multiplied by π while the denominator is multiplied by n/2. That is, Vn+2 = 2π/n Vn. Consequently, as soon as n exceeds 2π, the volume of the unit sphere will decrease. We conclude that the maximum value of Vn occurs either for n=5 or n=6. Calculating with (4) shows that V5 = 5.264 exceeds V6 = 5.168.

It is obvious (by comparison to a geometric sequence, for example) that both the surface area factors Sn and volume factors Vn approach zero as a limit. What is more interesting, though, is how these compare to the surface areas and volumes of other figures. We used the circumscribed cube for comparison.

For example, the hypercube circumscribed around a unit sphere has volume 2n and surface area n2n. The sphere therefore occupies Vn/2n = (π/4)n/2 / (n/2)! of the hypercube’s volume. This is less than 1 / (n/2)!, which is decreasing exponentially with n. The ratios of surface areas are the same. (Why is this obvious?)

Steve Wang encouraged us to find a fundamental, clear reason why this huge shrinkage occurs. Could we have anticipated this behavior before we did all that integration? We noted that the diagonals of a hypercube are getting arbitrarily large as n increases, because the Pythagorean Theorem tells us the unit hypercube has a diagonal of length √(12 + … + 12) = √n. However, by definition every point within a unit ball is within distance one of its center. This means there’s a lot of space left over in the corners of a


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hypercube after we stuff a ball into it! How much? A picture in two dimensions gives some hints:

The central ball has unit radius, so the side of its circumscribed hypercube is two and its diagonal is 2√n. There are four corners in two dimensions, 2n corners in n dimensions. A ball of radius r has been fit into each corner. Forgetting for the moment that we know Vn (because we won’t need this information), our elementary scaling argument from last week still holds and tells us the big ball has volume Vn and each little corner ball has volume Vnrn. How big is r? We had to scale the unit ball down from a radius of 1 to a radius of r in order to fit a little ball into the corner. At the same time, scaling by r relative to the upper left corner has to send point A, at distance 1+√n from the corner, to point B, at distance -1+√n from the corner. Therefore

r = (-1+√n) / (1+√n) = 1 – 2/(√n + 1).

For large n, this is close to the size of the original unit ball. Moreover, the total volume of the 2n corner balls equals

2n ⋅ Vnrn = 2n Vn[1 – 2/(√n + 1)]n.

A good way to estimate this is by taking logarithms and remembering that ln(1 – t) for small values of t is close to, but exceeds, –t itself. In the following calculation we will keep replacing nasty parts of the expression with simpler but slightly smaller values, so that we conservatively underestimate the total volume of the corner balls. Thus,

√n

1

2r

A

B

1


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ln(2n ⋅ Vnrn) = nln(2) + ln(Vn) + nln(r)

= nln(2) + ln(Vn) + nln(1 – 2/(√n + 1))

> nln(2) + ln(Vn) – n(2/(√n + 1))

> nln(2) + ln(Vn) – 2n / √n

= nln(2) + ln(Vn) – 2√n.

Exponentiating,

Total volume of corner balls > 2n Vn / e2√n. (5)

Compared to the volume of the original unit ball (Vn), these corner balls thereby occupy at least 2n / e2√n times the space. This expression becomes infinitely large as n grows large. (Eventually the nln(2) term in the logarithm dominates the 2√n term, showing that the rate of growth is almost exponential.) Consequently, as n increases, any ball occupies a vanishingly small amount of space within its circumscribed hypercube. There’s a lot of room in high dimensions!

Exercise

Determine what happens to the volume of the inscribed hypercube, relative to the ball’s volume, as n grows large.

Challenge

The formula for Vn tells us the ratio of volumes (of cube to ball) equals

/2

2 2

( / 2)!

n n

nnV

nπ

=

.

Because the cube includes space in addition to the central ball and corner balls, this ratio

must exceed the one we obtained at (5). Therefore /2

2

( / 2)!

nne

nπ

≤ , implying (upon

doubling n) that 2 2

!n

nn

eπ

≥ . This is a pretty lousy lower bound, because in higher

dimensions we left a lot of space unaccounted for. (For example, it tells us that 3,628,800 = 10! ≥ π10/e2√20 = 12.22.) Can we extend the geometric argument to improve this bound? How well can we do?

—Bill Huber, 14 October 2008

Documents

Notes and Solutions #5 - Haverford Collegeblogs.haverford.edu/mathproblemsolving/files/2010/05/S4-Sleeping... · Notes and Solutions #5 Meeting of 7 October 2008 Sleeping Beauty Michael