which we can use to find explicit formulae for I n , depending on whether n is even or odd.
For n even:
For n odd:
Now we use these to establish a connection between I n and I n −1 that is independent of the parity of n .
If n is even, it must be that n −1 is odd and we use the appropriate formulae to get
If n is odd, it must be that n − 1 is even and again we use the appropriate formulae to get
So, whatever the parity of n , I n I n −1 = 2 π / n .
Now recall that V n (1) = V n −1 (1) I n and reuse the formula once on itself to get
Table 12.1. Volumes of hyperspheres.
Now we can say that
We have a simple reduction formula for the V n (1) and we can chase it down, again depending on whether n is even or odd to get the answer:
Of course, this means that
Table 12.1 lists the volumes for small values of n and we can see that the volume of the unit hypersphere peaks when n = 5 and that this maximal volume is. A plot of V n (1) against n is shown in figure 12.5 , which indicates that the volume of the unit hypersphere is decreasing as n increases beyond 5, which appears very strange.
Figure 12.5. Volume of the unit hypersphere compared with integer dimensions.
Volume in Continuous Hyperdimensions
The plotted points in figure 12.5 are in such a regular pattern that it is natural to want to join them by a continuous curve, but doing so would force us to admit not only hyperdimensions but also nonintegral hyperdimensions. To approach this concept we need to rewrite the formulae for V n (1) as below:
If n is even, the formula can be written in terms of factorials as
If n is odd,is not defined, but its generalization − the Gamma function Γ ( x ) − is. This function’s somewhat strangedefinition is
which is defined for x > 0 and which has two particular properties:
and
Together they characterize the factorial function, since, if n is a positive integer,
So, this peculiar function is indeed an extension of the factorial function, which is defined only for positive integers, to all x > 0. In fact, the above relationship, when rewritten as
can be used to extend the idea of factorial to all numbers other than the negative integers and it is perfectly easy to extend the definition to complex numbers, but we will not concern ourselves with these excitements.
Note in particular that, if we accept a standard result that
we have
and using the substitution t = u 2 , we have that d t /d u = 2 u
and and this means that
It is then the case that for n even V n (1) can be rewritten yet again, this time as
The really nice thing is that this notation unifies the two formulae and it is easy to check that the Gamma function form of the formula holds whatever the parity of n .
For example,
Figure 12.6 is the plot of the continuous form of figure 12.5 , this time for n up to 20. It shows a little more clearly that the maximum occurs a little to the right of n = 5 and calculus should help us to find the coordinates of that point, provided that we can differentiate the components of
with respect to the continuous variable n .
The top of the fraction is easy to deal with using the fact that a b = e b ln a so that the formula becomes
Figure 12.6. Volume of the unit hypersphere compared with continuous dimensions.
The bottom of the fraction requires us to differentiate the Gamma function. We have no need to look closely at the implications of this and we will merely write the derivative in the usual way as Γ ′( x ); what we will need is the even more exotic Digamma function Ψ ( x ), which is defined by
and some powerful mathematical software to evaluate it.
Using the standard chain and quotient rules, the calculations are
The requirement that d V n (1)/d n = 0 means that
and so
Figure 12.7. The Digamma function
Figure 12.7 shows a plot of the Digamma function and the horizontal line at ln π . To findand hence n we need computational help to establish that n = 5.256 946 4 …
For n even:
For n odd:
Now we use these to establish a connection between I n and I n −1 that is independent of the parity of n .
If n is even, it must be that n −1 is odd and we use the appropriate formulae to get
If n is odd, it must be that n − 1 is even and again we use the appropriate formulae to get
So, whatever the parity of n , I n I n −1 = 2 π / n .
Now recall that V n (1) = V n −1 (1) I n and reuse the formula once on itself to get
Table 12.1. Volumes of hyperspheres.
Now we can say that
We have a simple reduction formula for the V n (1) and we can chase it down, again depending on whether n is even or odd to get the answer:
Of course, this means that
Table 12.1 lists the volumes for small values of n and we can see that the volume of the unit hypersphere peaks when n = 5 and that this maximal volume is. A plot of V n (1) against n is shown in figure 12.5 , which indicates that the volume of the unit hypersphere is decreasing as n increases beyond 5, which appears very strange.
Figure 12.5. Volume of the unit hypersphere compared with integer dimensions.
Volume in Continuous Hyperdimensions
The plotted points in figure 12.5 are in such a regular pattern that it is natural to want to join them by a continuous curve, but doing so would force us to admit not only hyperdimensions but also nonintegral hyperdimensions. To approach this concept we need to rewrite the formulae for V n (1) as below:
If n is even, the formula can be written in terms of factorials as
If n is odd,is not defined, but its generalization − the Gamma function Γ ( x ) − is. This function’s somewhat strangedefinition is
which is defined for x > 0 and which has two particular properties:
and
Together they characterize the factorial function, since, if n is a positive integer,
So, this peculiar function is indeed an extension of the factorial function, which is defined only for positive integers, to all x > 0. In fact, the above relationship, when rewritten as
can be used to extend the idea of factorial to all numbers other than the negative integers and it is perfectly easy to extend the definition to complex numbers, but we will not concern ourselves with these excitements.
Note in particular that, if we accept a standard result that
we have
and using the substitution t = u 2 , we have that d t /d u = 2 u
and and this means that
It is then the case that for n even V n (1) can be rewritten yet again, this time as
The really nice thing is that this notation unifies the two formulae and it is easy to check that the Gamma function form of the formula holds whatever the parity of n .
For example,
Figure 12.6 is the plot of the continuous form of figure 12.5 , this time for n up to 20. It shows a little more clearly that the maximum occurs a little to the right of n = 5 and calculus should help us to find the coordinates of that point, provided that we can differentiate the components of
with respect to the continuous variable n .
The top of the fraction is easy to deal with using the fact that a b = e b ln a so that the formula becomes
Figure 12.6. Volume of the unit hypersphere compared with continuous dimensions.
The bottom of the fraction requires us to differentiate the Gamma function. We have no need to look closely at the implications of this and we will merely write the derivative in the usual way as Γ ′( x ); what we will need is the even more exotic Digamma function Ψ ( x ), which is defined by
and some powerful mathematical software to evaluate it.
Using the standard chain and quotient rules, the calculations are
The requirement that d V n (1)/d n = 0 means that
and so
Figure 12.7. The Digamma function
Figure 12.7 shows a plot of the Digamma function and the horizontal line at ln π . To findand hence n we need computational help to establish that n = 5.256 946 4 …
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