Mathematical Tools for Physics

2—Infinite Series 44

The original function is a maximum when this denominator is a minimum. When the numbersN andkare big,
you can treatkas a continuous variable and differentiate with respect to it. Then set this derivative to zero and
finally, expand in a power series about that point.

d

dk

f(k) = lnk+ 1−ln(N−k)− 1 −lna+ lnb= 0

ln

k

N−k

= ln

a

b

,

k

N−k

=

a

b

, k=aN

At this point, what is the second derivative?

d^2

dk^2

f(k) =

1

k

+

1

N−k

whenk=aN,

1

k

+

1

N−k

=

1

aN

+

1

N−aN

=

1

aN

+

1

bN

=

1

abN

The power series forf(k)is

f(k) =f(aN) + (k−aN)f′(aN) +

1

2

(k−aN)^2 f′′(aN) +···

=NlnN+

1

2 abN

(k−aN)^2 +···

To substitute this back into Eq. ( 17 ), take its exponential. Then because this will be a fairly sharp maximum, only
the values ofknear toaNwill be significant. That allows me to use this central value ofkin the slowly varying
square root coefficient and also I can neglect higher order terms in the power series expansion. Letδ=k−aN.
The result is the Gaussian distribution.

1

√

2 π

√

N

aN(N−aN)

NN

NNeδ^2 /^2 abN

=

1

√

2 abNπ

e−δ

(^2) / 2 abN
(19)
Whena= 1/ 2 andb= 1/ 2 , this reduces to Eq. (1.16).
When you accumulateN trials at a time (largeN) and then look for the distribution in these cumulative
results, you will always get a Gaussian. This is the central limit theorem, which says that whatever set of
probabilities that you start with, not just a coin toss, you will always get a Gaussian by averaging the data.