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.