Mathematical Tools for Physics

2—Infinite Series 43

and the fractions of the time that you get each pair are respectively

a^2 ba ab b^2

This says that the fraction of the time that you get no heads, one head, or two heads are

a^2 , 2 ab, b^2 with total (a+b)^2 =a^2 + 2ab+b^2 = 1

ais the fraction for tails.
Generalize this to the case where you throwN coins at a time and determine how often you expect to see
0, 1,...,Nheads. Equation ( 14 ) says

(a+b)N=

∑N

k=0

(

N

k

)

akbN−k where

(

N

k

)

=

N!

k!(N−k)!

When you make a trial in which you tossNcoins, you expect that the first choice will come upN times only the
fractionaN of the trials. All tails and no heads. Compare problem 27.
The problem is now to use Stirling’s formula to find an approximate result for the terms of this series. This
is the fraction of the trials in which you turn upktails andN−kheads.

akbN−k

N!

k!(N−k)!

∼akbN−k

√

2 πN NNe−N

√

2 πk kke−k

√

2 π(N−k) (N−k)N−ke−(N−k)

=akbN−k

1

√

2 π

√

N

k(N−k)

NN

kk(N−k)N−k

(17)

The complicated parts to manipulate are the factors with all the exponentials ofkin them. Pull them out from
the denominator for separate handling.

kk(N−k)N−ka−kb−(N−k)

The next trick is to take a logarithm and to do all the manipulations on it.

ln→klnk+ (N−k) ln(N−k)−klna−(N−k) lnb=f(k) (18)