1—Basic Stuff 11

There are 8 possibilities here, 23 , so you expect (on average) one run out of 8 to give you 3 heads. Probability
1/8.

For the more interesting case of bigN, the exponent,e−^2 δ

(^2) /N
, varies slowly and smoothly asδchanges in
integer steps away from zero. This is a key point; it allows you to approximate a sum by an integral. IfN= 1000
andδ= 10, the exponent is 0.819. It has dropped only gradually from one.
FlipNcoins, then do it again and again. In what fraction of the trials will the result be betweenN/ 2 −∆
andN/2 + ∆heads? This is the sum of the fractions corresponding toδ= 0,δ=± 1 ,...,δ=±∆. Because
the approximate function is smooth, I can replace this sum with an integral.
∫∆
−∆
dδ

√

2

πN

e−^2 δ

(^2) /N
Make the substitution 2 δ^2 /N=x^2 and you have
√
2
πN

∫∆√ 2 /N

−∆

√

2 /N

√

N

2

dxe−x

2

=

1

√

π

∫∆√ 2 /N

−∆

√

2 /N

dxe−x

2

= erf

(

∆

√

2 /N

)

The error function of one is 0.84, so if∆ =

√

N/ 2 then in 84% of the trials heads will come up between
N/ 2 −

√

N/ 2 andN/2 +

√

N/ 2 times. ForN= 1000, this is between 478 and 522 heads.

1.5 Differentiating
When you differentiate a function in which the independent variable shows up in several places, how to you do
the derivative? For example, what is the derivative with respect toxofxx? The answer is that you treat each
instance ofxone at a time, ignoring the others; differentiate with respect tothat xand add the results. For
a proof, use the definition of a derivative and differentiate the functionf(x,x). Start with the finite difference
quotient:
f(x+ ∆x,x+ ∆x)−f(x,x)
∆x

f(x+ ∆x,x+ ∆x)−f(x,x+ ∆x) +f(x,x+ ∆x)−f(x,x)

∆x

=

f(x+ ∆x,x+ ∆x)−f(x,x+ ∆x)

∆x

+

f(x,x+ ∆x)−f(x,x)

∆x