Biological Physics: Energy, Information, Life

68 Chapter 3. The molecular dance[[Student version, December 8, 2002]]

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0.2

0.4

0.6

0.8

1.

P(

x)

x

x 0

Figure 3.2:(Mathematical function.) Unnormalized Gaussian distribution centered atx 0 =1withσ=1/√ 2 and
A=1(see Equation 3.6).

whereAandσare positive constants andx 0 is some other constant.

Your Turn 3a

Youcan quickly see what a function looks like with your favorite graphing software. For example,

in Maple writingplot(exp(-(x-1)ˆ2),x=-1..3);gives Figure 3.2. Try it, then play with the

constantsAandσto see how the figure changes.

The constantAisn’t free; it’s fixed by the normalization condition. This is such an important
and useful derivation that we should see how it works in detail.

Example Find the value ofArequired to normalize the Gaussian distribution.

Solution: First we need that

∫∞

−∞

dye−y

2

=

√

π. (3.7)

Youcan think of this as merely a mathematical fact to be looked up in an integral

table (or see the derivation given in Section 6.2.2′ on page 206). What’s more

important are a couple of easy steps from calculus. Equation 3.4 requires that we

choose the constantAso that

1=A

∫∞

−∞

dxe−(x−x^0 )

(^2) / 2 σ 2
.
Change variables toy=(x−x 0 )/(

√

2 σ), so dy=dx/

√

2 σ.Then Equation 3.7 gives

A=1/(σ

√

2 π).

In short, the Gaussian distribution is

P(x)=

1

√

2 πσ

e−(x−x^0 )

(^2) / 2 σ 2

. Gaussian distribution (3.8)

Looking at Figure 3.2, we see that it’s a bump function centered atx 0 (that is, maximum there).
The bump has a width controlled byσ. The largerσ,the fatter the bump, since one gets to go