250
13 Solutions for Selected Exercises
˜
R(k) =
1
a
a/2
−a/2
e
ikx
=
1
ika
e
ika/2
− e
−ika/2
=
sin(ka/2)
ka/2
.
(13.51)
Exercise 9.2
The cumulant expansion is defined in (9.16). Introducing y = ka/2 and d/dk =
(a/2)d/dy and defining f (y) = ln(sin(y)/y), we find the coefficients c m from
c m =
a
2
m (−i)
m d
m f (y)
dy m
y=0
.
(13.52)
We use the following code to symbolically differentiate f (y) and subsequently set
y to zero. Here n is the order of the derivative.
syms y
n=4;
% order of derivative
fun=(-i)ˆn*log(sin(y)/y)
f=diff(fun,y,n)
% symbolic differentiation
q=double(subs(f,eps))
% substitute y=0
[num,den]=rat(q,1e-8)
% write as fraction
Repeatedly running the code for different values of n, we find the cumulants to be
c 1 = 0, c 2 =
1
3
a
2
2 , c 3 = 0, and c 4 = −
2
15
a
2
4 .
(13.53)
Exercise 9.3
Convoluting a function with itself in real space is equivalent to transforming the
square of its Fourier transform back to real space. We note that the Fourier transform
˜
f (k) of the Cauchy distribution from (9.4) is given by (9.26) with μ = 1, such that
its square is
˜
f (k)
2
= e
−2a|k|
,
(13.54)
which is the Fourier transform of (9.4) with a replaced by 2a.
Exercise 9.4
he code below fills the array x with 10
4 groups of ten random numbers each. Then
it adds the ten numbers to produce an array y that is used to produce a histogram,
which shows a distribution close to a Gaussian, as predicted by (9.23). Then we
calculate the rms value of the histogram and find about 1.8, which is close to the
second cumulant c 2 = a
2
/12 with a = 2 from Exercise 2 multiplied by
√
N =
√
10.
This is a consequence of the central limit theorem, discussed in Sect. 9.7.
13 Solutions for Selected Exercises
˜
R(k) =
1
a
a/2
−a/2
e
ikx
=
1
ika
e
ika/2
− e
−ika/2
=
sin(ka/2)
ka/2
.
(13.51)
Exercise 9.2
The cumulant expansion is defined in (9.16). Introducing y = ka/2 and d/dk =
(a/2)d/dy and defining f (y) = ln(sin(y)/y), we find the coefficients c m from
c m =
a
2
m (−i)
m d
m f (y)
dy m
y=0
.
(13.52)
We use the following code to symbolically differentiate f (y) and subsequently set
y to zero. Here n is the order of the derivative.
syms y
n=4;
% order of derivative
fun=(-i)ˆn*log(sin(y)/y)
f=diff(fun,y,n)
% symbolic differentiation
q=double(subs(f,eps))
% substitute y=0
[num,den]=rat(q,1e-8)
% write as fraction
Repeatedly running the code for different values of n, we find the cumulants to be
c 1 = 0, c 2 =
1
3
a
2
2 , c 3 = 0, and c 4 = −
2
15
a
2
4 .
(13.53)
Exercise 9.3
Convoluting a function with itself in real space is equivalent to transforming the
square of its Fourier transform back to real space. We note that the Fourier transform
˜
f (k) of the Cauchy distribution from (9.4) is given by (9.26) with μ = 1, such that
its square is
˜
f (k)
2
= e
−2a|k|
,
(13.54)
which is the Fourier transform of (9.4) with a replaced by 2a.
Exercise 9.4
he code below fills the array x with 10
4 groups of ten random numbers each. Then
it adds the ten numbers to produce an array y that is used to produce a histogram,
which shows a distribution close to a Gaussian, as predicted by (9.23). Then we
calculate the rms value of the histogram and find about 1.8, which is close to the
second cumulant c 2 = a
2
/12 with a = 2 from Exercise 2 multiplied by
√
N =
√
10.
This is a consequence of the central limit theorem, discussed in Sect. 9.7.