each graphs will depend on the Gaussian function.
In-order for the program to work, we need to install pylab library into our computer, we can obtain the file from the link http://sourceforge.ent/projects/matplotlib/
Here is the code we had
_________________________________below is the code_________________________________
from pylab import*
harmonics= 20
center=harmonics/2
sigma=1
coeff=1/((sqrt(2*pi))*sigma)
gauss_list=[]
domain_list2=[]
L=10
knot= 2*pi/L
b_list=[]
kdomain=[]
##
##for k in arange(0,200,0.1):
## bofk=(1/(knot))
## b_list.append(bofk)
## kdomain.append(k)
####plot(kdomain,b_list)
##
for x in range(1,harmonics):
gauss=coeff*exp(-(x-center)**2/(2.*sigma**2))
gauss_list.append(gauss)
domain_list2.append(x)
##A_coeff_1=1
##wave_constant_1=1
##sine1_list=[]
##domain_list=[]
##for x in arange(-pi,pi,0.1):
## sine1=A_coeff_1*sin(wave_constant_1*x)
## sine1_list.append(sine1)
## domain_list.append(x)
w=1
Fourier_Series=[]
for i in range(1,harmonics):
sine_function=[]
x=[]
for t in arange(-pi,pi,0.01):
sine_f=gauss_list[i-1]*sin(i*w*t)
sine_function.append(sine_f)
x.append(t)
##plot(x,sine_function)
Fourier_Series.append(sine_function)
superposition=zeros(len(sine_function))
for function in Fourier_Series:
for i in range(len(function)):
superposition[i]+=function[i]
print kdomain
plot(x,superposition)
##plot(domain_list2,gauss_list)
##plot(domain_list,sine1_list)
show()
_________________________________End of the code_________________________________
Here are some of the results that we got
First this is the graph for jut plotting one sine function
After this is to plot the Gaussian function
At the end, we combine the since function with amplitude of the results from Gaussian function
Also, if we put more sine function (harmonics), the tails of the graph will be longer
After computing these results, we were ask few questions
Using the integral in
This represents an equal combination of all wave numbers between 0 and
a. Graph
b. Graph
c. Locate the two points closest to this maximum (one on each side of it) where
A:
=1.00 
d.Repeat part C for the case
.
e. Repeat part D for the case
.
f. The momentum
is equal to
, so the width of
in momentum is
. Calculate the product
for the case
A:
=
g. Calculate the product
for the case 
A:
=
h. Discuss your results in light of the Heisenberg uncertainty principle.
A: Since
=
and according to Heisenber uncertainty, which is
>=h/2π, the results is valid for any value of 
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