PI = 2.*np.arcsin(1)
a=0
b= 2*PI
N= 10
h = (a+b) / (N-1)
x = np.zeros(N)
y1 = np.zeros(N)
y2 = np.zeros(N)
y8 = np.zeros(N)
for i in range(N):
x[i] = a + h*i
y1[i] = np.sin(x[i])
y2[i] = np.sin(2*x[i])
y8[i] = np.sin(8*x[i])
print(intr.trap_d(x,intr.multiply(y1,y1)))
print(intr.trap_d(x,intr.multiply(y1,y2)))
print(intr.trap_d(x,intr.multiply(y2,y8)))
print(intr.trap_d(x,intr.multiply(y8,y8)))
#2nd problem
a2= 0
b2 = 1
N = [10, 100, 1000]
N2 = len(N)
for i in range(N2):
x = np.zeros(N[i])
y = np.zeros(N[i])
print("for N = " + str(N[i]))
print(intr.trap(intr.lin,a2,b2,N[i]))
print(intr.trap(intr.quad,a2,b2,N[i]))
PI = 2. * np.arcsin(1)
N = 1000
a = 0
b = 2. * PI
h = (b - a) / (N - 1)
x = np.zeros(N)
y1 = np.zeros(N)
y2 = np.zeros(N)
y3 = np.zeros(N)
for i in range(N):
x[i] = a + h * i
y1[i] = np.sin(x[i])
y2[i] = np.sin(2. * x[i])
y3[i] = np.sin(8. * x[i])
print(intr.trap_d(x,intr.multiply(y1,y1)))
print(intr.trap_d(x,intr.multiply(y1,y2)))
print(intr.trap_d(x,intr.multiply(y2,y3)))
print(intr.trap_d(x,intr.multiply(y3,y3)))
#We find that when m=n the scalar product is pi and when m=/n the scalar product
#is 0. This is very similar to the scalar product of the legendre functions.
def powr(x,n):
m = 1
if n == 0:
return 1
else:
for l in range(n):
m = m * x
return m
