I have the following code to compute a desired quantity:
import numpy as np
N = 2
lamda = 2
mu = 1
a = 0.5
St_Sp = np.arange(- N, N + 1)
Card = St_Sp.shape[0]
#%% Define infintesimal generator
def In_Ge(x, y):
if x == N or x == - N:
re = 0
elif x - y == - 1:
re = lamda
elif x - y == 1:
re = mu
elif x - y == 0:
re = - (mu + lamda)
else: re = 0
return re
x = St_Sp[0]
y = In_Ge(x, x) / (In_Ge(x, x) + np.log(a))
b = - 1 / y
print(b)
The result is inf. I checked and see that the value of y is non-zero, so I could not understand why such phenomenon happens. Could you elaborate on this issue?
#pinegulf's comment solves my question:
Your 'In_Ge(x,x)' retruns 0, thus y= 0 and 1/0 is pretty badly defined. Edit: You say it's not, but your x==--2 and functions first if is invoked.
Related
im trying to write a program that gives the integral approximation of e(x^2) between 0 and 1 based on this integral formula:
Formula
i've done this code so far but it keeps giving the wrong answer (Other methods gives 1.46 as an answer, this one gives 1.006).
I think that maybe there is a problem with the two for cycles that does the Riemman sum, or that there is a problem in the way i've wrote the formula. I also tried to re-write the formula in other ways but i had no success
Any kind of help is appreciated.
import math
import numpy as np
def f(x):
y = np.exp(x**2)
return y
a = float(input("¿Cual es el limite inferior? \n"))
b = float(input("¿Cual es el limite superior? \n"))
n = int(input("¿Cual es el numero de intervalos? "))
x = np.zeros([n+1])
y = np.zeros([n])
z = np.zeros([n])
h = (b-a)/n
print (h)
x[0] = a
x[n] = b
suma1 = 0
suma2 = 0
for i in np.arange(1,n):
x[i] = x[i-1] + h
suma1 = suma1 + f(x[i])
alfa = (x[i]-x[i-1])/3
for i in np.arange(0,n):
y[i] = (x[i-1]+ alfa)
suma2 = suma2 + f(y[i])
z[i] = y[i] + alfa
int3 = ((b-a)/(8*n)) * (f(x[0])+f(x[n]) + (3*(suma2+f(z[i]))) + (2*(suma1)))
print (int3)
I'm not a math major but I remember helping a friend with this rule for something about waterplane area for ships.
Here's an implementation based on Wikipedia's description of the Simpson's 3/8 rule:
# The input parameters
a, b, n = 0, 1, 10
# Divide the interval into 3*n sub-intervals
# and hence 3*n+1 endpoints
x = np.linspace(a,b,3*n+1)
y = f(x)
# The weight for each points
w = [1,3,3,1]
result = 0
for i in range(0, 3*n, 3):
# Calculate the area, 4 points at a time
result += (x[i+3] - x[i]) / 8 * (y[i:i+4] * w).sum()
# result = 1.4626525814387632
You can do it using numpy.vectorize (Based on this wikipedia post):
a, b, n = 0, 1, 10**6
h = (b-a) / n
x = np.linspace(0,n,n+1)*h + a
fv = np.vectorize(f)
(
3*h/8 * (
f(x[0]) +
3 * fv(x[np.mod(np.arange(len(x)), 3) != 0]).sum() + #skip every 3rd index
2 * fv(x[::3]).sum() + #get every 3rd index
f(x[-1])
)
)
#Output: 1.462654874404461
If you use numpy's built-in functions (which I think is always possible), performance will improve considerably:
a, b, n = 0, 1, 10**6
x = np.exp(np.square(np.linspace(0,n,n+1)*h + a))
(
3*h/8 * (
x[0] +
3 * x[np.mod(np.arange(len(x)), 3) != 0].sum()+
2 * x[::3].sum() +
x[-1]
)
)
#Output: 1.462654874404461
I have a list D containing 50 sub-lists. The number of elements in these sub-lists are decreasing. I visualize the list D by
for i, array in enumerate(D):
plt.scatter([i]*len(array), array)
I have 50 functions taking values from St_Sp, and Y is a list containing 50 elements, each of them is the output of each function. I visualize these functions
fig, ax = plt.subplots()
for i in range(len(Y)):
ax.plot(St_Sp, Y[i])
I found that too many colors are not easy to eyes. I would like to ask how to alternate color of the graphs between blue and white? I mean the color of the functions and dots in D are white > blue > white > blue ...
Could you please elaborate on how to do so?
##### Import packages
import numpy as np
import scipy.linalg as la
import time
import matplotlib
import matplotlib.pyplot as plt
##### Initial conditions
N = 100
lamda = 7
mu = 2
a = np.exp(-0.05)
r = - np.log(a).copy()
St_Sp = np.arange(- N, N + 1)
Card = St_Sp.shape[0]
##### Define infintesimal generator
def LL(x, y):
if x == N or x == - N: re = 0
elif x - y == - 1: re = lamda
elif x - y == 1: re = mu
elif x - y == 0: re = - (mu + lamda)
else: re = 0
return re
def L(x):
return - LL(x, x)
##### Define function Phi
def Phi(x):
return max(x, 0)
Phi = np.vectorize(Phi)
##### Define vector b
b = np.array(Phi(St_Sp))
##### Define function Psi
def Psi(x):
return L(x) / (L(x) + r)
Psi = np.vectorize(Psi)
##### Generate a Boolean vector whose all elements are False
d = np.array([0] * Card).astype(bool)
##### Define matrix A
A = np.zeros((Card, Card))
for i in range(Card):
for j in range(Card):
if (i != j) & (L(St_Sp[i]) != 0):
A[i, j] = LL(St_Sp[i], St_Sp[j]) / L(St_Sp[i])
elif (i != j) & (L(St_Sp[i]) == 0):
A[i, j] = 0
elif (i == j) & (Psi(St_Sp[i]) != 0):
A[i, j] = - 1 / Psi(St_Sp[i])
else: A[i, j] = 1
##### Row names of A
rows = np.arange(0, Card)
##### Define matrix B
B = np.zeros((Card, Card))
for i in range(Card):
for j in range(Card):
if i != j:
B[i, j] = LL(St_Sp[i], St_Sp[j])
else: B[i, j] = LL(St_Sp[i], St_Sp[j]) - r
start = time.time()
##### Generate I_0
I = [np.array([1] * Card).astype(bool), d.copy()]
Z = np.array(b.copy())
Z = Z.astype(float)
D = [St_Sp]
index0 = np.matmul(B, Z) <= 0
index1 = ~ index0
Y = [b.copy()]
##### Iterations
for i in range(1, Card):
I = [I[0] & index0, I[1] | index1]
Z = np.array(b.copy())
Z = Z.astype(float)
A1 = A[np.ix_(rows[I[1]], rows[I[1]])]
A2 = A[np.ix_(rows[I[1]], rows[I[0]])]
Z[I[1]] = la.solve(A1, - np.matmul(A2, Z[I[0]]))
Y = np.concatenate((Y, [Z]))
D.append(St_Sp[I[0]])
index = np.matmul(B[I[0]], Z) <= 0
index0, index1 = d.copy(), d.copy()
index0[I[0]], index1[I[0]] = index, ~ index
if (I[0] == index0).all() == True: break
for i, array in enumerate(D):
plt.scatter([i]*len(array), array)
fig, ax = plt.subplots()
for i in range(len(Y)):
ax.plot(St_Sp, Y[i])
The easiest approach is to set a custom color cycler. Instead of cycling between the 10 typical colors, the default colors for the plots will cycle through the given colors.
from cycler import cycler
custom_cycler = cycler(color=['white', 'blue'])
plt.gca().set_prop_cycle(custom_cycler)
for i, array in enumerate(D[:-1]):
plt.scatter([i] * len(array), array)
plt.scatter([len(D) - 1] * len(D[-1]), D[-1], color='crimson')
fig, ax = plt.subplots()
ax.set_prop_cycle(custom_cycler)
for i in range(len(Y) - 1):
ax.plot(St_Sp, Y[i])
ax.plot(St_Sp, Y[len(Y) - 1], color='crimson')
plt.show()
I have the following code. The beggining is quite long, but only serves to generate data. The problem happens with a few lines at the end.
##### Import packages
import numpy as np
import scipy.linalg as la
##### Initial conditions
N = 5
lamda = 7
mu = 2
a = 0.5
r = - np.log(a).copy()
St_Sp = np.arange(- N, N + 1)
Card = St_Sp.shape[0]
##### Define infintesimal generator
def LL(x, y):
if x == N or x == - N: re = 0
elif x - y == - 1: re = lamda
elif x - y == 1: re = mu
elif x - y == 0: re = - (mu + lamda)
else: re = 0
return re
def L(x):
return - LL(x, x)
##### Define function Phi
def Phi(x): return max(x, 0)
Phi = np.vectorize(Phi)
##### Define vector b
b = Phi(St_Sp).copy()
##### Define function Psi
def Psi(x): return L(x) / (L(x) + r)
Psi = np.vectorize(Psi)
##### Generate a Boolean vector whose all elements are False
d = np.array([0] * Card).astype(bool)
##### Define matrix A
A = np.zeros((Card, Card))
for i in range(Card):
for j in range(Card):
if (i != j) & (L(St_Sp[i]) != 0):
A[i, j] = LL(St_Sp[i], St_Sp[j]) / L(St_Sp[i])
elif (i != j) & (L(St_Sp[i]) == 0):
A[i, j] = 0
elif (i == j) & (Psi(St_Sp[i]) != 0):
A[i, j] = - 1 / Psi(St_Sp[i])
else: A[i, j] = 1
##### Row names of A
rows = np.arange(0, Card)
##### Define matrix B
B = np.zeros((Card, Card))
for i in range(Card):
for j in range(Card):
if i != j:
B[i, j] = LL(St_Sp[i], St_Sp[j])
else: B[i, j] = LL(St_Sp[i], St_Sp[j]) - r
##### Generate I_0
I = [np.array([1] * Card).astype(bool), d.copy()]
Z = b.copy()
index0 = np.matmul(B, Z) <= 0
index1 = ~ index0
##### Generate I_1
I = [index0, index1]
Z = b.copy()
if np.sum(I[1]) > 0:
order = np.concatenate((rows[I[1]], rows[~ I[1]]))
A1 = A[np.ix_(rows[I[1]], order)]
A2 = la.lu(A1)[2]
p = np.atleast_2d(A1).shape[0]
B1 = A2[:, range(p)]
B2 = - np.matmul(A2[:, p:], Z[I[0]])
print('Before being assigned new values, Z is \n', Z)
print('\n The index I[1] of elements of Z to be change \n', I[1])
M = la.solve_triangular(B1, B2, lower = False)
print('\n The values to be assigned to Z[I[1]] is \n', M)
Z[I[1]] = M
print('\n After being assigned new values, Z is \n', Z)
with result
Before being assigned new values, Z is
[0 0 0 0 0 0 1 2 3 4 5]
The index I[1] of elements of Z to be change
[False False False False False True True True True True False]
The values to be assigned to Z[I[1]] is
[2.08686055 2.88974949 3.40529229 3.88978577 4.41338306]
After being assigned new values, Z is
[0 0 0 0 0 2 2 3 3 4 5]
It's very weird to me that the command Z[I[1]] = M does not assign new values from M to the postion of Z indexed by I[1]. Could you please elaborate on why this problem arises and how to resolve it?
The datatype of your array Z is int, to the values are typecasted by python automatically, resulting in the interger values of int([2.08686055 2.88974949 3.40529229 3.88978577 4.41338306]) = [2 2 3 3 4 5].
If you want to change that behavour, you just need to add a line to change the type of your original array:
Z = Z.astype(float)
I have the following code that solves simultaneous linear equations by starting with the first equation and finding y when x=0, then putting that y into the second equation and finding x, then putting that x back into the first equation etc...
Obviously, this has the potential to reach infinity, so if it reaches +-inf then it swaps the order of the equations so the spiral/ladder goes the other way.
This seems to work, tho I'm not such a good mathematician that I can prove it will always work beyond a hunch, and of course some lines never meet (I know how to use matrices and linear algebra to check straight off whether they will never meet, but I'm not so interested in that atm).
Is there a better way to 'spiral' in on the answer? I'm not interested in using math functions or numpy for the whole solution - I want to be able to code the solution. I don't mind using libraries to improve the performance, for instance using some sort of statistical method.
This may be a very naive question from either a coding or maths point of view, but if so I'd like to know why!
My code is as follows:
# A python program to solve 2d simultaneous equations
# by iterating over coefficients in spirals
import numpy as np
def Input(coeff_or_constant, var, lower, upper):
val = int(input("Let the {} {} be a number between {} and {}: ".format(coeff_or_constant, var, lower, upper)))
if val >= lower and val <= upper :
return val
else:
print("Invalid input")
exit(0)
def Equation(equation_array):
a = Input("coefficient", "a", 0, 10)
b = Input("coefficient", "b", 0, 10)
c = Input("constant", "c", 0, 10)
equation_list = [a, b, c]
equation_array.append(equation_list)
return equation_array
def Stringify_Equations(equation_array):
A = str(equation_array[0][0])
B = str(equation_array[0][1])
C = str(equation_array[0][2])
D = str(equation_array[1][0])
E = str(equation_array[1][1])
F = str(equation_array[1][2])
eq1 = str(A + "y = " + B + "x + " + C)
eq2 = str(D + "y = " + E + "x + " + F)
print(eq1)
print(eq2)
def Spiral(equation_array):
a = equation_array[0][0]
b = equation_array[0][1]
c = equation_array[0][2]
d = equation_array[1][0]
e = equation_array[1][1]
f = equation_array[1][2]
# start at y when x = 0
x = 0
infinity_flag = False
count = 0
coords = []
coords.append([0, 0])
coords.append([1, 1])
# solve equation 2 for x when y = START
while not (coords[0][0] == coords[1][0]):
try:
y = ( ( b * x ) + c ) / a
except:
y = 0
print(y)
try:
x = ( ( d * y ) - f ) / e
except:
x = 0
if x >= 100000 or x <= -100000:
count = count + 1
if count >= 100000:
print("It\'s looking like these linear equations don\'t intersect!")
break
print(x)
new_coords = [x, y]
coords.append(new_coords)
coords.pop(0)
if not ((x == float("inf") or x == float("-inf")) and (y == float("inf") or y == float("-inf"))):
pass
else:
infinity_flag if False else True
if infinity_flag == False:
# if the spiral is divergent this switches the equations around so it converges
# the infinity_flag is to check if both spirals returned infinity meaning the lines do not intersect
# I think this would mostly work for linear equations, but for other kinds of equations it might not
x = 0
a = equation_array[1][0]
b = equation_array[1][1]
c = equation_array[1][2]
d = equation_array[0][0]
e = equation_array[0][1]
f = equation_array[0][2]
infinity_flag = False
else:
print("These linear equations do not intersect")
break
y = round(y, 3)
x = round(x, 3)
print(x, y)
equation_array = []
print("Specify coefficients a and b, and a constant c for equation 1")
equations = Equation(equation_array)
print("Specify coefficients a and b, and a constant c for equation 1")
equations = Equation(equation_array)
print(equation_array)
Stringify_Equations(equation_array)
Spiral(equation_array)
I need to find a way to write cos(1) in python using a while loop. But i cant use any math functions. Can someone help me out?
for example I also had to write the value of exp(1) and I was able to do it by writing:
count = 1
term = 1
expTotal = 0
xx = 1
while abs(term) > 1e-20:
print("%1d %22.17e" % (count, term))
expTotal = expTotal + term
term=term * xx/(count)
count+=1
I amm completely lost as for how to do this with the cos and sin values though.
Just change your expression to compute the term to:
term = term * (-1 * x * x)/( (2*count) * ((2*count)-1) )
Multiplying the count by 2 could be changed to increment the count by 2, so here is your copypasta:
import math
def cos(x):
cosTotal = 1
count = 2
term = 1
x=float(x)
while abs(term) > 1e-20:
term *= (-x * x)/( count * (count-1) )
cosTotal += term
count += 2
print("%1d %22.17e" % (count, term))
return cosTotal
print( cos(1) )
print( math.cos(1) )
You can calculate cos(1) by using the Taylor expansion of this function:
You can find more details on Wikipedia, see an implementation below:
import math
def factorial(n):
if n == 0:
return 1
else:
return n * factorial(n-1)
def cos(order):
a = 0
for i in range(0, order):
a += ((-1)**i)/(factorial(2*i)*1.0)
return a
print cos(10)
print math.cos(1)
This gives as output:
0.540302305868
0.540302305868
EDIT: Apparently the cosine is implemented in hardware using the CORDIC algorithm that uses a lookup table to calculate atan. See below a Python implementation of the CORDIS algorithm based on this Google group question:
#atans = [math.atan(2.0**(-i)) for i in range(0,40)]
atans =[0.7853981633974483, 0.4636476090008061, 0.24497866312686414, 0.12435499454676144, 0.06241880999595735, 0.031239833430268277, 0.015623728620476831, 0.007812341060101111, 0.0039062301319669718, 0.0019531225164788188, 0.0009765621895593195, 0.0004882812111948983, 0.00024414062014936177, 0.00012207031189367021, 6.103515617420877e-05, 3.0517578115526096e-05, 1.5258789061315762e-05, 7.62939453110197e-06, 3.814697265606496e-06, 1.907348632810187e-06, 9.536743164059608e-07, 4.7683715820308884e-07, 2.3841857910155797e-07, 1.1920928955078068e-07, 5.960464477539055e-08, 2.9802322387695303e-08, 1.4901161193847655e-08, 7.450580596923828e-09, 3.725290298461914e-09, 1.862645149230957e-09, 9.313225746154785e-10, 4.656612873077393e-10, 2.3283064365386963e-10, 1.1641532182693481e-10, 5.820766091346741e-11, 2.9103830456733704e-11, 1.4551915228366852e-11, 7.275957614183426e-12, 3.637978807091713e-12, 1.8189894035458565e-12]
def cosine_sine_cordic(beta,N=40):
# in hardware, put this in a table.
def K_vals(n):
K = []
acc = 1.0
for i in range(0, n):
acc = acc * (1.0/(1 + 2.0**(-2*i))**0.5)
K.append(acc)
return K
#K = K_vals(N)
K = 0.6072529350088812561694
x = 1
y = 0
for i in range(0,N):
d = 1.0
if beta < 0:
d = -1.0
(x,y) = (x - (d*(2.0**(-i))*y), (d*(2.0**(-i))*x) + y)
# in hardware put the atan values in a table
beta = beta - (d*atans[i])
return (K*x, K*y)
if __name__ == '__main__':
beta = 1
cos_val, sin_val = cosine_sine_cordic(beta)
print "Actual cos: " + str(math.cos(beta))
print "Cordic cos: " + str(cos_val)
This gives as output:
Actual cos: 0.540302305868
Cordic cos: 0.540302305869