What is A and B so that the line Ay = Bx + 1 passes through points (1, 3) and (5,13) in the Cartesian plane?
I have been trying to solve it using the slope intercept equation to no avail. This is taken from Dale Hoffman's Contemprary Calculus.
First, I would reorder to get canonical form,
y = (B/A) * x + (1/A) = m * x + b
Now we find slope (m):
m = dy / dx = (13 - 3) / (5 - 1) = 2.5
sub in to find b:
3 = 2.5 * 1 + b
b = 0.5
Now sub back to find the values you want,
b = 0.5 = 1 / A
A = 2
m = 2.5 = B / 2
B = 5
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
The following code generates numpy 2D lists of r and E values for the specified intervals.
r = np.linspace(3, 14, 10)
E = np.linspace(0.05, 0.75, 10)
r, E = np.meshgrid(r, E)
I am then using the following nested loop to generate output from the function ionisationGamma for each r and E interval value.
for ridx in trange(len(r)):
z = []
for cidx in range(len(r[ridx])):
z.append(ionisationGamma(r[ridx][cidx], E[ridx][cidx]))
Z.append(z)
Z = np.array(Z)
This loop gives me a 2D numpy array Z, which is my output and I am using it for a 3D graph. The problem with it is: it is taking ~6 hours to generate the output for all these intervals as there are so many values due to np.meshgrid. I have just discovered multi-threading in Python and wanted to know how I can implement this by using it. Any help is appreciated.
See below code for ionisationGamma
def ionisationGamma(r, E):
I = complex(0.1, 1.0)
a_soft = 1.0
omega = 0.057
beta = 0.0
dt = 0.1
steps = 10000
Nintervals = 60
N = 3000
xmin = float(-300)
xmax = -xmin
x = [0.0]*N
dx = (xmax - xmin) / (N - 1)
L = dx * N
dk = 2 * M_PI / L
propagator = None
in_, out_, psi0 = None, None, None
in_ = [complex(0.,0.)] * N
psi0 = [complex(0.,0.)] * N
out_ = [[complex(0.,0.)]*N for i in range(steps+1)]
overlap = exp(-r) * (1 + r + (1 / 3) * pow(r, 2))
normC = 1 / (sqrt(2 * (1 + overlap)))
gammai = 0.5
qi = 0.0 + (r / 2)
pi = 0.0
gammai1 = 0.5
gammai2 = 0.5
qi1 = 0.0 - (r / 2)
qi2 = 0.0 + (r / 2)
pi1 = 0.0
pi2 = 0.0
# split initial wavepacket
for i in range(N):
x[i] = xmin + i * dx
out_[0][i] = (normC) * ((pow(gammai1 / M_PI, 1. / 4.) * exp(complex(-(gammai1 / 2.) * pow(x[i] - qi1, 2.), pi1 * (x[i] - qi1)))) + (pow(gammai2 / M_PI, 1. / 4.) * exp(complex(-(gammai2 / 2.) * pow(x[i] - qi2, 2.), pi2 * (x[i] - qi2)))))
in_[i] = (normC) * ((pow(gammai1 / M_PI, 1. / 4.) * exp(complex(-(gammai1 / 2.) * pow(x[i] - qi1, 2.), pi1 * (x[i] - qi1)))) + (pow(gammai2 / M_PI, 1. / 4.) * exp(complex(-(gammai2 / 2.) * pow(x[i] - qi2, 2.), pi2 * (x[i] - qi2)))))
psi0[i] = in_[i]
for l in range(1, steps+1):
for i in range(N):
propagator = exp(complex(0, -potential(x[i], omega, beta, a_soft, r, E, dt, l) * dt / 2.))
in_[i] = propagator * in_[i];
in_ = np.fft.fft(in_, N)
for i in range(N):
k = dk * float(i if i < N / 2 else i - N)
propagator = exp(complex(0, -dt * pow(k, 2) / (2.)))
in_[i] = propagator * in_[i]
in_ = np.fft.ifft(in_, N)
for i in range(N):
propagator = exp(complex(0, -potential(x[i], omega, beta, a_soft, r, E, dt, l) * dt / 2.))
in_[i] = propagator * in_[i]
out_[l][i] = in_[i]
initialGammaCentre = 0.0
finalGammaCentre = 0.0
for i in range(500, 2500 +1):
initialGammaCentre += pow(abs(out_[0][i]), 2) * dx
finalGammaCentre += pow(abs(out_[steps][i]), 2) * dx
ionisationGamma = finalGammaCentre / initialGammaCentre
return ionisationGamma
def potential(x, omega, beta, a_soft, r, E, dt, l):
V = (-1. / sqrt((x - (r / 2)) * (x - (r / 2)) + a_soft * a_soft)) + ((-1. / sqrt((x + (r / 2)) * (x + (r / 2)) + a_soft * a_soft))) + E * x
return V
Since the question is about how to use multiprocessing, the following code will work:
import multiprocessing as mp
if __name__ == '__main__':
with mp.Pool(processes=16) as pool:
Z = pool.starmap(ionisationGamma, arguments)
Z = np.array(Z)
Where the arguments are:
arguments = list()
for ridx in range(len(r)):
for cidx in range(len(r[ridx])):
arguments.append((r[ridx][cidx], E[ridx][cidx]))
I am using starmap instead of map, since you have multiple arguments that you want to unpack. This will divide the arguments iterable over multiple cores, using the ionisationGamma function and the final result will be ordered.
However, I do feel the need to say that the main solution is not really the multiprocessing but the original function code. In ionisationGamma you are using several times the slow python for loops. And it would benefit your code a lot if you could vectorize those operations.
A second observation is that you are using many of those loops separately and it would be nice if you could separate that one big function into multiple smaller functions. Then you can time every function individually and speed up those that are too slow.
I am writing a function with a For loop, and ultimately the value of the function will depend on the output of the For loop. For now as a test, the value of the function is a constant. If the For loop is in the code, the function results in #Value!. If I remove the For loop, the output is the specified constant value. How does the For loop need to be specified to avoid this? Good values for Tc and Th as a test would be 100 and 300, respectively.
Function kndT(material As Integer, Tc As Double, Th As Double) As Variant
Dim x As Double
Select Case material
Case 4
If Th > 300 Then
Tmax = 300
Else
Tmax = Th
End If
A = 0.07918
b = 1.0957
c = -0.07277
D = 0.08084
e = 0.02803
f = -0.09464
g = 0.04179
h = -0.00571
i = 0
End Select
hh = (Tmax - Tc) / 999
fi = 0
nc = 1
For i = 1 To 999
Temp = (Tc + i * hh)
x = Log(Temp) / Log(10#)
y = A + b * x + c * x ^ 2 + D * x ^ 3 + e * x ^ 4 + f * x ^ 5 + g * x ^ 6 + h * x ^ 7 + i * x ^ 8
fn = 10 ^ y
If nc = 3 Then
fi = fi + 2 * fn
nc = 1
Else
fi = fi + 3 * fn
nc = nc + 1
End If
Next i
kndT = 2
End Function
I need to translate this formula into python:
x = b + (sqrt((b^2)-1)) / 2a
Can somebody please help me?
import math
x = b + (math.sqrt((b ** 2) - 1 )) / (2 * a)
Mathematically correct this solution is
import math
import cmath
def quadratic(a, b, c):
if b == 0:
f = (cmath.sqrt(b ** 2 - 1 )) / (2 * a)
else:
f = (math.sqrt(b ** 2 - 1 )) / (2 * a)
return b + f, b -f
print(quadratic(1, 2, 3))
This will give you the two roots.
You can access the roots by indexing:
my_roots = quadratic(1, 2, 3)
x_1 = my_roots[0]
x_2 = my_roots[1]
By using a Python function , you can re-use this code.
This should work as well. No need to import any library.
Assuming a != 0.
When b >= 1:
x = b + (((b ** 2) - 1 ) ** 0.5) / (2 * a)
NOTE: You should ideally have two roots for sqrt(b^2 - 1). Your function should return both values. Now, I am guessing that, under certain prior knowledge you have decided to only keep the positive root.
When b == 0, you get two roots. Note that sqrt(-1) = j or -j (complex roots).
x = (+1j) / (2 * a) # positive root
# OR
x = (-1j) / (2 * a) # negative root
More generally, when b**2 < 1, you get two complex roots.
x = b + (k*1j) ((1 - (b ** 2)) ** 0.5) / (2 * a)
# positive root: k = +1
# OR
# negative root: k = -1
If I have six variables representing two lines in general equation form (ax + by + c = 0). So for example:
ax + by + c = 0
jx + ky + l = 0
How can I find the intersection point's (x and y) [assuming there is one] from the six variables?
PS. Any recommendation of a good source for information on very simply computational geometry, like this, would be appreciated.
The intersection point satisfies both equations. So all you need is to solve them simultaneously:
ax + by + c = 0 (*j)
jx + ky + l = 0 (*a)
ajx + bjy + cj = 0 (-)
ajx + aky + al = 0
(bj-ak)y + cj - al = 0
y = (al-cj) / (bj-ak)
And similarly for x. (Or you can substitute the found value for y in any of the original equations and then find x):
x = (ck-bl) / (bj-ak)