Here's a solution that can handle x with single or multiple rows like scipy pdf:
from scipy.stats import multivariate_normal as mvn
# covariance matrix
sigma = np.array[[[2.3, 0, 0, 0],
[0, 1.5, 0, 0],
[0, 0, 1.7, 0],
[0, 0, 0, 2]
]]
# mean vector
mu = np.array[[2,3,8,10]]
# input
x1 = np.array[[2.1, 3.5, 8., 9.5]]
x2 = np.array[[[2.1, 3.5, 8., 9.5],[2.2, 3.6, 8.1, 9.6]]]
def multivariate_normal_pdf[x, mu, cov]:
x_m = x - mu
if x.ndim > 1:
sum_ax = 1
t_ax = [0]
t_ax.extend[list[range[x_m.ndim][:0:-1]]] # transpose dims > 0
else:
sum_ax = 0
t_ax = range[x_m.ndim][::-1]
x_m_t = np.transpose[x_m, axes=t_ax]
A = 1 / [ [[2* np.pi]**[len[mu]/2]] * [np.linalg.det[cov]**[1/2]] ]
B = [-1/2] * np.sum[x_m_t.dot[np.linalg.inv[cov]] * x_m,axis=sum_ax]
return A * np.exp[B]
print[mvn.pdf[x1, mu, sigma]]
print[multivariate_normal_pdf[x1, mu, sigma]]
print[mvn.pdf[x2, mu, sigma]]
print[multivariate_normal_pdf[x2, mu, sigma]]
This is a Python Program to read a number n and print and compute the series “1+2+…+n=”.
Problem Description
The program takes a number n and prints and computes the series “1+2+…+n=”.
Problem Solution
1. Take a value from the user and store it in a variable n.
2. Use a for loop where the value of i ranges between the values of 1 and n.
3. Print the value of i
and ‘+’ operator while appending the value of i to a list.
4. Then find the sum of elements in the list.
5. Print ‘=’ followed by the total sum.
6. Exit.
Program/Source Code
Here is the source code of the Python Program to read a number n and print and compute the series “1+2+…+n=”. The program output is also shown below.
n=int[input["Enter a number: "]] a=[] for i in range[1,n+1]: print[i,sep=" ",end=" "] if[i