I want to create a def that you provide a positive integer n and a number x, and it returns the n-th term of the following sequence:
x - x^3/3 + x^5/5 - x^7/7 + x^9/9 - x^11/11.....
where:
first term is: a1 = x
second term is : a2 = x - x^3/3
third term is: a3 = x - x^3/3 +x^5/5
etc
This is what i came up with, but it doesn't seem to return constistent results compared to calculating the values manually. Please, tell me if I'm missing something! Thank you very much!
def madh(n, x):
if n == 1:
return x
else:
result = (((-1)**n) * (x ** (2*n-1)) / (2*n - 1)) + madh((n - 1), x)
return result
Your mistake is in the sign of the element. For example, the second term is negative, but it is positive in your case ((-1) ** 2 = 1). So, the corrected version is:
def madh(n, x):
if n == 1:
return x
else:
# change n to (n+1) in the power of -1
result = (((-1)**(n+1)) * (x ** (2 * n - 1)) / (2 * n - 1)) + madh((n - 1), x)
return result
By the way, your function returns the sum of series up to n-th terms, not the n-th term of the series.
The nth term of the series is ((x^n)/n)*(-1^(n+1)). Simple function for that would be
def nth_num(n,x):
if n==1:
return x
return nth_num(n-1,x) + ((x**n)/n)*(-1**(n+1))
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
Is there a faster way to create an exponential moving average than for loop?
I am priming the first n values of exp_aves to be mean of first n obseravtions.
def exp_mov_ave(observations, n):
exp_ave = np.mean(observations[0:n])
lambdak = (n - 1) / (n + 1)
exp_aves = []
for i,element in enumerate(observations):
if (i >= n):
exp_ave = (lambdak * exp_ave) + ((1 - lambdak) * element)
exp_aves.append(exp_ave)
return np.array(exp_aves)
So I am supposed to find the sum of this series :
f(n) = 1 + (2*3) + (4*5*6) + .....n terms
I did this using recursion as follows:
def f(n):
if n == 1:
return 1
else:
product = 1
add = 0
s = (n * (n+1))/2
for i in range (0,n):
product = product * s
s = s - 1
add = product + f(n-1)
return add
Now please bear with me
I thought I could do this faster if I could use special series in linear algebra:
Here is what I attempted:
I found the nth term(through some vigorous calculations) : Tn =
Now is there a method I can use this formula to find sum of Tn and hence the series using python.
I also want to know whether we can do such things in python or not?
You can translate that product to Python using a for loop, analog to how you kept track of the product in your recursive function. So T(n) would be:
def T(n):
product = 1
for r in range(1, n+1):
product *= (n * (n - 1)) / 2 + r
return product
Now as you said, you need to find the sum of T(x) for x from 1 to n. In Python:
def f(n):
sum = 0
for i in range(1, n+1):
sum += T(i)
return sum
FYI:
a += x is the same as a = a + x,
analog a *= x is equal to a = a * x
How should I answer this-"Compute the total number of possible paths from (0,0) to (7,9) if the steps R (to the right) and U (up) are allowed, along with the diagonal step D:(x,y)→(x +1,y+ 1)"
Edit: added calculation for arbitrary cell, not only diagonal one.
The number of ways on square grid is known as Delannoy number, for (n,n) cell sequence is 1, 3, 13, 63, 321, 1683, 8989...
There is simple natural recurrence
D(m, n) = D(m-1, n) + D(m, n-1) + D(m-1,n-1)
that might be used to calculate values rather quickly for reasonable argument values (table approach, O(nm) operations including long summation).
"Closed formula"
D(m, n) = Sum[k=0..min(n, m)] {C(m + n - k, m) * C(m, k)}
for effective implementations requires table of binomial coefficients
#2D table quadratic approach
def PathsInSqGrid(n, m):
D = [[0 for x in range(m+1)] for y in range(n+1)]
for i in range(n+1):
D[i][0] = 1
for i in range(m+1):
D[0][i] = 1
for i in range(1, n+1):
for j in range(1,m+1):
D[i][j] = D[i][j-1] + D[i-1][j] + D[i-1][j-1]
return D[n][m]
def NCr(n, k):
result = 1
if k > n - k:
k = n - k
for i in range (1, k + 1):
result = (result * (n - i + 1)) // i
return result
#closed formula
def PathsCF(n, m):
#D(m, n) = Sum[k=0..min(n, m)] {C(m + n - k, m) * C(m, k)}
res = 0
for k in range(0, min(n, m) + 1):
res += NCr(m + n - k, m) *NCr(m, k)
return res
print(PathsInSqGrid(7, 9))
print(PathsCF(7, 9))
>>>
224143
224143
Wiki also shows two so-called "closed formulas" for central Delannoy numbers (while I believe that closed formula should be single expression without loop of length n):
D(n) = Sum[k=0..n]{C(n,k)*C(n+k,k)}
D(n) = Sum[k=0..n]{C(n,k)^2 * 2^n}
and recurrence (looks simple, linear complexity, but real implementation requires division of long number by short one)
n*D(n) = 3*(2*n-1) * D(n-1) - (n-1)*D(n-2)
and generating function
Sum[n=0..Inf]{D(n)*x^n} = 1/Sqrt(1 - 6 * x + x^2) = 1 + 3x + 13x^2 + 63x^3 +...
Code
#central Delannoy numbers
#2D table quadratic approach
#linear approach for diagonal cell (n,n)
def PathsInSqGridLin(n):
if n < 2:
return 2 * n + 1
A, B = 1, 3
for i in range(2, n + 1):
B, A = (3 * (2 * i - 1) * B - (i - 1) * A) // i, B
return B
print(PathsInSqGridLin(3))
print(PathsInSqGridLin(100))
>>
63
2053716830872415770228778006271971120334843128349550587141047275840274143041
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