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n=iterations
for some reason this code will need a lot more iterations for more accurate result from other codes, Can anyone explain why this is happening? thanks.
n,s,x=1000,1,0
for i in range(0,n,2):
x+=s*(1/(1+i))*4
s=-s
print(x)
As I mentioned in a comment, the only way to speed this is to transform the sequence. Here's a very simple way, related to the Euler transformation (see roippi's link): for the sum of an alternating sequence, create a new sequence consisting of the average of each pair of successive partial sums. For example, given the alternating sequence
a0 -a1 +a2 -a3 +a4 ...
where all the as are positive, the sequences of partial sums is:
s0=a0 s1=a0-a1 s2=a0-a1+a2 s3=a0-a1+a2-a3 s4=a0-a1+a2-a3+a4 ...
and then the new derived sequence is:
(s0+s1)/2 (s1+s2)/2 (s2+s3)/2 (s3+s4)/2 ...
That can often converge faster - and the same idea can applied to this sequence. That is, create yet another new sequence averaging the terms of that sequence. This can be carried on indefinitely. Here I'll take it one more level:
from math import pi
def leibniz():
from itertools import count
s, x = 1.0, 0.0
for i in count(1, 2):
x += 4.0*s/i
s = -s
yield x
def avg(seq):
a = next(seq)
while True:
b = next(seq)
yield (a + b) / 2.0
a = b
base = leibniz()
d1 = avg(base)
d2 = avg(d1)
d3 = avg(d2)
for i in range(20):
x = next(d3)
print("{:.6f} {:8.4%}".format(x, (x - pi)/pi))
Output:
3.161905 0.6466%
3.136508 -0.1619%
3.143434 0.0586%
3.140770 -0.0262%
3.142014 0.0134%
3.141355 -0.0076%
3.141736 0.0046%
3.141501 -0.0029%
3.141654 0.0020%
3.141550 -0.0014%
3.141623 0.0010%
3.141570 -0.0007%
3.141610 0.0005%
3.141580 -0.0004%
3.141603 0.0003%
3.141585 -0.0003%
3.141599 0.0002%
3.141587 -0.0002%
3.141597 0.0001%
3.141589 -0.0001%
So after just 20 terms, we've already got pi to about 6 significant digits. The base Leibniz sequence is still at about 2 digits correct:
>>> next(base)
3.099944032373808
That's an enormous improvement. A key point here is that the partial sums of the base Leibniz sequence give approximations that alternate between "too big" and "too small". That's why averaging them gets closer to the truth. The same (alternating between "too big" and "too small") is also true of the derived sequences, so averaging their terms also helps.
That's all hand-wavy, of course. Rigorous justification probably isn't something you're interested in ;-)
That is because you are using the Leibniz series and it is known to converge very (very) slowly.
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pip install sympy
import math as m
from sympy import *
a,b = symbols('a b')
x1, y1 = (2, m.radians(152.6))
x2, y2 = (4, m.radians(163.7))
result = solve((a*sin(b*x1)-y1, a*sin(b*x2)-y2), dict=True)
print(result)
result is like below
[{a: -3.15571732546928 + 1.66055281406852e-26I, b: -0.502333320617012 + 2.31610571514848e-22I}, {a: -3.15571732546928 + 1.66055281406852e-26I, b: 2.63925933297278 + 2.24993126614424e-22I}, {a: 3.15571732546928 + 1.66055281406852e-26I, b: -2.63925933297278 + 2.24993126614424e-22I}, {a: 3.15571732546928 + 1.66055281406852e-26I, b: 0.502333320617012 + 2.31610571514848e-22I}]
What does this number mean?, especially 'e-26*I'
I, in Sympy, is the imaginary number constant. That is, the number defined to be the square root of negative one. From Wikipedia
[A] complex number is an element of a number system that extends the real numbers with a specific element denoted i
In your case, Sympy has found complex-valued solutions to your trigonometry equation. It's possible to apply trigonometric functions to complex numbers, and indeed often the only solution to an equation like that is a complex number.
The e part is a common format for displaying numbers in computer terminals. It's similar to scientific notation. The number 3e+8 should be read as three times ten to the eighth power. Likewise, 2e-1 should be read as two times ten to the negative-first power, or 0.2.
So, to take your first example,
-3.15571732546928 + 1.66055281406852e-26I
This is a complex number whose real part is -3.1557 and whose imaginary part is, frankly, a really small number (a decimal point followed by 25 zeroes and then some nonzero quantities).
What this likely means is that Sympy could not algebraically solve the equation, so it resorted to some numerical methods which could only get close. It's possible a = -pi is a solution, since -3.155 is pretty close to pi and the imaginary part of that number might as well be zero.
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I already know how does abs function works.I mean it only shows you how far the number is from zero
The only thing i can't just get is with this example:
print(abs(3 + 4j)) # prints 5 !! Why ?
Because for complex numbers abs(number) will return magnitude of Complex Numbers.
Magnitude value will counted as :
√x2+y2 = √(3² + 4²) = √(9 + 16) = √(25) = 5.0
So abs with complex number will return magnitude of complex numbers.
for further reference you can use https://www.geeksforgeeks.org/abs-in-python/.
As the rest of the answers stated above, 3+4j is a complex number and the formula of calculating the absolute value of a complex number x+yi is sqrt( (x^2) + (y^2) ). In your case it's:
sqrt(3^2 + 4^2) = sqrt(9 + 16) = sqrt(25) = 5
Because in python, 3+4j is complex number and python calculates the absolute value or magnitude of the complex number when you do abs() on it. Magnitude of 3+4j is 5. Try this :
type(3+4j)
It should give <class 'complex'>.
Note : Magnitude of a complex number a+bj is ((a**2+b**2)**0.5)
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Is there a best practice to avoid ZeroDivisionErrorin python? For instance, consider the following function:
def division (x,y):
return x / y
In this function if y is set to zero, the ZeroDivisionError is raised. One strategy would be to replace the denominator of the division by y+1, so the function would be:
def division (x,y):
return x / (y+1)
or the other strategy can be returning nan value if denominator is zero:
def division (x,y):
if y == 0:
return np.nan
else:
return x / y
What are the cons of the above strategies? Is there any other approach to avoid ZeroDivisionError? Exception handling is also a popular solution for this problem, but I'm more looking for prevention strategy not intervention.
This follows the same conditioning strategy, but you can use this approach:
def division(x, y):
return x/y if y else 0
Con of your add one strategy is that if y is a small number (not -1) then your results will be very incorrect. Only sort of works when y is very large.
Con of returning NaN is that you need to document this clearly for the person calling your function. np.nan can surprise the caller. For example what is expected when you do comparison like division(1, 0) == division(2,0)
Raising an error is the right thing to do and let the caller handle according to their rules.
Keep in mind that numpy.true_divide() never returns ZeroDivisionError. Returns inf. instead.
i.e
a = [1,2,3]
b = [0,1,2]
np.true_divide(a, b)
RuntimeWarning: divide by zero encountered in true_divide
array([inf, 2. , 1.5])
It usually means that your algorithm is not carefully designed. Just taking care of +/- 0 is not going to help you in most cases. Almost every algorithm in literature deals with division by zero either by adding an ε to the absolute value or by thresholding the absolute value. One could also avoid such cases by limiting processing only on those values where denominator is significant.
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I have 2 questions I need to ask for help with. One question I don't entirely understand so if someone could help me to that would be great.
Question One (the one I don't entirely understand):
One definition of e is
Formula for e
This can be calculated as my_e = 1/math.factorial(0) + 1/math.factorial(1) + 1/math.factorial(2) + 1/math.factorial(3) + …
Let n be the input number of the math.factorial() function. n successively takes on 0, 1, 2, 3 and so on. Find the smallest n such that the absolute value of (my_e – math.e) is less than or equal to 10-10. That is, abs(my_e - math.e) <= (10 ** -10).
I just don't entirely understand what I am being asked to do. Clarification would be great. Thanks!
Question 2:
Ask the user to type in a series of integers. Sum them up and print out the sum and the number of integers the user has entered.
My code
So what should happen is after I enter the numbers I want to enter and hit the enter key, it should calculate and print out "sum = 25 count = 3". The screenshot shows what error message I am getting.
Any help you have is welcomed and greatly appreciated.
As far as you first question goes:
>>> import math
>>> math.e
2.718281828459045
>>> sum(1.0/math.factorial(i) for i in range(5))
2.708333333333333
>>> abs(sum(1.0/math.factorial(i) for i in range(5)) - math.e) < 10**-10
False
>>> abs(sum(1.0/math.factorial(i) for i in range(30)) - math.e) < 10**-10
True
So, somewhere between n == 5 and n == 30 you get 10 decimal places for e. Create the sum term by term in a while-loop (instead of by using the sum function as I have, since you are unlikely to have seen that syntax yet). At each pass through the loop, compare the sum with math.e. Stop when you get the target accuracy. Return the final n.
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I started programming in python not too long ago and I am having trouble with a part of a program. The program will ask for input from the user and he can input: A, B, C, M, or Q. I have completed the A, M, and Q part but I can't figure out how to do the parts for B (calculate the sine of the number you want) and C (calculate the sine).
All the information I was given was:
The power series approximation for the sine of X can be expressed as:
sine(X) = X – (X3/3!) + (X5/5!) – (X7/7!) + (X9/9!) .... Note that an
individual term in that power series can be expressed as: (-1)k *
X2k+1 / (2k+1)! where k = 0, 1, 2, 3, ….
Oooh, and (but for this a while loop should do right?):
When computing the sine of X or the cosine of X, the program will expand the power series
until the absolute value of the next term in the series is less than 1.0e-8 (the specified epsilon).
That term will not be included in the approximation.
And I can't use import math.
Can anyone give me an idea of how I can do this? I sincerely have no idea of where to even start hahaha.
Thanks in advance!
***Hey guys, I've been trying to do this for the last 3 hours. I'm really new to programming and some of yours answers made it a bit more understandable for me but my program is not working, I really don't know how to do this. And yes, I went to speak with a tutor today but he didn't know either. So yeah, I guess I'll just wait until I get the program graded by my teacher and then I can ask him how it was supposed to be done. Thank you for all the answers though, I appreciate them! :)
>>> e = 2.718281828459045
>>> X = 0.1
>>> (e**(X*1j)).imag # sin(X)
0.09983341664682815
>>> (e**(X*1j)).real # cos(X)
0.9950041652780258
Verify
>>> from math import sin, cos
>>> sin(X)
0.09983341664682815
>>> cos(X)
0.9950041652780258
You'll probably get better marks if you sum up the series explicitly though
result = 0
n = 1
while True:
term = ...
result += term
if term <= epsilon:
break
n += 2
It seems that you aren't supposed to import math because you are supposed to write your own function to compute sine. You are supposed to use the power series approximation.
I suggest you start by writing a factorial function, then write a loop that uses this factorial function to compute the power series.
If you still can't figure it out, I suggest you talk to your teacher or a teacher's assistant.
Since you have a condition to finish the loop last_term < 1.0e-8, you should use a while:
while last_term > 1.0e-8:
You will need a counter to keep the count of k (starting from 0) and a variable to keep the last term:
k = 10 # some initial value
last_term = 0
while ...:
last_term = ... # formula here
and also a result variable, let' say sin_x:
while ...:
...
sin_x += last_term
Note: In the formula you are using factorial, so will need to define a function that computes the factorial of a number, and use it properly.