# assignment 1a-d in Python. please explain your steps, thanks in advance. This is the only information...

assignment 1a-d in Python. please explain your steps, thanks in advance. This is the only information that we got for this practice exercise, my teacher said that we should try to get the output literally as in the example.  Assignment 1 (a: 1 point; b: 1 point; c: 2 points; d: 2 points) Your goal is to implement functions fac(n), cos1(x, n) and cos2 (x, n) and a small test program in the file A1.py and to test the functions in the Python shell and with your own test program. a) Implement a test program with the following output: x cos(x) cos(x,10) cos2 (x,10) 0.0 1.0 None None 0.1 0.9950041652780258 None None 0.2 0.9800665778412416 None None 0.3 0.955336489125606 None None 0.4 0.9210609940028851 None None 0.5 0.8775825618903728 None None 0.6 0.8253356149096783 None None 0.7 0.7648421872844885 None None 0.8 0.6967067093471654 None None 0.9 0.6216099682706644 None None 1.0 0.5403023058681398 None None 1.1 0.4535961214255773 None None 1.2 0.3623577544766736 None None 1.3 0.26749882862458735 None None 1.4 0.16996714290024104 None None 1.5 0.0707372016677029 None None 1.6 -0.029199522301288815 None None 1.7 -0.12884449429552464 None None 1.8 -0.2272020946930871 None None 1.9 -0.32328956686350335 None None 2.0 -0.4161468365471424 None None >>>
So, the main program / test program prints a header line and 21 lines with values for x, cos(x) (use the function in math) and function values of two functions that you will implement yourself. Because the two functions cosl and cos2 are still empty, the output will be None. Do not use any output formatting. The numbers do not need to be properly aligned. b) Implement a function fac (n) that returns n! for integer n. For negative n and for n - the returned value must be 1. It is not allowed to use the math.factorial function in your function. c) Implement a function cos1(x, n) that returns an approximation of cos(x) calculated with the following formula cos(x) (-1)' x x x x x x so for instance cos 1(x, (21)! Your 4! 6! 8! 10! cosi function must use the function fac. d) Implement a better function cos2(x, n). Start with a copy of cos1(x, n) and try to call the function in the Python shell with n=100, so type cos1(0.5, 100). Your function will probably crash with an error message like OverflowError: int too large to convert to float You can solve this problem by calculating the terms of the partial sum without calculating factorials. So, calculate the next term using the previous term. Intended output of your test program after implementing all functions: x cos(x) cos(x, 10) cos2(x,10) 0.0 1.0 1.0 1.0 0.1 0.9950041652780258 0.9950041652780258 0.9950041652780258 0.2 0.9800665778412416 0.9800665778412416 0.9800665778412416 0.3 0.955336489125606 0.955336489125606 0.955336489125606 0.4 0.9210609940028851 0.921060994002885 0.921060994002885 0.5 0.8775825618903728 0.8775825618903728 0.8775825618903728 0.6 0.8253356149096783 0.8253356149096783 0.8253356149096783 0.7 0.7648421872844885 0.7648421872844884 0.7648421872844884 0.8 0.6967067093471654 0.6967067093471654 0.6967067093471654 0.9 0.6216099682706644 0.6216099682706645 0.6216099682706645 1.0 0.5403023058681398 0.5403023058681397 0.5403023058681397 1.1 0.4535961214255773 0.45359612142557726 0.45359612142557726 1.2 0.3623577544766736 0.3623577544766735 0.3623577544766735 1.3 0.26749882862458735 0.26749882862458735 0.26749882862458735 1.4 0.16996714290024104 0.16996714290024104 0.169967142900241 1.5 0.0707372016677029 0.0707372016677029 0.0707372016677029 1.6 -0.029199522301288815 -0.029199522301288888 -0.029199522301288888 1.7 -0.12884449429552464 -0.12884449429552447 -0.12884449429552447 1.8 -0.2272020946930871 -0.22720209469308678 -0.22720209469308672 1.9 -0.32328956686350335 -0.32328956686350224 -0.32328956686350224 2.0 -0.4161468365471424 -0.41614683654713874 -0.41614683654713874

#source code:

import math
def fac(n):
if(n<=0):
return 1
else:
return n*fac(n-1)

print(fac(-1))

def cos1(x,n):
z=2
sum=1
for i in range(0,n):
if(i%2==0):
sum=sum-((x**z)/fac(z))
z=z+2
else:
sum=sum+((x**z)/fac(z))
z=z+2
return sum
def cos2(x,n):
return cos1(x,n)
i=0.0
print("x\tcos(x)\tcos1(x,10)\tcos2(x,10)\n")
while(i<=2.0):
print(str(i)+"\t"+str(math.cos(i))+"\t"+str(cos1(i,10))+"\t"+str(cos2(i,10)))
i=round(i+0.1,1) #output: #if you have any doubt comment below...

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