Question

python

b) Evaluating a polynomial derivative numerically For a function f(x), the derivative of the function at a value x can be found by evaluating f(x+2)-(*) and finding the limit as a gets closer and closer to 0. Using the same polynomial as the user entered in part (a), and for the same value of x as entered in part (a), compute the limit numerically. That is, start with an estimate by evaluating** 72 using a value for a such as 0.1. Then, repeatedly halve the value of a until the difference between successive evaluations of is less than some small value, such as 10%. Print the result, along with the number of evaluations it took. Calculate how close that result is to the actual answer, computed in part (a). Repeat the above activity by evaluating the limits of the following expressions: 17 2 or Try computing each of those, and calculate how many iterations you need to converge to the limit. Do you get different results with any of them, or does any of them take fewer steps to get an answer? (Put your answer as a comment in your code.)

the polynomial equation is Ax^3+Bx^2+Cx+D

Answer 1

#### while pasting indentations/tabs may get disturbed, please refer pic for correcttabs

##below is f(x)
def f_x(a,b,c,d,x):
return a*x**3+b*x**2+c*x+d;
  
def derivative_at_x(a,b,c,d,x):
a1=0.1;prev_d=1;d=0;ite=0
while(True):
ite=ite+1
d=(f_x(a,b,c,d,x+a1)-f_x(a,b,c,d,x))/a1;
#print(d)
diff=d-prev_d
prev_d=d
if(diff<10**(-6)):
break
a1=a1/2
#print(a1)
print("no of iterations:",ite)
print("derivative is:",d)

def method2_derivative_at_x(a,b,c,d,x):
a1=0.1;prev_d=1;d=0;ite=0
while(True):
ite=ite+1
d=(f_x(a,b,c,d,x+a1)-f_x(a,b,c,d,x-a1))/(2*a1);
diff=d-prev_d
prev_d=d
if(diff<=10**(-6)):
break
a1=a1/2
print("no of iterations:",ite)
print("derivative is:",d)

derivative_at_x(1,1,1,1,1)
method2_derivative_at_x(1,1,1,1,1)
  

  

1 ##below is f(x) 2- def f_x(a,b,c,d,x): return a x**3+b*x**2+<*x+d; 5. def derivative_at_x(a,b,c,d,x): a1=0.1;prev_d=1;d=0;ite=0 while(True): ite=ite+1 d=(f_x(a,b,c,d,x+al)-f_x(a,b,c,d,x))/a1; #print(d) diff=d-prev_d prev_dad if(diff<10**(-6)): break a1=a1/2 #print(al) print("no of iterations:",ite) print("derivative is:",d) 19 20. def method2 derivative_at_x(a,b,c,d,x): a1=0.1;prev_d=1;d=0;ite=0 while(True): ite=ite+1 d=(f_x(a,b,c,d,x+al)-f_x(a,b,c,d,x-a1))/(2*al); diff=d-prev_d prev ded 25 input no of iterations: 2 derivative is: 6.202500000000022 no of iterations: 2