My question is can I write this code in a different and easier way .? Python code: # there is only one root def fn(x): return x**3 - x 2 # define bisection method def bisection( eq, segment, app 0.3): a, b = segment['a'l, segment['b'1 Fa, Fb = eq(a), eq(b) if Fa Fb > 0: raise Exception( 'No change of sign bisection not possible') while( b a > app ): x = ( a + b) / 2.0 f eq(x) if f * Fa > 0: a x else: b = x %3D return x #test it print bisection ( fn, {'a':1, 'b':2}, 0.00003)
write this code in a different and easier way
![My question is can I write this code in a different
and easier way .?
Python code:
# there is only one root
def fn(x):
return x**3 - x - 2
# define bisection method
def bisection( eq, segment, app = 0.3):
a, b = segment ['a'1, segment['b']
Fa, Fb eq(a), eq(b)
if Fa * Fb > 0:
raise Exception('
while( b- a > app ):
x = ( a + b) / 2.0
f eq(x)
if f * Fa > 0: a x
else: b x
return x
#test it
print bisection (fn, {'a':1, 'b':2}, 0.00003)
%3D
change of sign - bisection not possible')](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0094fe34-92f6-406c-8255-2b6204c10f7f%2F68ff5824-ae89-4630-ad4f-39587c815ae1%2F47kse3p_processed.jpeg&w=3840&q=75)
The easier presentation of given Python code:
def fn(x):
return x**3 - x - 2
# define bisection method
def bisection(segment, app=0.3):
a, b = segment['a'], segment['b']
if (fn(a) * fn(b) >= 0):
raise Exception('No change of sign- bisection not possible')
x = a
while ((b-a) > app):
# compute middle point
x = (a+b)/2
# Check for middle point is root
if (fn(x) == 0.0):
break
# check to decide the side for iteration steps
if (fn(x)*fn(a) < 0):
b = x
else:
a = x
return x
# test it
print('The value of root is : %.5f'%bisection({'a':1, 'b':2}, 0.00003))
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