Solve polynomial between two values interval

Question

x = eval(input())
x = eval(input())
polinomio = 3*x**3- 5*x + 0.8
for x in range (x,x,0.5):
    print(polinomio)

Traceback (Most recent call last): File "C:/Users/User/Appdata/Local/Programs/Python/Python36-32/fdrew.py", line 4, in for x in range (x,x,0.5): Typeerror: 'float' Object cannot be Interpreted as an integer

Answer 1

We will solve your problem step by step.

First, let’s see the use of range. Let’s look at the documentation and... For life, it only accepts integer numbers! This function accepts up to 3 parameters (check that answer for more chewed details). How we will proceed then?

An alternative is to proceed arithmetically. If we want to take the values between m and M leaping from 0.5 in 0.5 (being i iterator), this is identical to picking the values between 2*m and 2*M leaping from 1 in 1, iterating on the variable i_dobro which can easily be transformed into i = i_dobro/2. But this only works in Python 3 (for Python 2 you have to divide by 2.0) and if m and M be whole...

So maybe we can use while? Let’s start with m, as small as M, increasing 0.5 every step... it seems, it seems reasonable.

Now, how about we find out who it is m and M that I so much quoted above? Well, m is the lower margin of the range, while M is the upper margin. As this data is read from the standard entry, and has no restriction or complementary text to this problem, I can only believe that the two entries will be numbers possibly out of order. Then let’s go n1 to the first number and n2 for the second number. To determine m and M that’s how it is:

(M, m) = (n1, n2) if n1 > n2 else (n2, n1)

Okay, so generally our iteration goes like this (I’ll keep your reading as much as I disagree with it):

n1 = eval(input())
n2 = eval(input())

(M, m) = (n1, n2) if n1 > n2 else (n2, n1)

i = m
while i < M:
    # ação interessante da iteração 
    i += 0.5

So, how to make the action interesting of iteration? The desired result is the impression of the evaluation of the polynomial before the i past. I see the following options:

• calculate directly in the print
• define a priori function and call it cheerfully
• create a lambda and call it even more cheerfully

For the first option, it is only necessary to change the commented section:

print(3* i**3 -5*i + 0.8)

For the second option, before taking the proper readings, set the function like this:

def polinomio(x):
    return 3* x**3 -5*x + 0.8

Then call inside the print:

print(polinomio(i))

For lambda alternative, before iteration, create lambda function and assign to variable polinomio:

polinomio = (lambda x: 3* x**3 -5*x + 0.8)

And call inside the print as if it were a traditional function:

print(polinomio(i))