Solve a polynomial between two values

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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 can not be interpreted as an integer

    
asked by anonymous 23.10.2017 / 05:21

1 answer

4

Let's solve your problem, step by step.

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

An alternative is to proceed arithmetically. If we want to get the values between m and M jumping from 0.5 to 0.5 (being i iterator), this is identical to getting values between 2*m and 2*M jumping from 1 in 1 , iterating in variable i_dobro that can easily be transformed into i = i_dobro/2 . But this only works in Python 3 (for Python 2 you need to split by 2.0 ) and if m and M are integers ...

So maybe we can use while ? Let's start from m , while less than M , increasing 0.5 at each step ... yeah, seems reasonable.

Now, how about finding out who is m and M that I quoted above? Well, m is the bottom margin of the range, while M is the top margin. As these data are read from the standard input, and have no restriction or supplementary text to this problem, I can only believe that the two inputs will be possibly out of order numbers. Then we read n1 for the first number and n2 for the second number. To determine m and M is as follows: 

(M, m) = (n1, n2) if n1 > n2 else (n2, n1)
Okay, so overall our iteration looks like this (I'll keep reading it 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 do you do the interesting iteration action? The desired result is the impression of the evaluation of the polynomial against the% of% past. I see the following options:

  • calculate directly on i
  • define a function a priori and call it cheerfully
  • create a lambda and call it even happier

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

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

For the second option, before doing the readings due, define the function like this: 

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

Then call within print : 

print(polinomio(i))

For the lambda-alternative, before the iteration, create the lambda function and assign it to the variable print : 

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

And call within polinomio as if it were a traditional function: 

print(polinomio(i))
    
23.10.2017 / 05:59
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