find x with f(x) = 0 (limited area of solution), Python 3

find x with f(x) = 0 (limited area of solution)

#!/usr/bin/python2

""" Exp:

find x with f(x) = 0
 f(x) = 2*x+3
solve for x 100
Trying...

---[ x = -2.000000000000000000 <=> f(x) = -1
---[ x = -1.499999999999998668 <=> f(x) = 2.6645352591003757e-15
---[ x = -1.499999999999998668 <=> f(x) = 2.6645352591003757e-15
---[ x = -1.499999999999999778 <=> f(x) = 4.440892098500626e-16
---[ x = -1.499999999999999778 <=> f(x) = 4.440892098500626e-16
---[ x = -1.499999999999999112 <=> f(x) = 1.7763568394002505e-15
---[ x = -1.500000000000000000 <=> f(x) = 0.0

x = -1.5
Done!

"""

def FloatRange(start, stop, step):
    while start < stop:
        yield start
        start += step

def FindxSmallest(expression, Start, Stop, Step):
    array = [abs(eval(expression)) for x in FloatRange(Start, Stop, Step)]
    return list(FloatRange(Start, Stop, Step))[array.index(min(array))]

def main():
    expression = input("find x with f(x) = 0\n f(x) = ")

# limited area of solution
    zone = int(input('solve for x '))
    NumStartScan = -abs(zone)
    NumStopScan = abs(zone)
    Step = 1  # default

print ("Trying...\n")
    while True:
        x = FindxSmallest(expression, NumStartScan, NumStopScan, Step)
        print ("---[ x = %-18.18f" % x, "<=> f(x) =", eval(expression))
        if round(eval(expression), 18)  == 0:
            print ("\nx =", x)
            break
        else:
            NumStartScan = x - Step
            NumStopScan = x + Step
            Step = Step / 10

if __name__ == "__main__":
    main()
    print ("Done!")


run \| edit \| history \| help	0

λ

                                      .NET NoSQL database for rapid development