问题
I am trying to write a program which calculates formula y=√x*x + 3*x - 500, Interval is [x1;x2] and f.e x1=15, x2=25.
I tried to use Exception Handling but it didn't help. And the code I try to use now gives me: ValueError: math domain error
import math
x1 = int(input("Enter first number:"))
x2 = int(input("Enter second number:"))
print(" x", " y")
for x in range(x1, x2):
formula = math.sqrt(x * x + 3 * x - 500)
if formula < 0:
print("square root cant be negative")
print(x, round(formula, 2))
The output should look like this:
x y
15 ***
16 ***
17 ***
18 ***
19 ***
20 ***
21 2.00
22 7.07
23 9.90
24 12.17
25 14.14
回答1:
The argument of square root mustn't be negative. It's perfectly fine to use exception handling here, see below:
Playground: https://ideone.com/vMcewP
import math
x1 = int(input("Enter first number:\n"))
x2 = int(input("Enter second number:\n"))
print(" x\ty")
for x in range(x1, x2 + 1):
try:
formula = math.sqrt(x**2 + 3*x - 500)
print("%d\t%.2f" % (x, formula))
except ValueError: # Square root of a negative number.
print("%d\txxx" % x)
Resources:
- wiki.python.org: Handling Exceptions
回答2:
You have to check whether the expression is < 0
before you take the square root. Otherwise you take the sqrt of a negative number which gives you a domain error.
回答3:
import math
x1 = int(input("Enter first number:"))
x2 = int(input("Enter second number:"))
print(" x", " y")
for x in range(x1, x2+1):
formula = x * x + 3 * x - 500
if formula < 0:
print (x, "***")
else:
formula = math.sqrt(formula)
print(x, round(formula, 2))
回答4:
You should try to modify your code as shown below. Compute the expression value before performing sqrt
print(" x", " y")
for x in range(x1, x2):
expres = x * x + 3 * x - 500
if expres >= 0:
formula = math.sqrt(expres)
print(x, round(formula, 2))
else:
print(x, "***")
# x y
# 15 ***
# 16 ***
# 17 ***
# 18 ***
# 19 ***
# 20 ***
# 21 2.0
# 22 7.07
# 23 9.9
# 24 12.17
# 25 14.14
来源:https://stackoverflow.com/questions/56313605/how-to-fix-valueerror-math-domain-error-in-python