Reputation: 111
I am trying to write some codes to find the global maximum of an equation, e.g. f = -x**4
.
Here is what I have got at the moment.
import sympy
x = sympy.symbols('x')
f = -x**4
df = sympy.diff(f,x)
ans = sympy.solve(df,x)
Then I am stuck. How should I substitute ans
back into f
, and how would I know if that would be the maximum, but not the minimum or a saddle point?
Upvotes: 1
Views: 2718
Reputation: 1370
If you are just looking for the global maximum and nothing else, then there is already a function for that. See the following:
from sympy import *
x = symbols('x')
f = -x**4
print(maximum(f, x)) # 0
If you want more information such as the x value that gives that max or maybe local maxima, you'll have to do more manual work. In the following, I find the critical values as you have done above and then I show the values as those critical points.
diff_f = diff(f, x)
critical_points = solve(diff_f, x)
print(critical_points) # x values
for point in critical_points:
print(f.subs(x, point)) # f(x) values
This can be extended to include the second derivative test as follows:
d_f = diff(f, x)
dd_f = diff(f, x, 2)
critical_points = solve(d_f, x)
for point in critical_points:
if dd_f.subs(x, point) < 0:
print(f"Local maximum at x={point} with f({point})={f.subs(x, point)}")
elif dd_f.subs(x, point) > 0:
print(f"Local minimum at x={point} with f({point})={f.subs(x, point)}")
else:
print(f"Inconclusive at x={point} with f({point})={f.subs(x, point)}")
To find the global max, you would need to take all your critical points and evaluate the function at those points. Then pick the max from those.
outputs = [f.subs(x, point) for point in critical_points]
optimal_x = [point for point in critical_points if f.subs(x, point) == max(outputs)]
print(f"The values x={optimal_x} all produce a global max at f(x)={max(outputs)}")
The above should work for most elementary functions. Apologies for the inconsistent naming of variables.
If you are struggling with simple things like substitution, I suggest going through the docs for an hour or two.
Upvotes: 3