Reputation: 607
I know how I can solve for a root in python
using scipy.optimize.fsolve
.
I have a function defined
f = lambda : -1*numpy.exp(-x**2)
and I want to solve for x
setting the function to a certain nonzero. For instance, I want to solve for x
using f(x) = 5
.
Is there a way to do this with fsolve
or would I need to use another tool in scipy? In other words, I'm looking for something analogous to Maple's fsolve
.
Upvotes: 0
Views: 493
Reputation: 309891
This is easy if you change your definition of f(x)
. e.g. if you want f(x) = 5
, define your function: g(x) = f(x) - 5 = 0
Upvotes: 4