numpy - Understanding the error for scipy.optimize.minimize() function

Question

I have a simple equation that I plotted via

def chernoff_bound(beta):
    return 0.5 * np.exp(-beta * (1-beta))

betas = np.arange(0, 1, 0.01)
c_bound = chernoff_bound(betas)

plt.plot(betas, c_bound)
plt.title('Chernoff Bound')
plt.ylabel('P(error)')
plt.xlabel('parameter beta')

plt.show()

Now, I want to find the value where P(error) is minimum. I tried to do it via the scipy.optimize.minimize() function (honestly, I haven't used it before and there is maybe some error of thought here ...)

from scipy.optimize import minimize

x0 = [0.1,0.2,0.4,0.5,0.9]
fun = lambda x: 0.5 * np.exp(-x * (1-x))
res = minimize(fun, x0, method='Nelder-Mead')

And the error I get is:

---------------------------------------------------------------------------
ValueError                                Traceback (most recent call last)
<ipython-input-25-2b04597c4341> in <module>()
      3 x0 = [0.1,0.2,0.4,0.5,0.9]
      4 fun = lambda x: 0.5 * np.exp(-x * (1-x))
----> 5 res = minimize(fun, x0, method='Nelder-Mead')
      6 print(res)

/Library/Frameworks/Python.framework/Versions/3.3/lib/python3.3/site-packages/scipy/optimize/_minimize.py in minimize(fun, x0, args, method, jac, hess, hessp, bounds, constraints, tol, callback, options)
    364 
    365     if meth == 'nelder-mead':
--> 366         return _minimize_neldermead(fun, x0, args, callback, **options)
    367     elif meth == 'powell':

    368         return _minimize_powell(fun, x0, args, callback, **options)

/Library/Frameworks/Python.framework/Versions/3.3/lib/python3.3/site-packages/scipy/optimize/optimize.py in _minimize_neldermead(func, x0, args, callback, xtol, ftol, maxiter, maxfev, disp, return_all, **unknown_options)
    436     if retall:
    437         allvecs = [sim[0]]
--> 438     fsim[0] = func(x0)
    439     nonzdelt = 0.05
    440     zdelt = 0.00025

ValueError: setting an array element with a sequence.

I would very much appreciate it if someone can point me to the right track here!

Answer 1

The second argument of optimize.minimize is an initial guess -- your guess for the minimum x-value that you wish optimize.minimize to find. So, for example,

import numpy as np
from scipy import optimize
x0 = 0.1
fun = lambda x: 0.5 * np.exp(-x * (1-x))
res = optimize.minimize(fun, x0, method='Nelder-Mead')
print(res)

yields

  status: 0
    nfev: 36
 success: True
     fun: 0.38940039153570244
       x: array([ 0.5])
 message: 'Optimization terminated successfully.'
     nit: 18

x0 need not always be a scalar. It could be an array -- it depends on fun. In the above example, x0 = np.array([0.1]) would also work. The key point is that for whatever you guess, fun(x0) should be a scalar.