square root in an optimization problem

[MarionaRieraGarrofe]

Hello,
I am trying to maximize this expression:
def output():
t = 0.7
R = 1.5
def Uc_1(x,t):
return \sqrt[n]{x/t}
def Uc_2(x,t,R):
return \sqrt[n]{((1-x)R)/(1-t)}
return optimize.fsolve(lambda x: tUc_1(x,t)+(1-t)*Uc_2(x,t,R),0.1)

When running the cell, it is giving an error that it is related with the utility function of the agent U(x)=square root of x.
How can I solve this?

[janboone]

There are a couple of problems with the screenshot of the code:

• there is no need to next function definitions; each def statement should start on the left
• variable definitions t and R can be global definitions (e.g. defined above the function output)
• \sqrt is latex notation; not python. I guess you want to use: https://numpy.org/doc/stable/reference/generated/numpy.sqrt.html
• function calls in python have arguments in parentheses "(" and ")" (not [] nor {})

Does this help?

[MarionaRieraGarrofe]

Our research paper is the Diamond and Dybvig model.
We want to derive with respect to x and = to 0 the following expression:

Given the values of t and R.

That is why we are using optimize.fsolve(lambda x : expression) . The expression contains a squared root since the utility function is u(x) = square root of (x) , and that is where we are having problems.

By using numpy.sqrt we would have to define an array first, but we don’t understand how that is feasible since x is the value that we are trying to maximise. About x we know that 0<x<1 since it is an assumption of this model.

How can we introduce the square root in the function in order to optimise it?

[janboone]

By using numpy.sqrt we would have to define an array first

I am not sure what you have in mind here: np.sqrt(9) gives the desired result, without defining an array first

If you know that $x \in [0,1]$, you can use: https://docs.scipy.org/doc/scipy/reference/generated/scipy.optimize.fminbound.html

We have also used this function in the lectures.

[MarionaRieraGarrofe]

Thanks!