lambdify | AWS Lambda Utility Library | Cloud Functions library

the code below is from the book "Python for Probability, Statistics, and Machine Learning. The clarification needed is for the plotting section. The problem is that "logJ" in the script is not-defined. However, the book provides this as the code to plot the graph. How do you correct (code) the plotting part of the script so it plots the output shown?

I would like to convert a SymPy expression in order to use as an objetive function in JuMP. Suppose my variable involves two variables

I just translated a set of scientific calculations involving matrices which elements are symbolic expressions which are differentiated and combined with various other mathematical expressions then numerically integrated. The pieces of code below constitute a minimal example for the sake of reproducing the performance gap I am experiencing. I understand that differentiating symbolically then integrating numerically does not make sense, but again, the point is about performance gap. It's important to note that importing libraries do not represent much time and do not explain the performance gap.

Julia code:

I have a function in sympy f(M,X) which is linear in V but nonlinear in X, both large n dimensional vectors. Give some X, how do I extract the linear vector that gives this function? At the moment I just lambdify and evaluate it for all the unit vectors in M, which seems very inefficient.

For a very simple example f([m1,m2],[x1,x2])= m1*(x1*x2) + m2*x2**2. Given x=[1,2] I want to get [2,4] since f([m1,m2],[1,2])= m1*2 + m2*4, without computing both f([1,0],[1,2]) and f([0,1],[1,2]) separately.

EDIT: added code below and fixed equations above.

Here's the code for the simple example:

I am trying to evaluate the derivative of the negative log likelihood functionin python. I am using sympy to compute the derivative however, I receive an error when I try to evaluate it. The example code below attempts to compute this for the lognormal function.

Through using the ways and obtaining help from Stackoverflow users, I could find half of the solution and I need to complete it. Through using Sympy I could produce my function parametrically and it became 100 different items similar to 0.03149536*exp(-4.56*s)*sin(2.33*s) 0.03446408*exp(-4.56*s)*sin(2.33*s). By using f = lambdify(s,f) I converted it to a NumPy function and I needed to do integral of in the different sthat I already have. The upper limit of the integral is a constant value and the lower limit must be done through afor loop`.

When I try to do, I get some error which I post below. The code that I wrote is below, but for being a reproducible question I have to put a generated data. TypeError: cannot determine truth value of Relational

I'm asking is it possible to get np.linalg.solve() out of the lambdify on an expression involving solving?

For example, let

I don't understand why the expression below is not being simplified. The example shows the bug:

I have a sympy symbolic expression that contains sin and cos. At the end of the code I use lambdify((Phi),function(Phi))(phi) but I'm getting many non-zero values because of the angle dependence which consists of several pi values, since sin(pi) doesn't return exactly 0. What I would like to do if possible, is to round while still in symbolic form. Something like sin(phi).round(5) where phi is just a symbol, so that when you lambdify the expression and sub in a value for phi the result of sin is automatically rounded to 5d.p. Is there any way to do so?

Thanks