n() aka numerical_approx() functionality in python?

[Jorge Garcia]

This is an off shoot of a discussion in sagecell, but it's not really sagecell specific, so here ya go:

"Is there a way to implement something similar to numerical_approx() in python, say with scipy or numpy?

My AP Calculus BC students are doing a post AP Exam final project related to solving free response questions from old AP exams using SAGE and also in plain python. For example they are writing their own (numerical) versions of find_root(), diff() and integrate(). I showed them a little bit of python and they wrote their own newton's method, difference quotient and Riemann sum algorithms. I set up an Ubuntu ssh server with python installed for them to use for this purpose.

All was going well until they asked me how to write their own n() method. This is where I got stumped."

Thanx in advance,

A. Jorge Garcia

[Luiz Felipe Martins]

If you are trying to have your students develop a symbolic library by themselves, programming something like n() would indeed be quite nontrivial, since you have to use a specific numeric algorithm to find an arbitrary precision approximation for any object that is in the library.

You could settle for a partial solution by casting to float, if you don't really need arbitrary precision. You could tell them to define the __float__ method for the objects in the module for which casting to float make sense. For things like integrals they could use a simple numerical method with error estimates. It would not be production code, but still a good exercise. 

If you want to use pure Python, you can "fake" it using sympy, which is a small symbolic library in Python. It has functionality analogous to n(), and has a nice interface to numpy/scipy (check the function lambdify). I use it with a numerical methods course where they have to learn pure Python (so I can't use Sage), to do things like computing the equations of motion from a Lagrangian and then solving the equations numerically.

Felipe Martins

Sent from my iPad

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[Jorge Garcia]

Thank you, sounds like sympy is the way to go!