atanh | A ponyfill for ES6 Math.atanh | Math library

Try initializing sum1 and sum2 to zero:

(This question has nothing to do with optimx, which is just a wrapper for R packages providing unconstrained optimization.)

The function in question is nmkb() in the dfoptim package for gradient-free optimization routines. The approach to transform bounded regions into unbounded spaces is a common one and can be applied with many different transformation functions, sometimes depending on the kind of the boundary and/or the type of the objective function. It may also be applied, e.g., to transform unbounded integration domains into bounded ones.

The approach is problematic if the optimum lies at the boundary, because the optimal point will be sent to (nearly) infinity and cannot ultimately be reached. The routine will not converge or the solution be quite inaccurate.

If you think the algorithm is not working correctly, you should write to the authors of that package and -- that is important -- add one or two examples for what you think are bugs or incorrect solutions. Without explicit code examples no one here is able to help you.

(1) Those transformations define bijective maps between bounded and unbounded regions and the theory behind this approach is obvious. You may read about possible transformations in books on multivariate calculus.

(2) The approach with penalties outside the bounds has its own drawbacks, for instance the target function will not be smooth at the boundaries, and the BFGS method may not be appropriate anymore.

(3) You could try the Hooke-Jeeves algorithm through function hjkb() in the same dfoptim package. It will be slower, but uses a different approach for treating the boundaries, no transformations involved.

EDIT (after discussion with Erwin Kalvelagen above)

There appear to be local minima (with some coordinates negative). If you set the lower bounds to 0, nmkb() will find the global minimum (1,1,1,1,1,1) in any case.
Watch out: starting values have to be feasible, that is all their coordinates greater 0.

You can use python's regular expression library to perform more advanced string matching and replacement than replace(). Specifically, re.sub(), which functions the same as replace() but takes regular expressions instead of simple strings.

Be sure to first import it with import re. You can then do the following:

The sibling answer shows the way to pass the extra parameter t to f. But, then you run into another problem. The equation seems to be complex, while fsolve only works for real functions.

One way to tackle these, is mpmath, Python's multiprecision library. mpmath has a function findroot that also works for complex numbers. Note that due to the multiprecision, it can be much slower than other libraries. I don't directly see a simple way to pass the t parameter, so I made use of a lambda function:

classy-prelude exports a MonoFoldable version of foldr, through the reexport of Data.MonoTraversable.Unprefixed:

https://hackage.haskell.org/package/mono-traversable-1.0.9.0/docs/Data-MonoTraversable-Unprefixed.html

Here are two alternatives for solving the problem on a Windows 10 machine:

Then, once jq has been installed (e.g. using linuxbrew), you could use this filter:

You'll be better off using numpy.arctan and numpy.arctanh instead of the math equivalents, as they can work with arrays:

You can write a function that recognizes certain names before calling float() to parse it normally.

The problem was a plugin that's not compatible with es2015. before (proberly with vue cli2) webpack was probably configured to transpile all node_modules, which is not necessarily common practice.

in this blog you can read how to publish es6 code to npm

https://xebia.com/blog/publishing-es6-code-to-npm/