3 years, 26 days

## @Carl Love I apologize —...

@Carl Love I apologize — I’ve been quite busy lately. I actually didn’t end up using the code since I’ve decided to go an alternative route. I do appreciate the time that you’ve spent on it though.

## @Carl Love Thank you for the packag...

@Carl Love Thank you for the package! Now how do I go about using it? (Sorry — I’m fairly new to Maple). Also, is it possible to modify the old code to so it does transform the subsequent variables?

## One Small Issue......

@Carl Love I got it working by using your code as a template. However, for some reason, when I do VuReduce(v[0]2v[1]u[0]) it doesn't run the procedure on v[1]*u[0]. It just outputs v[0]2v[1]u[0]. Other than that, it's working great!

## @Carl Love In practice, this is computin...

@Carl Love In practice, this is computing v[0]^2*u[0]^2 using the VuReduce procedure and manually replacing the other variables. Hopefully this gives a better idea at what I'm trying to aim for...

## @Carl Love Yes, they can be sp...

@Carl Love Yes, they can be split apart in that way so only 3 would be necessary. However, since I’ve been replacing manually, it’s been easier to use the six different procedures rather than split apart the variables.

So v[0]^3*u[0] = v[0]^2*v[0]*u[0] = v[0]^2*Vu(0,0,1). The value of Vu(0,0,1) is u[0]*v[0]+2w[0]. So we have v[0]^2*(u[0]*v[0]+2w[0]). Then distribute and repeat the process until the variables are in the order of u, v, w.

## @Carl Love All at one time if possible.....

@Carl Love All at one time if possible...

## Thank you!...

@Carl Love Thank you so much... This is exactly what I'm needing -- I'm speechless! There is one final caveat... I was hoping to be able to adapt this code and figure it out, but I want to make sure it's correct!

I have more than one procedure... So this works for Vu(a, b, c) which corresponds to variables of the form vac ub

However, the other procedures are vU(a, b, c) for variables like va ubc (Notice how the power appears on the variable that's capitalized in the procedure). And likewise Wv(a, b, c) for wac vb , wV(a, b, c) for wa vbc, Wu(a, b, c) for wac ub, and wU(a, b, c) for wa ubc

As you may notice, the procedure Vu(0,0,1) and uV(0,0,1) produce the same result since they're both v0u0. I might be dealing with an expression like this where I'm needing to replace multiple variables: w02u0v2u02w0v1. So the corresponding replacements would be:

Wu(0,0,2)vU(2,0,2)Wv(0,1,1). Does that make sense?

Thank you so much for your help!

## @Carl Love Thank you for your reply...

@Carl Love Thank you for your reply. Firstly, they are subscripted like v[a]. The exponent c is non-zero; it’s a natural number. I’m currently using the Physics package:

with(Physics)

setup(Noncommutativeprefix = {u, v, w})

This makes Maple view v[a] u[b] as different from u[b] v[a].

What you have on 4 is exactly what I’m needing. Essentially, for every occurrence of a variable of that form, I want to replace it with that procedure.

Thanks again for your help!

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