@Jimmy I am not sure if you are questioning that or just stating a fact. residualmeansquare will always be less than residualsumofsquares because residualmeansquare = residualsumofsquares / degreesoffreedom, and degreesoffreedom is an integer greater than 1 in any practical problem.

Are your "letters" actually entered as j[0], j[1], ..., j[15]? If they are not, would it be convenient to put them in that style?

Is it possible that any of the exponents are 0 or 1? Are the exponents all integers?

It is the expected behaviour: g(i) is 0 because i is not 1. The standard order of evaluation says that g(i) will be evaluated before being passed to the sum command. The i has no value at that time, so it is not 1.

It is the expected behaviour: g(i) is 0 because i is not 1. The standard order of evaluation says that g(i) will be evaluated before being passed to the sum command. The i has no value at that time, so it is not 1.

@Alejandro Jakubi There's also codegen:-cost and SoftwareMetrics:-HalsteadMetrics.

However, the complexity that I am talking about is something inherent to the polynomial rather than a measure of how the polynomial is written.

Specifically, we have a set of n polynomials {P[k] $ k= 1..n} and a set of n distinct names {v[k] $ k= 1..n}. We have a set E of very large polynomials on which we want to perform the simplify with side relations

simplify(E, {P[k] = v[k] $ k= 1..n})

However, that takes way too much time (when n is, say, 10). I have discovered that it is sometimes possible to achieve the same effect as the above command nearly instantly by applying the simplifications one at a time, but only if they are applied in a certain order. I call that order "complexity" because it seems related to the usual notions of being less or more complex. But it is a partial order, not a total order.

@Alejandro Jakubi There's also codegen:-cost and SoftwareMetrics:-HalsteadMetrics.

However, the complexity that I am talking about is something inherent to the polynomial rather than a measure of how the polynomial is written.

Specifically, we have a set of n polynomials {P[k] $ k= 1..n} and a set of n distinct names {v[k] $ k= 1..n}. We have a set E of very large polynomials on which we want to perform the simplify with side relations

simplify(E, {P[k] = v[k] $ k= 1..n})

However, that takes way too much time (when n is, say, 10). I have discovered that it is sometimes possible to achieve the same effect as the above command nearly instantly by applying the simplifications one at a time, but only if they are applied in a certain order. I call that order "complexity" because it seems related to the usual notions of being less or more complex. But it is a partial order, not a total order.

@casperyc "Complexity", in the sense that I am using it, is a partial order of polynomials, but not a total order. I said that the last two "cannot be ranked by complexity"; I did not say that you had them in the wrong order, because there is no right or wrong order. They have to be processed as a pair.

I do not have a precise definition of complexity; it is a heuristic concept. If A has higher powers than B but is otherwise identical, then I say that A is more complex than B.

You wrote:

Also, my current method
sskappa( sstest(4,2) , kappa );
ssTwoStage(%,4);
take less than 20 sec.

I thought the goal was to use the substitutions sstest3(4,2) instead of sstest(4,2).

@casperyc "Complexity", in the sense that I am using it, is a partial order of polynomials, but not a total order. I said that the last two "cannot be ranked by complexity"; I did not say that you had them in the wrong order, because there is no right or wrong order. They have to be processed as a pair.

I do not have a precise definition of complexity; it is a heuristic concept. If A has higher powers than B but is otherwise identical, then I say that A is more complex than B.

You wrote:

Also, my current method
sskappa( sstest(4,2) , kappa );
ssTwoStage(%,4);
take less than 20 sec.

I thought the goal was to use the substitutions sstest3(4,2) instead of sstest(4,2).

Note that in this thread we are talking about Newton's method for multivariate problems. I realize that that is not clear from the text.

Note that in this thread we are talking about Newton's method for multivariate problems. I realize that that is not clear from the text.

In all three of your examples, the result that "we are happy with" is the same result as returned by add. So what's the point of a new command?

@abbeykabir I can't fully describe it, because I know so little about the Document Mode. I just know that it is difficult for me to execute. The cursor does not jump to the next section. The value Digits gets reset in every section. When you create a new file, a submenu pops up with options "Worksheet Mode" and "Document Mode".

@abbeykabir I can't fully describe it, because I know so little about the Document Mode. I just know that it is difficult for me to execute. The cursor does not jump to the next section. The value Digits gets reset in every section. When you create a new file, a submenu pops up with options "Worksheet Mode" and "Document Mode".

The indicator that Mac Dude is taking about is available as

Statistics:-NonlinearFit(..., output= [residualmeansquare, ...])

@PatrickT So sorry. Please try again. I updated the code, changing expr to _E inside the `if`. I had made a mistake while transcribing the code from Maple by hand rather than cutting and pasting.