@Preben Alsholm You are passing arguments to numboccur in the wrong order; the first argument should be the container, and the second argument the item being counted.

I'd like to elaborate on what Mac Dude said is his P.S. The fact that a help page says that a command is intended for arguments of a certain type can't be used to infer that it won't also work for some arguments not of that type. I think that your type command returning false can be taken as definitive proof that the object is not type series.

The special parameter _rest can be used to "pass along" optional arguments without a procedure knowing what those arguments are or even whether they're present at all. I find this very useful for any procedure whose output is a plot.

waterfall:= proc(
    data::list(numeric),
    {colors::And(list({name, function, string}), 2 &under nops):= 
        ["green", "red"]
    }
)
uses P= plots, PT= plottools;
local d:= [0, data[]], c:= colors[(sign~(d[..-2] -~ d[2..]) +~ 3)/~2];
    d:= `[]`~([$0..nops(data)], d);
    P:-display(PT:-rectangle~(d[..-2], d[2..], ':-color'=~ c), _rest)
end proc
:
data := [6, 4, 4, 4, 7, 9, 12, 16, 25, 100, 105, 95, 90, 55, 45, 30]:
waterfall(data, colors= [purple, yellow], axes= boxed);

First, let's make sure that I've correctly guessed the package structure. You have a package that you didn't give a name to. I'll call it P. Are A and B both exports of P? Is B a module with a ModuleApply procedure?

Can you write C in such a way that it doesn't use any of P's module locals?

Is there a close relationship between a graph's crossing number and the least complicated surface (such as a torus) on which the graph can be embedded?

If you'd like, I could mark your account as a "spammer", which would delete (but not totally purge) everything related to your account (including this thread) and block the account's ability to make any future posts (thus, it'd be nearly impossible for a hacker to hijack the account and post in your name). I am by no means suggesting that you are a spammer! I only mention the possibility because it's something that I have the power to do. It'd only take about 20 seconds on my part, and you wouldn't need to do anything. So, if you'd like for me to do that, let me know.

@Christian Wolinski 

Your T is reminiscent to me of constructing a *sub*basis (rather than a basis) for a topology S.

[The next few definitions may already be familiar to you. If so, I include them for the benefit of other readers.] Let X be a set, and let S be a subset of the powerset of X containing at least the empty set and X itself. (For convenience, we define an intersection with zero operands to be X, which conforms with the usual rule that an associative operator applied to zero operands equals the identity for that operator.)

1. Definition of topology: Suppose that S contains (as elements) 1.1) all pairwise intersections of its elements and 1.2) all unions (not necessarily pairwise) of its subsets. Then S is called a topology on X and the elements of S are called the open sets. Note that 1.1 implies that S contains all intersections of its finite subsets.

2. Definition of basis (base is also commonly used) of a topology: Let S be a topology on X. Let B be a subset of S such every element of S (i.e., every open set of X) is a union of a subset of B. Then B is called a basis for S.

3. Definition of subbasis (or subbase) of a topology: Let S be a topology on X and let C be a subset of S such that every element of S is a union (not necessarily of a finite number of opetands) of intersections of finite subsets of C. Then C is called a subbasis for S.

Quick examples (that should be familiar to almost all readers): Let R be the real numbers. Let's declare a subset A of R to be open if for all x in A there is a d > 0 such that the open interval (x-d, x+d) is a subset of A. Then the set of all such A is a topology on R (indeed it's the standard topology). The set of all open intervals is a basis for that topology. The set of all open intervals that are unbounded at one end is a subbasis for that topology. (These things require some proof, which is essy, the subbasis part being trivial once the topology and basis are verified).

A place where this analogy between topology and your S and T breaks down is that the elements of a topological subbasis are not necessarily pairwise disjoint. However, it may be worthwhile to restrict your attention to cases where the union of S (in other words, X) is finite. 

@ehorta As far as I know, there is no mechanism for completely deleting an account. Whatever you did yesterday was sufficient to delete the personal info that you just mentioned. All that remains is the name "ehorta", the date that you joined, and (I guess) a password (not that I can see that part).

The only data associated to your account that I can see is the Question above. It looks like you've never posted anything else. Since you haven't acquired any reputation points, you don't even appear in the Users list.

Are you implying (and it may well be true, I don't know) that given S there is a unique T satisfying those conditions?

It is not necessary to transform it into a first-order system. The dsolve command does that automatically in the background (in a way that's transparent to the user). The transformation is unique in this case. If it weren't unique, you'd get an error message from dsolve.

Unfortunately, we also need to consider infinity as a parameter which is given a finite value. I'll guess that you're trying to duplicate numeric results from a paper where those results were obtained with RK4 and shooting. To exactly duplicate the results, you'll need to know what finite value the author used for infinity. I've seen some papers on boundary-layer problems (which this happens to be) where that detail unfortunately hasn't been included in the final paper. If that's the case, let me know, because that situation can still be handled.

Your handwriting is highly stylized but seems more suitable for artistic rather than mathematical work.

@ogunmiloro All 14 can be plotted with a single odeplot command like this:

plots:-odeplot(
    ans, 
    `[]`~(T, [indets(sys, typefunc(identical(T), Not(mathfunc)))[]]), 
    T= 0..12, 
    color= [seq]([red, blue, black, purple][1+irem(k,4)], k= 1..14)
);

A slightly more sophisticated approach, just as efficient, is as follows:

get:= module()
local 
    ST:= table((parse@lhs=rhs)~(S)),
    ModuleApply:= (index, field::name)-> ST[index][field]
;
end module
:
get(206, mu);
                                508.001018040

In all three methods that I've presented:

1. The table is only created once, yet can be accessed multiple times.
2. If a Record is changed, that change will be automatically reflected in the table. This is due to the mutability of Records; it's not a feature that I explicitly added.
3. Unlike Search, lookup times are not proportional to the number of records; the lookup times are *much* less than that.
4. The quotation marks are not needed.
5. There's no need to ever be aware of the position within the list of a particular record.

I don't understand what you don't like about my solution. Would you like it better if it were named get rather than ST?

get:= table((parse@lhs=rhs)~(S)):
get[206]:-mu, get[206]:-sigma;
                       508.001018040, 125.002863204708

Note that the above eliminates the need to use the quotation marks. That's what the parse does.

The repeated use of ListTools:-Search on the same list is very inefficient compared to creating a table. If you only wanted to look up one or two records (which seems unlikely), then Search might be worthwhile.

@Carl Love My final simplification above is superfluous. It says nothing more than h''(x) < 0.