Complex arithmetic allows for this transformation:

ln(a) - ln(b) = ln(a/b) = ln(-a/-b) = ln(-a) - ln(-b)

Now consider your eq16:

Arranging the ln terms on the left side and the non-ln terms on the right and factoring out hmax, we get on the left side

hmax*(ln(-hmax) - ln(h - hmax))

Applying what I said in the first sentence, this is equivalent to

hmax*(ln(hmax) - ln(hmax - h)),

where now all the arguments of ln are positive.

This is reminiscent of ?evalr and shake----range arithmetic.

A problem still arises if you want a local unevaluated name gamma. There is nothing that I can do to prevent evalf(gamma) from returning a number. Even `evalf/gamma`:= NULL doesn't seem to help.

A problem still arises if you want a local unevaluated name gamma. There is nothing that I can do to prevent evalf(gamma) from returning a number. Even `evalf/gamma`:= NULL doesn't seem to help.

@Preben Alsholm `?()` is the function invocation operator. Think of the expression f(x) as being a binary infix operation whose operands are f and x. If you want to treat that operator as a name, it is `?()`. The most documentation that I can find about it is at ?use , at the end of "Description" where the overloadable operators are described.

`?()`(f, [x,y]) is the same as f(x,y). I had not thought of apply. Yes, the are almost the same, the difference being that `?()` requires the arguments in square brackets. Also, `?()` is kernel and apply is library.

@Preben Alsholm `?()` is the function invocation operator. Think of the expression f(x) as being a binary infix operation whose operands are f and x. If you want to treat that operator as a name, it is `?()`. The most documentation that I can find about it is at ?use , at the end of "Description" where the overloadable operators are described.

`?()`(f, [x,y]) is the same as f(x,y). I had not thought of apply. Yes, the are almost the same, the difference being that `?()` requires the arguments in square brackets. Also, `?()` is kernel and apply is library.

@casperyc 

You can make it into column Vector by

V:= < v1, v2 >[..,1];

and the subs will work.

I don't think that it's strange that you want to make V a Vector. What's strange is that you want V1, ..., V4 to be Vectors also.

As to the "proper" way: I think it would be to define V first. Then if you really need V1, ..., V4 to be Vectors, create them from V like this:

s1:= a,b:
s2:= A,B:
s3:= seq(w[j],j=1..2):
s4:= w[3]*(x+y)^9:
V:= < s1, s2, s3, s4 >: or V:= Vector(7, [s1, s2, s3, s4]):
V1,V2,V3,V4:= V[1..2], V[3..4], V[5..6], V[7..7];

@casperyc 

You can make it into column Vector by

V:= < v1, v2 >[..,1];

and the subs will work.

I don't think that it's strange that you want to make V a Vector. What's strange is that you want V1, ..., V4 to be Vectors also.

As to the "proper" way: I think it would be to define V first. Then if you really need V1, ..., V4 to be Vectors, create them from V like this:

s1:= a,b:
s2:= A,B:
s3:= seq(w[j],j=1..2):
s4:= w[3]*(x+y)^9:
V:= < s1, s2, s3, s4 >: or V:= Vector(7, [s1, s2, s3, s4]):
V1,V2,V3,V4:= V[1..2], V[3..4], V[5..6], V[7..7];

Do you want to exclude linear combiations also? In order words, do you want your set to be linearly independent?

@tuGUTS Look at the output with lprint(%).

@tuGUTS Look at the output with lprint(%).

@erik10It makes perfect sense.

It should all be doable with the recursive procedure and within evalhf---assuming that your real functions can be computed with ordinary arithmetic and elementary functions and assuming that you don't go much bigger than (40,6).

Yes, we can return two matched Arrays of real numbers, in arbitrary order but with the pairs corresponding to a single combination.

@erik10It makes perfect sense.

It should all be doable with the recursive procedure and within evalhf---assuming that your real functions can be computed with ordinary arithmetic and elementary functions and assuming that you don't go much bigger than (40,6).

Yes, we can return two matched Arrays of real numbers, in arbitrary order but with the pairs corresponding to a single combination.

@roman_pearce Thanks, Roman, for the explanation. Your final result is the same as I what I got by applying the relations individually in reverse order.

So, I guess that you are saying that currently there is no way to do this with a single simplify command using only 0.140 seconds, right?

@roman_pearce Thanks, Roman, for the explanation. Your final result is the same as I what I got by applying the relations individually in reverse order.

So, I guess that you are saying that currently there is no way to do this with a single simplify command using only 0.140 seconds, right?