I am a beginner to Maple V rel. 4 and a novice on Maple User Group. I have two "assume" questions.

1) I think, that `"assume(...)"`

command has no inﬂuence on solving equations. Is it
true?

for example :

> assume(x,real);assume(m,real); > eq:=x^2-m*x+1=0; # quadratic equation with reals coefs. 2 eq := x~ - m~ x~ + 1 = 0 > ineqdiscr:=discrim(lhs(eq),x)>0; # for 2 real roots 2 ineqdiscr := 0 < -4 + m~ > solve(ineqdiscr,m); RealRange(-infinity, Open(-2)), RealRange(Open(2), infinity) > additionally(m,RealRange(Open(-2),Open(2))): #discrim < 0 => roots are complex, but it was assume(x,real) ... > solve(eq,x); 2 1/2 2 1/2 1/2 m~ + 1/2 (-4 + m~ ) , 1/2 m~ - 1/2 (-4 + m~ )

Why does Maple compute real roots? Assume(x,real) has no eﬀect ? Or where is the mistake?

2) I want to use "assume" command for variable x from `RealRange(-infinity, Open(-2))`

or from `RealRange(Open(2), infinity)`

(union intervals)

Is it possible?

| 1) I think, that "assume(...)" command has no influence on ...

It doesn’t restrict the values that "solve" will ﬁnd. I wouldn’t quite say it "has no inﬂuence", because it may in fact inﬂuence transformations of the equations that occur in the process of solving them. For example:

> assume(x > 0); assume(y < 0); > solve(sqrt(x^2)=1, x); 1 > solve(sqrt(y^2)=1, y); -1

This is because `sqrt(x^2)`

is simpliﬁed to x while `sqrt(y^2)`

is simpliﬁed to -y (actually,
before "solve" even starts).

| Why does Maple compute real roots? Assume(x,real) has no effect ?

It doesn’t cause Maple to check whether the results satisfy the assumptions.

Actually, Maple doesn’t even know whether or not `sqrt(-4+m^2)`

is real.

> assume(m, RealRange(Open(-2),Open(2))); > is(sqrt(-4+m^2),real); FAIL

although it does know that `-4+m^2 < 0:`

> is(-4+m^2 < 0); true > is(-4+m^2 >= 0); false

But even if it did know, it wouldn’t restrict the values obtained for x.

You can sometimes restrict the output of "solve" by including inequalities. For example:

> solve(x^4=1, x); 1, -1, I, -I > solve({x^4=1, x < infinity}, x); {x = 1}, {x = -1}

However, this usually doesn’t work if there are symbolic parameters.

> assume(m > 0); > solve({x^4 = m, x < infinity}, x);

(No result is returned)

And, strangely enough (still with the same assumption on m):

> solve({x^4 = m^4, x < infinity}, x); {x = -m~}

(Note to Maple developers: this is a bug)

(No result in Maple V Release 5, U. Klein)

| 2) I want to use "assume" command for variable x from ...

Yes.

> assume(x, OrProp(RealRange(-infinity, Open(-2)), RealRange(Open(2),infinity)));

Be aware that "assume" carries no weight with "type". In `BesselI(x,y)`

, for example,
the Maple routine requires y to be of type integer, and preceding its use with the
command `"assume(y,integer)"`

does not get past the typing check early in the
program.

Stupid!. I have complained to Maple about it without any observable improvement.

`Willard, Daniel (DUSA) <WillaD@hqda.army.mil> wrote:`

| Be aware that "assume" carries no weight with "type". ...

Please clarify your complaint. I would insist that "assume" should `_not_`

carry any weight
with "type": "type" must distinguish between, for example, variables and numbers. A variable
is still a variable, not a number, even when you assume its values are integers. If you
want to know whether the value of a quantity is an integer, you can use "is", not
"type".

As for BesselI, I presume you’re talking about the ﬁrst argument, not the second, where integers are not particularly special. AFAIK it does not "insist" that the ﬁrst argument is an integer either: Maple is perfectly happy to calculate, e.g.,

> BesselI(Pi, Pi); BesselI(Pi, Pi) > evalf(%); 1.011423336

I guess your complaint may be with some of the automatic simpliﬁcations, e.g.

> BesselI(3, 0); 0 > assume(x, posint); BesselI(x,0); BesselI(x~, 0) > simplify(%); BesselI(x~, 0)

In that case, you do have a point.