The purpose of this article is not to promote a particular software, but rather to be objective by showing that all calculator algebra systems react differently.

With an integer "n", one poses:

1-a): Calculate a2.

1-b): Show that using integration by parts.

1-c): By using the function , calculate a1.

2): Show that the sequence "an" is raised.

3-a): Show that

3-b): By using the variation of , show that for all x> 1 and one have

3-c): Find the limit of "an"

1-a): Calculate a2.

On the TI-92 Plus / Voyage 200, the function define enables us to store our function in order to use it throughout problem. a(2) is thus easily calculable.
	Simplification of a(n) when n=2.

On the HP49G, we can enter the expression in the editor of equations (EQW) for more convenience.
	Like the TI-92 Plus / Voyage 200, the function define enables us to store the expression in a(n)
	a(2) is obtained as on the TI-92 Plus / Voyage 200
	The primitive also has the same form.

On TI-92 Plus / Voyage 200 or HP49G, one will prefer to present the integral in the form 2/sqrt(3)

1-b): Show that using integration by parts.

: Let us integrate by parts

u(x) = : u' (x) =

v' (x) = 1 : v(x) = X

That is to say and

In this sequence of calculations, the TI-92 Plus / Voyage 200 and HP49G can help us evaluate u' and v (is it necessary?).

1-c): By using the function , calculate a1.

f is differentiable on [ 0,1 ] and is made up of differentiable functions.

Indeed, x->g(x)=1+x² is differentiable on [ 0,1 ] and strictly positive, and u->h(u)=v(u) is differentiable on [0,+8]

Therefore x->v(1+x²) = (hOg)(x) is differentiable on [ 0,1 ]. Consequently, x->x+v1+x² is also differentiable on [ 0,1 ] (sum of 2 funcs differentiable on [ 0,1 ]); moreover it is strictly positive on [ 0,1 ], therefore x->f(x)=ln(x+v(1+x²)) is differentiable on [ 0,1 ].

Therefore (*) = ln(1+sqrt(2))

Consequently, a1 =

On the TI-92 Plus / Voyage 200, derivative and integral are calculated and presented in the awaited form, simplified.
	(*) The series of calculations can be easily recreated with the calculator, thanks to, for example, the operator "Knowing that".

On the HP49G, it will be necessary for you to ask for a simplification.
	EVAL allows you to evaluate this expression. Let us note that EXPAND led to the same result. One could expect the same form returned by the TI-92 Plus / Voyage 200.
	(*) The series of calculations can be recreated with the calculator rather easily in this example, with command SUBST. Logic is the same one as with the TI-92 Plus / Voyage 200 because we only substitute one value for ONLY one variable. The operator "Knowing that" of the TI-92 Plus / Voyage 200 will prove to be much more powerful and practical in other problems, for example when you want to impose restrictions on the field of definition of variables. The second screen shows the use of lists. Once again, it will be necessary to ask a simplification to arrive at the desired result.

This simple example shows that the objectives of the developers (of the two softwares) are different.

The TI-89/92 Plus C.A.S. simplifies the results before displaying. This has two advantages: the expressions are adapted to the pupils, and the use of the calculator is much faster.

The HP49G return expressions in a "rough" state. It will be necessary for you most of the time to apply simplification functions to the expressions. This can be dangerous for an inexperienced user, and this logic in not (in my opinion) adapted for a college-level student, especially since the number handling is more important than on the TI-92 Plus / Voyage 200 (and just waste of time). On the other hand, this leaves more freedom to the user who is finally "the Master" of the expressions, which is really necessary in certain problems.

To each his own opinion...

2): Show that the sequence (an) is raised.

On [ 0,1 ], 1<1+x²< 2, therefore 1<sqrt(1+x²)<sqrt(2), for they are inequalities between positive numbers.

Therefore 1 < sqrt(1+x²)^n < sqrt(2)^n.

Let us integrate this inteval (framing) between 0 and 1:

The nth root is taken, and that gives 1 < an < sqrt(2)

The sequence an is raised by sqrt(2)

3-a): Show that

This demonstration is simple. It is enough to develop the right part of this inequality, and to compare:

, which is equivalent to ... what is exact.

On TI-92 Plus / Voyage 200, the function expand makes it possible to see the relationship to the left part of the inequality. Lines 2&3: a simple evaluation of the inequality does not make it possible to know if the direction of the inequality is good. This inequation should be solved. Solve is particularly well adapted. In general, to test an equality, use a nonpresent variable in the expression.

The test of inequality proceeds without problem. 1 means true and 0 means false

3-b): By using the variation , show that for all x>1 and one have

On R+, function Z is defined as z(x) = sqrt(1+x²) is increasing (z' >0), therefore this function is increasing for all since this interval is included in R+.

According to the 3-a question, we have :

because for n > 1.

Conclusion: for and n > 1, (Inequality 1)

The TI-92 Plus / Voyage 200 and HP49G can confirm this inequality using the same method.

3-c): In deducing the limit of the sequence an.

(Inequality 2)

Let us raise inequality 1 with power N and integrate the inequality obtained between 1-1/n and 1:

(Inequality 3)

By combining the inequalities (2) and (3) one obtains:

Let us raise this last inequality to the power 1/n.

We have :

However ; in addition it is known that ,

therefore by continuity of the exponential at 0.

One concludes

The theorem of framing makes it possible to conclude that lim an = sqrt(2)

No problem for this calculation of limit, in spite of a rather long calculating time.

Olivier Miclo, June 2000