The aim of this article is to show how the study of a function can be realised with the TI-92 Plus / Voyage 200.

Example with g(x) =

1 / Field definition of g(x). Show that G is prolonged by continuity into 0. Note F this prolongation.

2 / Show that F is derivable on ]-1,+inf[. Study the position of the curve C representative of F compared to its tangent at the point of X-coordinate x=0.

3 / Variations of F, graph of C.

4 / Find A, B, C such as:

5 / Calculate the surface of the portion of plan lain between the lines of equations x=-1 and x=0, the X-axis and the curve C.

1 / Field definition of g(x).

The denominator, x^3, must be different from 0, that is to say x/=0.

Also, the ln function is not defined for arguments < = 0. x+1 must be higher than 0, that is to say x+1>0, x>-1.

G is defined for x>1 and x/=0, so Dg = ]-1,0[U]0,+inf[

The limit of g(x) at -1 " + " is +inf because the numerator tends towards -inf and the denominator towards -1.

Show that G is prolonged by continuity into 0. Note F this prolongation.

The limit of G in 0 is unspecified " 0/0 ".

To raise this indetermination, let us carry out a TE (Taylor Expansion at the order 3) of ln(1+x).

The Taylor function of TI-92 Plus / Voyage 200 quickly gives this Taylor expansion. One will reproach the form in which the polynomial (decreasing powers) is given.

It is observed that the calculation of the limit does not pose a problem to the TI-92 Plus / Voyage 200. This lets predict that the software carried out an in-house simplification, transparent 2 the user.

T-E 3 with lim e(x) in 0, = 0.

So, lim g(x) in 0, = 1/3.

The function G is prolonged by continuity into 0 in an application F defined by :

2 / Show that F is derivable on ]-1,+inf[.

F is derivable on ]-1,0[U]0,+inf[, like quotient of derivable functions, whose denominator is not cancelled.

- Derivative : let us study the increase rate of T(x) of F.

One carries out this time a T-E at the order 4 because of the denominator.

What brings to:

The first line of this screen shows the automatic simplification of the TE4 of our expression. The calculation of the TE4 of the 2 polynomials does not have really interest, but shows how it is necessary to reason.

F is derivable into 0 and f'(0)=-1/4

The equation of the tangent in (0,1/3) is:

y-1/3=-1/4(x-0) --->>> y=-x/4+1/3

Study the position of the curve C representative of F compared to its tangent at the point of X-coordinate x=0.

We must study the sign of the quantity of D(x) defined by:

That is to say:

On the numerator, one recognizes the first four terms of the T-E of ln(1+x).

It is thus necessary to carry out a TE5 of ln(1+x) to be able to know the sign of this expression.

What brings to:

Because lim(1/5+j(x),x,0)=1/5, one can affirm that there is an interval I of center 0 such as for any x element of I, 1/5+j(x)>0.

Therefore in interval I, D(x) >=0.

In the vicinity of A(0,1/3) the representative curve of F is on the top of the tangent in A at C.

3 / Variations of F, graph of C.

For any X element of ]-1,0[U]0,+inf[, f' (x) =

f' (x) is sign of its numerator, because x^4 is positive on R.

Here the table of variations obtained starting from the downloadable FX program (via this site).

4 / Find A, B, C such as:

The equality is equivalent after reduction to the same denominator to the equality of the numerators, that is to say:

for any X element of ]-1,+inf[-{0}, 1=(a+c)*x² + (a+b)*x + b

These two polynomials are equal, therefore their coefficients are equal... what gives the system:

That is to say

The command comDenom (compared to x) of the TI-92 Plus / Voyage 200 enables us to see the equality of the numerators, as explained previously. Expand gives a simple elements decomposition.

5 / Calculate the surface between the lines of equations x=-1 and #x=0, the X-axis and the curve C.

This amounts calculating the integral of f(x) between -1 and 0. This integral is unsuitable into -1 and it is not wrongfully unsuitable because F (or G)can't be prolong by continuity in this point.

Let us check before calculation that the integral I= is convergent.

In -1, ln(1+x)-x+x²/ 2 is equivalent to ln(1+x) because ln(1+x) tends towards -inf, and - x+x²/ 2 towards 3/2

Also, x^3 tends towards -1. (limit in -1)

So f(x) is equivalent to - ln(1+x) in -1.

However, on the interval ]-1,0 ], a primitive of - ln(1+x) is x-(x+1)*ln(1+x) and lim(x-(x+1)*ln(1+x),x,-1) = -1.

Conclusion: the integral converges.

The downloadable function impropr (via this site) confirms it. (English version is available)

- Calculation of

Let us take x<y in ]-1,0[, and pose F(x,y)=

Then I=

F(x,y)=

F(x,y)=

F(x,y)= (1)

To calculate A(x,y), we process with an integration by parts, with:

u(t) = ln(1+t) ; u'(t) = 1/(1+t)

v'(t) = 1/t^3 ; v(t) = -1/(2t²)

A(x,y) =

A(x,y) =

A(x,y) =

A(x,y) =

A(x,y) =

A(x,y) = 1/2*G(y) - 1/2*G(x)

with G(t) =

Let us return to the expression (1) of F(x,y), one obtains:

F(x,y)=

F(x,y)=

F(x,y)=

H(t) =

H(t) =

H(t) =

The first result was obtained with command Factor. comDenom makes it possible to obtain a common denominator form, with a developed numerator.

- > Calculation of the limit of H(t) in 0 " - ".

Let us carry out a TE2 of the first ln - > -ln(1+t)

H(t) =

H(t) =

H(t) =

H(t) =

That is to say lim(H(t),t,0 -) = 1/2

- > Calculation of the limit of H(t) in -1 " + ".

H(t) =

H(t) =

However, lim((1+t)*ln(1+t), T, -1+) = 0, so lim H(t) in -1 " + " = -1.

To resume ...

F(x,y) = 1/2*(#H(y) - H(x)) and

=

=

However, H(y) is independent of X, so

H(x) is independent of y, so

Conclusion: = 1/4 + 1/2 = 3/4

OlivierMiclo, July 11 2000