I would choose the value of t where soln stops,

f(.90133);
                                             -5
                      0.727908994686135296 10

It is already close enough to 0 for the precision of your initial conditions. __________ Alec Mihailovs http://mihailovs.com/Alec/

The problem is that the soln doesn't continue after .90133000. For example, f(1) gives an error. So it doesn't reach 0. __________ Alec Mihailovs http://mihailovs.com/Alec/

The problem is that the soln doesn't continue after .90133000. For example, f(1) gives an error. So it doesn't reach 0. __________ Alec Mihailovs http://mihailovs.com/Alec/

If you are using Document mode, you might try it in the Worksheet mode (or in Classic Maple). If it won't work, that may mean that either the solution doesn't exist, or fsolve can't find it (in the latter case ranges may help.) __________ Alec Mihailovs http://mihailovs.com/Alec/

If you are using Document mode, you might try it in the Worksheet mode (or in Classic Maple). If it won't work, that may mean that either the solution doesn't exist, or fsolve can't find it (in the latter case ranges may help.) __________ Alec Mihailovs http://mihailovs.com/Alec/

If the solution exists, the following should work then, fsolve(f); And if you need a range, enter it as a range, 1..2, for example, without the variable name. __________ Alec Mihailovs http://mihailovs.com/Alec/

If the solution exists, the following should work then, fsolve(f); And if you need a range, enter it as a range, 1..2, for example, without the variable name. __________ Alec Mihailovs http://mihailovs.com/Alec/

Maybe you used sol instead of soln? Check the output of something like soln(0.1); and f(0.1); __________ Alec Mihailovs http://mihailovs.com/Alec/

Maybe you used sol instead of soln? Check the output of something like soln(0.1); and f(0.1); __________ Alec Mihailovs http://mihailovs.com/Alec/

That gives A^(-2) while she wants to get something like E^(1/2). The correct way is

solve({seq(i,i=A.A-E)});

  {d = 1, a = 1, b = 0, c = 0}, {d = -1, a = -1, b = 0, c = 0},

        {b = b, d = -1, a = 1, c = 0}, {d = 1, b = b, a = -1, c = 0},

                     2
               -1 + d
        {b = - -------, d = d, c = c, a = -d}
                  c

seq(eval(A,i),i=%);
                                                [              2]
    [1    0]  [-1     0]  [1    b ]  [-1    b]  [        -1 + d ]
    [      ], [        ], [       ], [       ], [-d    - -------]
    [0    1]  [ 0    -1]  [0    -1]  [ 0    1]  [           c   ]
                                                [               ]
                                                [c         d    ]

__________ Alec Mihailovs http://mihailovs.com/Alec/

That gives A^(-2) while she wants to get something like E^(1/2). The correct way is

solve({seq(i,i=A.A-E)});

  {d = 1, a = 1, b = 0, c = 0}, {d = -1, a = -1, b = 0, c = 0},

        {b = b, d = -1, a = 1, c = 0}, {d = 1, b = b, a = -1, c = 0},

                     2
               -1 + d
        {b = - -------, d = d, c = c, a = -d}
                  c

seq(eval(A,i),i=%);
                                                [              2]
    [1    0]  [-1     0]  [1    b ]  [-1    b]  [        -1 + d ]
    [      ], [        ], [       ], [       ], [-d    - -------]
    [0    1]  [ 0    -1]  [0    -1]  [ 0    1]  [           c   ]
                                                [               ]
                                                [c         d    ]

__________ Alec Mihailovs http://mihailovs.com/Alec/

First, about lower boundary of the projections of Itouch50. If you give only lower bound on h, i.e. h>0.25, then still the lower boundary of the projection will be y=0 and x=0, because when h is approaching infinity, x^2*y is approaching 0. If you give also a higher bound of h, such as h<100, for example, then the lower curve will be one of the curves with boundary values of h, i.e. either with h=0.25, or with h=100 - that curve that is lower. Now, about Itouch and Itouchi. The projections can be plotted the same way, maybe with different ranges of ts and h. If you use the display command, all of them can be plotted on the same plot. The procedure for finding the lower and the higher curves is slightly different,

a1:=implicitdiff(x=Itouch,ts,h):

fsolve(eval(a1,ts=100));

                            0.01000100025

fsolve(eval(a1,ts=100),h=2.);

                             2.547917145

As it can be seen from the plot

plot([eval(Itouch,h=0.01000100025),eval(Itouch,h=2.547917145)],
ts=0..10,color=[red,blue]);

the first value gives the lower boundary and the second value - upper boundary. I did calculations with ts=100, but any other value of ts could be chosen, with the same result since the curves with different values of h don't intersect. Similar calculation for Itouchi gives the value of h=1 for the upper curve and the lower curve in that case seems to be x=0. Again, as in the situation with Itouch50, if you have the boundaries on h, one of them would give the lower boundary curve. __________ Alec Mihailovs http://mihailovs.com/Alec/

First, about lower boundary of the projections of Itouch50. If you give only lower bound on h, i.e. h>0.25, then still the lower boundary of the projection will be y=0 and x=0, because when h is approaching infinity, x^2*y is approaching 0. If you give also a higher bound of h, such as h<100, for example, then the lower curve will be one of the curves with boundary values of h, i.e. either with h=0.25, or with h=100 - that curve that is lower. Now, about Itouch and Itouchi. The projections can be plotted the same way, maybe with different ranges of ts and h. If you use the display command, all of them can be plotted on the same plot. The procedure for finding the lower and the higher curves is slightly different,

a1:=implicitdiff(x=Itouch,ts,h):

fsolve(eval(a1,ts=100));

                            0.01000100025

fsolve(eval(a1,ts=100),h=2.);

                             2.547917145

As it can be seen from the plot

plot([eval(Itouch,h=0.01000100025),eval(Itouch,h=2.547917145)],
ts=0..10,color=[red,blue]);

the first value gives the lower boundary and the second value - upper boundary. I did calculations with ts=100, but any other value of ts could be chosen, with the same result since the curves with different values of h don't intersect. Similar calculation for Itouchi gives the value of h=1 for the upper curve and the lower curve in that case seems to be x=0. Again, as in the situation with Itouch50, if you have the boundaries on h, one of them would give the lower boundary curve. __________ Alec Mihailovs http://mihailovs.com/Alec/

Yes, it is a nice example. It can be done also as

1/(1/ln(1));
                                  0

It is consistent with the following calculations,

1/0.;

                           Float(infinity)

1/infinity;

                                  0

__________ Alec Mihailovs http://mihailovs.com/Alec/

Yes, it is a nice example. It can be done also as

1/(1/ln(1));
                                  0

It is consistent with the following calculations,

1/0.;

                           Float(infinity)

1/infinity;

                                  0

__________ Alec Mihailovs http://mihailovs.com/Alec/