In Mathematica, built-in functions start with a capital letter and function arguments or parameters are included in square brackets, so sin(x) should be Sin[x], and, say, cos(1) should be Cos[1].

Alec

At this moment, I have an access only to Maple 12.02.

Alec

Was that fixed in Maple 14?

Alec

If you mean a polynomial of t, it will never work for large t, because it will approach either infinity or minus infinity for large t, depending on the sign of the coefficient at the highest degree, while the solution is approaching 0.

The Pade approximation (a ratio of 2 polynomials of t) is also not good on infinite intervals, because it's asymptotic is C*t^(-k), while the solution decreases exponentially.

There is no other way around it as to use exponential terms, as in the asymptotic.

On the other hand, you may not need an infinite interval if you mean, say, precision up to 0.5e-4, in which case something like the following can be used in the example with 11,1,1

fsolve(s=0.000005,t);
                             135.2667941

appr:=numapprox:-minimax(s,t=0..135,[3,4],.5e-6);

  appr := (0.009791943912 + (-0.0002540483891

                           -5                  -8            /
         + (0.2258357652 10   - 0.6797441888 10   t) t) t)  /  (
                                                           /

        0.0097922193 + (0.000561098278 + (

                                         -6                  -8
        0.0000182807885 + (0.300056643 10   + 0.6569876340 10   t) t)

        t) t)

The smaller is the interval, the higher precision can be achieved

However, this procedure is numerical and can not give a symbolical answer depending on 3 parameters. Asymptotic can.

Alec

If you mean a polynomial of t, it will never work for large t, because it will approach either infinity or minus infinity for large t, depending on the sign of the coefficient at the highest degree, while the solution is approaching 0.

The Pade approximation (a ratio of 2 polynomials of t) is also not good on infinite intervals, because it's asymptotic is C*t^(-k), while the solution decreases exponentially.

There is no other way around it as to use exponential terms, as in the asymptotic.

On the other hand, you may not need an infinite interval if you mean, say, precision up to 0.5e-4, in which case something like the following can be used in the example with 11,1,1

fsolve(s=0.000005,t);
                             135.2667941

appr:=numapprox:-minimax(s,t=0..135,[3,4],.5e-6);

  appr := (0.009791943912 + (-0.0002540483891

                           -5                  -8            /
         + (0.2258357652 10   - 0.6797441888 10   t) t) t)  /  (
                                                           /

        0.0097922193 + (0.000561098278 + (

                                         -6                  -8
        0.0000182807885 + (0.300056643 10   + 0.6569876340 10   t) t)

        t) t)

The smaller is the interval, the higher precision can be achieved

However, this procedure is numerical and can not give a symbolical answer depending on 3 parameters. Asymptotic can.

Alec

In the examples with a negative t0, the asymptotic should work well on the entire interval from 0 to infinity. To increase the precision at 0, one can increase the number of terms in it. For example, in the same initial setup as above,

Km,S0,v:=11,1,1:
ser1:=convert(asympt(s,t,10),polynom);
plot(ser1-s,t=0..1);
evalf(eval(ser1,t=0));
                             1.000000235

Alec

In the examples with a negative t0, the asymptotic should work well on the entire interval from 0 to infinity. To increase the precision at 0, one can increase the number of terms in it. For example, in the same initial setup as above,

Km,S0,v:=11,1,1:
ser1:=convert(asympt(s,t,10),polynom);
plot(ser1-s,t=0..1);
evalf(eval(ser1,t=0));
                             1.000000235

Alec

It doesn't look as if that can be expressed in a closed form - as a finite combination of elementary or special functions with arithmetic operations and compositions. Since such an expression doesn't exist, Maple can't provide it.

Nevertheless, Maple can evaluate it for given x and plot,

evalf(eval(d,x=1));

                             3.469506315

plot(d,x=-1..1);

Alec

It doesn't look as if that can be expressed in a closed form - as a finite combination of elementary or special functions with arithmetic operations and compositions. Since such an expression doesn't exist, Maple can't provide it.

Nevertheless, Maple can evaluate it for given x and plot,

evalf(eval(d,x=1));

                             3.469506315

plot(d,x=-1..1);

Alec

1. "unlike in many other cases" - usually MapleNet simply doesn't work, at least when I am trying to see a file there.

2. "a lot of blue art" - that how it looked - the input at the beginning, then a lot of empty space, and then a lot of blue stains, which I don't know how to explain exactly, and then again some Maple stuff at the bottom.

3. "I couldn't find the answer" - yes, I couldn't find it there. I'll check again.

Does your answer coincide with Mathematica's answer in other cases?

Alec

1. "unlike in many other cases" - usually MapleNet simply doesn't work, at least when I am trying to see a file there.

2. "a lot of blue art" - that how it looked - the input at the beginning, then a lot of empty space, and then a lot of blue stains, which I don't know how to explain exactly, and then again some Maple stuff at the bottom.

3. "I couldn't find the answer" - yes, I couldn't find it there. I'll check again.

Does your answer coincide with Mathematica's answer in other cases?

Alec

Then, assuming that Mathematica's answer correct, the general answer would be

a*(a^2+3*b^2)*Pi*(cos(2*p)*(a^2+b^2)+2*I*a*b*sin(2*p))/(a^2-b^2)^2

Alec

Then, assuming that Mathematica's answer correct, the general answer would be

a*(a^2+3*b^2)*Pi*(cos(2*p)*(a^2+b^2)+2*I*a*b*sin(2*p))/(a^2-b^2)^2

Alec

I couldn't see that on MapleNet - the worksheet was displayed, unlike in many other cases, but with a lot of blue art in it, and I couldn't find the answer there.

Alec

I couldn't see that on MapleNet - the worksheet was displayed, unlike in many other cases, but with a lot of blue art in it, and I couldn't find the answer there.

Alec