OK, this actually makes sense. It is just not what is usually called "solve a system".

I didn't check the answer, but the following might work,

sol:=[solve(identity(sys[2],t))];

  sol := [{B = 0, a = 0}, {B = 0, a = 1/2}, {B = 1, a = 0},

        {B = -1/2, a = 1/2}]

sol1:=[seq(map2(`union`,i,[solve(identity(eval(sys[1],i),t),{g})])[],i=sol)];

  sol1 := [{B = 0, a = 0, g = 0}, {B = 0, a = 0, g = 1},

        {B = 0, a = 1/2, g = 0}, {B = 0, a = 1/2, g = -1/2},

        {B = 1, a = 0, g = 0}, {B = 1, a = 0, g = 1},

        {B = -1/2, a = 1/2, g = 0}, {B = -1/2, a = 1/2, g = -1/2}]

sol2:=[seq(map2(`union`,i,[solve(identity(eval(sys[3],i),t),{b,k,w})])[],i=sol1)];

  sol2 := [{B = 0, a = 0, b = b, g = 0, k = 0, w = w},

        {B = 0, a = 0, b = b, g = 1, k = 0, w = w},

                                            1/2
                                           3
        {B = 0, a = 1/2, b = b, g = 0, k = ----, w = 1/3},
                                            2

                                              1/2
                                             3
        {B = 0, a = 1/2, b = b, g = 0, k = - ----, w = 1/3},
                                              2

                                               1/2
                                              6
        {B = 0, a = 1/2, b = b, g = -1/2, k = ----, w = 1/3},
                                               2

                                                 1/2
                                                6
        {B = 0, a = 1/2, b = b, g = -1/2, k = - ----, w = 1/3},
                                                 2

        {B = 1, a = 0, b = b, g = 0, k = 0, w = w},

                                                1/2
        {B = 1, a = 0, b = b, g = 1, k = 1/2 I 2   , w = w},

                                                 1/2
        {B = 1, a = 0, b = b, g = 1, k = -1/2 I 2   , w = w},

        {B = -1/2, a = 1/2, b = 1, g = 0, k = 0, w = w}, {B = -1/2,

                          2                             2
                       4 k                           4 k
        a = 1/2, b = - ---- + 1, g = 0, k = k, w = - ---- + 2/3}]
                        3                             9

Now, is this the analytical solution or numerical?

Alec

Reading again,

"Hi
i want to solve this three equations as a function of ,t, (time). ..."

Alec

Reading again,

"Hi
i want to solve this three equations as a function of ,t, (time). ..."

Alec

What then (s)he wanted to find as a function of t? I reread it again, and it looks as if (s)he wants to find B, g and a as functions of t. Otherwise the problem doesn't seem to have sense.

This way it could be done at least numerically,

f:=x->fsolve(eval(sys,[b=1,k=1,w=1,t=x]));

f(1);

       {B = 0.07245435481, a = 0.4932406923, g = -0.1074762558}

f(1.1);

      {B = 0.04891094248, a = 0.4955740331, g = -0.06240010333}

f(1.2);

      {B = 0.02648671222, a = 0.4976846886, g = -0.03002731216}

Replacing fsolve with solve also works sometimes - for example, for t=1, but not for t=1.1.

Alec

What then (s)he wanted to find as a function of t? I reread it again, and it looks as if (s)he wants to find B, g and a as functions of t. Otherwise the problem doesn't seem to have sense.

This way it could be done at least numerically,

f:=x->fsolve(eval(sys,[b=1,k=1,w=1,t=x]));

f(1);

       {B = 0.07245435481, a = 0.4932406923, g = -0.1074762558}

f(1.1);

      {B = 0.04891094248, a = 0.4955740331, g = -0.06240010333}

f(1.2);

      {B = 0.02648671222, a = 0.4976846886, g = -0.03002731216}

Replacing fsolve with solve also works sometimes - for example, for t=1, but not for t=1.1.

Alec

Here is an example, B*t+a=0 has solution B=-a/t. It is not defined for t=0, but still gives a solution for B as a function of a and t.

Alec

Here is an example, B*t+a=0 has solution B=-a/t. It is not defined for t=0, but still gives a solution for B as a function of a and t.

Alec

In Maple,

sum(binomial(n, k)*x^(n-k)*y^k, k = 0 .. n);

                                     n  n
                            (y/x + 1)  x

So x can not be 0. For negative x, Maple gives

is(sum(binomial(n, k)*x^(n-k)*y^k, k = 0 .. n)=(x+y)^n) 
    assuming x<0;

                                false

This (with additional assumption that n is natural) may be considered as a bug.

Alec

In Maple,

sum(binomial(n, k)*x^(n-k)*y^k, k = 0 .. n);

                                     n  n
                            (y/x + 1)  x

So x can not be 0. For negative x, Maple gives

is(sum(binomial(n, k)*x^(n-k)*y^k, k = 0 .. n)=(x+y)^n) 
    assuming x<0;

                                false

This (with additional assumption that n is natural) may be considered as a bug.

Alec

Just tried to add a formula (in the recent coeff thread) using <math> and got the following:

Fatal error: Call to undefined function getgif() in C:\xampp\htdocs\modules\maplemath\maplemath.module on line 30
 

And the thread is unaccessible now - just showing that error page instead.

Alec

But that may be missing a lot. If a sum of 2 numbers is 0, that doesn't necessarily mean that each of them is 0.

Alec

But that may be missing a lot. If a sum of 2 numbers is 0, that doesn't necessarily mean that each of them is 0.

Alec

While adding a comma is more problematic, adding / to the letters may be not that difficult, is it?

Alec

The difference is more clear with using a(j) instead of a,

sum(a(j),j=5/2..17/3);

                          a(3) + a(4) + a(5)

add(a(j),j=5/2..17/3);

                  a(5/2) + a(7/2) + a(9/2) + a(11/2)

Alec

I don't remember the correct term - it is something like generalized sums, or extended sums. They are defined similarly to integration. It takes some time to give the exact definition, but it can be more easily seen in examples,

sum(a,j=0..N)=int(a,j=0..N+1);

                        (N + 1) a = (N + 1) a

sum(a,j=0..-2)=int(a,j=0..-1);

                               -a = -a
sum(a(j),j=0..-3);

                            -a(-2) - a(-1)

product(a(j),j=0..-3);

                                  1
                             -----------
                             a(-2) a(-1)

This is mentioned in sum,details

Alec