Inserting multiplication signs in your handwritten system, I get several sets of equations:

eq:={a=b, h=j, c=j, k=g, k*c-j*e=a*l-h*e-b*l+g*c,
 k*a-j*d=a*k-h*d-b*k+g*a,
 -k*b+j*b=-a*j+a*b+b*j+g*f,
 k*g-j*g=a*m-h*g-b*m+g*f,
 g*e=0, g*d=0,g*b=0 ,g*g=0 ,b*l-g*c=-k*c+j*e+a*l+m*e
 ,b*k-g*a=-k*a+j*d+a*k-m*d
 ,-b*j-g*f=-k*f-j*b-a*j+m*b
 ,b*m-g*h=-k*h+j*g+a*m-m*g
 ,a*l-h*e=k*c-j*e+b*l-g*c
 ,a*k-h*d=k*a-j*d+b*k-g*a
 ,-a*j+h*b=k*f+j*b-b*j-g*f
 ,a*m-h*g=k*h-j*g+b*m-g*h}:

solve(eq);
  {a = 0, b = 0, c = j, d = 0, e = 0, f = f, g = 0, h = j, j = j,

        k = 0, l = l, m = m}, {a = 0, b = 0, c = -m, d = 0, e = e,

        f = f, g = 0, h = -m, j = -m, k = 0, l = l, m = m}, {a = 0,

        b = 0, c = 0, d = d, e = e, f = f, g = 0, h = 0, j = 0, k = 0,

        l = l, m = 0}, {a = 0, b = 0, c = m, d = d, e = 0, f = f,

        g = 0, h = m, j = m, k = 0, l = l, m = m}, {a = m, b = m,

        c = m, d = d, e = 0, f = f, g = 0, h = m, j = m, k = 0, l = l,

        m = m}

The last one seems similar to your hand solution.

The answer goes by private mail as it is largely off topic.

The answer goes by private mail as it is largely off topic.

The technical issue is e.g. the trajectory of a mass point in the gravitational field of the Earth. That is mechanics.

But thinking about these objects as nuclear warheads and playing about destroying countries is not mechanics. Such movies on nuclear wars and war games involve a culture that is potentially dangerous. I find that this is a very important moral issue.

The technical issue is e.g. the trajectory of a mass point in the gravitational field of the Earth. That is mechanics.

But thinking about these objects as nuclear warheads and playing about destroying countries is not mechanics. Such movies on nuclear wars and war games involve a culture that is potentially dangerous. I find that this is a very important moral issue.

Europe  vs Africa? , US vs Russia? I find horrible that someone could consider cool thinking about a Global Thermonuclear War, even as a game. 

Europe  vs Africa? , US vs Russia? I find horrible that someone could consider cool thinking about a Global Thermonuclear War, even as a game. 

I am not familiar with those models, but as I see it, the sum form for K(delta) that I have presented above, if converges, it does for large delta. One form that I see it is from the expansion of its terms for delta -> infinity:

MultiSeries:-series(Kn,delta=infinity,1) assuming n>=0;

                  1/2     n /  1  \1/2     /  1  \(3/2)
              2 Pi    (-1)  |-----|    + O(|-----|     )
                            \delta/        \delta/
              ------------------------------------------
                                  delta
                              exp(-----)
                                    2

And most probably it holds as an asymptotic expansion. But, instead of that we can use this integral representation:

int(exp(-lambda*(u+1)),lambda=0..infinity) assuming positive;
                                  1
                                -----
                                u + 1

So, inserting it into K(delta) and integrating on u and then on lambda:

subs(1/(1+u)=exp(-lambda*(u+1)),K):
L:=(simplify@value)(%) assuming positive;

                                1/2                   1/2
                           delta    (delta + 4 lambda)       1/2
           2 exp(-lambda - ------------------------------) Pi
                                         2
      L := -----------------------------------------------------
                                      1/2
                                 delta

int(L,lambda=0..infinity) assuming positive;

       1/2                          1/2       1/2
  (2 Pi    + exp(delta) Pi erf(delta   ) delta

                              1/2        delta    /      1/2
         - Pi exp(delta) delta   ) exp(- -----)  /  delta
                                           2    /

If there is not typo or bug, it turns out not so ugly after all.

PS: A quick plot check seems to show that this result is OK:

K:=Int(1/u^(3/2)*exp(-1/4*delta*u)*exp(-1/4*delta/u)/(u+1),u = 0 .. infinity):
K1 := (2*Pi^(1/2)+exp(delta)*Pi*erf(delta^(1/2))*delta^(1/2)-
Pi*exp(delta)*delta^(1/2))*exp(-1/2*delta)/delta^(1/2):
plot(evalf(K)/K1,delta=0.1..3);

I.e. within this interval, it holds K/K1=1 plus numerical noise

I am not familiar with those models, but as I see it, the sum form for K(delta) that I have presented above, if converges, it does for large delta. One form that I see it is from the expansion of its terms for delta -> infinity:

MultiSeries:-series(Kn,delta=infinity,1) assuming n>=0;

                  1/2     n /  1  \1/2     /  1  \(3/2)
              2 Pi    (-1)  |-----|    + O(|-----|     )
                            \delta/        \delta/
              ------------------------------------------
                                  delta
                              exp(-----)
                                    2

And most probably it holds as an asymptotic expansion. But, instead of that we can use this integral representation:

int(exp(-lambda*(u+1)),lambda=0..infinity) assuming positive;
                                  1
                                -----
                                u + 1

So, inserting it into K(delta) and integrating on u and then on lambda:

subs(1/(1+u)=exp(-lambda*(u+1)),K):
L:=(simplify@value)(%) assuming positive;

                                1/2                   1/2
                           delta    (delta + 4 lambda)       1/2
           2 exp(-lambda - ------------------------------) Pi
                                         2
      L := -----------------------------------------------------
                                      1/2
                                 delta

int(L,lambda=0..infinity) assuming positive;

       1/2                          1/2       1/2
  (2 Pi    + exp(delta) Pi erf(delta   ) delta

                              1/2        delta    /      1/2
         - Pi exp(delta) delta   ) exp(- -----)  /  delta
                                           2    /

If there is not typo or bug, it turns out not so ugly after all.

PS: A quick plot check seems to show that this result is OK:

K:=Int(1/u^(3/2)*exp(-1/4*delta*u)*exp(-1/4*delta/u)/(u+1),u = 0 .. infinity):
K1 := (2*Pi^(1/2)+exp(delta)*Pi*erf(delta^(1/2))*delta^(1/2)-
Pi*exp(delta)*delta^(1/2))*exp(-1/2*delta)/delta^(1/2):
plot(evalf(K)/K1,delta=0.1..3);

I.e. within this interval, it holds K/K1=1 plus numerical noise

If you follow the calculation, you will see that the integral is the product of that sum, K(delta) say, times the prefactor 1/4*exp(delta)^(1/2)*delta^(1/2)*a^2/pi^(1/2), where a=(delta+1)/(delta*mu).

If you follow the calculation, you will see that the integral is the product of that sum, K(delta) say, times the prefactor 1/4*exp(delta)^(1/2)*delta^(1/2)*a^2/pi^(1/2), where a=(delta+1)/(delta*mu).

These maple tags should be inserted in source mode. I.e. press the "Source" button and then copy. It should work like this:

sum(cos(x^2)/sin(exp(y)),x=1..infinity)

The 'seq' command creates a sequence of expressions, while the 'for' - 'do' loop creates a repetition of statements, like the assignment statement here (cf ?assignment).

The 'seq' command creates a sequence of expressions, while the 'for' - 'do' loop creates a repetition of statements, like the assignment statement here (cf ?assignment).

Several file managers, like Total Commander (Windows) and Krusader (Linux) are useful for text searching within files, and this is what I use.