@Markiyan Hirnyk I didn't mean to imply equivalence. But from the differential inequation follows the integrated version.

@Markiyan Hirnyk I didn't mean to imply equivalence. But from the differential inequation follows the integrated version.

@Markiyan Hirnyk If N(t) is your proposed

N1:=t->-(-Pi+Pi*exp(-t))/exp(-t)-2-sin(exp(t^2));

then N1(0) = -2-sin(1). But you assumed N(0) = 0.

@Markiyan Hirnyk If N(t) is your proposed

N1:=t->-(-Pi+Pi*exp(-t))/exp(-t)-2-sin(exp(t^2));

then N1(0) = -2-sin(1). But you assumed N(0) = 0.

It is not clear to me what you are trying to do. What role does the variable 't' play?
Maybe you could tell us where this problem came up?

@samiyare Notice that I have added a faster version in my answer above.

@samiyare Notice that I have added a faster version in my answer above.

You will have a better chance of getting a response if you provide the code. You could upload a worksheet.

@samiyare You could try minimizing the sum of squares of pC and pT using LSSolve from the Optimization package.

Input something like

Optimization:-LSSolve([pC,pT]);

It worked for me in Maple 12, which is what I have available this week.

@samiyare You could try minimizing the sum of squares of pC and pT using LSSolve from the Optimization package.

Input something like

Optimization:-LSSolve([pC,pT]);

It worked for me in Maple 12, which is what I have available this week.

@samiyare The algorithm in the link you give could perhaps be adjusted to the present situation like this:

solpar:=dsolve(subs(f(x)=F(x),{eq2=0,eq3=0, C(0)=1,D(C)(0)=c1, T(0)=1, D(T)(0) = t1}), parameters=[c1,t1],numeric,output=listprocedure,known=F);
Cn,Tn:=op(subs(solpar,[C(x),T(x)]));
#Making procedure pC and pT which compute C(b) and T(b), i.e. Cn(b) and Tn(b)
pC:=proc(c1,t1) if not type([c1,t1],list(numeric)) then return 'procname(_passed)' end if;
solpar(parameters=[c1,t1]);
Cn(b)
end proc;

pT:=proc(c1,t1) if not type([c1,t1],list(numeric)) then return 'procname(_passed)' end if;
solpar(parameters=[c1,t1]);
Tn(b)
end proc;

#Testing on previously found
sol(0);
pC(-2.11068134505884,-.922230659910390);
pT(-2.11068134505884,-.922230659910390);

Then you would have to use fsolve on the system {pC(c1,t1),pT(c1,t1)}.

@samiyare The algorithm in the link you give could perhaps be adjusted to the present situation like this:

solpar:=dsolve(subs(f(x)=F(x),{eq2=0,eq3=0, C(0)=1,D(C)(0)=c1, T(0)=1, D(T)(0) = t1}), parameters=[c1,t1],numeric,output=listprocedure,known=F);
Cn,Tn:=op(subs(solpar,[C(x),T(x)]));
#Making procedure pC and pT which compute C(b) and T(b), i.e. Cn(b) and Tn(b)
pC:=proc(c1,t1) if not type([c1,t1],list(numeric)) then return 'procname(_passed)' end if;
solpar(parameters=[c1,t1]);
Cn(b)
end proc;

pT:=proc(c1,t1) if not type([c1,t1],list(numeric)) then return 'procname(_passed)' end if;
solpar(parameters=[c1,t1]);
Tn(b)
end proc;

#Testing on previously found
sol(0);
pC(-2.11068134505884,-.922230659910390);
pT(-2.11068134505884,-.922230659910390);

Then you would have to use fsolve on the system {pC(c1,t1),pT(c1,t1)}.

@AliKhan In the new version you changed the definition of C3. Unfortunately with the consequence that the formal variable n is not seen after the arrow. Since the body of the procedure is not evaluated at the time of definition this means that something like C3(k) or C3(4) does not result in the expected. The solution is simple, either keep the original definition or use unapply:

C3 := unapply(C1+C2, n):

Btotal can be defined like this:

Btotal:=add(add(Bar[m,n],n=1..5),m=1..5):

@AliKhan In the new version you changed the definition of C3. Unfortunately with the consequence that the formal variable n is not seen after the arrow. Since the body of the procedure is not evaluated at the time of definition this means that something like C3(k) or C3(4) does not result in the expected. The solution is simple, either keep the original definition or use unapply:

C3 := unapply(C1+C2, n):

Btotal can be defined like this:

Btotal:=add(add(Bar[m,n],n=1..5),m=1..5):

@Markiyan Hirnyk

limit(J,a=R/sqrt(2),left) assuming R>0;

Answer: 5/3*sqrt(2)*Pi*R^3-4/3*R^3*Pi

(Now done as it happens in Maple 16).