You are using 'a' as a function of two variables (x and v), but it is never defined as such. However, you also use it without arguments, so you are not being consistent. In any case for the algorithm to run you have to define 'a'.

You are using 'a' as a function of two variables (x and v), but it is never defined as such. However, you also use it without arguments, so you are not being consistent. In any case for the algorithm to run you have to define 'a'.

@Markiyan Hirnyk It was put there by hand. I was observing that the integral was bounded by 1/(2*(8+x^3)^2) thus is O(1/x^6)). Since the first term in res2 is also  O(1/x^6)), the result follows.

@Markiyan Hirnyk It was put there by hand. I was observing that the integral was bounded by 1/(2*(8+x^3)^2) thus is O(1/x^6)). Since the first term in res2 is also  O(1/x^6)), the result follows.

@Markiyan Hirnyk This is for getting the u-factor, i.e. the factor to be differentiated. The point is that we want to have cos(s) (and sin(s) later) as the factor to be integrated.

@Markiyan Hirnyk This is for getting the u-factor, i.e. the factor to be differentiated. The point is that we want to have cos(s) (and sin(s) later) as the factor to be integrated.

The system (as I wrote it above) is linear. It has exactly one solution, and Maple finds it. No RootOf is used.
So I wonder what you are doing.

The command

solve(eqns, {Citcyt, Citmt, Malmt, Pyrmt}, useassumptions) assuming Citcyt > 0, Citmt > 0, Malmt > 0, Pyrmt > 0;

also just gives one solution. In addition it gives conditions under which the unknowns are positive.

The system (as I wrote it above) is linear. It has exactly one solution, and Maple finds it. No RootOf is used.
So I wonder what you are doing.

The command

solve(eqns, {Citcyt, Citmt, Malmt, Pyrmt}, useassumptions) assuming Citcyt > 0, Citmt > 0, Malmt > 0, Pyrmt > 0;

also just gives one solution. In addition it gives conditions under which the unknowns are positive.

Notice that there is no difference between the name T and the name `T`.
However, something like `T ` (i.e. a space after T) would be different from T. That is why in my comment above I called the same quantity Tseq.

Notice that there is no difference between the name T and the name `T`.
However, something like `T ` (i.e. a space after T) would be different from T. That is why in my comment above I called the same quantity Tseq.

@J4James Changing slightly the answer by Kitonum you could do the following.

restart;
n:=3:
ode1:=diff(T(x),x$2)-T(x)^2=0;
ode2:=diff(S(x),x)=T(x); #To find the T average
bc:=T(0)=1,T(n)=0,S(0)=0;
V:=dsolve({ode1,ode2, bc}, numeric,output=listprocedure);
VT,VS:=op(subs(V,[T(x),S(x)]));
AverageT:=VS(n)/n;
dx:=1/1000;
Tseq:=seq(VT(i),i=0..n,dx):
N:=nops([Tseq]);
theta1:=sqrt(add((Tseq[i]-AverageT)^2, i=1..N)/(N-1))/AverageT;
#N is roughly
n/dx;
#so you could integrate instead of sum:
theta2:=sqrt( evalf(Int((VT(x)-AverageT)^2,x=0..n))/n)/AverageT;

@J4James Changing slightly the answer by Kitonum you could do the following.

restart;
n:=3:
ode1:=diff(T(x),x$2)-T(x)^2=0;
ode2:=diff(S(x),x)=T(x); #To find the T average
bc:=T(0)=1,T(n)=0,S(0)=0;
V:=dsolve({ode1,ode2, bc}, numeric,output=listprocedure);
VT,VS:=op(subs(V,[T(x),S(x)]));
AverageT:=VS(n)/n;
dx:=1/1000;
Tseq:=seq(VT(i),i=0..n,dx):
N:=nops([Tseq]);
theta1:=sqrt(add((Tseq[i]-AverageT)^2, i=1..N)/(N-1))/AverageT;
#N is roughly
n/dx;
#so you could integrate instead of sum:
theta2:=sqrt( evalf(Int((VT(x)-AverageT)^2,x=0..n))/n)/AverageT;

Playing a little with the two branches of the square root:

restart;
eq:=sqrt(x+3-4*sqrt(x-1))+sqrt(x+8-6*sqrt(x-1)) = 1;
plot([op(eq)],x=1..15,0..2,thickness=2); #Perfectly fine
R:=convert(lhs(eq),RootOf);
plot([R,1],x=1..15,0..2,thickness=2); #Still perfectly fine
subs((index=1)=NULL,R); #Now forgetting about the branch (index)
L:=[allvalues(%)];
plots:-display(seq(plot(L[k],x=-5..15),k=1..nops(L)),insequence=true);

#One of these (number 7 for me) is constantly 1.

Playing a little with the two branches of the square root:

restart;
eq:=sqrt(x+3-4*sqrt(x-1))+sqrt(x+8-6*sqrt(x-1)) = 1;
plot([op(eq)],x=1..15,0..2,thickness=2); #Perfectly fine
R:=convert(lhs(eq),RootOf);
plot([R,1],x=1..15,0..2,thickness=2); #Still perfectly fine
subs((index=1)=NULL,R); #Now forgetting about the branch (index)
L:=[allvalues(%)];
plots:-display(seq(plot(L[k],x=-5..15),k=1..nops(L)),insequence=true);

#One of these (number 7 for me) is constantly 1.

I think you need more than 5 terms due to the sine term in the series definition. 10 terms seems more like it.

JT1:= 2*q^(1/4)*Sum((-1)^n*q^(n*(n+1))*sin((2*n+1)*z),n=0..infinity);
evalc(abs(sin((2*n+1)*Pi*I*(a+b*I))));
simplify(%);
T:=(1/10)^(n*(n+1))*exp((2*n+1)*Pi*a);
seq(evalf[20](eval(T,a=4)),n=1..11);
diff(T,n);
solve(%=0,n);
evalf(eval(%,a=4));