restart;
with(GraphTheory):
A,B,C:=seq(RandomGraphs:-RandomGraph(8, 0.5), i=1..3);
plots:-display(DrawGraph~(<A | B | C>));

Obviously that the denominator is a periodic function (with respect to x-variable with period  2*Pi  and respect to t-variable with period  8*Pi ). Let us prove that the denominator of your fraction for any values of the variables is greater than 0. Due to the periodicity it is enough to check this on the rectangle  x=0..2*Pi , t=0..8*Pi . It's simplest to do this test geometrically. Here are two ways: using  plot3d  command and using  Explore command (we obtain slices of the graph  T(x,t)  for different  t ):

V := (x, t)->((5185/2592)*cos(x+(7/4)*t)+(5/72)*cos(3*x+(9/4)*t)+(1/18)*cos(2*x+(1/2)*t)+(1/72)*cos(4*x+4*t)+41489/20736)/``((5/36)*cos(x+(7/4)*t)+(1/32)*cos(2*x+(7/2)*t)+(5185/11664)*cos(3*x+(9/4)*t)+(1/2592)*cos(6*x+(9/2)*t)+(20737/41472)*cos(2*x+(1/2)*t)+(1297/10368)*cos(4*x+4*t)+29985553/26873856);
T:=expand(op(2,denom(V(x,t))));
plot3d(T, x = 0 .. 2*Pi, t = 0 .. 8*Pi, view = 0 .. 3, axes = normal);
Explore(plot(T, x=0..2*Pi, 0..3), t=0...evalf(8*Pi));

Analytically, we can establish the positivity of the denominator using  DirectSearch:-GlobalOptima  command:

DirectSearch:-GlobalOptima(T, [x = 0 .. 2*Pi, t = 0 .. 8*Pi]);

This package should be downloaded from the Maple Application Center and installed on your computer.

Use back quotes for this.

Example:

Addition. With back quotes you can make nonstandard names consisting of any symbols and even whole phrases, for example:

restart;
`My matrix`:=<1,2; 3,5>;
``('`My matrix`')^`-1`=`My matrix`^(-1);

I propose a solution based on two procedures  SetPartition  and  Partition . More about them you can read  here  and  here. For completeness, I included the texts of both procedures in the worksheet below. The example is considered for 12 books, the number of pages for which are randomly generated. This method gives an exact solution, because is based on consideration of all variants of partitions.

Download Division_of_books.mw

Instead of  test relation  command from the pop-up window, use  is  command. It makes easy to add other commands or assuming option:

R := sin(Pi*x/T)*cos(Pi*x/T) = 1/2*(sin(Pi*x/T+Pi*x/T)+sin(Pi*x/T-Pi*x/T)):
is(combine(R));
                                            true

Another example:

Addition. Now as for your first example. Maple is more suitable for calculations, and not for the proof of various symbolic identities. In your example, m is an arbitrary positive integer. If you specify its value, then there are no problems:

restart;
P := a[0]+Sum((2*(1-(-1)^k))/T*(Int(cos(2*Pi*k*x/T), x = 0 .. t)), k = 1 .. m);
Q := a[0]+Int(Sum(2*(1-(-1)^k)/T*cos(2*Pi*k*x/T), k=1..m), x=0..t);
is(value(eval(P=Q, m=10)));

1. Unfortunately in your code, the equations of the trajectory of motion as a function of time are not indicated. Then you could write a much more compact code that calculates the coordinates of the vectors you are interested in for any  t .

2. To plot vectors (arrows) in space, you can use  plots:-arrow  command.

See corrected code

Simplification1.mw

Edit.

We do not need ourselves to write the code for the solution by Runge-Kutta method. As it is written in the help by default  dsolve command with  numeric  option finds the numerical solution using a Fehlberg fourth-fifth order Runge-Kutta method (rkf45  method). Here is the solution of your problem with specific values of the parameters and the plot of this solution:

restart;
Sol:=dsolve({m*l*diff(theta(t),t,t)+b*diff(theta(t),t)+m*g*sin(theta(t))=A*cos(omega*t), theta(0)=1, D(theta)(0)=0}, theta(t), numeric, parameters=[m,l,g,b,A,omega]);
Sol(parameters=[1,2,9.8,1,4,0.3]):
plots:-odeplot(Sol, [t,theta(t)], t=0..40, size=[900,300]);

In Maple 2017.3 :

Eq:=diff(u(x,y),x,x)+diff(u(x,y),y,y)=0;
bc:=u(x,0)=0, u(x,a)=0, u(0,y)=sin(y), u(infinity,y)=0;
pdsolve({Eq, bc}, u(x,y)) assuming a>0;  # The symbolic result
F:=eval(rhs(%), [a=5, infinity=50]);  # An approximate result for plotting
plot3d(F, x=0..5, y=0..Pi);  # Visualization

I assumed that interest is charged at the beginning of each month, as well as at the moment (n-th day of the month) when an additional payment  R  is made. If  n = 1 , then we get the initial result.

Download Recurring_new.mw

restart;
F(x,y):=ln((1+x)*y) + exp(x^(2)*y^(2)) = x + cos(x):
L:=[1, seq(eval(implicitdiff(F(x,y), y, x$n), [x=0,y=1]), n=1..6)];
add(L[i]/(i-1)!*x^(i-1), i=1..7);

y:= x-> ((1 + a*x + b*x^2)/(1 + c*x + d*x^2))*(ln(sinh(x)^2 + cosh(x)^2));
Sys:={seq(coeff(series(y(x), x=0, 7), x^n)=0, n=3..6)};
solve(Sys);

                         {a = 0, b = 16/15, c = 0, d = 2/5}

plot([seq([PP[n], QQ[n]], n = 1 .. numelems(PP))], style = pointline, labels = ["PP", "QQ"]);

 
Edit.               

In addition to the unknown variables  {C, I, R, S, V} , your nonlinear system contains three more parameters (k , tau, Upsilon). If you want to get all the solutions expressed through these parameters, then use the options parametric=full  and  allsolutions. If you specify the values of these parameters, you can easily get all the solutions:

Sys := {eqn1 = 0, eqn2 = 0, eqn3 = 0, eqn4 = 0, eqn5 = 0};
Equilibria1 := solve(Sys, {C, I, R, S, V}, parametric=full, allsolutions);  # All the solutions expressed through 3 parameters
Equilibria2 := solve(eval(Sys, [k = 1, tau = 2, Upsilon = 3]), {C, I, R, S, V});  # All the solutions for specific parameters

If I correctly understood the problem, then the following code solves it. You can yourself specify the number of steps and the reduction factor of the height at each step (parameters  N  and  k):

restart;
L1:=s->[[0,0],[5,0],[2.5,s],[0,0]]:
L2:=s->[[2,0],[7,0],[4.5,s],[2,0]]:
L3:=s->[[4,0],[9,0],[6.5,s],[4,0]]:
A:=s->plot([L1(s),L2(s),L3(s)], color=red, scaling=constrained):
N:=15:  k:=0.8:
for n from 0 to N do
S[n]:=plots:-display([A(4), A(4*k^n)]):
od:
plots:-display(seq(S[n]$5, n=0..N), insequence);