Rouben Rostamian

MaplePrimes Activity


These are answers submitted by Rouben Rostamian

You have

fun := piecewise(x+y > 1, (x+y)^2, x-y);

It seems to me that you call that object fun because you are thinking of it as a function.  Maple, however, does not recognize that as a function.  It sees it just as an expression in x and y.  To define fun as a function, do:

fun := (x,y) -> piecewise(x+y > 1, (x+y)^2, x-y);

After doing that, you may replace the very ugly eval(5*fun, {:-x=x, :-y=y} with the very obvious 5*fun(x,y).

As to your problem with "cannot determine if this expression is true or false: 2 < z", in your proc change if x > y to if is(x > y).

Here is the modified code.

restart;

fun := (x,y) -> piecewise(x+y > 1, (x+y)^2, x-y);
temp_proc := proc(x, y)
        local out, ind:
        ind := 9:
        if is(x > y) then ind := 1 else ind := 0 end if;
        if ind = 1 then out := 5*fun(x,y) else out := -5*fun(x,y) end if:
        return(out);
end proc:

fun := proc (x, y) options operator, arrow; piecewise(1 < x+y, (x+y)^2, x-y) end proc

xt := z; yt := 2;

z

2

out1_fin := temp_proc(xt, yt);

out1_fin := `&ndash;`(5)*piecewise(0 < z+1, (z+2)^2, z-2)

plot(out1_fin, z=-2..3);

 

Download mw.mw

Things will work as you expect if you define fun as a function:

fun := (x,y) -> x^2+y^2;

Then within the proc, replace all fun with fun(x,y).  No other changes are necessary.

Why do you expect a^(2/3) to be the only solution?  Suppose a = -1.  Then we have

restart;

eq := x^(3/2) = -1;

x^(3/2) = -1

sol := solve(eq);

(1/4)*(I*3^(1/2)-1)^2, (1/4)*(1+I*3^(1/2))^2

simplify(subs(x=sol[1], eq));

-1 = -1

simplify(subs(x=sol[2], eq));

-1 = -1

If you are specifying the x-range as x=a..b,  then you want to include the point with coordinates (a,0) in your graph.  So something like this should work:

with(plots):
display(
   plot(10 + sin(x), x=3*Pi..4*Pi),
   pointplot([3*Pi,0], symbol=point)
);

 

Maple's native export to PDF is useless since it embeds the exported drawing within a letter-size (i.e., 8.5in by 11in) document.

Maple's export as EPS (Encapsulated PostScript) works fine with line drawings such as that you have shown.  Save your graphics as EPS and then apply a conversion utility, outside of Maple, to the saved file to produce the PDF equivalent.

There are many such conversions utilities available for free.  The choice depends on your operating system.  I use epstopdf on Linux. 

PS: I might have misunderstood your question.  If by "3D PDF" you mean a PDF document meant to be viewed with the help of 3D glasses, then my answer is totally irrelevant.  Ignore it.

Your function F is negative near phi=0 as we see in the graph below.  Then 1/sqrt(F) is complex and therefore your x is complex.  Is that what you expect?  If not, be sure that you have entered F correctly.

restart;

F:= phi -> 3.924999 - 0.024999/sqrt(1 - 2*phi) - 3.900/(1 - phi/6)^(3/2) - 1.648094618*10^(-14)*sqrt(3)*sqrt(1836)*(((1.3972 + sqrt(3)*sqrt(1836))^2 + 3672*phi)^(3/2) - (1.3972 + sqrt(3)*sqrt(1836))^3 - ((1.3972 - sqrt(3)*sqrt(1836))^2 + 3672*phi)^(3/2) + ((1.3972 - sqrt(3)*sqrt(1836))^2)^(3/2))-(1/18)*(sqrt(3)*sqrt(300)*(((1.4472 + sqrt(3)*sqrt(300)/300)^2 - 2*phi)^(3/2) - (1.4472 + sqrt(3)*sqrt(300)/300)^3 - ((1.4472 - sqrt(3)*sqrt(300)/300)^2 - 2*phi)^(3/2) + (1.4472 - sqrt(3)*sqrt(300)/300)^3));

proc (phi) options operator, arrow; 3.924999-0.24999e-1/sqrt(1-2*phi)-3.900/(1-(1/6)*phi)^(3/2)-0.1648094618e-13*sqrt(3)*sqrt(1836)*(((1.3972+sqrt(3)*sqrt(1836))^2+3672*phi)^(3/2)-(1.3972+sqrt(3)*sqrt(1836))^3-((1.3972-sqrt(3)*sqrt(1836))^2+3672*phi)^(3/2)+((1.3972-sqrt(3)*sqrt(1836))^2)^(3/2))-(1/18)*sqrt(3)*sqrt(300)*(((1.4472+(1/300)*sqrt(3)*sqrt(300))^2-2*phi)^(3/2)-(1.4472+(1/300)*sqrt(3)*sqrt(300))^3-((1.4472-(1/300)*sqrt(3)*sqrt(300))^2-2*phi)^(3/2)+(1.4472-(1/300)*sqrt(3)*sqrt(300))^3) end proc

plot(F(phi), phi=-1..1);

 

Download plot-of-F.mw

 

restart;

We wish to evaluate

Int(abs(sin(x^4))/(sqrt(x) + x^2), x=0..infinity);

Int(abs(sin(x^4))/(x^(1/2)+x^2), x = 0 .. infinity)

The change of variable x^4 = u converts the integral to

Int(abs(sin(u))/((u^(1/8) + u^(1/2))*(4*u^(3/4))), u=0..infinity);

Int((1/4)*abs(sin(u))/((u^(1/8)+u^(1/2))*u^(3/4)), u = 0 .. infinity)

The integrand is O(1/u^(5/4)) as proc (u) options operator, arrow; infinity end proctherefore the integral converges, albeit slowly.

This is what the integrand looks like:

plot(abs(sin(u))/(4*(u^(1/8) + sqrt(u))*u^(3/4)), u=0..10*Pi);

The abs(sin(u)) factor is Pi-periodic. Accordingly, we split the integration into a sum of

integrations over intervals of length Pi:

J := n -> int(abs(sin(u))/((u^(1/8) + u^(1/2))*(4*u^(3/4))),
        u = n*Pi .. (n+1)*Pi, numeric );

proc (n) options operator, arrow; int((1/4)*abs(sin(u))/((u^(1/8)+u^(1/2))*u^(3/4)), u = n*Pi .. (n+1)*Pi, numeric) end proc

J(0), J(1), J(2), J(3), J(1000);

.2311384906, 0.4737708191e-1, 0.2625581792e-1, 0.1783240068e-1, 0.2025655105e-4

Here is the result:

add(J(n), n=0..1000);

.5348206907

Take more terms for greater accuracy.

 

 

Aside: I attempted to do the change of variables through Maple's IntegrationTools:-Change

but the result was wrong.  Perhaps a bug?

Int(abs(sin(x^4))/(sqrt(x) + x^2), x=0..infinity);
IntegrationTools:-Change(%, x=u^(1/4));

Int(abs(sin(x^4))/(x^(1/2)+x^2), x = 0 .. infinity)

-(1/4)*(Int((-u^(11/4)+u^(7/8)+u^(13/8)+u^(19/8)-u^(1/2)-u^(5/4)-u^2+u^(1/8))*abs(sin(u))/(u*(u^3-1)), u = 0 .. infinity))

 

 

Download integration.mw

 

Try this (and also read the help page on plot,options).

plots:-display(
	plot(x^2, x=-1..1,   linestyle=dot),
	plot(x^2+1, x=-1..1, linestyle=spacedash),
	plot(x^2+2, x=-1..1, linestyle=dashdot),
thickness=3);

 

The construction `A=70` (with the back-quotes) builds a name consisting of those four characters, just like AB70 is a name consisting of four characters.  The equal sign has no special meaning in that name.  If instead you do

lst := [A=70, B=17, C=27];

(without the quotes,) then

tt := convert(lst, table);

will do what you want.

This problem is not well-posed.  See the discussion here.

 

What is the meaning of delta(x^2+y^2-1)?

The Dirac delta is defined through the property
int(delta(x)*f(x), x = -infinity .. infinity) = f(0)    for all continuous functions f.

 

Let's see if we can make sense of something simpler than your

exmaple, say delta(x^3).

Make a change of variables x^3 = u and calculate
"(&int;)[-infinity]^(infinity)delta(x^(3))f(x) &DifferentialD;x= (&int;)[-infinity]^(infinity)delta(u)f(u^(1/3)) (&DifferentialD;u)/(3 x^(2))=(&int;)[-infinity]^(infinity)delta(u)[(f(u^(1/3)) )/(3 u^(2/3))] &DifferentialD;u".

To continue, we need to evaluate the expression within the square brackets

at u = 0 which may not be feasible for all f and therefore delta(x^3) makes no

sense.

restart;
with(plots):
with(geom3d):
ball := draw(TruncatedIcosahedron(``)(color="Green"));

This worksheet shows how to calculate the first five terms of the desired series.  I have coded it as 5 similar blocks.  They may be combined into single for-loop to compute any number of terms.

The coefficient of u^3 in what you have shown seems to be in error.  The (a-1)(a+3) should be (a-1)(a+2)..

 

We have
z = int(sinh(a*sin(x*y)), y = 0 .. 1)

This defines z as a function of x but we want x as a function of z,

as in x(z) so we calculate the Maclaurin series of x(z).
Observe that x = 0 implies that z = 0, and therefore

x(0) = 0.  That gives us the first term of the series.

We find the remaining terms by repeated differentiations.

restart;

eq[0] := z = int(sinh(a*sin(x(z)*y)), y=0..1);

z = int(sinh(a*sin(x(z)*y)), y = 0 .. 1)

rels[0] := x(0)=0;

x(0) = 0

eq[1] := diff(eq[0], z);
convert(%, D);
eval(%, z=0);
eval(%, {rels[0]});
isolate(%, D(x)(0));
rels[1] := rels[0], simplify(%);

1 = int(a*(diff(x(z), z))*y*cos(x(z)*y)*cosh(a*sin(x(z)*y)), y = 0 .. 1)

1 = int(a*(D(x))(z)*y*cos(x(z)*y)*cosh(a*sin(x(z)*y)), y = 0 .. 1)

1 = int(a*(D(x))(0)*y*cos(x(0)*y)*cosh(a*sin(x(0)*y)), y = 0 .. 1)

1 = (1/2)*a*(D(x))(0)

(D(x))(0) = 2/a

x(0) = 0, (D(x))(0) = 2/a

eq[2] := diff(eq[1], z):
convert(%, D);
eval(%, z=0);
eval(%, {rels[1]});
isolate(%, (D@@2)(x)(0));
rels[2] := rels[1], simplify(%);

0 = int(a*((D@@2)(x))(z)*y*cos(x(z)*y)*cosh(a*sin(x(z)*y))-a*(D(x))(z)^2*y^2*sin(x(z)*y)*cosh(a*sin(x(z)*y))+a^2*(D(x))(z)^2*y^2*cos(x(z)*y)^2*sinh(a*sin(x(z)*y)), y = 0 .. 1)

0 = int(a*((D@@2)(x))(0)*y*cos(x(0)*y)*cosh(a*sin(x(0)*y))-a*(D(x))(0)^2*y^2*sin(x(0)*y)*cosh(a*sin(x(0)*y))+a^2*(D(x))(0)^2*y^2*cos(x(0)*y)^2*sinh(a*sin(x(0)*y)), y = 0 .. 1)

0 = (1/2)*a*((D@@2)(x))(0)

((D@@2)(x))(0) = 0

x(0) = 0, (D(x))(0) = 2/a, ((D@@2)(x))(0) = 0

eq[3] := diff(eq[2], z):
convert(%, D);
eval(%, z=0);
eval(%, {rels[2]});
isolate(%, (D@@3)(x)(0));
rels[3] := rels[2], simplify(%);

0 = int(a*((D@@3)(x))(z)*y*cos(x(z)*y)*cosh(a*sin(x(z)*y))-3*a*((D@@2)(x))(z)*y^2*(D(x))(z)*sin(x(z)*y)*cosh(a*sin(x(z)*y))+3*a^2*((D@@2)(x))(z)*y^2*cos(x(z)*y)^2*(D(x))(z)*sinh(a*sin(x(z)*y))-a*(D(x))(z)^3*y^3*cos(x(z)*y)*cosh(a*sin(x(z)*y))-3*a^2*(D(x))(z)^3*y^3*sin(x(z)*y)*cos(x(z)*y)*sinh(a*sin(x(z)*y))+a^3*(D(x))(z)^3*y^3*cos(x(z)*y)^3*cosh(a*sin(x(z)*y)), y = 0 .. 1)

0 = int(a*((D@@3)(x))(0)*y*cos(x(0)*y)*cosh(a*sin(x(0)*y))-3*a*((D@@2)(x))(0)*y^2*(D(x))(0)*sin(x(0)*y)*cosh(a*sin(x(0)*y))+3*a^2*((D@@2)(x))(0)*y^2*cos(x(0)*y)^2*(D(x))(0)*sinh(a*sin(x(0)*y))-a*(D(x))(0)^3*y^3*cos(x(0)*y)*cosh(a*sin(x(0)*y))-3*a^2*(D(x))(0)^3*y^3*sin(x(0)*y)*cos(x(0)*y)*sinh(a*sin(x(0)*y))+a^3*(D(x))(0)^3*y^3*cos(x(0)*y)^3*cosh(a*sin(x(0)*y)), y = 0 .. 1)

0 = -2/a^2+2+(1/2)*a*((D@@3)(x))(0)

((D@@3)(x))(0) = -2*(-2/a^2+2)/a

x(0) = 0, (D(x))(0) = 2/a, ((D@@2)(x))(0) = 0, ((D@@3)(x))(0) = (-4*a^2+4)/a^3

eq[4] := diff(eq[3], z):
convert(%, D);
eval(%, z=0);
eval(%, {rels[3]});
isolate(%, (D@@4)(x)(0));
rels[4] := rels[3], simplify(%);

0 = int(a*((D@@4)(x))(z)*y*cos(x(z)*y)*cosh(a*sin(x(z)*y))-4*a*((D@@3)(x))(z)*y^2*(D(x))(z)*sin(x(z)*y)*cosh(a*sin(x(z)*y))+4*a^2*((D@@3)(x))(z)*y^2*cos(x(z)*y)^2*(D(x))(z)*sinh(a*sin(x(z)*y))-3*a*((D@@2)(x))(z)^2*y^2*sin(x(z)*y)*cosh(a*sin(x(z)*y))-6*a*((D@@2)(x))(z)*y^3*(D(x))(z)^2*cos(x(z)*y)*cosh(a*sin(x(z)*y))-18*a^2*((D@@2)(x))(z)*y^3*(D(x))(z)^2*sin(x(z)*y)*cos(x(z)*y)*sinh(a*sin(x(z)*y))+3*a^2*((D@@2)(x))(z)^2*y^2*cos(x(z)*y)^2*sinh(a*sin(x(z)*y))+6*a^3*((D@@2)(x))(z)*y^3*cos(x(z)*y)^3*(D(x))(z)^2*cosh(a*sin(x(z)*y))+a*(D(x))(z)^4*y^4*sin(x(z)*y)*cosh(a*sin(x(z)*y))-4*a^2*(D(x))(z)^4*y^4*cos(x(z)*y)^2*sinh(a*sin(x(z)*y))+3*a^2*(D(x))(z)^4*y^4*sin(x(z)*y)^2*sinh(a*sin(x(z)*y))-6*a^3*(D(x))(z)^4*y^4*sin(x(z)*y)*cos(x(z)*y)^2*cosh(a*sin(x(z)*y))+a^4*(D(x))(z)^4*y^4*cos(x(z)*y)^4*sinh(a*sin(x(z)*y)), y = 0 .. 1)

0 = int(a*((D@@4)(x))(0)*y*cos(x(0)*y)*cosh(a*sin(x(0)*y))-4*a*((D@@3)(x))(0)*y^2*(D(x))(0)*sin(x(0)*y)*cosh(a*sin(x(0)*y))+4*a^2*((D@@3)(x))(0)*y^2*cos(x(0)*y)^2*(D(x))(0)*sinh(a*sin(x(0)*y))-3*a*((D@@2)(x))(0)^2*y^2*sin(x(0)*y)*cosh(a*sin(x(0)*y))-6*a*((D@@2)(x))(0)*y^3*(D(x))(0)^2*cos(x(0)*y)*cosh(a*sin(x(0)*y))-18*a^2*((D@@2)(x))(0)*y^3*(D(x))(0)^2*sin(x(0)*y)*cos(x(0)*y)*sinh(a*sin(x(0)*y))+3*a^2*((D@@2)(x))(0)^2*y^2*cos(x(0)*y)^2*sinh(a*sin(x(0)*y))+6*a^3*((D@@2)(x))(0)*y^3*cos(x(0)*y)^3*(D(x))(0)^2*cosh(a*sin(x(0)*y))+a*(D(x))(0)^4*y^4*sin(x(0)*y)*cosh(a*sin(x(0)*y))-4*a^2*(D(x))(0)^4*y^4*cos(x(0)*y)^2*sinh(a*sin(x(0)*y))+3*a^2*(D(x))(0)^4*y^4*sin(x(0)*y)^2*sinh(a*sin(x(0)*y))-6*a^3*(D(x))(0)^4*y^4*sin(x(0)*y)*cos(x(0)*y)^2*cosh(a*sin(x(0)*y))+a^4*(D(x))(0)^4*y^4*cos(x(0)*y)^4*sinh(a*sin(x(0)*y)), y = 0 .. 1)

0 = (1/2)*a*((D@@4)(x))(0)

((D@@4)(x))(0) = 0

x(0) = 0, (D(x))(0) = 2/a, ((D@@2)(x))(0) = 0, ((D@@3)(x))(0) = (-4*a^2+4)/a^3, ((D@@4)(x))(0) = 0

eq[5] := diff(eq[4], z):
convert(%, D);
eval(%, z=0);
eval(%, {rels[4]});
isolate(%, (D@@5)(x)(0));
rels[5] := rels[4], simplify(%);

0 = int(10*a^3*((D@@3)(x))(z)*y^3*cos(x(z)*y)^3*(D(x))(z)^2*cosh(a*sin(x(z)*y))-40*a^2*((D@@2)(x))(z)*y^4*(D(x))(z)^3*cos(x(z)*y)^2*sinh(a*sin(x(z)*y))+10*a^4*((D@@2)(x))(z)*y^4*cos(x(z)*y)^4*(D(x))(z)^3*sinh(a*sin(x(z)*y))-10*a^4*(D(x))(z)^5*y^5*sin(x(z)*y)*cos(x(z)*y)^3*sinh(a*sin(x(z)*y))-60*a^3*((D@@2)(x))(z)*y^4*(D(x))(z)^3*sin(x(z)*y)*cos(x(z)*y)^2*cosh(a*sin(x(z)*y))+a*((D@@5)(x))(z)*y*cos(x(z)*y)*cosh(a*sin(x(z)*y))-30*a^2*((D@@3)(x))(z)*y^3*(D(x))(z)^2*sin(x(z)*y)*cos(x(z)*y)*sinh(a*sin(x(z)*y))-10*a*((D@@3)(x))(z)*y^3*(D(x))(z)^2*cos(x(z)*y)*cosh(a*sin(x(z)*y))+a^5*(D(x))(z)^5*y^5*cos(x(z)*y)^5*cosh(a*sin(x(z)*y))-15*a*((D@@2)(x))(z)^2*y^3*(D(x))(z)*cos(x(z)*y)*cosh(a*sin(x(z)*y))+a*(D(x))(z)^5*y^5*cos(x(z)*y)*cosh(a*sin(x(z)*y))+5*a^2*((D@@4)(x))(z)*y^2*cos(x(z)*y)^2*(D(x))(z)*sinh(a*sin(x(z)*y))-45*a^2*((D@@2)(x))(z)^2*y^3*sin(x(z)*y)*(D(x))(z)*cos(x(z)*y)*sinh(a*sin(x(z)*y))+15*a^3*((D@@2)(x))(z)^2*y^3*cos(x(z)*y)^3*(D(x))(z)*cosh(a*sin(x(z)*y))+15*a^2*(D(x))(z)^5*y^5*sin(x(z)*y)*cos(x(z)*y)*sinh(a*sin(x(z)*y))-10*a^3*(D(x))(z)^5*y^5*cos(x(z)*y)^3*cosh(a*sin(x(z)*y))+15*a^3*(D(x))(z)^5*y^5*sin(x(z)*y)^2*cos(x(z)*y)*cosh(a*sin(x(z)*y))+10*a*((D@@2)(x))(z)*y^4*(D(x))(z)^3*sin(x(z)*y)*cosh(a*sin(x(z)*y))+30*a^2*((D@@2)(x))(z)*y^4*(D(x))(z)^3*sin(x(z)*y)^2*sinh(a*sin(x(z)*y))-5*a*((D@@4)(x))(z)*y^2*(D(x))(z)*sin(x(z)*y)*cosh(a*sin(x(z)*y))-10*a*((D@@3)(x))(z)*y^2*((D@@2)(x))(z)*sin(x(z)*y)*cosh(a*sin(x(z)*y))+10*a^2*((D@@3)(x))(z)*y^2*cos(x(z)*y)^2*((D@@2)(x))(z)*sinh(a*sin(x(z)*y)), y = 0 .. 1)

0 = int(10*a^3*((D@@3)(x))(0)*y^3*cos(x(0)*y)^3*(D(x))(0)^2*cosh(a*sin(x(0)*y))-40*a^2*((D@@2)(x))(0)*y^4*(D(x))(0)^3*cos(x(0)*y)^2*sinh(a*sin(x(0)*y))+10*a^4*((D@@2)(x))(0)*y^4*cos(x(0)*y)^4*(D(x))(0)^3*sinh(a*sin(x(0)*y))-10*a^4*(D(x))(0)^5*y^5*sin(x(0)*y)*cos(x(0)*y)^3*sinh(a*sin(x(0)*y))-60*a^3*((D@@2)(x))(0)*y^4*(D(x))(0)^3*sin(x(0)*y)*cos(x(0)*y)^2*cosh(a*sin(x(0)*y))+a*((D@@5)(x))(0)*y*cos(x(0)*y)*cosh(a*sin(x(0)*y))-30*a^2*((D@@3)(x))(0)*y^3*(D(x))(0)^2*sin(x(0)*y)*cos(x(0)*y)*sinh(a*sin(x(0)*y))-10*a*((D@@3)(x))(0)*y^3*(D(x))(0)^2*cos(x(0)*y)*cosh(a*sin(x(0)*y))+a^5*(D(x))(0)^5*y^5*cos(x(0)*y)^5*cosh(a*sin(x(0)*y))-15*a*((D@@2)(x))(0)^2*y^3*(D(x))(0)*cos(x(0)*y)*cosh(a*sin(x(0)*y))+a*(D(x))(0)^5*y^5*cos(x(0)*y)*cosh(a*sin(x(0)*y))+5*a^2*((D@@4)(x))(0)*y^2*cos(x(0)*y)^2*(D(x))(0)*sinh(a*sin(x(0)*y))-45*a^2*((D@@2)(x))(0)^2*y^3*sin(x(0)*y)*(D(x))(0)*cos(x(0)*y)*sinh(a*sin(x(0)*y))+15*a^3*((D@@2)(x))(0)^2*y^3*cos(x(0)*y)^3*(D(x))(0)*cosh(a*sin(x(0)*y))+15*a^2*(D(x))(0)^5*y^5*sin(x(0)*y)*cos(x(0)*y)*sinh(a*sin(x(0)*y))-10*a^3*(D(x))(0)^5*y^5*cos(x(0)*y)^3*cosh(a*sin(x(0)*y))+15*a^3*(D(x))(0)^5*y^5*sin(x(0)*y)^2*cos(x(0)*y)*cosh(a*sin(x(0)*y))+10*a*((D@@2)(x))(0)*y^4*(D(x))(0)^3*sin(x(0)*y)*cosh(a*sin(x(0)*y))+30*a^2*((D@@2)(x))(0)*y^4*(D(x))(0)^3*sin(x(0)*y)^2*sinh(a*sin(x(0)*y))-5*a*((D@@4)(x))(0)*y^2*(D(x))(0)*sin(x(0)*y)*cosh(a*sin(x(0)*y))-10*a*((D@@3)(x))(0)*y^2*((D@@2)(x))(0)*sin(x(0)*y)*cosh(a*sin(x(0)*y))+10*a^2*((D@@3)(x))(0)*y^2*cos(x(0)*y)^2*((D@@2)(x))(0)*sinh(a*sin(x(0)*y)), y = 0 .. 1)

0 = 16/3+(16/3)/a^4-(160/3)/a^2+10*(-4*a^2+4)/a^2-10*(-4*a^2+4)/a^4+(1/2)*a*((D@@5)(x))(0)

((D@@5)(x))(0) = -2*(16/3+(16/3)/a^4-(160/3)/a^2+10*(-4*a^2+4)/a^2-10*(-4*a^2+4)/a^4)/a

x(0) = 0, (D(x))(0) = 2/a, ((D@@2)(x))(0) = 0, ((D@@3)(x))(0) = (-4*a^2+4)/a^3, ((D@@4)(x))(0) = 0, ((D@@5)(x))(0) = (16/3)*(13*a^4-10*a^2+13)/a^5

Here is the series:

series(x(z), z);
eval(%, {rels[5]});

series(x(0)+(D(x))(0)*z+((1/2)*((D@@2)(x))(0))*z^2+((1/6)*((D@@3)(x))(0))*z^3+((1/24)*((D@@4)(x))(0))*z^4+((1/120)*((D@@5)(x))(0))*z^5+O(z^6),z,6)

series((2/a)*z+((1/6)*(-4*a^2+4)/a^3)*z^3+((2/45)*(13*a^4-10*a^2+13)/a^5)*z^5+O(z^6),z,6)

 

 

Download series.mw

 

You have four PDEs in the five unknowns 

C(X, R, t), R(X, R, t), T(X, R, t), U(X, R, t), V(X, R, t)

Additionally, U(X,R,t) appears just as U in the fourth PDE.  There may be other errors as well.

Suggestions:

  1. Post an actual worksheet rather than copy/paste.  That will help people that are going to help you.
  2. Before you do a composite computation in a for-loop, try just one case in order to detect where the error occurs.  You may add the for-loop afterward once you see that the basic code works.
restart;
ode1 := diff(x(t),t)=2*x(t)*y(t);
ode2 := diff(y(t),t)=1-x(t)^2-y(t)^2;
DEtools:-DEplot([ode1,ode2],[x(t),y(t)],
	t=0..1,x = -4..4, y = -4..4,
	arrows=curve, linecolor=red, arrowsize=1.5,
	axes=boxed, color='magnitude[legacy]');

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