## 380 Reputation

10 years, 240 days

## How to do a nested collect...

Maple 2019

Second Try:

f:= c/(c - 1) - c*(Pi^2 - 12*ln(c))*(1 + c)/(12*(c - 1)^3*n) + (((144*c^3 + 1584*c^2 + 1584*c + 144)*ln(c)^2 - 24*Pi^2*(1 + c)*(c^2 + 10*c + 1)*ln(c) + (-96*c^3 - 288*c^2 + 288*c + 96)*Zeta(3) + Pi^2*((Pi^2 + 24)*c^3 + (11*Pi^2 + 72)*c^2 + (11*Pi^2 - 72)*c + Pi^2 - 24))*c)/(288*(c - 1)^5*n^2)

This expression is already arranged with respect to n. However, the nominator of each term is not collected with respect to c. So naturally I thought

collect(f,[n,c]) or collect(f,[n,c],simplify)

would work. But now he messes up the nominator i.e. he can not factor and simplify. Note that the single variable case collect(f,n,simplify) works in not messing up, but this is not what I want, since the nominator of each n-term is not in c-collected form.

collect(f,[n],u->collect(u,[c],...))

also does not work, since he messes up again.

By c-collected form I mean the following:

The n^0 and n^{-1} term are actually fine. The factorization in the second term is ok. But for the third term

f2:=(144*c^3 + 1584*c^2 + 1584*c + 144)*ln(c)^2 - 24*Pi^2*(1 + c)*(c^2 + 10*c + 1)*ln(c) + (-96*c^3 - 288*c^2 + 288*c + 96)*Zeta(3) + Pi^2*((Pi^2 + 24)*c^3 + (11*Pi^2 + 72)*c^2 + (11*Pi^2 - 72)*c + Pi^2 - 24)

what I mean by c-collected is

collect(f2,c,simplify)

This is the nominator of the n^{-2} term.

I sort of managed by the following to procedings:

of:=[op(f)];
add(`~`[`/`](collect~(numer~(of), c, factor), ` \$`, denom(of)));

but the first one seems cumbersome for such a trivial thing that should be handled by collect. It also does not factor out the e.g. 12*ln(c)-Pi^2. This seems to be a general behaviour

g:=a*x*(x+1);
collect(g,x) does not factor out the a as in a*(x^2+x).

With the second method I'm more or less happy, but I didn't manage to completely collect the c terms i.e. the n^{-1} term is still c*(c+1) and not c^2+c.

## simple random number construction...

Maple 2019

Hello, Probably there is a way to do this easily but I do not quickly find it within the help.

I want

rand(0..1)

to give a true! random number and not always the same number; otherwise it should be called

predefinedlist()

Seed is deprecated, not sure it would help though. So how do I go abouts?

## Calculate Triple numeric integral...

Maple 2019

I want to calculate the below triple integral numerically. I tried different methods and reduced accuracy but nothing works.

Anyone capable?

Thanks

restart;
Digits := 15;
with(VectorCalculus);
v1 := <0, 0, 1>;
v2 := <sin(theta2), 0, cos(theta2)>;
v3 := <VectorCalculus:-`*`(sin(theta3), cos(phi3)), VectorCalculus:-`*`(sin(theta3), sin(phi3)), cos(theta3)>;
v1v2 := CrossProduct(v1, v2);
v2v3 := CrossProduct(v3, v2);
DotProduct(v1v2, v2v3);
(simplify(VectorCalculus:-`*`(%, VectorCalculus:-`*`(Norm(v1v2), Norm(v2v3))^VectorCalculus:-`-`(1))) assuming (0 < theta2, theta2 < Pi, 0 < theta3, theta3 < Pi));
evalf(VectorCalculus:-`*`(Int(VectorCalculus:-`*`(arccos(%), sin(theta3)), [theta2 = 0 .. Pi, theta3 = 0 .. Pi, phi3 = 0 .. VectorCalculus:-`*`(2, Pi)], epsilon = 0.001, method = _d01akc), VectorCalculus:-`*`(VectorCalculus:-`*`(VectorCalculus:-`*`(VectorCalculus:-`*`(VectorCalculus:-`*`(2, Pi), 2), Pi), 4), Pi)^VectorCalculus:-`-`(1)));

## simplifying an expression...

Maple 2019

Hey,

Is anyone of you capable of simplifying this expression

f1:=(-3*sin(8*x) + 3*sin(8*x + 2*y) - 3*sin(8*x + 6*y) + 3*sin(8*y + 8*x) + 3*sin(8*y + 6*x) + 3*sin(8*y) - 18*sin(8*y + 4*x) + 3*sin(8*y + 2*x) - 45*sin(6*y + 6*x) + 87*sin(4*y + 6*x) - 3*sin(6*x - 2*y) - 87*sin(6*x + 2*y) + 18*sin(4*x - 4*y) - 93*sin(4*x + 4*y) + 93*sin(4*x + 6*y) - 51*sin(2*x - 4*y) - 342*sin(2*x + 4*y) - 3*sin(-6*y + 2*x) + 51*sin(6*y + 2*x) - 93*sin(-2*y + 4*x) + 342*sin(-2*y + 2*x) + 639*sin(2*x + 2*y) - 639*sin(2*x) + 45*sin(6*x) + 93*sin(4*x) + 231*sin(4*y) - 225*sin(2*y) - 63*sin(6*y) - 57*sqrt(3)*cos(2*x) - 375*sqrt(3)*cos(2*y) + sqrt(3)*cos(8*y + 8*x) - 5*sqrt(3)*cos(8*x + 6*y) - 7*sqrt(3)*cos(8*y + 6*x) + sqrt(3)*cos(8*x) + 192*sqrt(3)*cos(2*y + 4*x) + 43*sqrt(3)*cos(-2*y + 4*x) - 7*sqrt(3)*cos(6*x + 2*y) + 7*sqrt(3)*cos(-6*y + 2*x) - 5*sqrt(3)*cos(6*y) - 149*sqrt(3)*cos(4*x + 4*y) - 149*sqrt(3)*cos(4*x) - 65*sqrt(3)*cos(6*y + 2*x) + 126*sqrt(3)*cos(2*x + 4*y) - 65*sqrt(3)*cos(2*x - 4*y) - 5*sqrt(3)*cos(8*x + 2*y) - sqrt(3)*cos(8*y) + 7*sqrt(3)*cos(8*y + 2*x) + 6*sqrt(3)*cos(8*x + 4*y) - 57*sqrt(3)*cos(2*x + 2*y) + 125*sqrt(3)*cos(4*y) + 126*sqrt(3)*cos(-2*y + 2*x) - 7*sqrt(3)*cos(6*x - 2*y) + 19*sqrt(3)*cos(6*x) + 43*sqrt(3)*cos(4*x + 6*y) + 19*sqrt(3)*cos(6*y + 6*x) - 7*sqrt(3)*cos(4*y + 6*x) + 246*sqrt(3))/(2*(-261*sin(4*x + y) - 297*sin(2*x + 3*y) - 48*sin(5*y + 6*x) + 126*sin(5*y + 2*x) + 9*sin(5*y + 8*x) + 12*sin(7*y + 6*x) - 9*sin(7*y + 4*x) - 36*sin(5*y + 4*x) + 261*sin(3*y + 4*x) + 9*sin(-3*y + 4*x) + 297*sin(-y + 2*x) - 135*sin(3*y) - 21*sin(5*y) - 147*cos(y)*sqrt(3) - 9*sqrt(3)*cos(7*y + 4*x) - 3*sqrt(3)*cos(5*y + 8*x) - 3*sqrt(3)*cos(3*y + 8*x) + 54*sqrt(3)*cos(6*x + 3*y) + 5*sqrt(3)*cos(-5*y + 2*x) + 5*sqrt(3)*cos(7*y + 2*x) - 2*sqrt(3)*cos(6*x - y) - 20*sqrt(3)*cos(6*x + y) - 69*sqrt(3)*cos(4*x + y) + 68*sqrt(3)*cos(4*x - y) + 2*sqrt(3)*cos(8*x + y) + 2*sqrt(3)*cos(7*y + 8*x) - 20*sqrt(3)*cos(5*y + 6*x) - 2*sqrt(3)*cos(7*y + 6*x) + 68*sqrt(3)*cos(5*y + 4*x) - 9*sqrt(3)*cos(-3*y + 4*x) - 69*sqrt(3)*cos(3*y + 4*x) - 171*sqrt(3)*cos(2*x + 3*y) - 35*sqrt(3)*cos(5*y) + 171*sqrt(3)*cos(3*y) - 171*sqrt(3)*cos(-y + 2*x) + 354*sqrt(3)*cos(2*x + y) + sqrt(3)*cos(7*y) + 639*sin(y) - 9*sin(3*y + 8*x) - 12*sin(6*x - y) + 3*sin(7*y) - 9*sin(7*y + 2*x) + 9*sin(-5*y + 2*x) + 48*sin(6*x + y) + 36*sin(4*x - y) - 126*sin(2*x - 3*y)))

into

cos(y-Pi/3).

PS: Actually I managed by expanding the thing out and converting to exp then expanding again and using radnormal. In essence I leave the question, because maybe somebody can explain to me why radnormal seems to be superior (sometimes) to simplify which I thought of as the USEALL choice. Thanks

## Simplification using a polylog(2,*) iden...

Maple 2019

Why doesn't

f:=ln(s + 2)^2 + 2*polylog(2, -1 - s) + 2*polylog(2, (1 + s)/(s + 2))

simplify to zero assuming s>0?

﻿