I want to divide every element of A1 list by every element of list A2. Lists are unequal length.

Example:

A1:= [a1, a2, a3];

A2:= [b1, b2];

Output would be:

[a1/b1, a2/b1, a3/b1, a1/b2, a2/b2, a3/b2]

I managed to do this by this means:

LinearAlgebra:-OuterProductMatrix([a1, a2, a3], [1/b1, 1/b2]); convert(%, list)

My questions are:

1. Is there a more succinct, practical method?
2. How about if I want to use a different operation like say adding or just use a function f?

Thank you all in advanced.

I've been using the following syntax to set boundary condition which is a derivative, when passing it to pdsolve. Say we want to set u(r,theta,t) to have insulated boundary conditions at r=1. So the BC will be

For example, to set derivative of u w.r.t. "r" to zero when r=1

       eval(  diff(u(r,theta,t),r), r=1) = 0;  #(1)

or using this syntax

       D[1](u)(1, theta, t) = 0;  #(2)

But now I find, on one example below, that the above no longer works.  I have to use this syntax (which I did not know about) for it to work

        D[1]*u(1, theta, t) = 0;  #(3)

Has something changed? why when using (3) pdsolve now gives result, but when using (2) or (1) it returns unevaluated? are the three semantically equivalent? when to use which syntax?

I am using Physics updates 265, Latest Maple 2018.2 

Here is an example showing the (1,2)  syntax no longer works, but the (3) syntax works

#articolo example 6.9.2
restart;

#using (1) syntax
pde := diff(u(r, theta, t), t) = (diff(u(r, theta, t), r)+r*(diff(u(r, theta, t), r, r))+(diff(u(r, theta, t), theta, theta))/r)/(25*r);
bc_on_r := eval(diff(u(r,theta,t),r), r=1) = 0;
bc_on_theta:= u(r,0,t)=0, u(r,Pi,t)=0;
ic := u(r,theta,0)=(r-1/3*r^3)*sin(theta);
pdsolve([pde, bc_on_r,bc_on_theta,ic], u(r, theta, t), HINT = boundedseries(r = [0]))

does not solve it.

restart;

#using (2) syntax
pde := diff(u(r, theta, t), t) = (diff(u(r, theta, t), r)+r*(diff(u(r, theta, t), r, r))+(diff(u(r, theta, t), theta, theta))/r)/(25*r);
bc_on_r := D[1](u)(1, theta, t) = 0; 
bc_on_theta:= u(r,0,t)=0, u(r,Pi,t)=0;
ic := u(r,theta,0)=(r-1/3*r^3)*sin(theta);
pdsolve([pde, bc_on_r,bc_on_theta,ic], u(r, theta, t), HINT = boundedseries(r = [0]))

does not solve it.

restart;

#using(3) syntax
pde := diff(u(r, theta, t), t) = (diff(u(r, theta, t), r)+r*(diff(u(r, theta, t), r, r))+(diff(u(r, theta, t), theta, theta))/r)/(25*r);
bc_on_r := D[1]*u(1,theta,t)=0;
bc_on_theta:= u(r,0,t)=0, u(r,Pi,t)=0;
ic := u(r,theta,0)=(r-1/3*r^3)*sin(theta);
pdsolve([pde, bc_on_r,bc_on_theta,ic], u(r, theta, t), HINT = boundedseries(r = [0]))

I've used syntax (1) before many times and it works. Here is an example where all three syntax work

pde:=diff(u(x,t),t)=k*diff(u(x,t),x$2);
bc:=eval(diff(u(x,t),x),x=0)=0,u(L,t)=0;
ic:=u(x,0)=f(x);
sol:=pdsolve([pde,bc,ic],u(x,t));

pdsolve gives

 Using syntax (2)

pde:=diff(u(x,t),t)=k*diff(u(x,t),x$2);
bc:=D[1](u)(0,t)=0,u(L,t)=0;
ic:=u(x,0)=f(x);
sol:=pdsolve([pde,bc,ic],u(x,t));

gives same answer as (1) and using syntax (3)

pde:=diff(u(x,t),t)=k*diff(u(x,t),x$2);
bc:=D[1]*u(0,t)=0,u(L,t);
ic:=u(x,0)=f(x);
sol:=pdsolve([pde,bc,ic],u(x,t));

The answer also looks like different and simpler, but I assume they are equivalent for now without looking too much into it.

Which syntax should one use as now I am really confused.

It looks like (3) is the one that should be used? Why the others did not work on first example? i.e. pdsolve did not give an answer at all?  And if (1,2,3) syntax are supposed to be equivalent, whysecond example gives slightly different looking answer when using one syntax vs. the other?

And finally to make things more confusing, here is an example where syntax (3) does not work, but syntax 1 and 2 work:

#articolo example 8.4.3
restart;
pde := diff(u(x, t), t) = (1/20)*(diff(u(x, t), x, x))+t;
bc := u(0, t) = 5, (u(1, t)+ D[1](u)(1, t)) = 10;
ic:= u(x, 0) = -40*x^2*(1/3)+45*x*(1/2)+5;
pdsolve([pde, bc,ic], u(x, t))

gives answer.

restart;
pde := diff(u(x, t), t) = (1/20)*(diff(u(x, t), x, x))+t;
bc := u(0, t) = 5, (u(1, t)+eval(diff(u(x,t),x),x=1))  = 10;
ic:= u(x, 0) = -40*x^2*(1/3)+45*x*(1/2)+5;
pdsolve([pde, bc,ic], u(x, t))

gives same answer. But

restart;
pde := diff(u(x, t), t) = (1/20)*(diff(u(x, t), x, x))+t;
bc := u(0, t) = 5, (u(1, t)+ D[1]*u(1, t)) = 10;
ic:= u(x, 0) = -40*x^2*(1/3)+45*x*(1/2)+5;
pdsolve([pde, bc,ic], u(x, t))

does not work.

Clearly there is something I do not understand between these 3 syntaxes and when to use which.

Using 265 version

Physics:-Version()
 "C:\Maple_updates\Physics+Updates.maple", 2018, December 22, 

    10:41 hours, version in the MapleCloud: 265, version 

    installed in this computer: not installed

downloaded today.

Download Code_1.mw

I have
a:=-(diff(x(t), t))^2*h^2/(x(t)^2-2*d*x(t)+d^2+h^2)^(3/2);
                       

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Hoping to get your very own copy of Maple for Christmas? 

No, I'm not about to announce a sale. No crass commercialism allowed on MaplePrimes!  But we have created this handy-dandy letter for Santa that students can use to try to convince him to put Maple under the tree this year*.

Happy holidays from Maplesoft!

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===

===

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