## Predict Perfect...

Hi

Initially I tried to find pemutations of 1..9, ie [[1,2,3],[4,5,6],[7,8,9]],[[1,2,4],[3,5,6],[7,8,9]],...etc, But maybe not all of them?

I wonder if someone out there can change my code to reflect the following:

1/According to the pdf excerpt, there are a total of 199 unique sums, the smallest sum is 774 and the largest 2556.(From displaying 280 outcomes, the code got the 774).

2/Show which sums have the greatest probability (specifically, 1566, 1575, 1638, 1656, 1674, 1692, 1755, and 1764, each of which can be shown with effort to have a 3/280, or 1.07%, probability – the calculations to determine this probability is a brute-force computation, and requires enumerating all outcomes of the sample space within a computational software package). We then finally reveal the prediction, which of course is correct..(I don't get it)

predict_perfect.mw

## How to apply a function to two unequal length list...

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?

## what is difference between the following syntax f...

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))
```

```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))
```

```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()

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

installed in this computer: not installed```

## Installing packages from the cloud hangs on window...

When I try to install (as an example) physics update using Maple 2018.2.1 on windows 10, it keeps hanging in the middle as shown above.

I closed Maple, starting again and tried again, same thing happens. I waited for more than 10 minutes and nothing happens.  I click on the install button in the clouds windows from Maple itself to install it as I always did.

Do others have problem installing this?  I'll try again in few hrs, it might be the Maple server is having some issues but thought to ask.

## Can't solve ODE...

I try to solve ODE with conditions, but it give answer only without conditions:

SOT.mw

Thank you.

## How to make a variable exchange of quadratic terms...

Hello,

Suppose we have a set of quadratic equations of the form:

U_11 * a * q + U_12 * b * q + U_13 * c * q + U_14 * d * q + U_15 * a * w + U_16 * b * w + U_17 * c * w + U_18 * d * w + .. = C_i

.

.

.

Where terms in uppercase means constant while lowercase letters correspond to unknowns.

Now, I want to make a change of variable, so instead of having non-linear term `a * q` we would have `s_ij`, meaning i-th equation and j-th unknown after the change was done.

I'm trying to do so, because in my case I will be left with a linear system with a small number of unknowns(relative to N) and N equations, and so I could then solve it easily.

Any help is appreciated, thank you!

## Bisection method...

I need Maple code of Bisection method.

## Mittag-Leffler Function

by: Maple 2018

Mittag-Leffler function generall simulation for fractional calculus

 >
 >
 >
 (1)
 >
 >
 >
 >
 >

 >
 >

## int - too many levels of recursion...

A simple (?)  int and too many levels of recursion

 > J := Int(cos(2*x)/(1+2*sin(3*x)^2), x = 0 .. Pi);
 (1)
 > evalf[30](J);
 (2)
 > value(J);
 >

The workaround I know (for J = 0) it is not very simple.
Can you find an easy one?

## Need help with Linear Transformations, standard ma...

Hello again, I posted a thread here earlier and received some great responses. I've made some progress in Maple since then but once again I've ended up on a question where I am completely stuck. I am only meant to solve this one in maple as the answer is written below the question. The question is:

https://imgur.com/rhFNh2l

I realise this is alot to ask for but I'm studying from afar and I don't have as many options for help as other students at the moment. I have managed to solve a), and I know how to solve b) the traditional way (pen and paper), but I have no idea how to do it in maple. My solution for a) is attached below. c) and d) strictly rely on the answer from b) so I'd greatly appreciate if I could have some help with it so I atleast could attempt the others on my own. This is the final question I have so once I'm done with this I'll pretty much be a master Maple... or not.. :P

I have commented the maple document so it is easier to understand what I've done and what I want help with. Also, I very much apologise if something that I write don't make sense, English is not my native language.

## Find all values of b, such that s is a linearly in...

Hello, I just started working in Maple and I am struggling with a question that I found in my book. The question goes as follows:

Consider the set of vectors T = {v1, v2, v3​​​​, v4} in ℝwith

v= (2,5,-2,0), v2 = (-2,-4,b,4), v3 = (-1,2,-2,-5), v= (b,2,5,3)

Find all values of b, such that T is a linearly independent set

I am supposed to first solve it by hand and then in maple. I can solve it by hand but I have no clue how to do it in maple.The only thing I've done so far is create a matrix that contains the vectors in the set S as column entries.

I know that the columns of A are linearly independent if  det A ≠ 0. This is where my near non-existent knowledge in Maple stops me...

The answer is for all {2}

I appreciate any tips...