## 6331 Reputation

7 years, 349 days

## A slight variant...

I feel it could be more efficient as it avoids using Explode (?)

```restart
A := {12, 17, 24, 28}:
B := {358, 568}:
TA:=table([map(a -> a=convert(a, base, 10), A)[]]):
TB:=table([map(b -> b=convert(b, base, 10), B)[]]):

{seq(seq(`if`(ormap(member,TA[a],TB[b]),
NULL,parse(cat(a,b))),b=B),a=A)};

{12358, 12568, 17358, 17568, 24358, 24568}
```

```expr1 := 4*(theta - 1/4)*gamma1^2:
# The representation you required:
#                     2
# (4 theta - 1) gamma1
simplify(expr1)
2
(4 theta - 1) gamma1

expr2 := (alpha + gamma1)^2*epsilon^2 + 2*(alpha + gamma1)*(beta + gamma1)*epsilon - 4*alpha*gamma1*theta + beta^2 + 2*beta*gamma1 - 4*(theta - 1/4)*gamma1^2:
# The representation you required:
#                                           2
# (beta + (alpha + gamma1) epsilon + gamma1)  - 4 gamma1 theta (alpha + gamma1)
Student:-Precalculus:-CompleteSquare(expr2, beta):
op(1, %) + factor(simplify(%-op(1, %)));
2
(beta + (alpha + gamma1) epsilon + gamma1)   - 4 gamma1 theta (alpha + gamma1)
```

solutions.mw

## @cq  Thanks for your return. Feel ...

@cq

Feel free to ask for more help.

## @dharr Infinitely more clever than ...

@dharr

Infinitely more clever than what I did.
I vote up.

For the record my first code was the one I sent in my reply.
But then I focused on the name Zeckendorf and I wonder myself "how could I modify my code to give the Zeckendorf's combination?" and this is the code I attached in my answer. ... but this obviously ended up in an unnecessary complex stuff.

## If so why do you speak about the Zeckend...

If you want ALL THE COMBINATIONS, there is no need to evoke the Zeckendorf's Theorem unless to look smart.

 > restart
 > GeneratorsOf := proc(n)   local N, F, G, s:   uses combinat:   N := 1:   while fibonacci(N) < n do N := N+1 end do: N := max(6, N-1):   F := {seq(fibonacci(i), i=2..N)};   G := convert(convert~(powerset(F) minus {{}}, list), list):   s := select((g -> add(g)=n), G); end proc:
 > GeneratorsOf(2);
 (1)
 > GeneratorsOf(3);
 (2)
 > GeneratorsOf(10);
 (3)
 > GeneratorsOf(100)
 (4)

## My observation...

Something I systematically observe is: if you edit your previous reply/answer within which you inserted the content of a worksheet and want to upload the content of a corrected/modified version of this same worksheet, the uploading time can take several dozens of seconds (but uploading the link alone is immediate).

Other observation: as for your using preview before submit seems to slowen dramatically the submission process (italic seems because this doesnt look as a systematic issue).

## The disappearance of min(m, n)...

@Dkunb

In the fike attached to my answer, formula (8) is the sum of two terms each involving thr hypergeom function.
One of them contains

`hypergeom([1, -m+1+min(m, n), -n+1+min(m, n)], [2+min(m, n)], 1/(2*a^2))`

This function is multiplied a term of the form

`something*binomial(m, min(m, n)+1)*binomial(n, min(m, n)+1)`

Let us focus on the product P(m, n) of the two binomial functions.

P(m, n) = 0 for every couple (m, n) of integers.

Then the only contributor to formula (8) is the term

`something*hypergeom([-m, -n], [], 1/(2*a^2))`

... and this contributor doesn't contain any term involving min(m, n).

So, if you look to expression (7) in the file I attached to my answer, you clearly see that the sum goes from 0 to min(m, n).
The evaluation of this sum (relation  (8) ) also contains the function min(m, n) (which also appears in the "LaguerreL" expression you give in your question).
But at the very end  min(m, n) disappears because binomial(m, min(m, n)+1)*binomial(n, min(m, n)+1) is always null.

This disappearance explain why the final result (equation (10) ) is that simple.

## @Preben Alsholm Thank you PrebenPS;...

Thank you Preben

PS; I like your "You don't need the assumption at the end, but it doesn't hurt."

## Can you please check this?...

@NIMA112I have corrected you document (Maple 2015 worksheet).
Please check it out and get back to me:

 > restart
 >
 (1)
 >
 (2)
 >
 >
 (3)
 > G := E/(2*(1+nu));
 (4)
 > Q[11] := E/(-nu^2+1); Q[55] := G: ([Q[110], Q[111], Q[112]] =~ Q[11] *~ int~([1, z, z^2], z=-h/2..h/2))[]; assign(%);
 (5)
 > ([Q[113], Q[114], Q[115]] =~ Q[11] *~ int~([Phi(z), Phi(z)^2, z*Phi(z)], z=-h/2..h/2))[]; assign(%);
 (6)
 > ([Q[550], Q[551], Q[552]] =~ (Q[55]/Q[11]) *~ [Q[110], Q[111], Q[112]])[]; assign(%);
 (7)
 > ([Q[553], Q[554], Q[555]] =~ (Q[55]/Q[11]) *~ [Q[113], Q[114], Q[115]])[]; assign(%);
 (8)
 > ([Q[556], Q[557], Q[558], Q[559]] =~ Q[55] *~ int~([diff(Phi(z), z), z*diff(Phi(z), z), Phi(z)*diff(Phi(z), z), diff(Phi(z), z)^2], z=-h/2..h/2))[]; assign(%);
 (9)
 > # check eval(Q)
 (10)
 > terms := [1, z, z^2, Phi(z), Phi(z)^2, z*Phi(z), diff(Phi(z), z), z*diff(Phi(z), z), Phi(z)*diff(Phi(z), z), diff(Phi(z), z)^2, diff(Phi(z), z\$2), diff(Phi(z), z\$2)*diff(Phi(z), z), diff(Phi(z), z\$2)^2]: seq(beta[k] = mu*int(terms[k], z=-h/2..h/2), k=1..nops(terms)); assign(%);
 (11)
 > # check eval(beta)
 (12)
 > N__T := 0; H__x := 0; N__0 := 0; K__2 := 0; K__1 := 0; DD := 0
 (13)
 > local I: seq(I[k-1] = rho*int(terms[k], z=-h/2..h/2), k=1..6); assign(%);
 (14)
 > eq1 := - Q[110] * diff(U(x, t), x, x)        + Q[111] * diff(W(x, t), x, x, x)        - Q[113] * diff(Upsilon(x, t), x, x)        - Q[110] * diff(W(x, t), x)/R        - Q[556] * Upsilon(x, t)/R        + Q[550] * U(x, t)/R^2        - Q[551] * diff(W(x, t), x)/R^2        + Q[553] * Upsilon(x,t)/(R^2)        + (beta[1]*l^2)/(4*R)  * diff(W(x,t), x, x, x)        - (beta[7]*l^2)/(8*R)  * diff(Upsilon(x,t), x, x)        - (beta[1]*l^2)/(8*R^2)* diff(U(x,t), x, x)        + (beta[2]*l^2)/(8*R^2)* diff(W(x,t), x, x, x)        - (beta[4]*l^2)/(8*R^2)* diff(Upsilon(x,t), x, x)        + I[0] * diff(U(x,t), t, t)        - I[1] * diff(W(x,t), t, t, x)        + I[3] * diff(Upsilon(x,t), t, t)
 (15)
 >

## @MaPal93 What I've done is tail...

What I've done is tailored to your initial problem and I've never claimed that it's a generic method for dealing with other problems.
It is quite easy to ask for the solution of some problem and say this solution doen't work for another problem.
I suggest you think carefully about the problem, or category of problems, you want to solve and Reply here with more precise specifications.

## @MaPal93 Screen capture + conversio...

Screen capture + conversion to png to be uploaded:

Direct export to png from the interactive graphic menu

There is no blurring here.

I use Maple 2015.2 but I don't think it is a version issue. If you cannot solve this problem by your own I suggest you to open a new thread and join my code, yours, and the images above: someone would likely have an idea of what happens on your side.

## @MaPal93 For your functions f__1 an...

For your functions f__1 and f__2, a simple solve is enough to find the locus o couples (sigma_-d, sigma__v) which verify f__1=f__2.

"For a more complex case ..." means essentially "whan solve is not enough".
Then implicitplot(....) or contourplot(..., contours=[0]) may be good alternatives.

"A minor follow-up: why is the plot significantly blurred ..." there is no such thing in the worksheet.
As a rule the quality of the plots is significantly altered when displayed on Mapleprimes

## @Carl Love  Thanks Carl for this e...

@Carl Love

Thanks Carl for this extremely detailed answer which seems to cover all my questions.
I'm not going to pretend I understood it in its entirety on the first sight, and I think a second (third?) reading will be necessary for me to get the right idea.

## @Axel Vogt  Thnk you Axel, this re...

@Axel Vogt

Thnk you Axel, this really helps.

Let's say we have some sample S, for instance S := Statistics:-Sample(Normal(0, 1), 10^3).
What is it that does not suit you in the command Statistics:-Histogram(S, opts), where opts is any sequence of options described in the help page?

Here are a few examples of histograms of samples drawn from a continuous or discrete random variable.
The last example is the histogram of the table you provided in your original question.

 > restart:
 > with(Statistics):

Histogram of a sample from a continuous random variable.

 > S := Sample(Uniform(-5, 5), 10^3): h0 := Histogram(S): h1 := Histogram(S, style=polygon): h2 := Histogram(S, maxbins=10): h3 := Histogram(S, frequencyscale=absolute): h4 := Histogram(S, binbounds=[seq(-5+k, k=0..10)]): h5 := Histogram(S, frequencyscale=absolute, transparency=0.8, color=red): plots:-display(Matrix(2, 3, [seq(h||k, k=0..5)])):

Histogram of a sample from a discrete random variable.

 > S := Sample(DiscreteUniform(-5, 5), 10^3): h0 := Histogram(S): h1 := Histogram(S, discrete=true): h2 := Histogram(         S         , discrete=true         , axis[1]=[tickmarks=[\$(-5..5)]]         , thickness=20         , style=polygon         , view=[-5.5..5.5, default]       ): plots:-display(Matrix(1, 3, [h0, h1, h2])):