sand15

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9 years, 178 days

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These are replies submitted by sand15

@dharr 

I voted up, but I believe there are a few points you could consider.

Firstly I don't thik writting K[3] = K[1]*k[3] is appropriate because neither K[1]nor K[3] intervene by themselves in the equation:  eqn only depends on the difference K[3]-K[1].
I think it would have been better to set K[3] = K[1] + k instead, which reduces the number ot parameters by 1.

Next, dimension analysis shows that gamma[1] and gamma[2] have the same dimension: this could lead to a simpler final equation with still one less parameter.

(last point: it is easy to show that theta is dimensionless).


Completing @acer's reply:

  • You meant probably k*alpha instead of ?
    Copy-pasting into a 1D mode worksheet mode
    `kα`
  • There is no need do declare GAMMA as local as soon as you define GAMMA... like the original (when its argument is an integer).
    By the way
    GAMMA(k*alpha+1) / GAMMA(k*alpha+alpha+1) = GAMMA(k+1) / GAMMA(k+2) = 1/(k+1)
    
  • Within your solve command, the first argument is
    1/(k+1)*(diff(U[k](x), x, x)+2*sum(U[r](x)*(diff(U[k-r](x), x)), r = 0 .. k)-(diff(sum(U[r](x)*U[k-r](x), r = 0 .. k), x)))
    

    (
    I used sum instead of add to "show" what this term [TERM] looks like

    )
    What result do you expect by doing

    solve(TERM, U[k+1](x))

    given TERM doesn't contain U[k+1](x) ?

@raj2018 

Integrating formally f1^4 is a dead end for the same reason that integrating f2 is.
f1^4 contains terms of the form:

(1-x/a)^(-kc+1/2)*(1-x/b)^(-kh+1/2)

So the only possibility is to use a numerical integration.



Several errors:

  1. No termination of the loop (end do is missing).
  2. What do you expectto do writting f[i] = f[i+1] ?
  3. n being undefined there is no chance that an hypothetical result matches the expected one (which doesn't contain n).
    Are those n typos?
     

Point 1 put apart, your problem has nothing to do with Maple: this is more a question about how you are capable or not to write correctly your recurrence relations.
Fix all these points and come again with something more consistent if you still get errors.

@Anthrazit 
Good luck

@Anthrazit 

I see...
I can't do more for you and I hope someone else will fix your problem.
As a last resort, even if I know this is a lot of work, could it be an option for you to use only Maplets?

@dharr 

For information:
Concerning the square root of symmetric positive definite matrices I submitted a question here
https://www.mapleprimes.com/questions/235834-MatrixPower-Doesnt--Give-The-Right-Answer


Extremely useful!
I agree with @C_R, a post would seem to me more appropriate and would give it more exposure.

@Preben Alsholm 

Given the definition of the integration over an infinite domain (simply definition in the sequel):

int(C, x=0..+infinity) = limit(int(C, x=0..a), a=+infinity);

one gets

limit(int(C, x=0..a), a=+infinity);
eval(%, C=0)
                       signum(C) infinity
                           undefined

In the other way

limit(int(0, x=0..a), a=+infinity);
                               0

Does this difference comes from the way eval acts?
---------------------------------------------------------------------------------------------------
Now consider this

int(0, y=0..1, x=0..+infinity);
                           undefined

But, applying the definition:

int(limit(int(0, x=0..a), a=+infinity), y=0..1);
limit(int(int(0, y=0..1), x=0..a), a=+infinity);
limit(int(0, y=0..1, x=0..a), a=+infinity);
                               0
                               0
                               0

So, do we have to consider that Maple doesn't apply the definition when it directly computes the integral?

@vv @Mahardika Mathematics

Even if the result is very pretty, it xwould have be nice to knowm how this (extremely complex) eqyation has beeen derived.

To @vv : give a look to this site
https://www.researchgate.net/publication/368438992_CREATING_3D_GRAPH_EQUATION_by_DHIMAS_MAHARDIKA
it contains a few other amazing images but the full article is not free.

To @Mahardika Mathematics : do you have a free paper about your work?

@Thiago_Rangel7 

If you prefer the way MMA simplifies expressions, why don't you use MMA instead of Maple?

I wonder if I'm not going to some Mapleprimes-like forum about MMA and ask
"Why doesn't MMA simplify it to (e + r)/cot(alpha/2)?", just to have an idea of what the answers will be.

The only"properties one can define for a NewDistribution those ones  List_of_properties.mw

It took me a lot of time to find where this information, and as some others about random variables, where hidden in the Statistics package.
I guess that the properties Parameters and ParentName, that any KnownDistribution has, are used elsewhere for some internal purpose.

@MaPal93 

Sorry for this late reply but I won't have much time to spend with you from now on as I have just returned to work.

Concerning your not that naive question: "by hand" simplification often involve non explicit knowledge (for instance a variance is a strictly positive quantity, a correlation coefficient rho verifies abs(rho) <= 1, and so on).
If you want Maple to simplfy an expression must set explicitely all the assumtions which seems implicit to you.

Here is your file with simplified expression for the lambdas.
(NOTE that I didn't run your code for I don't have time: I just copy-paste some expresssions into a variable Z ans simplified Z).

Analytical_SOL.mw

@RezaZanjirani 


 

restart

Z := (beta__c*y__c^2 + beta__m*y__c*y__n + beta__n*y__n^2)/(2*h*(-r__c*y__c - r__n*y__n + h)) + (beta__c*(1 - y__c)^2 + beta__m*(1 - y__c)*(1 - y__n) + beta__n*(1 - y__n)^2)/(2*h*(h - (1 - y__c)*r__c - (1 - y__n)*r__n)):

sys := {diff(Z, y__c) = 0, diff(Z, y__n) = 0}:

solve(sys, [y__c, y__n]);

[[y__c = 1/2, y__n = 1/2]]

(1)

infolevel[solve]:=100:
solve(sys, [y__c, y__n])

Main: Entering solver with 2 equations in 2 variables

Main: attempting to solve as a linear system
Main: attempting to solve as a polynomial system
Main: Polynomial solver successful. Exiting solver returning 1 solution

 

[[y__c = 1/2, y__n = 1/2]]

(2)

SolveTools:-PolynomialSystem(sys, {y__c, y__n});

Main: polynomial system split into 1 parts under preprocessing

Main: trying resultant methods
PseudoResultant: normalize equations, length= 4401
PseudoResultant: 15047473 [1 7000392304 y__c] 2 2 4401 2 107 0
PseudoResultant: factoring all equations length = 4401
PseudoResultant: factoring done length = 4401
PseudoResultant: normalize equations, length= 5467
PseudoResultant: 3917948 [1 4033335347 y__n 1] 2 2 5467 2 107 0
PseudoResultant: normalize equations, length= 4898
PseudoResultant: normalize equations, length= 5452
PseudoResultant: 592024 [1 100005376 y__n] 2 2 4898 2 107 1
PseudoResultant: normalize equations, length= 1300
PseudoResultant: 408832 [1 200169041 y__c] 1 1 1276 2 78 1
PseudoResultant: normalize equations, length= 3
PseudoResultant: -10 [] 0 0 3 0 3 1
PseudoResultant: 4229639 [1 4233342435 y__n 2] 2 2 5452 3 123 0
PseudoResultant: normalize equations, length= 5060
PseudoResultant: normalize equations, length= 5271
PseudoResultant: 1672501 [1 700025440 y__n] 2 2 5060 3 123 1
PseudoResultant: normalize equations, length= 8883
PseudoResultant: 1953068 [1 700028504 y__n] 2 2 5271 4 301 1
PseudoResultant: normalize equations, length= 15387
PseudoResultant: 2065476 [1 700026140 y__n] 2 2 8883 3 123 1
PseudoResultant: normalize equations, length= 8228
PseudoResultant: 2918733 [1 700042247 y__n] 2 2 15387 4 301 0
PseudoResultant: normalize equations, length= 14354

PseudoResultant: 1 solutions found, now doing backsubstitution
PseudoResultant: backsubstitution of y__c
PseudoResultant: backsubstitution of y__n

 

{y__c = 1/2, y__n = 1/2}

(3)

 


 

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