To complete @Christian Wolinski 's reply: f can be integrated provided alpha is an integer <= 1.
I guess that the reason why the WhittakerM function

In the attached file you will see that the integration can be done only (that's my hunch, not a mathematical proof) when
WhittakerM(a, b, 3/2*t) can be converted into elementary functions.
(
Maybe there exist some recurrence relation to integrate int(h(t)*WhittakerM(a, b, 3/2*t) , t), but I wasn't capable to find any.
More precisely I thought that one could take advantage that

diff(WhittakerW(p, q, 3/2*t), t);
                                                  /          3  \
                                       WhittakerW |p + 1, q, - t|
 3 /1   2 p\             /      3  \              \          2  /
 - |- - ---|  WhittakerW |p, q, - t| - -------------------------
 2 \2   3 t/             \      2  /               t            
          

).

When alpha is an integer > 1, the denominator GAMMA(2-alpha)=0 and f is undefined.

Whittaker.mw

@mz6687 

I'm sorry, but I don't understand a single word of what you're saying.
What is this 8x8 matrix you are now talking about?

BTW: is it you who deleted your last question or was it moved by someone else?

Whatever, is this new thread isrelated to this one https://www.mapleprimes.com/questions/237066-Determinant-The-System-Hangs, you would better keep this discussion goin in this latter thread.

@mz6687 

 

Download Are_you_ok_with_that.mw

 

@mz6687 

Sorry, I don't understand what you say

@mz6687 


Your initial question was "Can Maple solve the determinant of the 9x9 matrix? "
I showed you the answer was YES.

Now you are talking about something else.
But you missed a point: in the reference you mention the matrix had a particular shape and depended only on 6 quantities

{v__11, v__12, v__21, v__22, lambda__1, lambda__2}

So the expression of its determinant was potentially simplifiable (and indeed was).

Your matrix B is dense and depends on 81 indeterminates (not only 6) meaning  there is no possible simplification between terms.
Using simplify or collect, for instance, will be a dead end.

Things could be completely different if all the m[i, j] were expressions involving a few number of independent quantities.


BTW: what does "solve the determinant" mean to you?

@acer 

You're right, I edited my answer consequently.

@Traruh Synred 

I know how to use Histogram, but I want to plot binned data!

Can you please give an example of those binned data and of the original data instead of speaking in the void?

For one thing, a Histogram requires a large list of each data point and is thus a memory hog, and each time you make a histogram Histogram loops through the data for the quantity, and if you make it for a related quantity you have to loop again. 

That is totally unclear, can you provide an example?

@AHSAN 

Your claim that "each curve have their maximum point at some vale of x" is obviously wrong (last figure on my last file on forst figure in For_reference.mw ... soom around the peaks).
and not consistent with your need "to obsrvse that difference in percentage at the maximum value of x" at the same time (if x is a constant as you before claimed it was it' difference in percentage is 0).

I hope you'll find someone better able than me to understand the subtleties of your question.
For my part, I'm done with you.

@AHSAN 


I still don't understand what is your criterion to assess this relative increase/decrease:

• Is this criterion the height of the peak?
   
• You taught before of "enlargement"; is your criterion the peak width at some conventional height?
   
• Is this criterion one of those I use in my first file?
   

Whatever, the case here is an example based on the peak height
Could_It_Be_This_2.mw

@AHSAN 

My mistake, this phrase about "a loop" is indeed an error.

It's likely  I didn't understand correctelyyour question.
I understood you wanted to compite the decrease/increase of the eights of the peaks.

What do you mean by "the decrease or incease in percentage between curves" ?
If you this

A_ref := eval(A, lambda = 1.3015):
data_1 := [beta=0.1, Q=1.3015, lambda=0.9986];
B_1    := eval(B, data_1):
plot((B_1-A_ref)/A_ref*100, x=-4..0)

you get a curve which represents the relative variation of B_1 wrt A_ref, but this is a function x.
So do you want a number or an algebraic expression or a curve

Could_It_Be_This.mw

@dharr 

Right, I didn't pay attention to that.

type "SIR  model" and click on Search

@Carl Love 

Using fsolve(..., complex) as you suggested gives a solution close to yours to within  10^-11 (or less) which verifies rel to within 10^-42.
Given that diff(Re(rel[1]), A) is infinite at this point, the slightest variation in A makes "huge" differences in the values of rel[1].

2023_vs_2015.mw

So you're right, there is indeed a second real solution

@Carl Love 

Thanks Carl... but this is not the result I get with Maple 2015:

eval(rel, {
    A = -2.7553365135418814642586082436429575890825402826031,
    B = -0.70285804987973303586180028708027467941012949957141
})

[                                                     -43    
[0.00004788232651393033381187767465396229938775 - 2 10    I, 

                                                        -40
  1.8649856419668477410903757547200476549949700479584 10   

                                                           -41  ]
   - 1.2309622539151182340327668587211822233603338843467 10    I]

A version issue?

@NIMA112 


Are there other real roots than the one @Rouben Rostamian got?
I don't think so (the couple (A, B) @Carl Love found doesn't verify the equations with an enough small error to be considered, IMO, as a solution).

Some details are given here Fung_sand15.mw