@maplelearner 

solve(% = 0, alpha0);

                           1           1     
                  0, - ----------, ----------
                            (1/2)       (1/2)
                       (-k2)       (-k2)     

@Markiyan Hirnyk Yes, I noticed. I also notice that your C is exactly 1.0, whereas mine is 0.99997. The smallest value of x is 1.0, so it makes sense that it would be 1.0. The reason that mine is 0.99997 is that z = 0.00003 is nearly the smallest positive z for which Maple will numerically evaluate the infinite sum. Unfortunately, `evalf/Sum` does not have any option equivalent to the digits or epsilon option of `evalf/Int`, so there's no way to get extra guard digits and changing Digits won't help that. This is a drawback of my method.

evalf(Sum(exp(-.00003*n^2)/n^2, n= 1..infinity));
                        1.63524093728545

evalf(Sum(exp(-.00002*n^2)/n^2, n= 1..infinity));
                    infinity                
                     -----                  
                      \        /          2\
                       )    exp\-0.00002 n /
                      /     ----------------
                     -----          2       
                     n = 1         n        

evalf(%% - Pi^2/6);
                       -0.00969312956278

@Markiyan Hirnyk Yes, I noticed. I also notice that your C is exactly 1.0, whereas mine is 0.99997. The smallest value of x is 1.0, so it makes sense that it would be 1.0. The reason that mine is 0.99997 is that z = 0.00003 is nearly the smallest positive z for which Maple will numerically evaluate the infinite sum. Unfortunately, `evalf/Sum` does not have any option equivalent to the digits or epsilon option of `evalf/Int`, so there's no way to get extra guard digits and changing Digits won't help that. This is a drawback of my method.

evalf(Sum(exp(-.00003*n^2)/n^2, n= 1..infinity));
                        1.63524093728545

evalf(Sum(exp(-.00002*n^2)/n^2, n= 1..infinity));
                    infinity                
                     -----                  
                      \        /          2\
                       )    exp\-0.00002 n /
                      /     ----------------
                     -----          2       
                     n = 1         n        

evalf(%% - Pi^2/6);
                       -0.00969312956278

@mriedel To cut-and-paste mono-spaced content, first place your text cursor at the beginning of a blank line where you want the content to start. Then use the pull-down menu labelled "Paragraph" from the top toolbar of the MaplePrimes editor. Select the third item, "Preformatted". Then paste the content. The content will looked messed up in the editor, but Preview will show it properly spaced.

This process will automatically convert a "cut" of Maple 2d output (prettyprint=2) to mono-spaced 2d form (prettyprint=1).

The problem doesn't seem fully specified. Perhaps other details were given in the class notes? maybe a diagram?

If I were to write code for this I'd need to know

1. Is the reactor considered a single point? Can we think of all the neutrons as emanating from the origin (0,0) with an initial direction up the y-axis? or all four axial directions?
2. Where is the shield? I see that its size is specified, but not its location. Is the shield immediately surrounding the reactor? Or is there empty space in between?
3. What shape is the shield? Is it a circle around the reactor?

@Markiyan Hirnyk Plots show that all forms of the function presented in this thread---LommelS1, both hypergeoms, conversion to rational, and the results of simplifying those to BesselJ times Gamma---are equivalent and that the limit is 1/4/a ~ 0.06188. Series expansions at z=0---when they work at all---also show equivalence. So I think that the bug is in limit.

@Markiyan Hirnyk Plots show that all forms of the function presented in this thread---LommelS1, both hypergeoms, conversion to rational, and the results of simplifying those to BesselJ times Gamma---are equivalent and that the limit is 1/4/a ~ 0.06188. Series expansions at z=0---when they work at all---also show equivalence. So I think that the bug is in limit.

@Markiyan Hirnyk 

eval(%, [sc= 1, t= 1, a= 1, k1= 1]);
  1       6   2       5         4            2                 
- - lambda  - - lambda  + lambda  + 32 lambda  + 64 lambda + 96
  9           9    
                                            
galois(%, lambda);
        "6T16", {"S(6)"}, "-", 720, 

          {"(1 6)", "(2 6)", "(3 6)", "(4 6)", "(5 6)"}


The galois group being S(6), the polynomial is not solvable.

@Markiyan Hirnyk 

eval(%, [sc= 1, t= 1, a= 1, k1= 1]);
  1       6   2       5         4            2                 
- - lambda  - - lambda  + lambda  + 32 lambda  + 64 lambda + 96
  9           9    
                                            
galois(%, lambda);
        "6T16", {"S(6)"}, "-", 720, 

          {"(1 6)", "(2 6)", "(3 6)", "(4 6)", "(5 6)"}


The galois group being S(6), the polynomial is not solvable.

Why don't you try it yourself first? Then, if you have any problems, post a Reply.

By the way, you should remove with(linalg). That is a very old package that is not meant to be used anymore.

There's no need to apologize. I realize that there's no easy way to avoid multiple answering, and I enjoy seeing a different presentation of essentially the same answer. Anyway, you have the additional information about Explore, which was educational for me.

If someone simul-posts an Answer that seems significantly better than mine, I usually delete mine.

There's no need to apologize. I realize that there's no easy way to avoid multiple answering, and I enjoy seeing a different presentation of essentially the same answer. Anyway, you have the additional information about Explore, which was educational for me.

If someone simul-posts an Answer that seems significantly better than mine, I usually delete mine.

Warning: The worksheet attached to the Question has a single expression that is 3753 pages long.

@maplelearner Note that your variable RootOfLambda has two parts. One of the roots is 0. Your eval statement should be eval(P, lambda= RootOfLambda[2]) so that you select only the second root. At this point, you can continue working with the P expression, even though it contains RootOfs. For example, you can set it to 0 and solve it, getting three simple solutions with no RootOfs.

Numerical solution will not be possible because of your four symbolic constants.

The RootOf is a degree 6 polynomial with symbolic coefficients. Since it's polynomial, it is not considered transcendental, for whatever that's worth. Still, I don't have much hope for simplifying it.

@maplelearner Note that your variable RootOfLambda has two parts. One of the roots is 0. Your eval statement should be eval(P, lambda= RootOfLambda[2]) so that you select only the second root. At this point, you can continue working with the P expression, even though it contains RootOfs. For example, you can set it to 0 and solve it, getting three simple solutions with no RootOfs.

Numerical solution will not be possible because of your four symbolic constants.

The RootOf is a degree 6 polynomial with symbolic coefficients. Since it's polynomial, it is not considered transcendental, for whatever that's worth. Still, I don't have much hope for simplifying it.