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4

Warning: long answer with little practical use. As of Mathematica 10, there're no completely symbolic ways to define radial functions $\operatorname{Mc}^{(j)}$ and $\operatorname{Ms}^{(j)}$. Still, I'll show how to define them using as few numerics as I could achieve. In particular, here all the dependence of the functions on $z$ as in e.g. ...

4

You can use a plot to find the roots and then follow up with FindRoot to get precise solutions. Here I'll use ContourPlot to plot where the real part vanishes and use MeshFunctions to find where the imaginary part is simultaneously zero. cp = Show[ Table[ ContourPlot[ Evaluate[Re[WhittakerW[1, (I α)/2, 10] /. α -> a + b I] == 0], {a, a1, a1 ...

3

Sometimes it is just a matter of adjusting the options. Here is a modification of your code that gives an answer fairly quickly and without any messages, using NIntegrate and some of its options: P = 0; l = 0; x = 4; κ = 0.01`20; (* note increased precision *) n = 5; q = (κ*n)/2; Do[Q = Exp[((-I)*(MathieuCharacteristicA[ν, q] - x^2/4)*τ/2)]* ...

0

Solve's core functionality is symbolic algebraic (polynomial) equations. Your equation is far from algebraic, and Solve can't find a transformation that would make it algebraic. Since you seem to want a numerical answer, a numerical method is probably more appropriate: FindRoot[Sum[(R[k, i, .00001]^1 + L[k, 1, i, .00001]^2)* Binomial[40, ...

5

For large $x$, the value of $\operatorname{erf}(x)$ approaches $1$, so even if you were able to evaluate it you would encounter a catastrophic loss of precision when you subtracted $1$ from it. For this reason, implementations typically also provide the complementary function $$\operatorname{erfc}(x)=1-\operatorname{erf}(x)$$ designed to provide better ...

1

Here's the answer to your question, how Mathematica can give you a function wich calculates the sum for all combination of the parameters n, m, and z. And you almost have found it! Simply define the function a[n_, m_, z_] := Sum[(j + n - 2)!^2/((j + n - m - 1)!*(j - 1)!), {j, 1, z - n}] Now you can calculate the correct result of your example a[7, 4, ...

1

It's being caused by a misplaced ')'. Nested tables are frequently a bad idea. Here: Table[C'[i][t] == (B.Table[C[i][t], {i, 1, 10}])[[i]], {i, 2, 9}] + ... You can't do that with the plus on the end. In the equation you're feeding into NDSolve, essentially you have (stuff == moreStuff) + extra which makes no sense. Once I correct for that the error goes ...

0

Note function LegendreP gives normalized orthogonal polynomials. Approaching your recursion by nesting can reproduce these (though not proving where coefficients vanish). So for illustrative purposes (changing l to j): f[n_, j_] := - (j + n + 1) (j - n)/((n + 1) (n + 2)); es[a_, j_, n_] := First@Nest[{f[#[[2]], j] #[[1]], #[[2]] + 2} &, {a, 0}, n/2] ...

2

I think you need to use this form (subscripts removed because they're messy): FullSimplify[RSolve[a[n + 2] == ((n + k + 1) (n + k) - l (l + 1))/((n + k + 1) (n + k + 2)) a[n], a[n], n]] (* you should get...*) result = (2^n (C[1]+(-1)^n C[2])Gamma[1+k]Gamma[1/2 (k-l+n)]Gamma[ 1/2 (1+k+l+n)])/(Gamma[(k-l)/2] Gamma[1/2 (1+k+l)]Gamma[1+k+n]) Using this, ...

2

If you plot the real parts of each half of your integral together Plot[{Re[-2^p Beta[2/3, -p, 1 + p]], Re[((-1)^p Hypergeometric2F1[1, 1, 1 - p, 2])/p]}, {p, 0, 10}] you'll see that each half goes to infinity in opposite directions, so the two singularities must cancel each other. Mathematica is stumbling over combining the two separate singularities ...

1

In version 8 the integral is very simply solved without any explicit assuptions: \$Version (* Out[6]= "8.0 for Microsoft Windows (64-bit) (October 7, 2011)" *) Integrate[Sqrt[a + b Cos[q]], q] (* Out[5]= (2 Sqrt[a + b Cos[q]] EllipticE[q/2, (2 b)/(a + b)])/Sqrt[(a + b Cos[q])/(a + b)] *)

2

You really need to know the mathematics you are dealing with as well as the related areas of Mathematica to understand what is going on. If you had read the MathWorld article carefully and looked up Laguerre in the Documentation Center (as I did) you would realize that you have done nothing wrong; you code is finding what you are looking for. You just don't ...

2

Mathematica does not like your h[i]s. If you introduce shrt[head_, indices_] := ToExpression[StringJoin[ ToString /@ Flatten[ {head, indices}]]] so that e.g. shrt[h,1] becomes h1 or shrt[h,{i,j}] hij Then this works: S = 10^-9.2; cl = 3*10^8 ; f = 5.9*10^9; w = cl/f ; A = (4*π/w)^(2/1); β = 2.5 ; m ...

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