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5

One can make use of Simplify with Assumptions I. Compute the sum s=Sum[HarmonicNumber[n,5]/n^8,{n,1,Infinity}] (* -(1/63) π^6 Zeta[7]-13/15 π^4 Zeta[9]-55 π^2 Zeta[11]+644 Zeta[13] *) II. Make a table of Zeta-functions with even arguments t=Flatten[Table[{ζ[2n]==Zeta[2n]},{n,0,6}]] (* {ζ[0]==-(1/2),ζ[2]==π^2/6,ζ[4]==π^4/90,ζ[6]==π^6/945,ζ[8]==π^8/9450,ζ[...


1

It seems the problem came from defining the function in terms of k while the mathematical formulation is defined in terms of k+1. When the two were combined, it introduced the problem of including the input term in the output, which caused an infinite loop. For example d[2] -> With[{k = 2}, Table[d[k + 1 - i], {i, k + 1}]] {d[2], d[1], d[0]} would ...


2

I'd proceed as follows: delta[0, alpha_, lambda_] = 1; delta[k_, alpha_, lambda_] := delta[k, alpha, lambda] = alpha/k Sum[Sum[(1 - lambda[[1]]/lambda[[j]])^i, {j, 1, Length[lambda]}] delta[k - i, alpha, lambda], {i, 1, k}] To test it with your values: list = {1.2, 2.4, 3.3}; a = 2.3; delta[1, a, list] delta[2, a, list] delta[3, a, list] 2.61364 ...


2

I think you have to be mindful of the radius of convergence of the series. From the Mejer reference you cited, if $f(z)=x+x^p$, then define $$ h(z)=\sum_{k=0}^\text{kMax} \frac{(-1)^k}{pk-k+1}\binom{pk}{k}z^k$$ However, convergence of the series is restricted to the region of convergence $$ R=\frac{(p-1)^{p-1}}{p^p}$$ so that if $0\leq y\leq R^{\frac{1}{p-1}...


2

Hope this is what you're looking for: F[s_, l_, p_] := Sum[((-1)^k (1/2 Sinh[s])^(2 k))/(k! (p - 2 k)! (k + (l - p)/2)!), {k, (p - l)/2, p/2}] A[q_, s_, b_, p_] := Sum[Sqrt[q! p! (1/2 Tanh[s])^((l - p)/2)]/(Cosh[s])^(p + 1/2) b^(l - q) E^(-1/2 b^2) F[s, l, p] LaguerreL[q, l - q, b^2] Cos[\[Pi]/2 (p - l)], {l, q, Infinity}]^2 q = 2; s = 1; b = 5 (Cosh[s] + ...


4

You can use a recursive definition Clear["Global`*"] d[0] = f[x]; d[n_Integer?Positive] := d[n] = f[x]*D[d[n - 1], x] // Simplify Column[d /@ Range[0, 4], Dividers -> All]


1

The Null appears because two-argument If[] produces Null if the condition in the first argument is not satisfied. Αλέξανδρος shows one possibility, but you can fix your original code by recalling that $1$ is the identity element for multiplication; thus, you can implement Lagrangian interpolation like so: With[{x1 = {1, 2, 3, 4}, y1 = {3, 4, 2, 5}}, ...


1

What you want is already built-in as SymmetricPolynomial[]: SymmetricPolynomial[3, Array[f, 4]] f[1] f[2] f[3] + f[1] f[2] f[4] + f[1] f[3] f[4] + f[2] f[3] f[4] SymmetricPolynomial[4, Array[f, 4]] f[1] f[2] f[3] f[4] but otherwise, Bill's suggestion can be vastly simplified using Sum[] and Product[]'s ability to take a list of indices: Sum[Product[...


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