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Designating your matrix as m, I ran the following: fsteig= Eigenvalues[m, 1]; Normal[Series[ fsteig /. {t1 -> a t1, t2 -> a t2, xi -> a xi, J -> a J}, {a, 0, 3}]] // Simplify Although I received a warning message, Root::sbr: Because of branch cuts, the series may represent a different root of ... I received no error messages. How does ...


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To solve this problem under the assumption that the stated quantities are small, you can insert a perturbation expansion for the unknown phi and solve for its expansion coefficients in the smallness parameter: First I'm copying the definitions from the question to be self-contained: Ecp = (s1*s3*Abs[J]/DE^2/2 - 1/DE*s2*s3*KroneckerDelta[1, ...


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This might be along the lines of what you want. I expand the terms before trying the replacement, and do that in two steps so I don't have to deal separately with all manner of products. Df = f'[x]; Dhf = (f[x + h] - f[x])/h; ss = Expand[Normal[Series[-Df/Dhf, {h, 0, 3}]]] (* Out[79]= -1 + (h*Derivative[2][f][x])/(2*Derivative[1][f][x]) - ...


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Clear[delta] delta[\[Alpha]_?NumericQ, q_?NumericQ, n_?NumericQ] := Module[ {z = I 2.0` l \[Pi]/Log[q]}, Re[2 NSum[Gamma[\[Alpha] - z] n^z, {l, 1, Infinity}]]]; delta[1, 10, 10] 0.0818184


1

Using Mathematica 10.0.2.0, I do not receive the error message you did. However, Sum returns unevaluated, which happens when Sum cannot do the summation. Since you are seeking a numerical answer, I suggest that you use a large upper bound on your Sum instead of Infinity. For instance, delta[\[Alpha]_, q_, n_] := 2 Sum[Re[Gamma[\[Alpha] - I 2 l ...


4

If you hit = in a clean cell twice the cell will turn into a Wolfram Alpha cell, and you can ask it using the English language. It will give you the series expansion as well as a series representation:


5

I think SeriesCoefficient is what you want. Then you can use it to display formatted formulas series[expr_, x_, x0_] := Defer[expr = Sum[#, {n, 0, ∞}]] &[ FullSimplify@SeriesCoefficient[expr, {x, x0, n}, Assumptions -> {n >= 0}] (x - x0)^n] series[Sin[x] Cos[x]/x, x, 0]


3

If you are interested in the analytic form for the series coefficient $c[n]$ of $x^{-n}$, it is (I think) $c[n]=-\frac{4~\Gamma[n/2]}{n~(n+2)~\sqrt{\pi}~\Gamma[(n+1)/2]}$, for $n=1,3,5,7,...$. (Gory details available on request). Finding the sum of coefficients from $n=1$ to arbitrary odd upper limit $m$ has an analytic form in terms of $m$ returned by the ...


3

It looks like you're trying to approximate the result using a partial sum of the first 30 Laurent coefficients. But you can actually get the infinite sum with far less code by substituting $x=1$: Limit[Integrate[2*x*ArcSin[a/x] - 2*a*Sqrt[1 - (a/x)^2], x], x -> 1] (*-(π/2)*)


4

Simply replace Table by Sum: a = 1; s = Series[ Integrate[2*x*ArcSin[a/x] - 2*a*Sqrt[1 - (a/x)^2], x], {x, Infinity,29}] Sum[SeriesCoefficient[s, i], {i, 0, 29}] Or if you want to keep the table, use Total: a = 1; s = Series[ Integrate[2*x*ArcSin[a/x] - 2*a*Sqrt[1 - (a/x)^2], x], {x, Infinity,29}] t = Table[SeriesCoefficient[s, i], {i, 0, 29}] ...



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