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5

Here a one-liner twitterable solution: Expand@Normal@Series[Hypergeometric2F1[1, 1 - eps/2, 3 - eps, x], {eps,0,0}]/.x->(1/2) giving 4 - 4 Log[2]


4

Your reply to Jens's answer suggests that you do not want to manually replace variables. This function just automates the method in Jens answer... cme[hgm_, symb_: c] := Module[ { places, symbs }, places = Position[hgm, eps]; symbs = Array[symb, Length[places]]; FullSimplify[Normal[Series[ ReplacePart[hgm, (Rule @@ # &) /@ Transpose[{places, ...


6

This does indeed work flawlessly with Mathematica version 8. But in version 10, I had to resort to the following workaround: AbsoluteTiming[ Series[Normal[ Series[Hypergeometric2F1[1, 1 - eps1, 3 - eps, 1/2], {eps1, 0, 1}, {eps, 0, 1}]] /. eps1 -> (eps/2), {eps, 0, 0}]] (* ==> {0.016225, SeriesData[eps, 0, {2 (EulerGamma + PolyGamma[0, ...


1

If you'd checked the document of shooting method carefully, you would notice that a Method can be added inside Method: eqn1 = t x'[t] - (-x[t] + y[t]); eqn2 = t y'[t] - (-5 t^2/x[t]^2 + x[t] - y[t]); sol = Quiet@ NDSolve[{eqn1 == 0, eqn2 == 0, x[0] == y[0], x[1] == 1}, {x, y}, {t, 0, 1}, Method -> {"Shooting", Method -> ...


0

I don't know any way to tell Integrate that 1/z[x]^2 has a pole at zero. However, we can integrate your series expression: series = Assuming[{z[0] == 0}, Series[1/z[x]^2, {x, 0, 1}]]; 1/(2 Pi I) Integrate[Normal[series] * I x /. {x -> Exp[I t]}, {t, 0, 2 Pi}] This gives us -z''[0]/z'[0]^3 This makes sense, since Cauchy's integral formula is: $$ ...


1

f=Normal[# + O[x]^(#2[[2]] + 1)] - Normal[# + O[x]^(#2[[1]])] &; ser = Series[Exp[x], {x, 0, 10}]; f[ser, {2,5}] (* x^2/2 + x^3/6 + x^4/24 + x^5/120 *) This handles series where powers are not sequential/predictable, where the nice and compact use of Part in other answers fails/becomes difficult to use: ser = Series[Exp[2 x^2]*x, {x, 0, 10}] f[ser, ...


3

ser=Series[Exp[x], {x, 0, 10}] $ 1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\frac{x^5}{120}+\frac{x^6}{720}+\frac{x^7}{5040}+\frac{x^8}{40320}+\frac{x^9}{362880}+\frac{x^{10}}{3628800}+O\left(x^{11}\right)$ Normal[ser][[3;;5]] $ \frac{x^4}{24}+\frac{x^3}{6}+\frac{x^2}{2} $ Or spartsF = FromDigits[Reverse[#[[3, #2 + 1 ;; #3 + 1]]], #[[1]]] ...


1

expr = Series[E^Sin[x], {x, 0, 10}] // Normal; Cases[expr, a_.*x^n_?(2 <= # <= 5 &)] // Total x^2/2 - x^4/8 - x^5/15


0

mySeries = Series[Exp[Sin[x]], {x, 0, 10}] $1+x+\frac{x^2}{2}-\frac{x^4}{8}-\frac{x^5}{15}-\frac{x^6}{240}+\frac{x^7}{90}+\frac{31 x^8}{5760}+\frac{x^9}{5670}-\frac{2951 x^{10}}{3628800}+O\left(x^{11}\right)$ SeriesCoefficient[mySeries, #] & /@ Range[2, 5] $\left\{\frac{1}{2},0,-\frac{1}{8},-\frac{1}{15}\right\}$


1

SeriesCoefficient[mySeries, n] For instance: mySeries = Series[Exp[Sin[x]], {x, 0, 10}] $ 1+x+\frac{x^2}{2}-\frac{x^4}{8}-\frac{x^5}{15}-\frac{x^6}{240}+\frac{x^7}{90}+\frac{31 x^8}{5760}+\frac{x^9}{5670}-\frac{2951 x^{10}}{3628800}+O\left(x^{11}\right) $ SeriesCoefficient[mySeries, 8] $\frac{31}{5760}$


1

You can use Piecewise to filter out any components you do not desire: (* your exclusion conditions *) conds={2-n/2<0, (* possibly more *) }; filter[c_]:=Piecewise[{{1,And@@c}},0] and then Sum[filter[conds] c[n]r^(2-n/2) + filter[conds] d[n]r^(-n/2),{n,Infinity}] This can easily be expanded for any conditions you might encounter.


1

It might help you to look at each side separately. Starting with the left-hand side lhs = Sum[(p/(1 - p))^s*(q/(1 - q))^s* Binomial[n, s]*(Binomial[m - 1, s]*(p*q*(m + n) + (2*m - 1)*(-p - q + 1))), {s, 0, n}] -Hypergeometric2F1[1 - m, -n, 1, (pq)/((-1 + p)(-1 + q))] + 2*mHypergeometric2F1[1 - m, -n, 1, (pq)/((-1 + ...


3

There are excellent answers of kguler and belisarius. Not to compete with them, but to offer a different view, you might operate within a different paradigm. Here it is: Step 1: Let us first find the solution of your equation for J: sl = Solve[16 x^3 - 4 x^2 + J^2/64 == 0, J] (* {{J -> -16 Sqrt[x^2 - 4 x^3]}, {J -> 16 Sqrt[x^2 - 4 x^3]}} *) ...


2

Perhaps Series[ToRadicals@Root[j^2 - 256 #1^2 + 1024 #1^3 &, 2], {j, 0, 4}]


2

How about this: Clear[x, t, X, Y, X0, δ, ϵ]; With[ {functions = {X, Y}, equilibrium = {X0, 0} }, Normal@Series[ D[X[t, x], t] + D[Y[t, x], x] + X[t, x] D[Y[t, x], x] + X[t, x] Y[t, x] == 0 /. Thread[functions -> Map[Function[{t, x}, #] &, equilibrium + ϵ Through[ ...


1

Beginning with D[X[t, x], t] + D[Y[t, x], t] + X D[Y[t, x], x] + X Y == 0 First replace the derivative by some other function to simplify manipulations. %[[1]] /. Derivative[z1__][z2__][z3__] -> W[{z1}, z2, {z3}] Make the first order substitution. %/. {X -> X0 + d X1, Y -> Y0 + d Y1} Expand the second argument of W. (Depending on the ...



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