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Answer revised for Abs[R] > 1 The code for sigx in the question contains a typo, [-1 + Sqrt[R^2]] instead of (-1 + Sqrt[R^2]). With it fixed and R > 1 chosen, consistent with the code in the question, the following returns an answer in just a few minutes for Version 12.1.1 Integrate[x*sigx, x, Assumptions -> 0 < x < 1 && R > 1]; ...


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From your question, we have num = 3 X = Table[Symbol["x" <> ToString[i]], {i, 1, num}]; Y = Table[Symbol["y" <> ToString[j]], {j, 1, num}]; and we have two lists of values of the X's and Y's, {xvals, yvals} = RandomReal[{-10, 10}, {2, num}]; We can use Thread to create our replacement rules like this rules = Join[Thread[X -&...


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When I rationalize eq, AsymptoticDSolveValue returns a solution for me: eq = FullSimplify[r1 + r2 + r3 - l] 1 + (-1 + x + x^2 + x^3) y[x] + x^3 (1 + 2 x) y'[x] + 1/2 x^5 y''[x] seq = CoefficientList[ AsymptoticDSolveValue[{eq == 0, y[0] == 1}, y, {x, 0, 30}], x] (* {1, 1, 2, 5, 14, 46, 166, 652, 2780, 12644, 61136, 312676, 1680592, \ 9467680, 55704104, ...


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Error Corrected. Because neither DSolve nor AsymptoticDSolveValue produced a solution, I expanded the ODE itself and computed the coefficients of each term. nmax = 10; Normal@Series[x^5 y''[x]/2 + (2 x + 1) x^3 y'[x] + (x^3 + x^2 + x - 1) y[x] + 1, {x, 0, nmax}] CoefficientList[%, x] /. Derivative[n_][y][0] -> c[n] /. y[0] -> c[0] Flatten@Solve[...


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How about this? f[x_, a_, b_, c_] = a PDF[NormalDistribution[b, c], x] Then ContourPlot3D[f[x, 1, b, c], {x, -3, 3}, {b, -1, 1}, {c, 1, 2}] Note that I am not claiming it is particularly informative! This might be more useful? Plot[Table[f[x, 1, b, 1], {b, -1, 1, 1/4}] // Evaluate, {x, -2, 2}, PlotStyle -> ColorData[10]] Plot[Table[f[x, 1, 1/2, c], {...


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You can get the sum for 169 values of $(i,j)$ by taking the limit of the what you get from Sum: expr = FullSimplify[Sum[F[m, k, i, j], {m, 0, ∞}, {k, 0, ∞}]]; ss2[iv_, jv_] := ss2[iv, jv] = Limit[expr, {i, j} -> {iv, jv}] mat = Table[ss2[iv, jv], {iv, 0, 12}, {jv, 0, 12}]; From these values FindSequenceFunction can suggest a closed form that works on ...


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$Version (* "12.1.1 for Mac OS X x86 (64-bit) (June 19, 2020)" *) Clear["Global`*"] F[m_, k_, i_, j_] := (-1)^(m + k)/(m!*k!)*2^m*Binomial[m, i]*Binomial[k + 1, j] sum[i_, j_] = Sum[F[m, k, i, j], {m, 0, Infinity}, {k, 0, Infinity}] // FullSimplify (* -2^i E^(-3 + I i π) (-1 + j) Binomial[0, i] Binomial[1, j] (Gamma[1 - i] + i ...


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I tried to Plot3D it. Plot3D[Evaluate@ Sum[(-1)^(m + k)/(m!*k!)*2^m*Binomial[m, i]*Binomial[k + 1, j], {m, 0, Infinity}, {k, 0, Infinity}], {i, -5, 5}, {j, -5, 5}, PlotPoints -> 20, PlotRange -> All] Looks like a lot of discontinuity. But there are some points return values. /. {i -> 51/10, j -> 45/10} -(1/(E^3))(-((544 (-2)^(1/10) E Binomial[...


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If I understand correctly, this should be easy. First, as μ is a constant, it does not need an index and I do not write it in the following. Second, I think it is a bad idea to write a arguments like your "f" and "s" as sub- and superscript. They are arguments, so treat them like arguments. The derivative of B[x] can simply be written as ...


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If you replace a few of the complex terms with constants the simplification is done within 15 seconds. reps = {E^(-j k (d - t)) -> c[1], E^(-h1 j t) -> 1/c[2], E^(h1 j t) -> c[2], E^(-h2 j t) -> 1/c[3], E^(h2 j t) -> c[3], E^(-h3 j t) -> 1/c[4], E^(h3 j t) -> c[4], E^(-h4 j t) -> 1/c[5], E^(h4 j t) -> c[5], E^(-j k t) ->...


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Cs = {1, 2, 4, 5}; Ss = {1, -1, 1, -1}; Ss1 = Ss /. {1 -> "+", -1 -> "-"}; Thread[Subsuperscript["c", Cs, Ss1]] You can also Apply (@@@) the function Subsuperscript["c", ##]& to pairs of values from Cs and Ss1 (that is, to Transpose[{Cs, Ss1}}]): Subsuperscript["c", ##] & @@@ Transpose[{Cs, ...


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You may use, e.g., MapThread[ SubsuperscriptBox["c", ##] &, {Cs, Ss1} ]


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GeneratedParameters should resolve Your problem. GeneratedParameters is an option to such functions as DSolve, RSolve, and Reduce.


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