# Tag Info

0

I'd recommend doing the whole calculation with symbols and substituting only at the end: ft[t1_] = a*Exp[-t1/d] - b*Exp[-t1/e]; g[t1_, t2_, t3_] = ft[t1]*ft[t2]*ft[t3]; factor[x_, y_] = factor[x, y] = g[0, x*dt, y*dt]; sumfactor[h_, g_] = Sum[factor[x, y], {x, 0, h}, {y, 0, g}] // FullSimplify (* (1/((-1 + E^(dt/d))^2 (-1 + E^(dt/e))^2))(a - b) E^(-((dt (...

2

In version 12.1, you can speed up the accepted answer from Szabolcs with FunctionCompile: cf = FunctionCompile[ Function[{Typed[k, "MachineInteger"]}, Module[{a, b, c, d, tau, dt, e, ft, g, factor, xx, yy}, a = -0.07; b = -0.45; c = 2.0; d = 3.0; tau = 2.0; dt = 0.01; e = c*tau; xx = ConstantArray[N@Range[0, k], ...

3

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 ...

5

$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 ... 2 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[... 0 Binom is not valid Mathematica syntax. Try Binomial instead. Sum[(2n-1)*(Sum[Binomial[6, k]*Binomial[10, 2 n-1-k], {k,0,n-1}])/(Binomial[16, 2 n-1]), {n,1,4}] 12245/1001 1 ip = {{17, 3}, {13, 7}, {11, 3, 3, 3}, {7, 7, 3, 3}, {7, 5, 5, 3}, {5, 5, 5, 5}}; Total[Times @@ Values[1 / Factorial @ #] & /@ Counts /@ ip] 71/24 Also Query[Total, Times @@ Values[1/Factorial @ #] &] @ Dataset[Counts /@ ip] 71/24 Dataset[Counts /@ ip][Total, Times @@ Values[1/Factorial @ #] &] 71/24 2 fn=Tr[1/Times @@@ (Tally[#][[All, 2]] & /@ #!)]&; fn@{{17, 3}, {13, 7}, {11, 3, 3, 3}, {7, 7, 3, 3}, {7, 5, 5, 3}, {5, 5, 5, 5}} 71/24 2 There seems to be a weakness in series expansions of QPochhammer in Mathematica: Series[QPochhammer[q, q, 2], {q, 0, 1}] //TeXForm$1+q \left(\text{QPochhammer}^{(0,1,0)}(0,0,2)+\text{QPochhammer}^{(1,0,0)}(0,0,2)\right)+O \left(q^2\right)$You can fix this by using FunctionExpand (essentially Vaclav Kotesovec's comment to Bill Watt's answer): Series[... 0 I'm not sure how you get all the terms, but if you want to look at a finite number you can do something like this: First get a series in q from fn = q^(1/2 n (n + 1))/QPochhammer[q, q, n] Series[fn, {q, 0, 100}] // Normal Then apply the first few n Plus @@ (Table[%, {n, 0, 100}] // Expand) That should give you several coefficients to play with. If you ... 3 Clear["Global*"]$Assumptions = a > 0 && ρ > 0; z[ρ_, ϕ_, a_ : 1, nmax_ : 40] := 50 + 200/Pi + Sum[(ρ/a)^(2 n + 1) Sin[(2 n + 1) ϕ]/(2 n + 1), {n, 0, nmax}]; Plot3D[z[ρ, ϕ], {ρ, 0, 1}, {ϕ, 0, Pi}, WorkingPrecision -> 20, PlotPoints -> 75, MaxRecursion -> 5] // Quiet If the number of terms is infinite f = z[ρ,...

1

As @I.M. suggested FrobeniusSolve[ConstantArray[1,m],n] with give all m tuples of indices at least 0 summing to n. However, you have added the extra constraint that each index should be at least 3. Subtracting 3 from each index, we get that the sum should be n-3m instead and can construct your indices as f[m_,n_] := FrobeniusSolve[ConstantArray[1,m],n-3m]+...

0

This example builds a homogeneous polynomial with $m$ variables and total degree $n$. $$p(m,n) = \sum_{i_1 + \dots + i_m = n} x_{1}^{i_1} \dots x_{m}^{i_m}$$ Power, can be converted to a two-indexed form. ClearAll[index] ; index[m_,n_] := FrobeniusSolve[ConstantArray[1,m],n] ; (* m - number of variables *) (* n - total monomial power *) m = 2 ; n = 6 ; vars ...

Top 50 recent answers are included