15

Consider: sum = Inactive[Sum][(1 - a j + m)^n, {j, 1, m - 1}]; and differentiate with respect to a: D[sum, a] Inactive[Sum][-j (1 - a j + m)^(-1 + n) n, {j, 1, -1 + m}] Up to a factor of -n (and the presence of the parameter a), this is the sum you're trying to evaluate. So, you're desired result is: r[m_, n_] = -D[Activate @ sum, a]/n /. a->1 (...


8

Workaround: $$\Gamma \left(\frac{k}{n}\right)=\frac{\Gamma \left(\frac{k}{n}+1\right)}{\frac{k}{n}}$$ $Version (* "12.0.0 for Microsoft Windows (64-bit) (April 6, 2019)" *) Product[Gamma[k/n + 1]/(k/n), {k, 1, n - 1}] (* (2 \[Pi])^(1/2 (-1 + n))/Sqrt[n] *)


6

You can use Optional in your pattern, which has the short form of .: arr = Table[i*j a^i b^j, {i, 1, 5}, {j, 1, 5}]; arr /. a^i_. b^j_. :> RuleCondition[0, i + j > 5] //TeXForm $\left( \begin{array}{ccccc} a b & 2 a b^2 & 3 a b^3 & 4 a b^4 & 0 \\ 2 a^2 b & 4 a^2 b^2 & 6 a^2 b^3 & 0 & 0 \\ 3 a^3 b & 6 a^3 b^2 &...


5

This post contains several code blocks, you can copy them easily with the help of importCode. As already mentioned in the comment above, the deduction of $(1)$ is incorrect because OP forgot $A$ cannot be treated as constant when solving ODE $(3)$, so it doesn't make much sense to continue discussing the Laplace inversion of $(1)$. Since OP's target is just ...


3

DSolve gives incorrect answer With Hom, InHom, and myeq as defined in the question, DSolve indeed gives an incorrect answer, sol = DSolveValue[myeq == 0, f0, z]; FullSimplify[(myeq == 0) /. f0 -> %] (* ((-1 + z) (z^2 - 5 (-2 + z) z Log[1 - z] + (14 + z (-14 + 3 z)) Log[1 - z]^2))/z == 0 *) which is not, in general, True, as can be seen by evaluating ...


3

If you don't specify which of the variables that you want excluded, Mathematica will use which ever ones fall out of the algorithms naturally. To preclude solving for c[14] use Drop[Flatten[mat], {14}] for the variable list. mat = Partition[Array[c, 25], 5]; equations = {Equal @@ Join[Total[mat], Total[mat, {2}], {Tr[mat], Tr[Reverse[mat]]}], ...


3

You can do an algebraic replacement for that position. Start as above: mat = Partition[Array[c, 25], 5]; equations = {Equal @@ Join[Total[mat], Total[mat, {2}], {Tr[mat], Tr[Reverse[mat]]}], Total[mat, 2] == Total[Range[25]]}; sol = Solve[equations, Flatten[mat], Integers][[1]]; mat2 = mat /. sol /. ConditionalExpression[a_, __] :> a (* Out[...


3

Here's a simple algebraic solution. The trick is to consider a factor (1-j+m) instead of just a factor j, and then subtract the access part (1+m). Here we go The sum being immediately evaluated is s0[n_, m_] := Sum[(1 - j + m)^(n - 1), {j, 1, m - 1}] s0[n, m] (* Out[6]= -0^(-1 + n) + (-1)^(1 + n) HurwitzZeta[1 - n, -m] *) The problem of the OP is the ...


3

You can define something like the following numberq[_] := False numberq[_?NumericQ] := True f[u_Plus] := f /@ u f[a_?numberq x_] := a f[x] You then specify that a and b are to be treated as numeric constants numberq[a | b] = True; You then get the simplification you requested f[3 x^2 - 8 a y^2 + 4 z b] (* 3 f[x^2] - 8 a f[y^2] + 4 b f[z] *)


3

Interpreting your $a^+$ and $a^-$ as limits from above and below, (and in one dimension), this function takes the limit from above and from below and then takes the difference: jump[q_, x_, val_] := Limit[q[x], x -> val, Direction -> "FromAbove"] - Limit[q[x], x -> val, Direction -> "FromBelow"] So for example, say we ...


3

Does this do what you want? Are you looking for additional formatting of input or output? Attributes[jump] = HoldFirst; jump[fn_[args__]] := fn[args] - fn @@ -{args} jump[f[a, b, c]] -f[-a, -b, -c] + f[a, b, c] Following your update please try this and report its utility. Parameters f and g can be changed to whatever symbol modifier you please. ...


3

The indeterminate can be overcome using the full identity for $\Gamma(nz)$: $$\Gamma(nz)=(2\pi)^{(1-n)/2}n^{nz-1/2}\prod_{k=0}^{n-1}\Gamma(z+\frac{k}{n})$$ and taking the limit as $z\rightarrow 0$: $Version (* "12.0.0 for Linux x86 (64-bit) (April 15, 2019)" *) Limit[Product[Gamma[z + k/n], {k, 1, n - 1}], z -> 0] (* (2 \[Pi])^(1/2 (-1 + n))/Sqrt[n] *)...


3

The integral doesn't seem to have any closed form solution. You can get an approximation to it by building an interpolating function and integrating it. Like so: Clear[f, pts, ff, int] f[τ_][u_] := Exp[Csch[(u - τ)]^2]; ff[τ_, x_, dx_] := Interpolation[Table[{u, f[τ][u]}, {u, 0, x, dx}]]; int[x_, dx_][τ_] := Function[u, Integrate[ff[τ, x, dx][uu], {uu, 0, ...


2

This looks like a better start: Ee[x_, y_, z_] := Sqrt[x^2 + y^2 + z^2]; η[Ee_, z_] := Sqrt[(Ee - z)/(Ee + z)]; α[y_, Ee_, z_] := ArcTan[y/Sqrt[Ee^2 - y^2 - z^2]]; κ[y_, z_, Ee_, V_] := Sqrt[y^2 + z^2 - (Ee - V)^2]; ξ[κ_, Ee_, V_, z_, y_] := (κ + y)/(Ee - V + z); Φ[η_, α_, ξ_] := 2 ArcTan[η Cos[α/(ξ - η Sin[α]) - α]]; dΦ = D[Φ[η[Ee[x, y, z], z], α[...


1

Think, you can only solve it numericaly. Define the inner integral and solve the differential equation for t,a,b values of interest. (takes a few seconds). nint1[ts_?NumericQ, a_?NumericQ, b_?NumericQ] := NIntegrate[ a (-1 + Exp[ b (-1/(ts - \[Tau])^2 - 1/(I + ts - \[Tau])^2 + \[Pi]^2 Csch[\[Pi] (ts - \ \[Tau])]^2)]), {\[Tau], 0, ts}] ...


1

As suggested by @JM, you can use ParametricNDSolveValue to obtain a numerical approximation to your integral: pf = ParametricNDSolveValue[ { int'[u] == Exp[Csch[u-τ]^2], int[0] == 0 }, int, {u, 0, τ-.1}, (* avoid singularity at τ *) τ ]; Visualization (using @m_goldbergs settings): Plot[{Exp[Csch[x-2]^2], pf[2][x]}, {x, 0, 1.25}...


1

You can use PiecewiseExpand to make sure the conditions are disjoint. For your example: pw = Piecewise[{{x, 3*x > y && x < 2*y}, {y, 2*x > y && x < 3*y}}, 0]; disjoint = PiecewiseExpand[ pw, Method -> {"OrderlessConditions"->True, "ConditionSimplifier"->FullSimplify} ]; disjoint //TeXForm $\begin{cases} x &...


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