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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. ...


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}...


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, ...


0

Probably not complete answer, but some (clumsy) thoughts. The idea is to introduce a dummy variable instead of y (=0) so that Mathematica can properly compute partial derivatives: expr0 = Normal@ Series[expr, {y, 0, 0}] /. {(t : (T | S))[x, 0] :> t[x, z], Derivative[n_, m_][t : (T | S)][x, 0] -> Derivative[n, m][t][x, z]} {-4*(-1 + 3*x^2)*T[...


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 ...


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 &...


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[...


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 &...


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] *)...


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

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 ...


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] *)


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 (...


0

You could define a function that generates the explicit components for you: ClearAll[p] p[i_] := Array[HoldForm[p][i], 4] so you can use p[2] for your $p_2$ vector: p[2] (* Out: {p[2][1], p[2][2], p[2][3], p[2][4]} *) With that in hand we can try your simple equality; perhaps I am misunderstanding your notation here, but I am not sure how your equality ...


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] *)


1

Evidently it's having trouble with the transcendental equation. Plug in values and it can do it numerically. d = .5 h = .5 θn = 45 ° ϕn = 30 ° Minimize[U, {θ, ϕ}] (*{-0.899519, {θ -> 0.523599, ϕ -> 0.523599}}*) You need to test your own values, since I don't know what reasonable input value are.


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