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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