11
List available as EntityList[EntityClass["WolframLanguageSymbol", "Experimental"]]
9
_ matches one thing.
__ matches one or more things.
___ matches zero or more things.
8
The answer is simple: First, you really need to look up the documentation of ##. It means "take all arguments". But more importantly, you can use the property that a > b stays unevaluated when neither true or false can be computed. So check this out:
If[#1 > #2, ##] &[a, b]
(* If[a > b, a, b] *)
Do you see how ## converts this construct into a ...
7
f[a, l[[1]],l[[2]]] is already concise. You may save a few key strokes using:
f[x, ##] & @@ l
f[x, b, c]
Also
f[x, ## & @@ l]
f[x, b, c]
and
f[x, Sequence @@ l]
f[x, b, c]
f @@ Prepend[l, x]
f[x, b, c]
6
To understand how the SlotSequence (##) works, use Trace
Trace[If[#1 > #2, ##] &[22, 21]]
Trace[If[#1 > #2, ##] &[21, 22]]
6
Use ComplexExpand and then Simplify:
Simplify[ComplexExpand[ReIm[(-I a - m^2) Log[(I m^2)/(2 a)]]], a > 0]
Simplify[ComplexExpand[ReIm[(-I a - m^2) Log[(I m^2)/(2 a)]]], a < 0]
{(a π)/2 - m^2 Log[1/(2 a)] - m^2 Log[m^2], -((m^2 π)/2) -
a Log[1/(2 a)] - a Log[m^2]}
{-((a π)/2) - m^2 Log[-(1/(2 a))] - m^2 Log[m^2], (m^2 π)/2 -
a Log[-(...
6
My first guess when I see parenthesized exponents is that this represents a derivative of a multivariate expression. To check this guess I try
FullForm[%]
and look at what that shows me. That confirms there is a Derivative wrapped around your HypergeometricPFQ
If I use NIntegrate instead of Integrate to get an approximate result instead of an exactly ...
5
Using pts from mikado's answer:
DelaunayMesh[pts]["Faces"]
{{3, 4, 1}, {1, 4, 5}, {1, 2, 3}, {1, 5, 6}}
Grid @ %
4
You might find the following works. I'll create some points
pts = Table[(5 + t) {Cos[t], Sin[t]}, {t, 0, 5}] // N
(* {{5., 0.}, {3.24181, 5.04883}, {-2.91303, 6.36508}, {-7.91994,
1.12896}, {-5.88279, -6.81122}, {2.83662, -9.58924}} *)
and extract the cells of the triangulation
MeshCells[DelaunayMesh[pts], 2]
(* {Polygon[{3, 4, 1}], Polygon[{1, 4, 5}],...
4
a = 30;
f[x_] := Sqrt[2/a] (3/5 Sin[2 \[Pi] x/a] + 4/5 Sin [(9 \[Pi] x)/a])
Plot[f[x], {x, 0, a}, AxesLabel -> {x, "f(x)"},
PlotStyle -> {Black, Dashed}]
Note that you forgot the closing ) in the AxesLabel expression f(x).
When you actually type the input, you may want to take advantage of the Mathematica front end by typing Esc p Esc (with the ...
4
With f[x__Integer] := ... you can use
{x}[[i]]
to get the ith argument.
4
eqn = ψ[x] (k x^2/2 - En) - ℏ^2 ψ''[x]/(2 m) == 0;
sol = DSolve[eqn, ψ, x][[1]]
(* {ψ ->
Function[{x},
C[2] ParabolicCylinderD[(-2 En Sqrt[m] - Sqrt[k] ℏ)/(
2 Sqrt[k] ℏ), (I Sqrt[2] k^(1/4) m^(1/4) x)/Sqrt[ℏ]] +
C[1] ParabolicCylinderD[(2 En Sqrt[m] - Sqrt[k] ℏ)/(
2 Sqrt[k] ℏ), (Sqrt[2] k^(1/4) m^(1/4) x)/Sqrt[ℏ]]]} *)
Verifying ...
3
If you want to determine a symmetric matrix $A$ such that if $x=(1,x_1,x_2,x_3,x_4)$ and
$$x^T A x = 3 + x_1 - 2 x_4 + 2 x_3 x_4 + x_2 (-x_3 + x_4)$$
then the following (based on the comment by @yarchik) should do that:
(* Create a symbolic symmetric matrix *)
A = Table[a[Min[i, j], Max[i, j]], {i, 0, 4}, {j, 0, 4}];
(* Quadratic form *)
y = 3 + x1 - 2 ...
3
Szabolcs's answer (wrapping the sequence x with List and using Part) is the way.
An alternative is to use Indexed
ClearAll[f2]
f2[x__Integer] := Indexed[{x}, 1] + Indexed[{x}, 2]
f2[1, 2, 3, 4, 5]
3
You can also use Slot (#):
ClearAll[f3, f4]
f3[x__Integer] := Slot[1] + Slot[2] &[x]
f3[1, 2, 3, 4, 5]
3
f4[x__Integer] := #1 + #2 &[x]
f4[...
3
The O representation of an expansion point of Infinity is obtained with:
O[x, Infinity]
(see this part of the documentation for O).
So, you just need to do:
M = {
{1+5/s+6/s^3+O[s,Infinity]^4,1+8/s+4/s^2+O[s,Infinity]^4},
{1+2/s+2/s^3+O[s,Infinity]^4,1-1/s+8/s^3+O[s,Infinity]^4}
};
Det[M] //TeXForm
$-\frac{6}{s}-\frac{25}{s^2}+\frac{4}{s^3}+O\...
2
Why not just use Solve?
Solve[rho[r] == z && r>0, r, Reals]
{{r -> ConditionalExpression[Sqrt[1 + z^2], z > 0]}}
For values of q other than -1 you will also need to specify the value of z. For example:
q = 1/2;
Quiet @ Solve[rho[r] == 2 && r>0, r, Reals] //N
{{r -> 1.45108}}
You can use ConditionalExpression with ...
2
Change a+1to a++
For[m = 0, m < 0.05, m++, For[a = 5, a < 20, a++, a1 = a*(a/2);x[m] = a1]]
x[0]
(*361/2*)
2
symbols = Array[Symbol["x" <> ToString@#] &, 5];
ClearAll[f];
f[## & @@ (Pattern[#, Blank[]] & /@ symbols)] := Evaluate[Sin[Times @@ symbols]];
f[a, b, c, d, e]
Sin[a b c d e]
f @@ {a, b, c, d, e}
Sin[a b c d e]
2
As suggested, I'm expanding my comment into an answer
The problem here is the HoldAll attribute of ContourPlot. Like Plot and similar functions the process goes something like this:
Look at the first argument, and decide what form it has:
If it's a list, the user wants to plot multiple functions
If it's an equation with ==, the user wants to plot the ...
1
As with most *Q functions, CoprimeQ will evaluate to True or False immediately:
CoprimeQ[a,b,c]
False
You could instead use GCD:
FindInstance[
a + b == c && And @@ Thread[GCD @@@ Subsets[{a, b, c}, {2}] == 1],
{a, b, c},
Integers
]
{{a -> 1, b -> 0, c -> 1}}
although this approach won't work to find your example.
1
Clear[f, lhs, rhs]
{lhs, rhs} = Transpose@ToExpression[
{StringJoin["x", #, "_"], StringJoin["x", #]} & /@
Array[ToString, 4]];
Evaluate[f @@ lhs] := Evaluate[Sin[Times @@ rhs]]
f[a, b, c, d]
(* Sin[a b c d] *)
1
In the definition of modes, you set a new value for U here:
U = (U /. Table[var2[[i]] -> nn[[i]], {i, 1, Length[var2]}]);
Any subsequent calls to mode will simply return this value. Try instead
modes[j_] := Module[{nn},
nn = Last[
Transpose[
Last[SingularValueDecomposition[Rarz /. \[Omega] -> s2[[j]]]]]];
U /. Thread[var2 -> nn]
]
...
1
Try ?NumericQ and assign the NSolve result to h
h[r_?NumericQ, R_?NumericQ, l_?NumericQ, n_?NumericQ] :=h /. NSolve[
l^2 == r^2 + R^2 - 2 h^2 -2 Sqrt[(r^2 - h^2) (R^2 - h^2)] Cos[2 Pi/n] && h >= 0, h,Reals][[1]]
DensityPlot[h[r, 1, l, 4], {r, 0, 1}, {l, 0, 1}]
1
If HeavisideTheta[0] = 1, is HeavisideTheta in v11.3 equal to UnitStep in v5.0?
For calculus purposes, sure.
Can I replace UnitStep with HeavisideTheta?
HeavisideTheta[] is what you should be using in version 6 and later versions, because it's the one now supported extensively by the calculus functions. In earlier functions, you have to settle for ...
answered Sep 21 at 5:15
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