7
Try the following
Series[f[InverseFunction[g][y]],{y,0,10}]
5
Try Map[(If[# <= 0.4, 1, 0]) &, RandomReal[{0, 1}, 20]]!
4
@LeslieChiu
Follow @kglr suggestion
Clear["`*"];
f[x_] = Piecewise[{{1, x <= 0.4}, {2, 0.4 < x <= 0.7}, {3,
0.7 < x <= 1}}];
f /@ RandomReal[{0, 1}, 20]
4
SeedRandom[1]
x = RandomReal[{0, 1}, 20];
t = .4;
You can also use UnitStep or Boole or Clip as follows:
result1 = UnitStep[t - x];
result2 = Boole[# <= t] & /@ x;
result3 = Round @ Clip[x, {t, t}, {1, 0}];
Grid[Prepend[Transpose[{x, result1, result2, result3}],
{"x", "UnitStep[t-x]", "Boole[#≤t]&/@x", "...
4
In an even more trivial case. It appears that Module is not willing to recognize such indexed variables as symbols:
f2[x_] := Module[{out[1]}, out[1] = x; out[1]]
f2[5]
Module::lvsym: Local variable specification {out[1]} contains out[1], which is not a symbol or an assignment to a symbol
Module[{out[1]}, out[1] = 5; out[1]]
3
According to the Help, ''Every time a module is evaluated, a new temporary symbol is created'' and ''Block localizes values only; it does not create new symbols''
Clear["`*"];
f1[x_] := Block[{out}, Table[out[i], {i, 1, Length[x]}, 42]]
f1[{1, 2, 3}]
3
You can get the row sums using {2} as the second argument of Total:
Total[{{1, 2}, {2, 0}, {3, 4}}, {2}]
{3, 2, 7}
Then you can use Ordering and Sort to get the ordering of sums and the sorted sums, respectively:
Through @ {Ordering, Sort} @ %
{{2, 1, 3}, {2, 3, 7}}
and Transpose the resulting pair of lists:
Transpose @ %
{{2, 2}, {1, 3}, {3, 7}}
...
3
Does it work?
{f[4] - g[-1], h[2], 2*g[2] + h[-1]} /. y_[x_ /; x < 0] -> c*y[x]
2
Integrate[(λ0/Sqrt[2])*Sech[((t - τ) - 1/2*T)/σ] (1 +
Tanh[((t - τ) - 1/2*T)/σ]), τ]
gives
(*1/Sqrt[2]λ0 σ (ArcTan[Sinh[(-2 t+T+2 τ)/(2 σ)]]+Sech[(-2 t+T+2 τ)/(2 σ)]) *)
then enter the limits.
Numerical evaluation may be difficult for your parameter values
2
I have rewritten your code somewhat. It now completes in python and it works in Mathematica. Please tell me if the outputs are reasonable.
solution3d[edges_, vd_, vl_, ew_] := Module[{
uedges = UndirectedEdge @@@ edges,
vcoords = List @@ vd,
ew2 = KeyMap[UndirectedEdge @@ # &, ew],
vars3d, \[Lambda], lbnd, ubnd, obj3d, err, sol, edgeLengths3d}...
2
Are you sure you have typed your function as you intended? I can't make it work, so I think you mean:
f[x : {{_, _} ...}] := f /@ x;
f[{a_, b_}] := a + b;
f[{{1, 2}, {2, 0}, {3, 4}}]
Which outputs
{3,2,7}
You could also use (amongst many other possibilities):
{{1, 2}, {2, 0}, {3, 4}} /. {x_, y_} -> x + y
I think from your example output {{2,2},{1,3},{...
2
MapIndexed work fine.
list = {{1, 2}, {2, 0}, {3, 4}};
result=SortBy[First]@MapIndexed[{Total@#1, #2} &][list]
(* {{2, {2}}, {3, {1}}, {7, {3}}} *)
and then adjust the appearance.
Flatten[#, 1] & /@ Reverse /@result
(* {{2, 2}, {1, 3}, {3, 7}} *)
2
f[list_] := (
k = 1;
Sort[Map[{k++, #[[1]] + #[[2]]} &, list], #1[[2]] < #2[[2]] &]
)
f[{{1, 2}, {2, 0}, {3, 4}}]
(* Result of calling f *)
(* {{2, 2}, {1, 3}, {3, 7}} *)
2
Integrate[
E^(-\[Alpha] \[Sqrt](x^2 +
y^2)), {x, -\[Infinity], \[Infinity]}, {y, -\[Infinity], \
\[Infinity]}, Assumptions -> a > 0]
(ConditionalExpression[(2 \[Pi])/\[Alpha]^2, Re[\[Alpha]] > 0])
There is an internal parameter to the notebooks called $Assumption.
There is the documentation page for Assumptions for the built-ins Simplify, ...
2
Do integration first and then take the limit k-> inf.
Use indefinite integration. Integrate only finds a solution for m==0.
g[x_, m_, k_] = 1/((2*(x - m))^(2*k) + 1)
mint[x_, k_] = Integrate[g[x, 0, k], x]
(* x Hypergeometric2F1[1, 1/(2 k), 1 + 1/(2 k), -4^k x^(2 k)] *)
Use a trick. Tell Limit that x^(2 k) is always positive. (Valid for Integer k)...
2
Does this help?
f2[x_?NumericQ, m_] := Limit[1/((2*(x - m))^(2*k) + 1), k -> Infinity];
Module[{m = 0}, NIntegrate[f2[x, m], {x, m - 1/2, m + 1/2}]]
(* 1. *)
1
f[x_, y_, z_] := x + y + z;
Try @@@ (Apply at Level 1):
f[#,#2, 1] & @@@ {{1, 2}, {3, 4}}
{4, 8}
1
The equations depend on the variables Hb , Db , kb , Ks , Gb and two parameters K,Ho
For given parameters (examplary {K -> 1, Ho -> 1}) NMinimize might solve your problem:
eqn = {Hb/Db == K,Ks == Sqrt[(Coth[kb Db])/(1 + Gb)],Gb == (2 kb Db)/(Sinh[2 kb Db]) Hb/Ho == Ks (Ho)^2 == Hb^2 (1 + Gb) Tanh[kb Db]}
NMinimize[{1, eqn} /. {K -> 1, Ho -> 1}...
1
Clear["Global`*"]
Rather than using If, I recommend that you use Piecewise
area[J_, L_] =
Piecewise[{{Abs[L*Integrate[Max[J x, x^2], {x, 0, 1}]], L < 0}},
Abs[L*Integrate[Min[J x, x^2], {x, 0, 1}]]]
Vol[J_] = Integrate[area[J, L], {L, -1, 1}] // Simplify
Vol[-J] == Vol[J] // Simplify
(* True *)
Plotting,
Plot[Vol[J], {J, 0, 5}, ...
1
This is not an answer but rather something I do not understand. This seems quite Bizarre to me the more I think about it.
pts = RandomPoint[Ball[], 10000];
pts2 = Map[RandomReal[{0, 0.5}, 3] + {1.5, 0, 0} + # &, pts]; (* points with noise and translation *)
pts3 = RandomPoint[Ball[{1.5, 0, 0}, 1], 10000]; (* points sampled from another ball that is only ...
1
Your two functions do not share the same formula. However they share the same denominator:
numerator[r_, q_, d_] := numerator[r, q, d] = (2 \[Pi] (1 + r q - r q E^(-2 q d)))
(* there was a product r1 r1 that I wrote as r1^2, hope it's ok *)
denominator[r1_, r2_, q_, d_] := \[Epsilon] q ((1 + r1 q ) (1 + r2 q) - r1^2 q^2 E^(- 2 q d)))
The uV1 function ...
1
You can use Manipulate, PopupMenu and InputField as follows:
functions = {Total, Min, Max, MinMax, Mean, Median, Commonest, Quartiles};
flabels = {"Suma Datos", "Dato Minimo", "Dato Maximo", "Rango", "Media Aritmetica",
"Mediana", "Moda", "Cuartiles"};
fl = Thread[functions -...
1
You can use .5 {-y, x}/(x^2 + y^2) // TraditionalForm to see that .5 {-y, x}/(x^2 + y^2) can be use to express the vector value function.
$$\left(\frac{-0.5y}{x^2+y^2},\frac{0.5x}{x^2+y^2}\right)$$
and actually,{-y, x}/(x^2 + y^2) is the gradient of the function ArcTan[y/x] or ArcTan[x,y]
Grad[ArcTan[x, y], {x, y}]
Grad[ArcTan[y/x], {x, y}] // Simplify
(* {-(...
1
Another realization of ybeltukov's code.
start=1/2;
f[x_] = x + Sin[x];
Manipulate[
Show[Plot[{x, f[x]}, {x, 0, Pi}],
NestList[{Last@#, f[First@# ]} &, {start, f[start]}, n] // ListLinePlot
], {n, 1, 10, 1}
]
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