9
data = ConstantArray[Range[20], 5];
MatrixForm[data]
You can use Delete:
Map[Delete[{{9}, {11}}]] @ data // MatrixForm
Alternatively:
data[[All, DeleteCases[9 | 11] @ Range[20]]]
data[[All, Delete[{{9}, {11}}] @ Range[20]]]
MapAt[Nothing &, data, {All, {9, 11}}]
Module[{d = data}, d[[All, {9, 11}]] = Nothing; d]
Transpose @ ReplacePart[ {{9}, {11}} ...
5
You can use the assumptions that a^2 == a to do this:
Simplify[F, a^2==a && b^2==b]
(-1 + a) (-1 + b)
5
$n = 0$ is a removable singularity. If you take the limit of the result of the integral, you will get the same as if you put $n = 0$ before taking the integral:
result = Integrate[Exp[I n x], {x, 0, Pi}]
Limit[result, n -> 0]
(* Pi *)
Integrate[Exp[I 0 x], {x, 0, Pi}]
(* Pi *)
Addendum
Re @cvgmt on Integrate[x^a, x] for a = -1: that also is a removable ...
5
B = {{17/3, 1/3, 1/3, 1/3}, {1/3, 1/3, 1/3, 1/3}, {1/3, 1/3, 1/3, 1/3}, {1/3,
1/3, 1/3, 82/3}};
A = {{5/3, 0, 1/3, 0}, {0, 5/3, 0, 1/3}, {1/3, 0, 10/3, 0}, {0, 1/3, 0,
10/3}};
val = First /@ Eigensystem[{B, A}, 1];
All of the values are real
valR = val // RootReduce
Element[valR, Reals]
(* True *)
valR // N
(* {8.33542, {0.0311157, -0.179391, ...
5
Chop[N[First /@ Eigensystem[{B, A}, 1]]]
returns
{8.33542,{0.0311157,-0.179391,0.00719285,1.}}
5
There are many identities reminding one in the question (unclear what kind of relation has been intended), e.g. expressing FresnelS in terms of Erfi
FullSimplify[-1/4 (1 + I)( Erfi[(1 + I)Sqrt[Pi]z/2]- I Erfi[(1 - I)Sqrt[Pi]z/2])]
FresnelS[z]
Let's demonstrate other relations:
FullSimplify[{-1/4 (1+I)(I Erfi[(1+I)Sqrt[Pi]z/2]- Erfi[(1-I)Sqrt[Pi]z/2]),
...
4
This takes a long while, but it works if you just want to prove that the imaginary part is $0$.
B = {{17/3, 1/3, 1/3, 1/3}, {1/3, 1/3, 1/3, 1/3}, {1/3, 1/3, 1/3, 1/
3}, {1/3, 1/3, 1/3, 82/3}};
A = {{5/3, 0, 1/3, 0}, {0, 5/3, 0, 1/3}, {1/3, 0, 10/3, 0}, {0, 1/3,
0, 10/3}};
result = First /@ Eigensystem[{B, A}, 1]
Im[result] // FullSimplify
(*{0,{0,0,...
4
Apart from FullSimplify in my comment, you can also use SolveAlways to get the constants that make the two forms match:
eqn = 29/36 - x/2 + x^2/4 + y/9 + y^2/36 - (4 z)/9 + z^2/9 == 1;
otherform = a (b + x)^2 + c (d + y)^2 + e (f + z)^2 == 1;
sol = SolveAlways[First[eqn] == First[otherform], {x, y, z}]
(* {{a -> 1/4, c -> 1/36, e -> 1/9, f -> -2,...
3
U = Erfi[((1/2 + I/2) (R - z))/Sqrt[k R]] +
Erfi[((1/2 + I/2) (R - Sqrt[D^2 + z^2]))/Sqrt[k R]];
Use ComplexityFunction to penalize the use of Erfi
U2 = FullSimplify[U,
ComplexityFunction -> (LeafCount[#] +
1000 Count[#, _Erfi, {0, Infinity}] &)]
(* (1 + I) (FresnelC[(R - z)/(Sqrt[π] Sqrt[k R])] +
FresnelC[(R - Sqrt[D^2 + z^2])/(...
3
We had to manual D the condition :)
u[t_] = Through[{u1, u2, u3}[t]];
Simplify[u[t].u'[t], Assumptions -> D[u[t].u[t] == 1, t]]
3
When numbers plugged in - this expression evaluates to zero, but otherwise will not evaluate to zero
Since a1,a2,b1,b2 are all positive, we can substitute them with other numbers $x^2, y^2, z^2, w^2$.
FullSimplify[d2 - d1 /. {a1 -> x^2, a2 -> y^2, b1 -> z^2, b2 -> w^2},
Element[{x,y,z,w}, PositiveReals]]
(* result: 0 *)
3
If you have a Dataset with capital D:
Using Drop:
dataset = Dataset[{
<|"a" -> 1, "b" -> "x", "c" -> {1}|>,
<|"a" -> 2, "b" -> "y", "c" -> {2, 3}|>,
<|"a" -> 3, "b" -> "z", "c" -> {3}|&...
3
You could use ReplaceRepeatedand completing squares:
29/36 - x/2 + x^2/4 + y/9 + y^2/36 - (4 z)/9 + z^2/9 == 1 //.
a_ *s_^2 + b_ *s_ + rest__ :> a (s + b/(2 a))^2 - b^2/(4 a) + rest
(*1/4 (-1 + x)^2 + 1/36 (2 + y)^2 + 1/9 (-2 + z)^2 == 1*)
3
Try numerical evaluation with Mathematica's N[]
B = {{17/3, 1/3, 1/3, 1/3}, {1/3, 1/3, 1/3, 1/3}, {1/3, 1/3, 1/3,
1/3}, {1/3, 1/3, 1/3, 82/3}};
A = {{5/3, 0, 1/3, 0}, {0, 5/3, 0, 1/3}, {1/3, 0, 10/3, 0}, {0, 1/3,
0, 10/3}};
First /@ Eigensystem[{N@B, N@A}, 1]
The result should be
{8.33542, {0.0306117, -0.176485, 0.00707634, 0.983802}}
2
To piggyback off Bill's answer, one can just use CountRoots[] on the characteristic polynomial of the given matrix pencil, if one only wishes to show that the eigenvalues are all real:
CountRoots[CharacteristicPolynomial[{B, A}, x], x]
4
One can then use RootIntervals[] to find brackets for the roots:
RootIntervals[CharacteristicPolynomial[{B, A}, x], ...
answered Sep 12 at 5:55
2
Let us first introduce a function I will use:
factorMinus[expr_] := Module[{}, (-1)*HoldForm[Evaluate[-expr]]]
Here is your expression (I write it down in a simplified notation to make it more visible):
expr = (z2 - z2*z3)^-e;
Try this:
MapAt[factorMinus, MapAt[Factor, expr, {1}], {1, 3}] //
PowerExpand // ReleaseHold
(* z2^-e (1 - z3)^-e *)
Have ...
2
ReplaceAll[Power -> List] @ Replace[SortBy[Length] /@ (List @@@ List @@ k1),
{a_, b_, c___} :> If[a === -1, {{-b}, c}, {{a}, b, c}], 2]
{{{k}, {a + b, -1 - e}, {x + y, 2 + e}},
{{-z}, {a + b, -1 - e}, {x + y, 2 + e}}}
Use ReplaceAll[Power | Plus -> List] to get
{{{k}, {{a, b}, {-1, -e}}, {{x, y}, {2, e}}},
{{-z}, {{a, b}, {-1, -e}}, {{x, ...
2
Add /. Abs -> RealAbs // Simplify at the end of every line.
m = {{0, WP[t], 0}, {WP[t], 0, WS[t]}, {0, WS[t], 0}};
q = Eigenvectors[m];
p1 = Normalize[q[[1]]] /. Abs -> RealAbs // Simplify
p2 = Normalize[q[[2]]] /. Abs -> RealAbs // Simplify
p3 = Normalize[q[[3]]] /. Abs -> RealAbs // Simplify
{-(WS[t]/(Sqrt[1 + Abs[WS[t]/WP[t]]^2] WP[t])), 0,
...
2
I think you must have made a mistake when entering the expression in typeset form. When I pasted the LaTeX form
$\frac{x^3 - y^3}{x-y}$
After entering
Simplify[
and pasting the LaTeX form, I got a dialog pane asking me if wanted the front-end to translate from LaTeX to Mathematica typesetting. I said yes, and then adding the closing bracket and evaluating ...
1
The simplify work for Sqrt if we assumption that all the variables is positive real numbers.
Simplify[Sqrt[a b + x y z] Sqrt[-(1/(-a b - x y z))],
Assumptions -> AllTrue[{a, b, x, y, z}, Element[#, PositiveReals] &]]
1
Maybe the following is not exactly what you want, but may be useful for larger problems.
First, convert your function F to a boolean function G:
F[a_, b_] = (1 - a - b)^3 (1 - b);
# -> F @@ # & /@ Tuples[{0, 1}, 2]
(* {{0, 0} -> 1, {0, 1} -> 0, {1, 0} -> 0, {1, 1} -> 0} *)
G = # -> F @@ # & /@ Tuples[{0, 1}, 2] /. {0 -> ...
1
The earlier Q/A to which you linked provides a simpler form for F than what you proposed.
Clear["Global`*"]
binarySimplify[eq_, vars_] := Module[{rels, gb}, rels = (#^2 - # &) /@ vars;
gb = GroebnerBasis[Join[{eq /. Equal -> Subtract}, rels], vars];
Simplify@Thread[Complement[gb, rels] == 0]]
bs = binarySimplify[F == (1 - a - b)^3 (1 - ...
1
That's where I can bring it:
expr1 = 2^(1 - 2*n^2)*(3 - Sqrt[5])^(-1 - n)*(-1 +
Sqrt[5])^(n*(1 + n))*(1 + Sqrt[5])^n^2;
expr2 = expr1 /. (-1 + Sqrt[5])^(n (1 + n)) -> a*(-1 + Sqrt[5])^n^2 //
Simplify;
expr3 = expr2 /. a -> (-1 + Sqrt[5])^n;
expr4 = expr3 /. (3 - Sqrt[5])^(-1 - n) -> a*(3 - Sqrt[5])^-n //
Simplify[#, n > 0] &;
...
1
You can also use Internal`SyntacticNegativeQ with GeneralUtilities`SelectDiscard, GroupBy, Cases, DeleteCases, Select or Pick as follows:
expr = x1^d1*x2^d2*x3^d3
x1^(4 d) x2^(-5 e) x3^(-6 d)
exponents = Exponent[expr, {x1, x2, x3}]
{4 d, -5 e, -6 d}
{neg, pos} = GeneralUtilities`SelectDiscard[Internal`SyntacticNegativeQ] @ exponents
{neg, pos} = ...
1
If you write aϕ, then this is interpreted as one single variable. You need to put a space between a and ϕ. With:
DB = σ*σ*ϕ*ϕ
BB = Sqrt[DB] // FullSimplify // PowerExpand
g = 1/DB
αI = a ϕ
α = αI - 0.5*BB*D[BB, ϕ]
αs = α - ((1/2)*BB*D[BB, ϕ])
UG = 0.375 σ^2
Phi = UG - ((1/2)*(g*αs*αs)) // FullSimplify // PowerExpand
you then get:
a - (0.5 a^2)/σ^2 - 0....
1
With DSolve you get y[t] as an InverseFunction , diffucult to understand. Invert this function again with Solve to get t[y] in simple form you can plot with ParametricPlot .
f = q2[t] - q3[t] // Together // Numerator
dsol1 = DSolve[f == 0, y, t,
GeneratedParameters -> (ToExpression[
StringJoin["c", ToString[#]]] &)]
(* {{y -> ...
1
In light of the recent edit to the question, consider a slightly more difficult case of two Powers:
test2 = (z2p - z2p z3)^-ϵ + (z2p1^2 - z2p1^2 z31^3)^-ϵ1
Define the function,
m[z1_, z2_] := Module[{fac}, fac = List @@ Factor[z1];
If[fac[[1]] < 0, fac[[1 ;; 2]] = -fac[[1 ;; 2]]]; Times @@ (fac^z2)]
Then
test2 /. z1_^z2_ :> m[z1, z2]
(* (-z2p)^-...
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