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

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], ...


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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