# Tag Info

4

Sort @ Select[Exponent[# /. y -> x, x] < 5 &] @ MonomialList[ser] TeXForm @ % $\left\{1,-\frac{x}{2},-\frac{x^2}{8},-\frac{x^3}{16},\frac{y}{2},\frac{3 x y}{4},\frac{23 x^2 y}{16},\frac{27 x^3 y}{32},-\frac{y^2}{8},-\frac{31 x y^2}{16},-\frac{127}{64} x^2 y^2,\frac{y^3}{16},\frac{35 x y^3}{32}\right\}$

2

You could also use CoefficientArrays v = {x, y}; c = CoefficientArrays[ser, v] result = {c[[1]], c[[2]].v, c[[3]].v.v, c[[4]].v.v.v, c[[5]].v.v.v.v} // Expand $$\left\{1,\frac{y}{2}-\frac{x}{2},-\frac{x^2}{8}+\frac{3 x y}{4}-\frac{y^2}{8},\frac{23 x^2 y}{16}-\frac{x^3}{16}-\frac{31 x y^2}{16}+\frac{y^3}{16},-\frac{127 x^2 y^2}{64}+\frac{27 x^3 y}{32}+\... 2 You can adapt Yode's method from his answer to: Removing terms of certain degree in multivariable polynomial var={x,y}; Sort@Select[MonomialList[ser], Tr[Exponent[#, var]] < 5 &]$$\left\{1,-\frac{x}{2},-\frac{x^2}{8},-\frac{x^3}{16},\frac{y}{2},\frac{3 x y}{4},\frac{23 x^2 y}{16},\frac{27 x^3 y}{32},-\frac{y^2}{8},-\frac{31 x y^2}{16},-\frac{...

2

You can use CoefficientList and select necessary items in the matrix: ser = 1 - x/2 - x^2/8 - x^3/16 + y/2 + (3 x y)/4 + (23 x^2 y)/16 + (27 x^3 y)/32 - y^2/8 - (31 x y^2)/16 - (127 x^2 y^2)/64 - (351 x^3 y^2)/128 + y^3/16 + (35 x y^3)/32 + (407 x^2 y^3)/128 + (1915 x^3 y^3)/256; PolynomialQ[ser, x] (* True *) coef = CoefficientList[ser, {x,...

5

You can use Jens's method from this answer: expr[x_, y_] := 1 - x/2 - x^2/8 - x^3/16 + y/2 + (3 x y)/4 + (23 x^2 y)/16 + (27 x^3 y)/32 - y^2/8 - (31 x y^2)/16 - (127 x^2 y^2)/64 - (351 x^3 y^2)/128 + y^3/16 + (35 x y^3)/32 + (407 x^2 y^3)/128 + (1915 x^3 y^3)/256 expr2[x_, y_] = Normal[Series[expr[x t, y t], {t, 0, 4}]] /. t ...

0

One possible approach is to write out a list of possible invariants that could be involved in the result, e.g. v2 = v1.v1 w2 = w1.w1 vw = v1.w1 ctr = Tr[cM] c2tr = Tr[cM.cM] vcv = v1.cM.v1 {...} As far as I can tell, your expression only contains terms that are quadratic & quartic in the components of $\bf{v}$, $\bf{w}$, and $\bf{c}$, so there will be ...

2

An expression like: Root[1 + 2*#1 + #1^5 & , 1, 0] is simply an exact representation of an algebraic number. Radicals like Sqrt[2] are more familiar, but cannot express every algebraic number. You can treat such things pretty much as any other exact representation of a number in Mathematica. You can get numeric approximations using N[]. However, ...

0

The following works correctly. f = Function[x, 15 + 2 x - 0.100000000000000000000* x^2]; DifferenceDelta[f[i], i] // FullSimplify (*1.900000000000 - 0.2000000000000 i *) The same phenomenon appears for other commands, e.g. Residue. PS. To be clear, g = Function[x, 15 + 2 x - 0.123456789101121314* x^2]; DifferenceDelta[g[i], i] // FullSimplify (*1....

6

This answer is for free! ;-) (Sorry, didn't understand the purpose of your comment "By the way, I paid nearly \$2000 for Mathematica.") Try NMinimize: NMinimize[{Grad[Func, Variables[Func]].Grad[Func, Variables[Func]],a^2 <= 1 && 1 + 2 a b f >= a^2 + b^2 + f^2}, Variables[Func]] (*{4.20993*10^-18, {a -> -0.754063, b -> -0.0172513, c ...

4

How about InternalRationalFunctionQ[a + 1/a - k, a] (*True*)

0

I suspect that the problem is undecidable, except for particulars classes of expressions, e. g., rational expressions.

3

You essentially want to test whether a given expression is rational in some given variable. The general case is probably going to be subtle, but the naive approach works for your particular example: And @@ { PolynomialQ[Numerator[#], a], PolynomialQ[Denominator[#], a] } &@ Together[a + 1/a - k] If you want the expanded form, use ...

3

I do not recall offhand how these are computed in general. For the case of distinct roots I can illustrate one method. I'll take the example in that Wikipedia article. We have polynomials as below. polys = {x^2-1, (x-1)*(y-1), y^2-1}; We take as separating element t = (x-y)/2. seppoly = t-(x-y)/2; First compute the polynomial referred to as h(t). Then ...

2

This is doable in a much more compact way than the previous answer: Collect[x^3 + C[2] x^2 + C[1] x + C[0] /. SolveAlways[{x^3 + C[2] x^2 + C[1] x + C[0] == (x - a) (x - b) (x - c), a^2 == b, b^2 == c, c^2 == a}, x], x, FullSimplify] // Union {x^3, -1 + 3 x - 3 x^2 + x^3, -1 - 1/2 I (-I + Sqrt[7]) x + 1/2 (1 - I Sqrt[7]) ...

2

The polynomial $$p(x,y,z) = 15 (x+y+2)^2 (x+z+2)^2 (y+z+2)^2-32 (x+y+z+3)^3$$ has a Newton polytope hull which is NP = {{4, 2, 0}, {4, 0, 2}, {4, 0, 0}, {0, 4, 2}, {0, 4, 0}, {0, 2, 4}, {0, 0, 4}, {0, 0, 0}, {2, 4, 0}, {2, 0, 4}} has all the powers even. Follows a picture for the Newton polytope. In black, the hull. The monomials that can generate ...

1

Root cann't handel the additional parameter t! Try g[z_, t_] := 1 + t z + z^3 root[t_] := Root[Function[z, g[z, t]], 1] Plot[Evaluate[Residue[1/g[z, t], {z, root[t] }]], {t, 0, 10} ]

3

You need to indicate the variable and the root number. By way of an example: Residue[1/(z^3 + p z + q), {z, Root[#^3 + p # + q &, 1]}] gives 1/(p + 3 Root[q + p #1 + #1^3 &, 1]^2) and then Plot3D[1/(p + 3 Root[q + p #1 + #1^3 &, 1]^2), {p, -3, 3}, {q, -3, 3}, PlotPoints -> 75, MaxRecursion -> 5, MeshFunctions -> {#3 &}] ...

5

Try IrreduciblePolynomialQ[x^5+2x+1] which returns True and Factor[x^5+2x+1] returns the original polynomial while IrreduciblePolynomialQ[x^5+x+1] returns False and Factor[x^5+x+1] returns (1+x+x^2)(1-x^2+x^3) The documentation for IrreduciblePolynomialQ https://reference.wolfram.com/language/ref/IrreduciblePolynomialQ.html shows how you provide ...

1

Use Plot[Re[Root[H[1, y, #] &, 1]], {y, 0, 10}] or Plot[Re[Root[Function[x, H[1, y, x]], 1]], {y, 0, 10}] to get Note: The first argument of Root should be a pure function.

3

This site is about Mathematica and the Wolfram language, not WolframAlpha. eqns = {a^2 == c, b^2 == a, c^2 == b}; The solutions are solns = Solve[eqns, {a, b, c}] Verifying the solutions (And @@ eqns) /. solns (* {True, True, True, True, True, True, True, True} *) The polynomials (with duplicates removed) (polys = Times @@ (x - {a, b, c}) /. solns //...

3

You can use Chop: expr = expr = {(-0.12952142851754841 + 6.938893903907228*^-18 I) - (17.525908000000005 - 4.975740737167372*^-17 I) k1^2, (-2.914139090711302 - 1.3833455871105996*^-17 I) - (3.6056850228887516*^-15 + 1.7532479660249581`*^-15 I) k1^2}; Chop[expr] {-0.129521 - 17.5259 k1^2, -2.91414}

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