# 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[], c[].v, c[].v.v, c[].v.v.v, c[].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 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 x^2 + C x + C /. SolveAlways[{x^3 + C x^2 + C x + C == (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) x + 1/2 (1 - I Sqrt) ...

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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