# Tag Info

4

Here's another way, using TagSetDelayed: Irrationals /: Element[x_, Irrationals] := Element[x, Reals] && ! Element[x, Rationals]; Element[Sqrt[2], Irrationals] (* True *) This should work for elementary calculation. However, I don't think there is way to fully incorporate Irrationals as a domain into Mathematica (e.g., into Reduce[expr, ...

4

This code should give you some insight as to why you are seeing this behavior: Manipulate[Plot[(a - Sqrt[a^2 + x])/(a^2 - a*Sqrt[a^2 - x]), {x, -1, 1}], {a, -3, 3}] When Assumptions -> {a > 0} is used, you get the correct limit. But when no assumptions are placed, Mathematica tries to evaluate the limit for a general complex $a$. This second ...

4

The original question is not clear. There is a wide range of functionality to work with polynomials symbolically e.g. PolynomialReduce, SymmetricReduction, etc. however it is not clear why one just assumes that expr1 and expr2 are as given above and if they are unique representants of possible reductions in terms of third order polynomials of two variables, ...

3

The issue we encounter here is an apparent incompleteness of the recent updates in the system, we should remember that Solve has been updated in the recent versions of Mathematica and although documentation pages say "last modified in 8", one can distinguish various different issues between ver.8 and ver.9, it's just a state of art. In ver. 8 we get: ...

3

I suspect you want the principal value, since the integral is divergent. Integrate[ z Exp[I z r]/Sqrt[z^2 + m^2], {z, -∞, ∞}, PrincipalValue -> True, Assumptions -> m > 0 && r ∈ Reals] (* 2 I m BesselK[1, m Abs[r]] Sign[r] *) If r > 0, then it agrees with your expected answer. Another derivation, although it seems more ...

3

The usual way is to make a Groebner basis from the polynomials that define the relationships you intend to apply ("reduce with", if you will), and then do that with PolynomialReduce. poly = x^2 + 2 x^3 + 3 x^2 y; polys = {poly - e1, (poly /. {x -> y, y -> x}) - e2}; gb = GroebnerBasis[polys, {x, y}]; Now for the example in question. expr = x^2 + ...

2

Using knowledge from CoefficientList docs (Properties and Relations section) of how to recreate multivariate polynomial from coefficient list: coe = CoefficientList[expr, {x, y}][[ All , ;; 2]]; expr1 = Fold[FromDigits[Reverse[#1], #2] &, coe, {x, y}] // Expand expr2 = expr1 /. {x -> y, y -> x} x^2 + 2 x^3 + 3 x^2 y y^2 + 3 x y^2 + 2 y^3

1

One way to do that is : expr = x^2 + y^2 + 2*x^3 + 2*y^3 + 3 x^2 y + 3 y^2 x; expr1 = x^2 + 2 x^3 + 3 x^2 y; expr2 = expr1 /. {x -> y, y -> x} expr1 + expr2 == expr Another way is to define functions : expr1[x_,y_]:=x^2 + 2 x^3 + 3 x^2 y; expr2[x_,y_]:=expr1[y,x];

1

After your comment showed that I misunderstood your problem, I cannot reproduce the issue you have. Let me copy and evaluate your complete example and show you that I get the correct behavior: f[v_]:=Sqrt[(1-2 v+8 v^2+Sqrt[-3+12 v+4 v^2-32 v^3+64 v^4])/(-1+2 v)]/Sqrt[2] g[v_]:=3/8+1/8 Sqrt[25+16 v]+Sqrt[5+8 v-25/Sqrt[25+16 v]-(16 v)/Sqrt[25+16 v]]/(4 ...

1

Define \[Beta] = 2; Solve[y*(((y*x)/(\[Beta]*b))^(1/(\[Beta] - 1)) - v) - c*\[Alpha] == 0 && ((x/\[Alpha]))*(((y*x)/(\[Alpha]*\[Beta]*b))^(1/(\[Beta] - 1)) - v) + (((y*x)/\[Alpha]) - 2*\[Alpha]*((yx)/(2*\[Beta]*b))^(1/(\[Beta] - 1)))*(1/(\[Beta] - 1))*(x/(\[Alpha]*\[Beta]* b))*((y*x)/(\[Alpha]*\[Beta]* b))^((2 - ...

1

It seems to me that this is probably a transcendental set of equations, with no algebraic solution. You've got x and y raised to a bunch of incompatible powers (functions of beta). There's little hope of ever isolating x and y. My advice is to defer to a numerical solution: With[ {v = 0.1, b = 0.2, alpha = 0.3, beta = 0.4, c = 0.5}, NSolve[{y (-v + ((x ...

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