Using the output of Reduce

Question

β = 100;
v0 = 5*10^-9;

 Reduce[{a/b^2 + 1/a == (2 b)/a^2 + 2/3 β b && (3 v0)/(
      4 π a^2) < b < (3 v0)/(2 π a^2)}, a, Reals] // N

This gives me the output

0.00106077 < b < 0.00133648 && 
 a == Root[-6 b^3 + 3 b^2 #1 - 200 b^3 #1^2 + 3 #1^3 &, 1]

I would like to put the minimum and miximum values of b into bmin and bmax and set a function a(b) i.e.

bmin=0.00106077 
bmax=0.00133648 
a[b_]:=Root[-6 b^3 + 3 b^2 #1 - 200 b^3 #1^2 + 3 #1^3 &, 1]

The thing is I don't want to copy and paste like this because next, I want to be using manipulate to change the values of beta and v0 and I want bmin, bmax, a[b_] to change accordingly.

Answer 1

You might approach the problem something like this.

There exists a function ToRules that converts some Reduce output but it doesn't know what to do with inequalities. Disregarding open vs closed intervals we can create a replacement rule to transform these inequalities into Interval expressions:

ineqToIntv = HoldPattern[
  Inequality[m_, Less | LessEqual, s_Symbol, Less | LessEqual, M_] | 
  Inequality[M_, Greater | GreaterEqual, s_Symbol, Greater | GreaterEqual, m_]] :>
    (s == Interval[{m, M}]);

Now:

β = 100;
v0 = 5*10^-9;

sol = Reduce[{a/b^2 + 1/a == (2 b)/a^2 + 2/3 β b && (3 v0)/(4 π a^2) < 
      b < (3 v0)/(2 π a^2)}, a, Reals] // N;

rules = sol /. ineqToIntv // ToRules

{b -> Interval[{0.00106077, 0.00133648}], 
 a -> Root[-6 b^3 + 3 b^2 #1 - 200 b^3 #1^2 + 3 #1^3 &, 1]}

I recommend against defining a since it can conflict with the lines above, so I shall use afn instead:

afn[b_] = a /. rules
{bmin, bmax} = First[b /. rules];

afn[bmin]
afn[bmax]

0.00106079

0.00133652

In this case Interval might be seen as an inconvenience as it simply gets stripped with First, but in other cases it may be useful as arithmetic will operate over the expression, e.g.

b^-1 /. rules

Interval[{748.235, 942.71}]

Answer 2

This will get the desired expressions:

β = 100;
v0 = 5*10^-9;

red = Reduce[{a/b^2 + 1/a == (2 b)/a^2 + 
       2/3 β b && (3 v0)/(4 π a^2) < b < (3 v0)/(2 π a^2)}, a, Reals];

N[red] /. {HoldPattern[And[_[b1_, ___, b2_], a == a0_]] :> {a0, {b1, b2}}}
(*  {Root[-6 b^3 + 3 b^2 #1 - 200 b^3 #1^2 + 3 #1^3 &, 1], {0.00106077, 0.00133648}}  *)

As for defining the function a, I would rather not have a be both a variable and a function. To evaluate a at b == b0, use a /. b -> b0. Or define

a[b0_] := a /. b -> b0