0

This is my code

Manipulate[
 Reduce[{a/b^2 + 1/a == (2 b)/a^2 + 2/3 \[Beta] b && (3 v)/(
      4 \[Pi] a^2) < b < (3 v)/(2 \[Pi] a^2)}, a, Reals] // N,
 {\[Beta], 50, 200}, {v, 1*10^-9, 10*10^-9}
 ]

And this is the outcome i got for a certain set of beta and v

0.00110566 < b < 0.00139303 && 
 a == Root[-30. b^3 + 15. b^2 #1 - 1214. b^3 #1^2 + 15. #1^3 &, 1]

What does the # mean in this outcome?Also what I would like is something like for example;

a = Sqrt[b^2 + 30 b - 20]

So the program automaticall puts an expression of b in the place of a in another inequality for example;

3/(2 \[Pi]a^2) v < c < b

So I can find the range of c too just by selecting the values of beta and v. So what I ultimately want is the outcome

sth < b < sth
sth < c < sth

just by using one manipulate

Jun
  • 93
  • 5
  • Do read the linked thread. Briefly: read a as "the 1-st Root[] of the polynomial -30. b^3 + 15. b^2 x - 1214. b^3 x^2 + 15. x^3 in x". – J. M.'s missing motivation Jan 08 '17 at 14:28
  • Thank you now I get what # means. But I still don't know the next part of my question. How can I use that outcome to get the range of c? – Jun Jan 08 '17 at 15:02
  • I don't understand what you mean. If specific values of b are plugged into that Root[] expression, you get a number that is usable for subsequent bounds. – J. M.'s missing motivation Jan 08 '17 at 15:11
  • hmm okay so what I want mathematica to give me is sth like this. If 0 < x <2, x< y <2x I want 0 < y < 4. So I guess what I want is the range of y for all the possible x connected by or – Jun Jan 08 '17 at 16:00

0 Answers0