0
$\begingroup$

I am considering the function

f[T_, c1_, c2_] := \[Pi] (rh[T]^2 (L^2 - rh[T]^2))/(G (L^2 + 3 rh[T]^2)) (1 + (64 \[Pi] G (c2 - c1))/L^2) + 64 \[Pi]^2 c1

with $G=L=1$, and

rh[T_] := 1/3 (Sqrt[L^2 (4 \[Pi]^2 L^2 T^2 - 3)] + 2 \[Pi] L^2 T)

I have plotted this function as a function of $T$ ($T \geq 0$) for several values of the coefficients $c_1$ and $c_2$. For some of these coefficients (for example, $c_1=0.006, c_2=0.001$), the function $f$ is always positive. I wanted to know for what region of the parameters $c_1$ and $c_2$ the function $f[T,c_1,c_2]$ is always positive. For this, I have used to following function:

Reduce[ForAll[T, T >= 0, f[T, c1, c2] < 0], {c1, c2}, Reals]

This yields 'False' although I have plotted some values of the coefficients for which, clearly, the function $f$ never crosses the axis and is always positive. I have also tried using Resolve instead of Reduce, with the same outcome. What am I doing wrong, and what can I do to make this right?

Improve this question
$\endgroup$
6
  • 2
    $\begingroup$ The function f is neither positive nor negative. For small T, f[T, 6/1000, 1/1000] is complex, e.g. f[1/5, 6/1000, 1/1000] // N yields 3.78632 - 0.00166533 I and so Reduce is right. $\endgroup$ Commented Jun 8, 2022 at 13:29
  • $\begingroup$ @Artes You are right. Thank you very much for pointing this out, I know this is only real for $T>=\sqrt{3}/2\pi$, but somehow I didn't include it in my code... my bad! $\endgroup$ Commented Jun 8, 2022 at 13:34
  • 1
    $\begingroup$ For general T >= Tmin red = Reduce[ForAll[T, T >= Tmin, f[T, c1, c2] < 0], {c1, c2}, Reals] $\endgroup$ Commented Jun 8, 2022 at 14:01
  • $\begingroup$ @Akku14: This produces Tmin == Sqrt[3]/(2 \[Pi]) .... $\endgroup$ Commented Jun 8, 2022 at 14:14
  • 1
    $\begingroup$ @creidhne: It is not a good practice to essentially modify a question, not indicating the changes. $\endgroup$ Commented Oct 31, 2023 at 21:22

1 Answer 1

Reset to default
0
$\begingroup$

The answer to the corrected question ( Sqrt[3]/2/Pi instead of 0) is as follows.

L = 1; G = 1; f[T_, c1_, c2_] := \[Pi] (rh[T]^2 (L^2 - rh[T]^2))/(G (L^2 + 3 rh[T]^2)) (1 + (64 \[Pi] G (c2 - c1))/L^2) + 64 \[Pi]^2 c1; rh[T_] := 1/3 (Sqrt[L^2 (4 \[Pi]^2 L^2 T^2 - 3)] + 2 \[Pi] L^2 T); Resolve[ForAll[T, T >= Sqrt[3]/2/Pi, f[T, c1, c2] < 0], {c1, c2}, Reals]

c1 < 0 && (-1 + 64 c1 \[Pi])/(64 \[Pi]) <= c2 < (-1 - 512 c1 \[Pi])/( 64 \[Pi])

Improve this answer
$\endgroup$

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.