0
$\begingroup$

I want to smoothen the following contour.

ContourPlot[
 10^b - (10^180/(10^m  10^23)^6)^(1/4)/Sqrt[ (((10^36)^(3/2))/(
   10^m  10^23 Sqrt[ 100]))] == 0, {m, -3, 15}, {b, -25, -2}, 
 ContourStyle -> {{Purple, Thickness[.01]}}]

Is there a way to do so?

$\endgroup$

2 Answers 2

3
$\begingroup$

Try the option MaxRecursion

ContourPlot[
 10^b - (10^180/(10^m 10^23)^6)^(1/4)/Sqrt[(((10^36)^(3/2))/(10^m 10^23Sqrt[100]))] == 0, 
 {m, -3, 15}, {b, -25, -2}, 
 ContourStyle -> {{Purple, Thickness[.01]}}, 
 MaxRecursion -> 5]

enter image description here

$\endgroup$
4
$\begingroup$
10^b - (10^180/(10^m 10^23)^6)^(1/4)/Sqrt[(10^36)^(3/2)/(
   10^m 10^23 Sqrt[100])] == 0 // FullSimplify

get

10^b == (10^(-6 (-7 + m)))^(1/4)/Sqrt[10^(30 - m)]

Solve m

m /. Solve[10^b == (10^(-6 (-7 + m)))^(1/4)/Sqrt[10^(30 - m)], {m}]

get

{(-9 Log[10] - 2 Log[10^b])/(2 Log[10])}

Plot it

Plot[(-9 Log[10] - 2 Log[10^b])/(2 Log[10]), {b, -25, -2}]

enter image description here

This is m=f[b] , change to b=f[m]

Plot[InverseFunction[(-9 Log[10] - 2 Log[10^#])/(2 Log[10]) &][
  m], {m, -3, 15}]

enter image description here


And after simplify, we get

Plot[1/2 (-9 - 2 m), {m, -3, 15}]

😂,such a simple expression.

$\endgroup$
1
  • 3
    $\begingroup$ Simplify[10^b - (10^180/(10^m 10^23)^6)^(1/4)/ Sqrt[(((10^36)^(3/2))/(10^m 10^23 Sqrt[100]))], Assumptions -> -25 <= b <= -2 && -3 <= m <= 15] $\endgroup$
    – cvgmt
    Commented Apr 12, 2021 at 8:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.