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?
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2 Answers
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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] 
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1 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}] 
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}] 
And after simplify, we get
Plot[1/2 (-9 - 2 m), {m, -3, 15}] 😂，such a simple expression.
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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– cvgmt2021-04-12 08:40:52 +00:00Commented Apr 12, 2021 at 8:40