# Improve the accuracy of calculations of a quadratic cosine?

Posted 1 year ago
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 There is a formula Plot[ArcSin[Cos[x]]/ArcCos[Abs[Sin[Abs[x]]]] Made calculations with increased accuracy Block[{$MinPrecision = 100000000,$MaxPrecision = 100000000}, Plot[ArcSin[Cos[x]]/ArcCos[Abs[Sin[Abs[x]]]], {x, N[Pi, 100000000]/2 - 0.0000001, N[Pi, 100000000]/2 + 0.0000001}]]  Can someone calculate with more accuracy? I do not understand whether there is a gap there or not.Sorry my English.
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Posted 1 year ago
 Is this what you're after?: Plot[ArcSin[Cos[x]]/ArcCos[Abs[Sin[Abs[x]]]], {x, Pi/2 - 0.0000001, Pi/2 + 0.0000001}, WorkingPrecision -> 16] (Note: N[Pi, 100000000]/2 - 0.0000001 is just a long way to compute Pi/2 - 0.0000001, since Mathematica converts N[Pi, 100000000]/2 to machine precision when -0.0000001 is added to it.)
Posted 1 year ago
 Plot[ArcSin[Cos[x]]/ArcCos[Abs[Sin[Abs[x]]]], {x, Pi/2 - 0.0000001, Pi/2 + 0.0000001}, WorkingPrecision -> 16] this does not workI do not understand this line is vertical and there is no gap or there is a slope and there is a gap?
Posted 1 year ago
 The default symbolic analysis fails to detect the discontinuity. Compare Plot[ArcSin[Cos[x]]/ArcCos[Abs[Sin[Abs[x]]]], {x, Pi/2 - 0.0000001, Pi/2 + 0.0000001}, WorkingPrecision -> 16, Exclusions -> ArcCos[Abs[Sin[Abs[x]]]] == 0] with Plot[ArcSin[Cos[x]]/ArcCos[Abs[Sin[Abs[x]]]], {x, Pi/2 - 0.0000001, Pi/2 + 0.0000001}, WorkingPrecision -> 16, Exclusions -> Flatten@Values@ Solve[ArcCos[Abs[Sin[Abs[x]]]] == 0 && Pi/2 - 1*^-7 < x < Pi/2 + 1*^-7]] There should be a gap.
Posted 1 year ago
 Aleksey,There is a discontinuity there: In[20]:= Limit[ArcSin[Cos[x]]/ArcCos[Abs[Sin[Abs[x]]]], x -> Pi/2, Direction -> "FromAbove"] Out[20]= -1 In[21]:= Limit[ArcSin[Cos[x]]/ArcCos[Abs[Sin[Abs[x]]]], x -> Pi/2, Direction -> "FromBelow"] Out[21]= 1 You can evaluate the expression with arbitrary accuracy by specifying the accuracy. Michael is correct that as soon as you add the 0.0000001 to Pi/2 you use machine precision. The syntax is this: ArcSin[Cos[x]]/ArcCos[Abs[Sin[Abs[x]]]] /. x -> Pi/2 + .00000000000000000000000000000000150 to get -1.0000000000000000000000000000000000000000000000000 (Note: in this case I specified 50 digits of precision. You can go out to whatever you want.)
Posted 1 year ago
 just about the limits and forgot. And this proves that at the point Pi / 2 - the vertical line graphic and there is no gap or not? And how to prove it?
Posted 1 year ago
 Aleksey,I am not sure exactly what you are asking. The vertical line should not really be on the plot. It is a discontinuity (gap). The function approaches -1 from the positive side and 1 from the negative side. The limit is your proof of a discontinuity -- if the limit is different from each side, the function has a gap or discontinuity at that point. Plot connects all points so it will show the discontinuity as a vertical line.If I am not understanding your question and if your native language is Russian, please post it in Russian and my daughter will translate it for me.Regards,Neil
Posted 1 year ago
 Aleksey,I was able to understand the Russian post (before it was removed -- sorry - -I forgot about the rules). Many of the issues you raised are philosophical so I am not qualified to give you an opinion. I am also an engineer and not a pure mathematician. That being said, I think that the limit calculation proves that the function is not continuous and does not connect with a vertical line. The value at Pi/2 is undefined because it depends on the direction from which you approach Pi/2. Michael's plot with the Exclusions is the correct way to handle this plot as far as I know. I believe that you should exclude that point and the plot should not be connected with a line. Again, this is my opinion and you can ask a mathematician who would know more about the issues you raise. I hope this helps. Regards,Neil
 This is Not the answer for Yours question only another forumla for Quadratic cosine:  HoldForm[ArcSin[Cos[x]]/ArcCos[Abs[Sin[Abs[x]]]] == HeavisideTheta[x] + 2 Sum[(-1)^k*HeavisideTheta[\[Pi]/2 - k \[Pi] + x], {k, 1, Infinity}] == -1 + (-1)^Floor[1/2 + x/\[Pi]] + HeavisideTheta[x]] // TraditionalForm ` $$\frac{\sin ^{-1}(\cos (x))}{\cos ^{-1}(\left| \sin (\left| x\right| )\right| )}=\theta (x)+2 \sum _{k=1}^{\infty } (-1)^k \theta \left(\frac{\pi }{2}-k \pi +x\right)=-1+(-1)^{\left\lfloor \frac{1}{2}+\frac{x}{\pi }\right\rfloor }+\theta (x)$$Regards,Mariusz