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Improve the accuracy of calculations of a quadratic cosine?

Posted 1 year ago
9 Replies
7 Total Likes

There is a formula


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.

9 Replies

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]

enter image description here

(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 work

I do not understand this line is vertical and there is no gap or there is a slope and there is a gap?

The default symbolic analysis fails to detect the discontinuity. Compare

 {x, Pi/2 - 0.0000001, Pi/2 + 0.0000001},
 WorkingPrecision -> 16,
 Exclusions -> ArcCos[Abs[Sin[Abs[x]]]] == 0]


 {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.


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 + .000000000000000000000000000000001`50

to get


(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?


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.




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.



Posted 1 year ago

red line is True or false?

red vertical line is True or false?



in fact, here the horizontal line is the same vertical line in the previous graph How to calculate the length of this horizontal section? This can be very important

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

$$\frac{\sin ^{-1}(\cos (x))}{\cos ^{-1}(\left| \sin (\left| x\right| )\right| )}=\theta (x)+2 \sum _{k=1}^{\infty } (-1)^k \theta \left(\frac{\pi }{2}-k \pi +x\right)=-1+(-1)^{\left\lfloor \frac{1}{2}+\frac{x}{\pi }\right\rfloor }+\theta (x)$$


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