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MathGroup Archive 2004

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Re: Re: x-ArcSin[Sin[x]]

  • To: mathgroup at smc.vnet.net
  • Subject: [mg50091] Re: [mg50064] Re: x-ArcSin[Sin[x]]
  • From: DrBob <drbob at bigfoot.com>
  • Date: Sat, 14 Aug 2004 01:51:04 -0400 (EDT)
  • References: <200408110953.FAA04108@smc.vnet.net> <cfffgn$ilf$1@smc.vnet.net> <200408130956.FAA03696@smc.vnet.net>
  • Reply-to: drbob at bigfoot.com
  • Sender: owner-wri-mathgroup at wolfram.com

One typical situation is that you know Sin[theta] AND Cos[theta]. In that case, ArcTan[Cos[theta],Sin[theta]] is more likely to come out in the proper quadrant. I don't know if that will help in your problem.

Bobby

On Fri, 13 Aug 2004 05:56:38 -0400 (EDT), Paul Muller <paul.muller-at-epfl.ch at sicinfo2.epfl.ch> wrote:

> Thanks for your answer. Just to make clear: no book stated that ArcSin[Sin[x]]=x. The
> books give the magnetic field in the center of the spire, and that's where my integral
> gives a wrong value. Now this is probably due to my too wide interpretation of the
> ArcSin at the place it gets into the calculation, messing up the final result.
>
> Unfortunately, at some point in my calculation, I have sin(theta)= something and I need
> to extract theta from this. It seemed obvious to me to use ArcSin, but according to the
> whole discussion, this is what I'm not supposed to assume...
>
> Thanks again
>
> Paul
>
> DrBob wrote:
>>
>> There is not, and cannot be, an implementation of ArcSin[Sin[x]]. Sin and ArcSin are separate functions, separately implemented.
>>
>> Moreover, Sin is periodic on the real line, and hence it cannot have a left inverse for all x. Therefore, as explained at the link, ArcSin[Sin[x]]==x is false in general.
>>
>> If you want it to be true for x in a certain interval, you may be able to define an appropriate branch of ArcSin FOR THAT INTERVAL. That is, it wouldn't be a function of x alone, but also a function of the interval.
>>
>> As x varies from 0 to 2Pi, the quantity Pi/2-x sweeps out the following interval:
>>
>> Pi/2 - Interval[{0, 2*Pi}]
>> Interval[{-((3*Pi)/2), Pi/2}]
>>
>> Max[%] - Min[%]
>> 2*Pi
>>
>> Now, on any interval of that length, Sin takes on all values between -1 and 1 TWICE. So again, on an interval like that, Sin has no left inverse.
>>
>> In summary: the value you give for that integral is wrong, no matter how many books agree with you.
>>
>> Bobby
>>
>
>
>



-- 
DrBob at bigfoot.com
www.eclecticdreams.net


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