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Re: x-ArcSin[Sin[x]]
*To*: mathgroup at smc.vnet.net
*Subject*: [mg50064] Re: x-ArcSin[Sin[x]]
*From*: Paul Muller <paul.muller-at-epfl.ch at sicinfo2.epfl.ch>
*Date*: Fri, 13 Aug 2004 05:56:38 -0400 (EDT)
*References*: <200408110953.FAA04108@smc.vnet.net> <cfffgn$ilf$1@smc.vnet.net>
*Sender*: owner-wri-mathgroup at wolfram.com
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
>
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