Re: Circular inverse function of the hyperbola

From: N:dlzc D:aol T:com \(dlzc\) (net_at_nospam.com)
Date: 03/27/05 

Date: Sun, 27 Mar 2005 08:55:36 -0700

Dear Tom Capizzi:

"Tom Capizzi" <etianshrdlu@verizon.net> wrote in message
news:Nwz1e.19352$jt6.17502@trndny07...
>
> "N:dlzc D:aol T:com (dlzc)" <N: dlzc1 D:cox T:net@nospam.com>
> wrote in message news:q4i1e.6263$Mt5.4379@fed1read01...
>> Dear DavidBowman:
>>
>> "DavidBowman" <dt041054@yahoo.com> wrote in message
>> news:1111861567.202679.49900@l41g2000cwc.googlegroups.com...
>>>> simply resolve it into sin(x)/cos(x), and realize that
>>>> cos goes from very slightly positive, through zero, to
>>>> very slightly negative.
>>>
>>> yeah, but what i'm trying to figure out is whether + oo
>>> and -oo can be in the same place. and i think the
>>> TAN fn shows us that they can be.
>>
>> The COS fn shows that they are. But only for that one
>> relation.
>>
>>> dave, aren't you the one who first told me that the
>>> hyperbola is connected through hyperpace? is so,
>>> why do you think so?
>>
>> The hyperbola approaches a pair of non-vertical asymptotes.
>> You may have "heard" me say that, but I might have been
>> playing with the prefix "hyper-". I do not recall having said
>> this exact thing.
>>
>> If you include both hyperbolae (y^2 = 1 - x^2, both + and -
>
> Just a second. What you have written is the equation of a
> circle, not a
> hyperbola - it is, after all, equivalent to x^2 + y^2 = 1, the
> unit circle.

Doh!

y^2 = 1 + x^2

>> solutions), you could imagine leaving on the outgoing +y & +x
>> asymptote and arriving on the -y & -x asymptote. You could
>> even do a 45 deg rotation of coordinates, to make this more
>> clear.
>>
>>> I've had two threads going to find out if that's real.
>>
>> Oscillating (periodic) solutions are never entirely real.

Thanks Tom!

David A. Smith

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