Re: i^7 or up on TI83???
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Re: i^7 or up on TI83???
well, i have a simple method to determine it.
since
i^0=1
i^1=i
i^2=-1
i^3=-i
and after that is just cyclic (e.g. i^4=1 i^5=i.....)
so, you can just have:
for i^n
i^(mod(n,4))
it will always bring n down to 0<=n<=3, so it will work well.
kaus
----- Original Message -----
From: Kirk Lane <snake64@geocities.com>
Newsgroups: bit.listserv.calc-ti
Sent: Friday, March 12, 1999 6:21 PM
Subject: i^7 or up on TI83???
>All right. My calc does i to any power from 1 to 6 perfectly fine (i^2
>is -1, i^3 is -i, i^4 is 1, i^5 is i, and i^6 is -1) BUT....i^7
>= -3E-13-i!!! It should be -i! i^101 spits out 4.9E-12+i, but ((i^4)^25)*i
=
>i, like it should! What is wrong, and is there any program that can do
this
>(take the number, divide it down so that it can be i^2, 3, 4, 5, or 6 and
>that to whatever power to make it i^x...)
>
>Thanks
>
>--
>Kirk Lane
>snake64@geocities.com
>ICQ: 28171652
>
>
>
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