Re: How to prove the following for Φ^n
- From: Ray Vickson <RGVickson@xxxxxxx>
- Date: Sat, 28 Mar 2009 09:02:44 -0700 (PDT)
On Mar 28, 8:17 am, Alpha <vcpan...@xxxxxxxxx> wrote:
On Mar 28, 7:51 pm, achille <achille_...@xxxxxxxxxxxx> wrote:
On Mar 28, 10:18 pm, Alpha <vcpan...@xxxxxxxxx> wrote:
How to prove the following:
---------------------------------------
Lim [n→∞] Φ^n = Decimal part converges to 0 i.e. we have +ve INTEGER
Infinity
Nope, the decimal part of Φ^n does not converges to 0,
only the sub-sequence for odd n converges to 0,
the sub-sequence for even n converges to 1.
Here goes my Idea which may or may not be useful:
For an even power : Φ^2k + (1/Φ)^2k = a Whole Number
For an odd power : Φ^(2k+1) - (1/Φ)^(2k+1) = a Whole Number
Hint: To see how to use the two ideas above.
Compute the first few Φ^n and (1/Φ)^n. Can you
observe the pattern between decimal part of Φ^n
and (1/Φ)^n?
achille:
Thanks for your reply,
Yes I know that for odd n , decimal converges to 0 & for even it
converges to 1 but both of them will mean that result → +ve INTEGER
Infinity, don't you agree?
And Sorry I could not get your hint :( BTW this is not a homework
question, it's just something which I found interesting which I am not
able to prove.
Your original message did not say what phi was. I can't type the greek
letter phi on my keyboard, so I will use p instead. If p = 1, p^n = 1
for all integer n >= 1, so p^n does NOT approach + infinity. If 0 < p
< 1, p^n --> 0 as n --> infinity. If p > 1, p^n --> infinity as n -->
infinity, and since the fractional part of p^n stays between 0 and 1,
this automatically implies that the integer part of p^n also goes to
infinity.
R.G. Vickson
.
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