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interests / rec.puzzles / Re: A = (5 + 2 (sqrt 5)) and i do : A^2, A^3, A^4, . . .

SubjectAuthor
* Re: A = (5 + 2 (sqrt 5)) and i do : A^2, A^3, A^4, . . .Ilan Mayer
`* Re: A = (5 + 2 (sqrt 5)) and i do : A^2, A^3, A^4, . . .henh...@gmail.com
 +- Re: A = (5 + 2 (sqrt 5)) and i do : A^2, A^3, A^4, . . .Ilan Mayer
 `- Re: A = (5 + 2 (sqrt 5)) and i do : A^2, A^3, A^4, . . .Phil Carmody

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Re: A = (5 + 2 (sqrt 5)) and i do : A^2, A^3, A^4, . . .

<833c0065-59b1-4a2f-8d4f-127805a3f072n@googlegroups.com>

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Subject: Re: A = (5 + 2 (sqrt 5)) and i do : A^2, A^3, A^4, . . .
From: ilan_no_...@hotmail.com (Ilan Mayer)
Injection-Date: Wed, 22 Jun 2022 03:43:35 +0000
Content-Type: text/plain; charset="UTF-8"
 by: Ilan Mayer - Wed, 22 Jun 2022 03:43 UTC

On Sunday, June 19, 2022 at 9:33:32 PM UTC-4, henh...@gmail.com wrote:
> where
> A = 9.47213595499958...
> or
> A = (5 + 2 (sqrt 5))
>
> and i do : A^2, A^3, A^4, . . .
>
> by A^8 (or so), the accumulation of 9's is obvious....
> (as ........49 . 999.......... )
>
>
> -- is there a simple explanation for why this happens?
>
> -- Does this happen with (sqrt 2) ? (sqrt 3) ?
>
> -- Do we ever get similar accumulations of 2's, 3's, ... ?
>
>
> ( i wish i knew this in high school ! )

Explanation for the 9s:

A^1 = 5+2*sqrt(5)
A^2 = 45+20*sqrt(5)
A^3 = 425+190*sqrt(5)
A^4 = 4025+1800*sqrt(5)
A^5 = 38125+17050*sqrt(5)
A^6 = 361125+161500*sqrt(5)
A^7 = 3420625+1529750*sqrt(5)
A^8 = 32400625+14490000*sqrt(5)
....
A^n = a_n+b_n*sqrt(5)
a_n = ((5-2*sqrt(5))^n+(5+2*sqrt(5))^n)/2
b_n = (1/10)*sqrt(5)*((5+2*sqrt(5))^n-(5-2*sqrt(5))^n)
As n increases, a_n/b_n approaches sqrt(5) from above
Therefore b_n*sqrt(5) approaches a_n from below, and hence more and more 9s after the decimal point

Please reply to ilanlmayer at gmail dot com

__/\__
\ /
__/\\ //\__ Ilan Mayer
\ /
/__ __\ Toronto, Canada
/__ __\
||

Re: A = (5 + 2 (sqrt 5)) and i do : A^2, A^3, A^4, . . .

<e50bd759-b0bd-4a86-883f-0efbfdf90f40n@googlegroups.com>

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Subject: Re: A = (5 + 2 (sqrt 5)) and i do : A^2, A^3, A^4, . . .
From: henha...@gmail.com (henh...@gmail.com)
Injection-Date: Wed, 22 Jun 2022 19:34:32 +0000
Content-Type: text/plain; charset="UTF-8"
 by: henh...@gmail.com - Wed, 22 Jun 2022 19:34 UTC

On Tuesday, June 21, 2022 at 8:43:37 PM UTC-7, Ilan Mayer wrote:
> On Sunday, June 19, 2022 at 9:33:32 PM UTC-4, henh...@gmail.com wrote:
> > where
> > A = 9.47213595499958...
> > or
> > A = (5 + 2 (sqrt 5))
> >
> > and i do : A^2, A^3, A^4, . . .
> >
> > by A^8 (or so), the accumulation of 9's is obvious....
> > (as ........49 . 999.......... )
> >
> >
> > -- is there a simple explanation for why this happens?
> >
> > -- Does this happen with (sqrt 2) ? (sqrt 3) ?
> >
> > -- Do we ever get similar accumulations of 2's, 3's, ... ?
> >
> >
> > ( i wish i knew this in high school ! )
> Explanation for the 9s:
>
> A^1 = 5+2*sqrt(5)
> A^2 = 45+20*sqrt(5)
> A^3 = 425+190*sqrt(5)
> A^4 = 4025+1800*sqrt(5)
> A^5 = 38125+17050*sqrt(5)
> A^6 = 361125+161500*sqrt(5)
> A^7 = 3420625+1529750*sqrt(5)
> A^8 = 32400625+14490000*sqrt(5)
> ...
> A^n = a_n+b_n*sqrt(5)
> a_n = ((5-2*sqrt(5))^n+(5+2*sqrt(5))^n)/2
> b_n = (1/10)*sqrt(5)*((5+2*sqrt(5))^n-(5-2*sqrt(5))^n)
> As n increases, a_n/b_n approaches sqrt(5) from above
> Therefore b_n*sqrt(5) approaches a_n from below, and hence more and more 9s after the decimal point
>
> Please reply to ilanlmayer at gmail dot com
>
> __/\__
> \ /
> __/\\ //\__ Ilan Mayer
> \ /
> /__ __\ Toronto, Canada
> /__ __\
> ||

thank you for the help !


( i wish i knew this in high school ! )

i don't know why, but a similar pattern happens with (3 + 2 * sqrt(2)) ^ k

5.82842712474619.....................

33.9705627484771394478719195161

197.994949366116643469193236470189742320118659

1153.99913344822270261118589603047721268684116683595605815921

67259.998513232194628793508006917181208468837660012895581285831254498609099

392019.99974491093437509592477268670631674440337231921276927110823440631471667911839958281

2284859.9999562333744856481681683435615568663093769138768431377543192559229850045968629747114467231369939

Re: A = (5 + 2 (sqrt 5)) and i do : A^2, A^3, A^4, . . .

<b51adbc4-4b58-43fc-8c9c-a09cfc59f820n@googlegroups.com>

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Subject: Re: A = (5 + 2 (sqrt 5)) and i do : A^2, A^3, A^4, . . .
From: ilan_no_...@hotmail.com (Ilan Mayer)
Injection-Date: Thu, 23 Jun 2022 02:59:17 +0000
Content-Type: text/plain; charset="UTF-8"
 by: Ilan Mayer - Thu, 23 Jun 2022 02:59 UTC

On Wednesday, June 22, 2022 at 3:34:34 PM UTC-4, henh...@gmail.com wrote:
> On Tuesday, June 21, 2022 at 8:43:37 PM UTC-7, Ilan Mayer wrote:
> > On Sunday, June 19, 2022 at 9:33:32 PM UTC-4, henh...@gmail.com wrote:
> > > where
> > > A = 9.47213595499958...
> > > or
> > > A = (5 + 2 (sqrt 5))
> > >
> > > and i do : A^2, A^3, A^4, . . .
> > >
> > > by A^8 (or so), the accumulation of 9's is obvious....
> > > (as ........49 . 999.......... )
> > >
> > >
> > > -- is there a simple explanation for why this happens?
> > >
> > > -- Does this happen with (sqrt 2) ? (sqrt 3) ?
> > >
> > > -- Do we ever get similar accumulations of 2's, 3's, ... ?
> > >
> > >
> > > ( i wish i knew this in high school ! )
> > Explanation for the 9s:
> >
> > A^1 = 5+2*sqrt(5)
> > A^2 = 45+20*sqrt(5)
> > A^3 = 425+190*sqrt(5)
> > A^4 = 4025+1800*sqrt(5)
> > A^5 = 38125+17050*sqrt(5)
> > A^6 = 361125+161500*sqrt(5)
> > A^7 = 3420625+1529750*sqrt(5)
> > A^8 = 32400625+14490000*sqrt(5)
> > ...
> > A^n = a_n+b_n*sqrt(5)
> > a_n = ((5-2*sqrt(5))^n+(5+2*sqrt(5))^n)/2
> > b_n = (1/10)*sqrt(5)*((5+2*sqrt(5))^n-(5-2*sqrt(5))^n)
> > As n increases, a_n/b_n approaches sqrt(5) from above
> > Therefore b_n*sqrt(5) approaches a_n from below, and hence more and more 9s after the decimal point
> >
> > Please reply to ilanlmayer at gmail dot com
> >
> > __/\__
> > \ /
> > __/\\ //\__ Ilan Mayer
> > \ /
> > /__ __\ Toronto, Canada
> > /__ __\
> > ||
> thank you for the help !
> ( i wish i knew this in high school ! )
> i don't know why, but a similar pattern happens with (3 + 2 * sqrt(2)) ^ k
>
>
> 5.82842712474619.....................
>
> 33.9705627484771394478719195161
>
> 197.994949366116643469193236470189742320118659
>
> 1153.99913344822270261118589603047721268684116683595605815921
>
> 67259.998513232194628793508006917181208468837660012895581285831254498609099
>
> 392019.99974491093437509592477268670631674440337231921276927110823440631471667911839958281
>
> 2284859.9999562333744856481681683435615568663093769138768431377543192559229850045968629747114467231369939

This happens for a similar reason:
A^1 = 3+2*sqrt(2)
A^2 = 17+12*sqrt(2)
A^3 = 99+70*sqrt(2)
A^4 = 577+408*sqrt(2)
A^5 = 3363+2378*sqrt(2)
....
A^n = a_n+b_n*sqrt(2)
a_n = ((3+2*sqrt(2))^n+(3-2*sqrt(2))^n)/2
b_n = ((3+2*sqrt(2))^n-(3-2*sqrt(2))^n)/(2*sqrt(2))
As n increases, a_n/b_n approaches sqrt(2) from above
Therefore b_n*sqrt(2) approaches a_n from below, and hence more and more 9s after the decimal point

Please reply to ilanlmayer at gmail dot com

__/\__
\ /
__/\\ //\__ Ilan Mayer
\ /
/__ __\ Toronto, Canada
/__ __\
||

Re: A = (5 + 2 (sqrt 5)) and i do : A^2, A^3, A^4, . . .

<87fsjve5r7.fsf@zotaspaz.fatphil.org>

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From: pc+use...@asdf.org (Phil Carmody)
Newsgroups: rec.puzzles
Subject: Re: A = (5 + 2 (sqrt 5)) and i do : A^2, A^3, A^4, . . .
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 by: Phil Carmody - Thu, 23 Jun 2022 15:53 UTC

"henh...@gmail.com" <henhanna@gmail.com> writes:
> On Tuesday, June 21, 2022 at 8:43:37 PM UTC-7, Ilan Mayer wrote:
>> On Sunday, June 19, 2022 at 9:33:32 PM UTC-4, henh...@gmail.com wrote:
>> > where
>> > A = 9.47213595499958...
>> > or
>> > A = (5 + 2 (sqrt 5))
>> >
>> > and i do : A^2, A^3, A^4, . . .
>> >
>> > by A^8 (or so), the accumulation of 9's is obvious....
>> > (as ........49 . 999.......... )
>> >
>> >
>> > -- is there a simple explanation for why this happens?
>> >
>> > -- Does this happen with (sqrt 2) ? (sqrt 3) ?
>> >
>> > -- Do we ever get similar accumulations of 2's, 3's, ... ?
>> >
>> >
>> > ( i wish i knew this in high school ! )
>> Explanation for the 9s:
>>
>> A^1 = 5+2*sqrt(5)
>> A^2 = 45+20*sqrt(5)
>> A^3 = 425+190*sqrt(5)
>> A^4 = 4025+1800*sqrt(5)
>> A^5 = 38125+17050*sqrt(5)
>> A^6 = 361125+161500*sqrt(5)
>> A^7 = 3420625+1529750*sqrt(5)
>> A^8 = 32400625+14490000*sqrt(5)
>> ...
>> A^n = a_n+b_n*sqrt(5)
>> a_n = ((5-2*sqrt(5))^n+(5+2*sqrt(5))^n)/2
>> b_n = (1/10)*sqrt(5)*((5+2*sqrt(5))^n-(5-2*sqrt(5))^n)
>> As n increases, a_n/b_n approaches sqrt(5) from above
>> Therefore b_n*sqrt(5) approaches a_n from below, and hence more and
>> more 9s after the decimal point
>
> thank you for the help !
>
> ( i wish i knew this in high school ! )

Look up the derivation for the formula for the fibonnaci numbers:
F(n) = A^n + B^n
where A and B are conjugates (they differ only in the signs of the
part that's a radical)

It's the same principle here, you're looking at the (P+sqrt(Q))^n
part of (P+sqrt(Q))^n + (P-sqrt(Q))^n, where that sum is an integer,
and (P-sqrt(Q)) < 1, so (P-sqrt(Q))^n -> 0 as n increases.

Phil
--
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