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

From henhanna@gmail.com@21:1/5 to All on Sun Jun 19 18:33:29 2022
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 ! )

--- SoupGate-Win32 v1.05
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• From Ilan Mayer@21:1/5 to henh...@gmail.com on Tue Jun 21 20:43:34 2022
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

__/\__
\ /
__/\\ //\__ Ilan Mayer
\ /
/__ __\
||

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From henhanna@gmail.com@21:1/5 to Ilan Mayer on Wed Jun 22 12:34:31 2022
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

__/\__
\ /
__/\\ //\__ Ilan Mayer
\ /
/__ __\
||

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

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Ilan Mayer@21:1/5 to henh...@gmail.com on Wed Jun 22 19:59:16 2022
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

__/\__
\ /
__/\\ //\__ Ilan Mayer
\ /
/__ __\
||
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

__/\__
\ /
__/\\ //\__ Ilan Mayer
\ /
/__ __\
||

--- SoupGate-Win32 v1.05
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• From Phil Carmody@21:1/5 to henh...@gmail.com on Thu Jun 23 18:53:00 2022
"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

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
--
We are no longer hunters and nomads. No longer awed and frightened, as we have gained some understanding of the world in which we live. As such, we can cast aside childish remnants from the dawn of our civilization.
-- NotSanguine on SoylentNews, after Eugen Weber in /The Western Tradition/

--- SoupGate-Win32 v1.05
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• From Jonathan Dushoff@21:1/5 to henh...@gmail.com on Thu Jun 23 09:24:34 2022
Another way to look at it is that it happens because the conjugate 5-2√5 has absolute value <1.

If we let A, B be the solutions to x² - 10x + 5 (equivalent to x² = 10x - 5) (equivalent to we can write any series that follows the recursion

s(n+2) = 10 s(n+1) - 5 s(n)

as a sum of powers of A and B.

If we start such a recursion with A⁰+B⁰, A¹+B¹ = 2, 10, we can continue it as 2, 10, 90, 850, ... The key point is that in each case A^n + B^n has to be an integer (since the recursion multiplies and adds integers). Since B^n keeps getting smaller,
A^n has to keep getting closer and closer to the integer.

There is also a parallel series: if you subtract instead of add then the first two terms of the series are 0, 4√5. Thus, if you take the series A^n/√5. You will get a similar convergence.

This is a a particularly fun game to play with the roots of x²-x-1 or x²-2x-1.

Jonathan

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))

9.47213595499958 * (10 - 9.47213595499958) ## 4.99999999999999456920

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 ! )

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)