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 ! )
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
/__ __\
||
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
__/\__thank you for the help !
\ /
__/\\ //\__ Ilan Mayer
\ /
/__ __\ Toronto, Canada
/__ __\
||
( 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
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:
whereExplanation for the 9s:
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 ! )
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 ! )
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 ! )
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