• 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
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • 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

    Please reply to ilanlmayer at gmail dot com

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

    --- 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

    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

    --- 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

    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
    /__ __\
    ||

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