       RE: Re: : huge number, ciphers after decimal point?

• To: mathgroup at smc.vnet.net
• Subject: [mg36605] RE: [mg36537] Re: : huge number, ciphers after decimal point?
• From: "DrBob" <drbob at bigfoot.com>
• Date: Fri, 13 Sep 2002 01:16:14 -0400 (EDT)
• Reply-to: <drbob at bigfoot.com>
• Sender: owner-wri-mathgroup at wolfram.com

```I puzzled over this and came up with an explanation.  The phenomenon
somebody noticed is that the fractional part of

(Sqrt + Sqrt)^(2*n)

approaches 1 from below as n (integer) gets large.  Hence, for instance,

Mod[ N[(Sqrt + Sqrt)^2002, 2000], 100]

has a string of 996 nines after the decimal point.  Well, here's the
reason for that.

Let x and y be the square roots of any integers and let n be an even
integer power.  Then

(x + y)^n == Sum[Binomial[n, m]*y^m*x^(n - m), {m, 0, n}];

and

(x - y)^n == Sum[(-1)^m*Binomial[n, m]*y^m*x^(n - m), {m, 0, n}];

so, adding the two equations, we have

(x + y)^n + (x - y)^n == Sum[Binomial[n, m]*y^m*x^(n - m)*
(1 + (-1)^m), {m, 0, n}];

1 + (-1)^m is 0 when m is odd and 2 when m is even, so

(x + y)^n + (x - y)^n ==
2*Sum[Binomial[n, 2*k]*y^(2*k)*x^(n - 2*k), {k, 0, n/2}];

Since n is even, all powers of x and of y are even in this expression.
Since x and y are the square roots of integers, it follows that all the
terms are integers.  Hence

(x + y)^n + (x - y)^n

is integer.  Now we assume, in addition to our other conditions, that

0 < x - y < 1

In that case, (x - y)^n approaches 0 from above as n gets large, and

FractionalPart[(x + y)^n] == 1 - (x - y)^n

which approaches one from below as n gets large.  The number of nines
after the decimal point can be calculated as

Floor[(-n)*Log[10, Sqrt - Sqrt]]

For n=2002 (the example we started with, this is indeed)

Floor[-2002*Log[10, Sqrt - Sqrt]]

996

Bobby Treat

-----Original Message-----
From: larry [mailto:goldbach at charter.net]
To: mathgroup at smc.vnet.net
Subject: [mg36605] [mg36537] Re: : huge number, ciphers after decimal point?

Fred Simons wrote:

> Stefan,
>
> To find n digits before the decimal point of a number, we can proceed
in the
> following way. We compute the number in sufficiently many digits, then
take
> the Floor of the result (i.e. we round it downwards to an integer) and
> finally take the result modulo 10^n.
>
> Mod[Floor[ N[(Sqrt + Sqrt)^2002 , 1000]], 10^2]
>
> results in 9, so the last two digits before the decimal point are 09.
>
> With a slight modification we can find the first n digits after the
decimal
> point. Simply find the last n digits before the decimal point of 10^n
times
> the number.
>
> Mod[Floor[ N[10^2 (Sqrt + Sqrt)^2002 , 1000]], 10^2]
>
> results in 99, so these are the digits you are interested in.
>
>
> Mod[Floor[N[10^1000 (Sqrt + Sqrt)^2002 , 2000]], 10^1000]
>
> results in 996 digits 9 followed by 7405.
>
> You can also play with the following command, resulting in the digits
around
> the decimal point:
>
> Mod[ N[(Sqrt + Sqrt)^2002, 2300], 10^6]
>
> The decimal expansion of (Sqrt+Sqrt)^2002 contains a sequence of
> 997 consecutive digits 9. Do you have any idea why?

Not unusual. Try other even exponents and there should be large blocks
of 9s . Smaller number for small exponents. Some kind of propagation of
9s as the exponent grows. Try other odd values for 3 and the same thing
happens for some values. Also for other values for 2.
Larry

>
> Fred Simons
> Eindhoven University of Technology
>
>
>
>
>

```

• Prev by Date: Re: creating adjacency matrices
• Next by Date: RE: Uneven FrameTicks with ExtendGraphics
• Previous by thread: Re: : huge number, ciphers after decimal point?
• Next by thread: Solve the oscillating function