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MathGroup Archive 2002

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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[3] + Sqrt[2])^(2*n)

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

Mod[ N[(Sqrt[2] + Sqrt[3])^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[3] - Sqrt[2]]]

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

Floor[-2002*Log[10, Sqrt[3] - Sqrt[2]]]

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[2] + Sqrt[3])^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[2] + Sqrt[3])^2002 , 1000]], 10^2]
> 
> results in 99, so these are the digits you are interested in.
> 
> But there is something curious about this number.
> 
> Mod[Floor[N[10^1000 (Sqrt[2] + Sqrt[3])^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[2] + Sqrt[3])^2002, 2300], 10^6]
> 
> The decimal expansion of (Sqrt[2]+Sqrt[3])^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
> 
> 
> 
> 
> 






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