| >For any finite sequence of base 10 digits there exists a
>power of 2 whose writing in base 10 begins precisely with
>those digits.
>
>Generalizations?
It would seem that this would be true for any integer base b > 1 and any x > 1
such that log[base b](x) is irrational, based on a fairly easy to prove theorem
that says if a is irrational there are infinitely many (p,q) such that
abs(a - p/q) < 1/q^2.
Since there are infinitely many such (p,q), there must be some with q larger
than any given finite bound. Pick such a (p,q) with q > ln(b)*(b^(n+1) + 1),
where n >=1 is the number of digits you want to match.
By the theorem, -1/q^2 < log[base b](x) - p/q < 1/q^2.
Multiply by q,
-1/q < q*log[base b](x) - p < 1/q
Since b^x is monotonic increasing we can exponentiate base b,
b^(-1/q) < (x^q)/(b^p) < b^(1/q)
For all y, b^(y/ln(b)) = (b^(1/ln(b)))^y = (b^(log[base b](e)))^y = e^y, so
this is equivalent to
e^(-1/(b^(n+1)+1)) < (x^q)/(b^p) < e^(1/(b^(n+1)+1))
For all y > 1, e^(1/y) < 1/(1-1/y) = 1 + 1/(y-1) by comparing series
expansions, so
e^(1/(b^(n+1)+1)) < 1 + 1/b^(n+1) and similiarly
e^(-1/(b^(n+1)+1)) > 1 - 1/b^(n+1)
i.e. (x^q)/(b^p) is within b^-(n+1) of 1.
So for any number z, the first n digits base b of z * (x^q)/(b^p) should either
be the same as the first n digits of z, or different by one (wrapping around
from the largest n-digit number to 10000... and vice versa), and the difference
will always be in the same direction (positive or negative).
Therefore, the powers of (x^q)/(b^p) should cover all possible combinations of
first n significant digits base b, and therefore so should the powers of x^q,
since all the b^p factor does is fiddle with the "decimal point".
You could generalize this to x < 1 as well if you didn't include leading zeros
(and minus signs) as digits.
Note that this means when b = 10 the leading n digits of the powers of any
integer that isn't itself a power of 10 will assume all possible values.
|
| Nice stab, Richie! I didn't know this proof - here is mine:
Proposition. Let a,b > 0 . Then the set
{m*a - n*b / m,n in N}
is dense in R if and only if a/b is irrational.
Corollary. Let a,b > 0 . Then the set
{(a^m) / (b^n) / m,n in N}
is dense in (0,oo) if and only if log[base b](a) is irrational.
(I have my own proof for the classical proposition above).
Generalization of .0:
Let p,q be natural numbers, p,q >= 2 , such that log[base q](p) is
irrational and consider a finite sequence of base p digits. Then there is
a power of q whose writing in base p begins precisely with those
digits.
Proof:
Let c(1),...,c(s) be the finite sequence of base p digits and let A be
the number whose base p writing digits are precisely c(1),...,c(s) .
The corollary above tells us that the set {(q^n) / (p^m) / m,n in N} is
dense in (0,oo) . Hence there are natural m,n such that
A * p^s <= (q^n) / (p^m) <= A * p^s + 1 - 1/(p^m)
Then
0 <= q^n - A * p^(m + s) <= p^m - 1
Let x = q^n - A * p^(m + s) , that is q^n = A * p^(m + s) + x . We also
have x < p^m , hence x is written in base p using m base p digits
x(1),...,x(m) (add zeros at the beginning if necessary).
As A is written c(1),...c(s) in base p , q^n is written in base p
using the digits c(1),...,c(s),x(1),...,x(m) , q.e.d.
Mihai.
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