Playing with numbers: 1/998001 and others

A post in response to Colin’s “Patterns in numbers”:

1/998001 = 0.000 001 002 003 004 005 006 007…

goes through all integers 000 through 999, skipping only 998. Maple confirms this:

1/998001

This fraction made rounds on the internet a while ago.

I begin with two general claims:

Integer numbers are more complicated than polynomials
Decimal fractions are more complicated than power series

To illustrate the first one: multiplying 748 by 367 takes more effort than multiplying the corresponding polynomials,

$(7x^2+4x+8)(3x^2+6x+7) = 21x^4+54x^3+97x^2+76x+56$

The reason is that there’s no way for the product of $4 x$ and $3x^2$ to “roll over” into $x^4$ . It’s going to stay in the $x^3$ group. Mathematically speaking, polynomials form a graded ring and integers don’t. At the same time, one can recover the integers from polynomials by setting $x=10$ in $21x^4+54x^3+97x^2+76x+56$ . The result is $274516$ .

Moving on to the second claim, consider the power series

$\displaystyle (*) \qquad \frac{1}{1-t}=1+t+t^2+t^3+\dots = \sum_{n=0}^{\infty} t^n$

I prefer to use $t$ instead of $x$ here, because we often want to replace $t$ with an expression in $x$ . For example, setting $t=2x$ gives us

$\displaystyle \frac{1}{1-2x}=1+2x+4x^2+8x^3+\dots = \sum_{n=0}^{\infty} 2^n x^n$

which is nothing surprising. But if we now set $x=0.001$ (and, for neatness, divide both sides by 1000), the result is

$\displaystyle \frac{1}{998}=0.001\ 002\ 004\ 008\ 016\ 032 \dots$

which looks like a magical fraction producing powers of 2. But in reality, setting $x=0.001$ did nothing but mess things up. Now we have a complicated decimal number, in which powers of 2 break down starting with “513″ because of the extra digits rolled over from 1024. In contrast, the neat power series keeps generating powers of 2 forever.

By the way, $\displaystyle \frac{1}{1-2x}$ is the generating function for the numbers 1,2,4,8,16…, i.e., the powers of 2.

So, if you want to cook up a ‘magical’ fraction, all you need to do is find the generating function for the numbers you want, and set the variable to be some negative power of 10. E.g., the choice $x=0.001$ avoids digits rolling over until the desired numbers reach 1000. But we could take $x=10^{-6}$ and get many more numbers at the cost of a more complicated fraction.

For example, how would one come up with 1/998001? We need a generating function for 1,2,3,4,…, that is, we need a formula for the power series $\sum nt^n$ . No big deal: just take the derivative of (*):

$\displaystyle \frac{1}{(1-t)^2}=\sum_{n=0}^{\infty} nt^{n-1}$

and multiply both sides by $t$ to restore the exponent:

$\displaystyle (**) \qquad \frac{t}{(1-t)^2}=\sum_{n=0}^{\infty} nt^{n}$

Now set $t=0.001$ , which is easiest if you expand the denominator as $1-2t+t^2$ and multiply both the numerator and denominator by $1000000$ . The result is

$\displaystyle \frac{1000}{998001} = \sum_{n=0}^{\infty} n\,0.001^{n} = 0.001002003\dots$

Dropping 1000 in the numerator is a matter of taste (cf. xkcd 163).

Let’s cook up something else. For example, again take the derivative in (**) and multiply by $t$ :

$\displaystyle \frac{t(1+t)}{(1-t)^3}=\sum_{n=0}^{\infty} n^2t^{n}$

Now set $t=0.001$ to get

$\displaystyle \frac{1001}{997002999} = 0.000\ 001\ 004\ 009\ 016\ 025\ 036\ 049\dots$

Admittedly, this fraction is less likely to propagate around the web than 1/998001.

For the last example, take the Fibonacci numbers 1,1,2,3,5,8,13,… The recurrence relation $F_{n+2}=F_n+F_{n+1}$ can be used to find the generating function, $\displaystyle \frac{1}{1-t-t^2} = \sum F_n t^n$ . Setting $t=0.001$ yields

$\displaystyle \frac{1}{998999} = 0.000\ 001\ 002\ 003\ 005\ 008\ 013\ 021\ 034\ 055\dots$

Update: this post was picked by @mathblogging

Calc VII / @lkovalev generating patterns in decimal expansions of fractions like 1/998001 lkovalev.wordpress.com/2012/02/25/pla… @mathblogging #picks—
Peter Krautzberger (@pkrautz) February 29, 2012

12 Responses to Playing with numbers: 1/998001 and others

colindcarroll says:

2012-02-25 at 15:07

Fantastic explanation. So I guess the answer to the question “why was 998 skipped?” is really that it was not, but there was a domino effect with one of the digits from 1000 “rolling over” onto the 999, which picked up an extra digit and “rolled over” to 998, making it look like 999.

- Leonid Kovalev says:
  
  2012-02-25 at 19:08
  
  Exactly. There’s another observation that did not make the post: if a function f is represented by a power series with bounded integer coefficients, then f can be determined by evaluating it at a single point. Indeed, define g(x)=f(x)+M/(1-x) where M is a sufficiently large integer so that the coefficients of g are positive. Then evaluate $g(10^{-d})$ where d is sufficiently large so there’s no rolling over, and read off the coefficients.
  
  - Leonid Kovalev says:
    
    2012-02-26 at 19:45
    
    Oh, and another illustration of the 1st claim would be the abc conjecture and FLT: both are easier for polynomials.
Lianxin says:

2012-12-30 at 00:43

It is easy to remove the domino effects of 1/998001. Notice we want to subtract $1000\sum_{n=1001}^2000 0.001^n$ , $2000\sum_{n=2001}^3000 0.001^n$ , and so on. Upon simplification this becomes $\sum_{n=1}^\infty n [10^{-3000}]^{n-1} \frac{1-10^{-3000}}{0.999 \times 10^{3000}}$ . Exchanging derivative and limit, we have the result as exactly $\frac{1}{0.999 \times 10^{3000}}$ . So the “ideal” faction should be 1/998001- $\frac{1}{0.999 \times 10^{3000}}$

- Leonid Kovalev says:
  
  2012-12-30 at 01:56
  
  You are right: $\displaystyle \frac{1}{998001}-\frac{1}{999\cdot 10^{2997}}$ has the period that ends with 997998999, going back to 000001002003…
  
  That said, part of what made 1/998001 interesting is that a 6-digit input (denominator) produces a nearly perfect 3000-digit output. After correction, one gets an absolutely perfect 3000-digit output, but from a fraction with over 3000 digits in the denominator.
  
Lianxin says:

2012-12-30 at 00:46

As another remark we should not pre-assume M and d discussed in the reply of the first comment; we have no idea of the magnitude of the coefficients of the power series. Thus additional computations should be used to determine M and d first.

Clint says:

2013-04-14 at 17:20

how do get a generating function for even numbers up to 100

- lkovalev says:
  
  2013-04-14 at 17:33
  
  Sounds like a question for Math.StackExchange… Take a generating function for 1,2,3,4,5, … and multiply it by 2.
  
  - Clint says:
    
    2013-04-14 at 19:48
    
    Now why did we deriviate to get a generating Function for 1/998001
  - lkovalev says:
    
    2013-04-14 at 20:24
    
    We know a formula for $\sum_{n=0}^\infty x^n$ and want a formula for $\sum_{n=1}^\infty n x^n$ . How to create this coefficient $n$ ? Taking the derivative, of course: $(x^n)'=nx^{n-1}$ . Then multiply by $n$ to restore the exponent to $n$ .
    
    There is an approach without derivative, but it requires an algebraic trick. Observe that $\sum_{n=1}^\infty n x^n$ is equal to the double series $\sum_{m=1}^\infty\sum_{n=m}^\infty x^n$ . Use the geometric series formula on the inner sum to get $\sum_{m=1}^\infty\frac{x^m}{1-x}$ , and then use the formula again to get $\frac{x}{(1-x)^2}$ .
Clint says:

2013-04-14 at 22:21

What is the geometric series formula

- lkovalev says:
  
  2013-04-14 at 22:22
  
  See Wikipedia: http://en.wikipedia.org/wiki/Geometric_series

Playing with numbers: 1/998001 and others

12 Responses to Playing with numbers: 1/998001 and others

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