Finite string of numbers

**redsoxfan325** · November 17th 2009, 12:35 AM

Question: For any pair $(x,n)$ , where $x$ is an irrational number such that $0<x<1$ , so $x=\sum_{i=1}^{\infty}\frac{x_i}{10^i}$ , and $n$ is a natural number, so $n=\sum_{i=0}^m 10^i n_i$ , what's the probability that the finite string of digits in $n$ will be contained in the decimal expansion of $x$ ? In other words, what's the probability that $\exists N$ such that $x_{N+k}=n_k$ for all $k\in[0,m]$ ?

Extra Questions: 1.) What's the probability that the string of digits in $n$ occurs in $x$ twice? $r$ times? Infinitely many times?
2.) How does the answer change (if at all) if we allow $x$ to be any real number?

Example: Say $x=\sqrt{2}-1$ and $n=1356237$ . Then you have $\sqrt{2}-1=0.4142\textbf{1356237}31...$ so the string of digits of $n$ does occur in $x$ .

Any ideas?

**aman_cc** · November 17th 2009, 12:53 AM

Originally Posted by redsoxfan325

If you take the decimal expansion of any irrational number less than one, i.e. $0.a_1a_2a_3a_4...$ and any finite string of digits $n_0n_1n_2...n_m$ (i.e. a natural number), will you always find that exact string somewhere in the decimal expansion of the irrational number. In other words, does there exist $N$ such that $a_{N+k}=n_k$ for all $k\in[0,m]$ ?

For instance, say you are given the irrational number $\sqrt{2}-1$ and a string of numbers $1356237$ . Then you have $\sqrt{2}-1=0.4142\textbf{1356237}31...$

Intuitively, it seems like you should be able to, but I have no idea how to justify that claim.

Any ideas?

Hi - A very interesting question indeed.
My thoughts:
We know that any rational has a repeating decimal representation.
So consider
x=0.121221222122221222221222222122222221......
This can never be repeating hence x is an irrational number.
Now consider, say, 3. You will never find it in the decimal expansion of x. So I would say claim is false.

Does it sound good?

**redsoxfan325** · November 17th 2009, 12:57 AM

Originally Posted by aman_cc

Hi - A very interesting question indeed.
My thoughts:
We know that any rational has a repeating decimal representation.
So consider
x=0.121221222122221222221222222122222221......
This can never be repeating hence x is an irrational number.
Now consider, say, 3. You will never find it in the decimal expansion of x.

Does it sound good?

Ah yes, true. I should refine my question then.

See above for refined question.

I feel like this is one of those questions where the answer is either zero or one hundred percent.

**aman_cc** · November 17th 2009, 01:05 AM

Originally Posted by redsoxfan325

Ah yes, true. I should refine my question then.

If you pick an irrational number (less than one) out of a (large) hat and a natural number out of a hat, what's the probability that the finite string will be contained in the decimal expansion?

I feel like this is one of those questions where the answer is either zero or one hundred percent.

Sorry but I guess your question is not defined well. What is 'large' hat? And I feel this is going to be one difficult question.

An answer based on pure/raw/unsupported gut feel is 0.

**redsoxfan325** · November 17th 2009, 01:08 AM

Originally Posted by aman_cc

Sorry but I guess your question is not defined well. What is 'large' hat? And I feel this is going to be one difficult question.

A answer based on pure/raw/unsupported gut feel is 0.

The 'large' part was a joke.

**gdmath** · November 17th 2009, 02:38 AM

Hi redsoxfan325,

I work on arbitary arithmetic patterns.

One of my concerns is if an irrational number, defined in the following form
$<br /> r=\sum_{i=1}^{\propto}a_i<br />$

Does have "random distribution" (or more proper "normal distribution") on its decimal digits.

I think that your question does have deeper relation to this condition

However, as aman_cc noticed, we must be very carefull to the mathematic definitions that we use to define the problem. In fact this is one of my biggest concerns in my work.

We keep in touch.

**redsoxfan325** · November 17th 2009, 03:05 AM

I wasn't trying to be mathematically precise when I worded the second question, but I have edited so that I think it is.

**gdmath** · November 17th 2009, 05:26 AM

Additionaly take a look at
Pi - Wikipedia, the free encyclopedia
and
Normal number - Wikipedia, the free encyclopedia
and
Normal number - Wikipedia, the free encyclopedia

**aman_cc** · November 17th 2009, 05:34 AM

Originally Posted by gdmath

Additionaly take a look at
Pi - Wikipedia, the free encyclopedia
and
Normal number - Wikipedia, the free encyclopedia
and
Normal number - Wikipedia, the free encyclopedia

The question posed appears pretty tough. I don't even know how to start.

**Drexel28** · November 17th 2009, 06:48 AM

Isn't this a very tough question? It depends on whether or not the digits of the number are random. Suppose that $x$ has random digits. Define $\sigma_n=\frac{\text{number of }\ell\text{s in the first n digits}}{n}$ where $0\le \ell \le 9$ . Then $\lim_{n\to\infty}\sigma_n=\frac{1}{10}$ . So we must theoretically only look at the first ten numbers to find an occurence of $\ell$ . If $\ell$ were a string of $k$ digits we would theoretically have to only look ath the first $10^k$ digits to find an occurence of that string.

**chiph588@** · November 17th 2009, 08:35 AM

It's unfortunate, but mathematicians don't know that much about normal numbers and hence I believe your question can't be answered...

**redsoxfan325** · November 17th 2009, 09:05 AM

It was just something I was thinking about last night, so I figured I'd post it and see what people thought. It's certainly possible that it's too difficult to solve.

Though the fact that almost all real numbers are normal seems to indicate (intuitively, not mathematically) that there's a high chance of a string of digits appearing in the expansion.

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