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Question 1: Let $a$ be a real number with a base-10 decimal representation $a_1a_2\ldots a_n \ldots$ Denote the number of ways to write $a_n$ as the sum of positive integers as $p(a_n)$ - also called the partition of $a_n$. I write $A(n)=\sum_{k=1}^n p(a_k)$ and $B(n)= \sum_{k=1}^n a_k$ and set $$\beta(a)=\lim_{n\to \infty}{A(n)\above 1.5 pt B(n)}$$ If $a$ is a base-10 simply normal number is it true that $\beta(a)={97 \above 1.5 pt 45}$ ?

Attempted Solution: Let $a$ be a base-10 simply normal number with decimal representation $a_1a_2\ldots a_n \ldots$. Explicitly we can write \begin{equation}\lim_{n\to \infty}{A(n)\above 1.5 pt B(n)} =\lim_{n\to \infty}{p(a_1)+p(a_2)+\ldots +p(a_n)\above 1.5pt a_1+a_2\ldots +a_n }\end{equation} Surely $a_k \in [0,9]$ for every $1\leq k \leq n$. Now let $\nu_n(a_k)$ count occurrences of the digit $a_k$ up to and including the n-th decimal place of $a$. Observe the following: the numerator in the above equation can be written explicitly as:

$$ \underbrace{(p(0)+\ldots +p(0))}_{\nu_n(0)\text{ times}}+\ldots+\underbrace{(p(9)+\ldots + p(9))}_{\nu_n(9)\text{ times}}$$ and similarly the denominator may be written as

$$\underbrace{(0+\ldots +0)}_{\nu_n(0)\text{ times}}+\ldots+\underbrace{(9+\ldots + 9)}_{\nu_n(9)\text{ times}}$$ Combining those two observation together we can rewrite the RHS of our first equation as

$$ \lim_{n \to \infty}{\nu_n(0)\times p(0)+\nu_n(1)\times p(1)+\ldots + \nu_n(9)\times p(9)\above 1.5pt \nu_n(0)\times 0+ \nu_n(1)\times1+\ldots + \nu_n(9)\times9} $$

Recall that every digit of $a$ occurs with equal frequency. In particular since $a$ is simply normal we have that $\lim_{n \to \infty}\nu_n(a_j)=\lim_{n \to \infty}\nu_n(a_k)$ for every $j$ and $k$. With out loss to generality we can substitute $\nu_n(j)$ with $\nu_n(1)$ in the numerators and denominators of the equation above and evaluate the limit,

$$\lim_{n \to \infty}{\nu_n(1)\times p(1)+\ldots + \nu_n(1)\times p(9) \above 1.5pt \nu_n(1)\times1+\ldots + \nu_n(1)\times9}$$

We can then pull out the term $ \nu_n(1)$, noticing those terms cancel out in the limit, and make a final simplification to show that $$\lim_{n \to \infty}{\nu_n(1) \above 1.5pt \nu_n(1)}{p(1) +\ldots +p(9)\above 1.5 pt 1+ \ldots +9}={1+1+1+2+3+5+7+11+15+22+30 \above 1.5pt 0+1+2+3+4+5+6+7+8+9}={ 97 \above 1.5pt 45}.$$

I believe the above solution to Question 1 is correct? Now my understanding is that we have the following implication: base-10 normal $\implies$ base-10 simply normal. So the follow up question seems natural.

Question 2: If $a$ is a base-10 normal is it true that $\beta(a)={97 \above 1.5 pt 45}$ ?

The answer to Question 2 should be yes by implication alone. Moreover it should be true for both algebraic and transcendental base-10 normal numbers. Since I believe I have shown every base-10 simply normal number converges to ${97 \above 1.5 pt 45}$ under $\beta$ and every base-10 normal number is simply normal then so to will every base-10 normal number converge to ${97 \above 1.5 pt 45}$ under $\beta$. It should be clear that the converse two Question 2 is false !

If we believe Questions 1 and 2 are true then we would expect several well known provably base-10 normal constants to converge to ${97 \above 1.5pt 45}$ under $\beta$. So I ran an experiment using data from Sloan's Database on the Erdos-Copeland constants which is proven to be normal in base-10. Using the first $2\text{ }000$ digits I plotted the values of $\beta(\text{Erdos-Copeland})$. The graph below illustrates the convergence. The blue line is the value of ${97 \above 1.5 pt 45}$ and the green graph are the values of the Erdos-Copeland constant under $\beta$. Up to the $2\text{ }000$-th digit $\beta(\text{Erdos-Copeland})=2.12999037999\ldots$ enter image description here

I believe given enough terms one can plot a similar graph for the Champernowne constant and one should expect the convergence to be significantly slower.

Finally it is not known if $\pi$ is normal in base-10. Out of curiosity I computed $\beta(\pi)$ up to the first $500\text{ }000$ digits of $\pi$. See the graph below. Again the blue line is the value of ${97 \above 1.5 pt 45}$ and the orange graph are the values of the $\pi$ under $\beta$. Up to the $500\text{ }000$-th digit $\beta(\pi)=2.153781\ldots$. $\pi$ appears to converge faster under $\beta$ than any other mathematical constant I have looked into. enter image description here

I have numerous plots for zeta values and square roots and they all appear to approach ${97 \above 1.5 pt 45}$ which can be written numerically as $2.15555\ldots$. I have not attempted to plot numbers that are known to be abnormal in base-10.

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    $\begingroup$ Instead of replacing $\nu_n(d)$ with $\nu_n(1)$, you might rather divide top and bottom by $n$ and observe that $\frac1n\nu_n(d)\to \frac19$. $\endgroup$ – Hagen von Eitzen May 26 '17 at 17:15
  • $\begingroup$ @HagenvonEitzen I cannot quite follow you ? I thought my substitution move was intuitive ? $\endgroup$ – Antonio Hernandez Maquivar May 26 '17 at 17:17
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    $\begingroup$ I think that would make the argument more stringent. But of course the result remains: For simply normal numbers, $$\lim_{N\to\infty}\frac{\sum_{n=1}^N f(a_n)}{\sum_{n=1}^N a_n}= \frac{\sum_{d=0}^9f(d)}{\sum_{d=0}^9d}$$ $\endgroup$ – Hagen von Eitzen May 26 '17 at 17:20
  • $\begingroup$ @HagenvonEitzen I think I now understand your comment. $\endgroup$ – Antonio Hernandez Maquivar May 26 '17 at 17:26
  • $\begingroup$ What exactly is your question? $\endgroup$ – orlp May 26 '17 at 18:23
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For simply normal numbers in base $b$ we have $\displaystyle \lim_{n\rightarrow\infty}\sum_{k=1}^n f(a_k) = \frac{n}{b}\sum_{d=0}^{b-1} f(d),$ as for simply normal numbers the digits in the limit all have frequency $\frac{1}{b}$. With that we have

$$\displaystyle \lim_{n\rightarrow\infty}\frac{A(n)}{B(n)} = \frac{\sum_{d=0}^{b-1}p(d)}{\sum_{d=0}^{b-1}d},$$

which for base 10 does result in $\frac{97}{45}$.

Since base-10 normal implies base-10 simply normal, yes, this also holds for base-10 normal numbers.

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  • $\begingroup$ I think I was looking for affirmation to my proof - and maybe this is not the right forum for that. However both you and @HagenvonEitzen have given succinct and more stringent proofs as better alternatives. $\endgroup$ – Antonio Hernandez Maquivar May 26 '17 at 19:41
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    $\begingroup$ @AnthonyHernandez Generally that is fine, but it should be clear that you're looking for verification. $\endgroup$ – orlp May 26 '17 at 19:58
  • $\begingroup$ Very random comment but a Chaitin's constant $\Omega$ (also called the halting probability) is apparently a normal number that is uncomputable - only a finite number of digits can be written. On the other hand $\beta(\Omega)={97 \above 1.5 pt 45}$ over all it's digits ? $\endgroup$ – Antonio Hernandez Maquivar May 26 '17 at 20:01
  • $\begingroup$ @AnthonyHernandez If it's proven to be normal in base 10, then yes. $\endgroup$ – orlp May 26 '17 at 20:24

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