19

I recently saw the identity $$ \frac{1}{1} - {1 \over 2} +{1 \over 3} - {1 \over 4} + {1 \over 5} - {1 \over 6} \dotsb = \log(2) $$ which I found rather interesting. I was intrigued by the way a transcendental number like $\log(2)$ could be expressed by using the reciprocals of every integer once. I then wondered what other real numbers could be represented this way, by using the infinite harmonic series, but changing the $+$ or $-$ signs before each fraction.

I soon realized that every real number could be represented this way, although not necessarily with such a clear pattern to the signs. Since + and - are binary options, I realized that the pattern of $+$ or $-$'s in the series could be represented by using a binary decimal number between 0 and 0.11111...

I called this number the "harmonic sign number" (or "HSN" for short). For instance, an example of a harmonic sign number for $\log(2)$ would be $0.\overline{10}$, because it alternates between $+$ and $-$, with the 1's representing a + and the 0's representing a -. However, $\log(2)$ is probably a special case, as it has a harmonic sign number that is rational.

It seems like most numbers could only be represented by irrational harmonic sign numbers, however that is still up for debate.

Another example of a rational HSN would be $0.\overline{1100}$, which is the HSN for ${\pi \over 4} + {\log(2) \over 2}$

Also, every real number has an infinite amount of HSN's, but every HSN corresponds to only one real number (or not, if it doesn't converge).

Some questions that I have thought about:

  1. Which numbers have rational harmonic sign numbers? Can only trancendental numbers have rational HSN's?
  2. If not, is there a quick way to tell whether a number has a rational HSN?
  3. Are there any rational numbers that have rational or at least algebraic HSN's?
  4. Does there exist a number which is it's own HSN? Are there more than one? Infinite?
  5. Which HSN's actually converge? I would suggest that they would have to be normal.

Please note, this is done purely out of interest, so only respond if you are actually interested! If you would like clarification also please ask in comments and I will edit. Also, I'm not looking for full concrete answers. Feel free to answer with anything you have found or is of interest. It does not need to answer the questions posed above.

volcanrb
  • 2,796
  • 9
  • 23
  • 1
    Interesting. Since the harmonic series diverges, $0.111\ldots$ doesn't represent anything. That suggests asking which HSBs actually converge. That probably depends on the density of $1$'s in some sense. – Ethan Bolker Jul 31 '17 at 18:32
  • One would need "about half" the digits to be 1s to get convergence, although I'm not going to try to make that formal. – Michael Lugo Jul 31 '17 at 18:33
  • I'd recommend you ask one question at a time. And everybody who doesn't just play around with things he didn't half understand knows that every single one is more complicated than some known, unsolved problems. The fact that every real number can be represented that way is known though, that's nearly trivial. –  Jul 31 '17 at 18:36
  • Do you think this question would be better in a different part of the site? I'm not really looking for a concrete answer, just some interesting discussion. – volcanrb Jul 31 '17 at 18:39
  • 2
    Note that your HSN representation is not unique -- in fact for every real number $x$, there are $2^{\aleph_0}$ different HSNs for $x$, which form a dense subset of $(0,1)$. (So in particular the answer to your (3) is "yes every real has irrational, transcendental, and non-computable HSNs"). – hmakholm left over Monica Jul 31 '17 at 18:39
  • Sorry, the 'irrational' was a typo, I meant rational – volcanrb Jul 31 '17 at 18:42
  • 1
    This site is about answers, not "interesting discussions" about nothing. And you won't get answers, not even if you formulate your questions rigorously. –  Jul 31 '17 at 18:44
  • @HenningMakholm to make sure I'm getting your idea, essentially we can do whatever we want with the first $N$ terms (for whatever finite $N$ we want), and then use essentially the Riemann rearrangement trick with terms $N+1$ and beyond, correct? – TomGrubb Jul 31 '17 at 18:51
  • If a series converges "conditionally", that is the series itself converges but the series of absolute values, then it can be rearranged to sum to any number. – user247327 Jul 31 '17 at 19:00
  • @ThomasGrubb: Yes, and we can get continuum many choices for the remaining terms too, say, by choosing each time the partial sum exceeds our target number whether we switch to negative terms immediately or only after an additional positive term. – hmakholm left over Monica Jul 31 '17 at 19:14
  • Have you thought about this $s_{13}= \sum {n=1}^{\infty } \dfrac{(-1)^{((n+1) \bmod 13)}}{n}$ where modulus is odd prime, or simply odd. $Mathematica$ and Inverse Symbolic calculator don't know closed forms. $s{13}\approx 0.779941;;s_5\approx 1.218754$. It's fun because $s_{10001}\approx \log 2$. Quite obviously $s_n\to \log 2$ as $n\to\infty$ – Raffaele Jul 31 '17 at 19:47
  • It's fun this other double harmonic series $H=\sum _{n=1}^{\infty } \frac{(-1)^{\left\lfloor \frac{n-1}{2}\right\rfloor }}{n}\approx 1.14705571938$ It's this is $1+\frac{1}{2}-\frac{1}{3}-\frac{1}{4}+\frac{1}{5}+\frac{1}{6}-\frac{1}{7}-\frac{1}{8}+\frac{1}{9}+\frac{1}{10}-\frac{1}{11}-\frac{1}{12}+\frac{1}{13}+\frac{1}{14}-\ldots$ – Raffaele Jul 31 '17 at 20:18
  • @volcanrb $0.\overline{1100}$ should be the series in my previous comment, with pattern $ + + - - + + - -$. But it doesn't converge to ${\pi \over 4} + {\log 2 \over 2}$ – Raffaele Jul 31 '17 at 20:21
  • Wouldn't $\sum _{n=0}^{\infty}\frac{\left(-1\right)^n}{2n+1}+\sum _{n=1}^{\infty}\frac{-\left(-1\right)^n}{2n}$ which is equal to ${\pi \over 4} + {\log(2) \over 2}$, also represent that series? – volcanrb Jul 31 '17 at 21:18
  • 1
    All 'rational series' that converge should converge to sums of digamma functions of rational arguments (plus some multiple of $\gamma$) by the usual partial-fraction decomposition techniques. This is probably enough to show that they're all trascendental. – Steven Stadnicki Jul 31 '17 at 21:27
  • Also, it's essentially trivial that 'most' numbers can only be represented by irrational HSNs, by the standard counting argument; there are only countably many rational HSNs but uncountably many real numbers, so the former have to be a set of measure zero in the latter. – Steven Stadnicki Jul 31 '17 at 21:29
  • 1
    @Henning Makholm (and others interested): Stronger statements than "continuum many rearrangements for each $x$" can be made. See this 18 September 2000 sci.math post. – Dave L. Renfro Jul 31 '17 at 21:40
  • 6
    @ProfessorVector Mathematics itself is driven more by questions than answers. And I'd like to think this site puts mathematics first than merely answers to math questions. Because otherwise, we're nothing more than an over-glorified tutoring lab. –  Jul 31 '17 at 21:45
  • You may find this post of mine relevant: https://math.stackexchange.com/a/2347650/166353 – Akiva Weinberger Aug 01 '17 at 13:33
  • 2
    The best way to explore these might be to substitute in $\frac1n=\int_0^1x^{n-1}dx$ and use the geometric series. – Akiva Weinberger Aug 01 '17 at 13:36

1 Answers1

2

Some notation first: $$\mbox{HSN}(z)=H_z=\{a_0,a_1,\dots\}=0.a_0a_1\dots, \; \forall a_i. |a_i|=1$$ $$ z = \sum_{i=0}^\infty{\frac{c_i}{i+1}}, \; c_i = \begin{cases} +1 & \mbox{if } a_i=1 \\ -1 & \mbox{if } a_i=0 \end{cases} $$

Now we can write $z$ as, $$(1) \Rightarrow z =\int_0^1P_z(x)dx,\; P_z(x) = {\sum_{i=0}^\infty{c_ix^i}dx}$$

Rational HSN

define $R_{z} = \{p\in \mathbb{N}_0|\forall a_i \in H_z. a_i=a_{i+p}\}$ and $r_z = min(R_z)$ then is $H_z$ rational $\Leftrightarrow R_z \ne \varnothing$

$$ \begin{align*} H_z \text{ is rational} \Rightarrow & P_z(x)&& = {\sum_{i=0}^\infty{c_ix^i}} \\ & && = {\sum_{i=0}^\infty{c_{(i\mbox{ mod }r_z)}x^i}} \\ & && = {\sum_{i=0}^{r_z-1}{c_ix^i}} + {\sum_{i=r_z}^\infty{c_{(i\mbox{ mod }r_z)}x^i}} \\ & && = {\sum_{i=0}^{r_z-1}{c_ix^i}} + x^{r_z}{\sum_{i-r_z=0}^\infty{c_{(i-r_z\mbox{ mod }r_z)}x^{i-r_z}}} \\ & && = p_z(x) + x^{r_z}P_z(x), \; p_z(x)={\sum_{i=0}^{r_z}{c_ix^i}}\\ \Leftrightarrow& P_z(x) && = \frac{p_z(x)}{1-x^{r_z}}\\ \Leftrightarrow& z && = \int_0^1\frac{p_z(x)}{1-x^{r_z}}dx\\ \end{align*}$$

convergence

$p_z(1) \neq 0$ will cause $\frac{p_z(x)}{1-x^{r_z}}$ to converge to $\frac1x$ which has infinite area in $[1-\epsilon,1]$. So in order for $z$ to have a finite value $ p_z(1) $ must be $0$, or $\sum_{i=0}^{r_z-1}{c_i} = 0$. This also means that $r_z$ must be even.

closed form

we can solve the integral for $z$ using the partial fraction decomposition of $\frac{p_z(x)}{1-x^{r_z}}$

$$\begin{align*} (2) \Rightarrow && \frac{p_z(x)}{1-x^{r_z}} & = \sum_{t=1}^{r_z}{\frac{p_z(\alpha_t)\alpha_t}{r_z(x-\alpha_t)}}, \; \alpha_t = e^{2\pi ti/{r_z}}\\ \Rightarrow && z &= \int_0^1{\sum_{t=1}^{r_z}{\frac{p_z(\alpha_t)\alpha_t}{r_z(x-\alpha_t)}}dx}\\ && z &= \sum_{t=1}^{r_z}{\frac{p_z(\alpha_t)\alpha_t}{r_z}\cdot\ln\left(\frac{1-\alpha_t}{0-\alpha_t}\right)} \\ && z &= \sum_{t=1}^{r_z}{\frac{p_z(\alpha_t)\alpha_t}{r_z}\cdot\ln(1-\bar\alpha_t)} \\ && z &= \frac{-p_z(-1)}{r_z}\cdot\ln(2) + \sum_{t=1}^{r_z/2-1}{\frac{p_z(\alpha_t)\alpha_t}{r_z}\cdot\ln(1-\bar\alpha_t)+\frac{p_z(\bar\alpha_t)\bar\alpha_t}{r_z}\cdot\ln(1-\alpha_t)} \\ && z &= \frac{-p_z(-1)}{r_z}\cdot\ln(2) + \sum_{t=1}^{r_z/2-1}{\frac{p_z(\alpha_t)\alpha_t}{r_z}\cdot\overline{\ln(1-\alpha_t)}+\frac{\overline{p_z(\alpha_t)\alpha_t}}{r_z}\cdot\ln(1-\alpha_t)} \\ && z &= \frac{-p_z(-1)}{r_z}\cdot\ln(2) + \sum_{t=1}^{r_z/2-1}{\frac{p_z(\alpha_t)\alpha_t}{r_z}\cdot\overline{\ln(1-\alpha_t)}+\overline{\frac{p_z(\alpha_t)\alpha_t}{r_z}\cdot\overline{\ln(1-\alpha_t)}}} \\ && z &= \frac{-p_z(-1)}{r_z}\cdot\ln(2) + \frac2{r_z}\sum_{t=1}^{r_z/2-1}{Re\left(p_z(\alpha_t)\alpha_t\cdot\overline{\ln(1-\alpha_t)}\right)} \\ \end{align*}$$ (!!) because we assumed $z$ is finite the partial fraction for $\alpha_0$ can be omitted

Question 1;

I suspect that only transcendental numbers have a rational HSN, because they can be writen as a sum of natural logaritm of algebraic numbers, so the sum of transcendental numbers. However not all sums of transcendental numbers are transcendental, you'd have to prove that ${\frac{p_z(\alpha_t)\alpha_t}{r_z}\cdot\ln(1-\bar\alpha_t)}$ are algebraically independent.

Question 2:

as far as i found not, the opposite would be easier (if Question 1)

Question 3:

If Question 1 is correct then no algebraic number could have a rational HSN

Question 4:

If Question 1 is correct, such a number would have to be irrational.

Question 5:

For rational HSN see above, for irrational HSN i suspect the same must be true.

$(1)$ $$\frac1n=\int_0^1{x^{n-1}dx}$$ $(2)$ $$\frac{P(x)}{Q(x)}=\sum_{i=0}^n{\frac{P(\alpha_i)}{Q'(\alpha_i)}\frac1{x-\alpha_i}}$$

Sam Coutteau
  • 357