# Is this number algebraic or transcendental?

1. May 13, 2010

### Dickfore

When we first encounter irrational numbers, we usually define them as decimal numbers with infinite number of decimals that do not repeat in a periodic fashion. The canonical example one sees is the number:

$$\beta = 0.1010010001\ldots$$

i.e. there is one more zero between each consecutive pair of ones.

Another way of representing this number is by a series. Let us find the positions of the ones in this representation. Let us denote by $n_{k}$ the decimal position of the k-th one. Obviously, $n_{1} = 1$.

The number of zeros between the second and first one is 1.

The number of zeros between the third and second one is 2.

...

The number of zeros between the (k + 1)-st and the k-th one is k.

However, the number of decimal places between a figure in the n-th and a figure in the m-th place is equal to $n - m - 1$ (not counting the end decimals themselves). So, we can write the recursion.

$$n_{k+1} - n_{k} - 1 = k, \ k \ge 1$$

$$n_{k+1} - n_{k} = k + 1, \ k \ge 1$$

$$n_{k} - n_{k - 1} = k, \ k \ge 2$$

which, together with the initial condition has the solution:

$$n_{k} = \sum_{j=1}^{k} {j} = \frac{k(k + 1)}{2}.$$

As a sanity check, we confirm that this formula gives 1, 3, 6, 10 as the decimal places where the first, second, third and fourth one occur, respectively. The value of a one in the n-th decimal place is:

$$\frac{1}{10^{n}}$$

Combining these results, we see that we can write $\beta$ as:

$$\beta = \sum_{k = 1}^{\infty} {\frac{1}{10^{\frac{k (k+1)}{2}}}}$$

The question is: Is this number algebraic or transcendental?

2. May 13, 2010

### mathman

Without further analysis, it looks transcendental to me. I can't imagine a polynomial with integer coefficients having that number as a root.

3. May 13, 2010

### CRGreathouse

I agree with mathman:
$$\beta = \sum_{k = 1}^{\infty} 10^{-k(k+1)/2}$$
is almost surely transcendental. Actually, the form is unusual enough that ot may even be provable, though I can't think of a way at the moment.

4. May 13, 2010

### Dickfore

Well, I found this link http://sprott.physics.wisc.edu/pickover/trans.html". In it, one of the numbers, called Morse-Thue's number $x = 0.01101001\ldots$, which has a simple relation to my $\beta$ as:

$$100 \, x = 1 + \beta$$

So, if $\beta$ were an alberaic number, then any rational function of it is still an algebraic number. Since it says x is transcendental (without proof), we can conclude that $\beta$ cannot be algebraic, i.e. it must be transcendental.

Last edited by a moderator: Apr 25, 2017
5. May 13, 2010

### D H

Staff Emeritus
The number 0.11010010001... is obviously related to your number and to your newly-found Morse-Thue's number.

Let s1=0.1, s2=0.11, s3=0.1101, s4=0.1101001, ... Writing sn as a rational sn=pn/qn, then

\aligned p_1 &= 1 \\ q_1 &= 10 \\ p_{n+1} &= 10^np_n+1 \\ q_{n+1} &= 10^nq_n \endaligned

Suggestion: Try expressing it as a continued fraction. (Not a hint; when I say "hint" it means I know the solution.)

6. May 13, 2010

### Dickfore

I found the first 100 terms in the continued fraction of $\beta$ (not the number you suggested):

Code (Text):

{0,9,1,9,11,9,11,90,1,9,1,90,110,9,12,4,1,3,1,4,2,1,4,4,10,1,10,9,1,907,2,1,2,6,3,1,4,1,9,1,1,4,2,1,2,1,1,1,8,5,10,1,1,3,1,1,4,2,1,1,6,3,1,6,2,1,3,1,4,6,2,6,1,39,2,1,7,1,1,2,11,2,2,1,8,1,1,13,28,89,1,1,5,1,26,1,3,2,1,8,2}

If you make a scatter plot of these values, you can't see any regularity:

I tried to see if there are some hidden periodicities in the sequence by performing the discrete Fourier transform. Here is the power spectrum:

It looks pretty much like white noise to me.

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7. May 13, 2010

### D H

Staff Emeritus
Finally! Sometimes LaTeX can be a beotch -- and I've been using it for decades.

Not quite canonical form, but

$$1.1010010001\cdots = \cfrac{1}{1-\cfrac{1}{11-\cfrac{10}{101-\cfrac{100}{1001-\cfrac{1000}{10001-\dotsb}}}}}$$

Last edited: May 13, 2010
8. May 13, 2010

### Dickfore

Yeah, it might be true. I didn't check it. Nevertheless, is this a proof of transcendence? Because the continued fraction of $\sqrt[3]{2}$ has no periodicity or regularity as far as I know.

EDIT:

Mathematica gives the result of the series in terms of elliptic theta function:

$$\frac{\sqrt[8]{10}}{2} \vartheta_{2}(0; \frac{1}{\sqrt{10}}) - 1$$

Last edited: May 13, 2010
9. May 14, 2010

### RedGolpe

Be aware that http://en.wikipedia.org/wiki/Thue%E2%80%93Morse_sequence" [Broken] is not $\frac{(1+\beta)}{100}$ as it goes on .01101001100101101001011001101001... so even if you found a proof for it, that wouldn't apply to your number anyway.

Last edited by a moderator: May 4, 2017
10. May 14, 2010

### Dickfore

Which is exactly what I quoted as Morse-Thue's number. Nevertheless, even if it is connected through some transformation involving algebraic operations only, the proof still holds. BTW, Wikipedia gives an article for a sequence, not a number.

Last edited by a moderator: May 4, 2017
11. May 14, 2010

### Petek

Assuming D H's continued fraction expansion in post #7, I think I can prove that $\beta$ is irrational. Not what the OP was looking for, but it's a partial result. I use the following result, taken from p. 512 of Vol. II of Chrystal's Textbook of Algebra, AMC Chelsea edition, 1999:

17. If $a_2, a_3, ..., a_n, b_2, b_3, ..., b_n$ be all positive integers, then

(...)

II. The infinite continued fraction

$$\cfrac{b_2}{a_2-\cfrac{b_3}{a_3-\cfrac{b_4}{a_4-\cfrac{b_5}{a_5-\cfrac{b_6}{a_6-\dotsb}}}}}$$

converges to an incommensurable limit provided that after some finite value of n the condition $a_n \geq b_n + 1$ be always satisfied, where the sign > need not always occur but must occur infinitely often.

(Chrystal uses the word incommensurable to mean what we now call irrational.)

The continued fraction in post #7 satisfies these conditions. Thus $\beta$ is irrational.

Petek

12. May 15, 2010

### Martin Rattigan

Isn't that already obvious because it doesn't recur?

13. May 15, 2010

### Petek

Yes, you're right.

14. May 22, 2010

### MathPhd

The Thue number is not clearly related to your number. The decimals with 1s in positions that are n(n+1)/2 or n^2, etc., are related to the Jacobi identity and, in particular,

Product [0 to infinity] {(1+x^n)(1-x^(2n+2))} = 2Sum [x^(n(n+1)/2)]

{I am having a little trouble seeing the product at n=0 and the 2 coefficient in the sum! Possibly, 1+x^n is not done at n=0 and there is no 2 on the right}

For x=.1, this gives (1+1)*(1-.1^2)*(1+.1)*(1-.1^4)*(1+.1^2)*(1-.1^6)*(1+.1^3)*(1-.1^8)*(1+.1^4)*(1-.1^10)*(1+.1^5)*(1-.1^12)*(1+.1^6)*(1-.1^14)... = 2( 1.1010010001...)

This does not prove transcendentalness, but gives new details.