Let's define $\rho=1.1010010001\dots$ which can be expressed by: $$\rho=\frac{1}{10^{0}}\underbrace{+\frac{1}{10^{1}}}_{\text{power}+1}\underbrace{+\frac{1}{10^{3}}}_{\text{power}+2}\underbrace{+\frac{1}{10^{6}}}_{\text{power}+3}\underbrace{+\frac{1}{10^{10}}}_{\text{power}+4}+\dots$$ That is to say: $$\rho=\sum_{i=0}^{\infty}10^{-i(i+1)/2}$$ Can we express $\rho$ in a simple form? I think it's not a rational but I don't know how to prove it, any ideas? Thanks.
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1It is equal to $\frac{\sqrt[8]{10}}{2} \theta_2\left(\frac{1}{\sqrt{10}}\right)$ where $\theta_2(q)$ is a Jacobi theta series. This number is a transcendental number, see this answer. – achille hui Oct 21 '14 at 12:45
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Exactly what I needed, thanks. – A. Breust Oct 21 '14 at 12:48
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Every rational number has a repeating pattern in it's decimal notation. Yours does not, since the subpaterns of zeroes become arbitrarily large. Therefore, your number is not rational.
A more formal proof:
Suppose $\rho$ is a rational number. Then $$\rho = 1.a_1a_2\dots a_n \overline{b_1b_2\dots b_m}$$ for some numbers $a_i, b_j$. There are two possibilities:
- All values of $b_i$ are equal to $0$. This is not possible because your definition of $\rho$.
- There exists a nonzero value $b_i$. Then the larges subsequence of zeroes that can repeat in $\rho$ can be at most $\max\{m,n\}$ zeroes long. But $\rho$ has infinitely many longer subsequences of zeroes, so this is not possible.
In both cases, you reach an inconsistency, meaning $\rho$ is not rational.
Edit:
How do we know that every rational number has a repeating pattern? Well, a rational number is equal to $\frac pq$ for some numbers $p,q$. Think about how you calculate the decimal expansion of $\frac pq$:
- The first number in the expansion is the whole part of $\frac pq$, i.e. $p = k\cdot q + n$ for some $0\leq n_1<q$ ($n_1$ is the remainder here)
- The second number is the whole part of the division $n_1:q$, i.e. $n_1 = k_2 q + n_2$ for some $0\leq n_2<q$.
- The third number is the whole part of the division $n_2:q$...
In this generation of $n_1,n_2,n_3$, it is clear that if $n_i = n_j$, then $n_{i+1} = n_{j+1}$. It is also clear that since all numbers $n_1,n_2,\dots$ are whole numbers between $0$ and $q-1$, then at least two numbers in the sequence $$n_1,n_2,\dots n_q, n_{q+1}$$ are equal. Let's say that $n_i = n_j$ for $i<j$. Then
$$n_1,n_2,n_3,\dots = n_1, n_2, \dots, n_i, n_{i+1}, \dots, n_j (=n_{i}),\dots $$ and the sequence is repeating itself.
5xum
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@A.Breust Apply long division to a fraction $\tfrac 1q$ - there can be at most $(q-1)$ different remainders, so after at most $q$ steps digits of the result will repeat. EDIT Strange, the comment which I tried to answer disappeared before I posted the answer... – CiaPan Oct 21 '14 at 12:50
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Because if you try to find the decimal expansion the rational $r=\frac{m}{n}$, you divide $m$ by $n$, take the remainder, multiply by $10$, then divide by $n$,... This procedure has to repeat at some point as you only have finite remainders when dividing by $n$. – Quang Hoang Oct 21 '14 at 12:50
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@A.Breust I edited my answer with an explanation about why all rationals have a repeating pattern. – 5xum Oct 21 '14 at 12:53
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Mathematica says that your sum is equal to $\frac{\sqrt[8]{5} \vartheta _2\left(0,\frac{1}{\sqrt{10}}\right)}{2^{7/8}}$
Dr. Sonnhard Graubner
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