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I completed the first part of this question, which is stated [here] Series convergence for sequence of zeroes and ones

I showed that the sequence of partial sums $S_n$ is bounded $0\le S_n\le1$, so as n goes to infinity, the sequence of partial sums converges to some value between 0 and 1. This is the second part of the question: Let $x\in[0,1]$. Prove that there exists a sequence $(a_n)_{n\in N}$ such that $$x=\sum_{n=1}^\infty \frac{a_n}{2^n}$$ This is the binary expansion of $x$. Isn't this part of the question basically stating what I proved in the first, i.e. I showed that the sequence of partial sums is between 0 and 1. Or am I misunderstanding the question?

Any clarification is appreciated.

Thank you.

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Hint: If you have defined $a_1,a_2,\dots,a_n$, let $a_{n+1}=1$ if $\frac{a_1}{2}+\cdots+\frac{a_n}{2^n}+\frac{1}{2^{n+1}}\le x$, and $0$ otherwise. – André Nicolas Nov 12 '12 at 7:01
This is clearly the way to go, via an inductive definition of the $a_k$. Should also throw the bounds into the inductive definition of sum-so-far < x < sum-so-far +1/2^n or the like. +1 for comment. – coffeemath Nov 12 '12 at 7:16
Thank you, so I use induction? Is this similar to Cantor's diagonal argument? – Alti Nov 12 '12 at 20:46
It's not like the diagonal argument. What Andre's remark is about is how to construct the sequence of $a_n$ from the number $x$. If you do some cases, you'll see that (given $a_1,...,a_n$ are correct), the definition gives the correct next base 2 digit $a_{n+1}$. – coffeemath Nov 12 '12 at 23:53
Can you give me an example? I'm sorry, I feel it's not clicking yet. – Alti Nov 13 '12 at 1:08
up vote 1 down vote accepted

In this part you are given x and have to show the existence of the sequenced $a_n$.

So this means, for each fixed $x$ in $[0,1]$, you have to do two things:

1) show how to construct the sequence $a_n$ from the number $x$, and

2) show that the resulting sum $\sum a_n/2^n$ actually converges to $x$.

In the previous part of the question you showed such sums converge, but in this latter part of the question you are to show for each $x$ you can find an appropriate sequence which converges with its sum being that particular $x$.

EDIT: The OP Alti has asked for details on the construction of the $a_n$ from the number $x \in [0,1]$. First note that the particular number $x=1$ has the expression in which all $a_n=1$, i.e. $1=1/2+1/4+1/8+...$, so that we may assume in fact that $X \in [0,1)$, the half-open interval where $0 \le x < 1.$

To get the construction started, we use that $[0,1)=[0,1/2) \cup [1/2,1)$ where the union is disjoint. We let $a_1=0$ if $x \in [0,1/2)$ and $a_1=1$ if $x \in [1/2,1)$. Note for this "base case" of the construction that we have $a_1/2 \le x < a_1/2+1/2$, which may be restated as $x \in [a_1/2,a_1/2+1/2)$ To construct the next $a$, which is $a_2$, we use that $$[a_1/2,a_1/2+1/2)=[a_1/2,a_1/2+1/4) \cup [a_1/2+1/4,a_1/2+1/2),$$ the union again being disjoint. We then define $a_2=0$ if $x$ is in the first half of the above disjoint union, and $a_1=1$ if $x lies in the second half of the above union.

For notation of left and right endpoints, let $v_n=a_1/2+a_2/4+...+a_n/2^n$, so that $v_n$ is the $n^{th}$ partial sum of the series we are constructing. Then provided we have inductively constructed each of $a_1,a_2,...,a_n$ we have at that stage that $x \in [v_n,v_n+1/2^n).$ For constructing $a_{n+1}$ we use the disjoint union $$[v_n,v_n+1/2^n)=[v_n,v_n+1/2^{n+1}) \cup [v_n+1/2^{n+1},v_n+1/2^n).$$ This is another disjoint union, and we let $a_{n+1}=0$ if $x$ lies in the left half and $a_{n+1}=1$ if $x$ lies in the right half.

Convergence of the partial sums to $x$ can be based on the nested interval theorem, or on using the partial sums of the series and also the lengths of the constructed half-open intervals in the proof.

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So I would let $x$ be in $[0,1]$, and let $(a_n)$ be a sequence such that $a_n$ is in {0,1} for all n. I'm not sure what you mean by constructing the sequence $a_n$ from $x$, though. $x$ is a number, so I would construct a number of zeroes and ones? – Alti Nov 12 '12 at 6:53
Yes, each particular $x$ would have its own particular sequence of 0's and 1's, which are obtained from the base two decimal version of $x$. For example I think that 1/3 has the base two expansion $0.01010101...$ where the digits repeat in blocls of two. – coffeemath Nov 12 '12 at 7:11
Thank you for all your help! – Alti Nov 14 '12 at 3:49
Alti: An interesting topic to look up is "bisection method" or some title like that. Basically one takes an interval and keeps cutting it in half, and focussing on where your point $x$ is, then cutting that part in half, and so on. For $x \in [0,1]$ you can keep track of the bisections and get base 2 digits of $x$. – coffeemath Nov 14 '12 at 4:08

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