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From the literature, I have found the following: Any real number A (say) can be expressed as

$ A = a_1 + (1/a_1) + (1/a_2) + (1/a_3) +\ldots$ Where $a_1\ge2$ and the recurrence relation $a_{i+1}\ge a_i(a_i - 1) + 1$ for $i \ge 1$.

I could not understand this statement due to the fowling reason:

I fixed $a_1$ = 2 then, $A = 2 + ½ + (1/a_2) + (1/a_3) + \dots$ ---------(i)

We can find $a_2$ by recurrence relation; $a_2 \ge a_1(a_1 -1) + 1 = 2(2-1) + 1 = 3$

i.e., $a_2\ge 3$ and $a_3\ge4$ and so on… Now, by our (i), $A = 2 + 1/2 + 1/3 + 1/4 + \dots$

How any real number A is equal to (i)?

I could not understand this statement.

Also, the same A can be expressible in other two ways:

$A = a_0 + (1/a_1) + (1/a_1)(1/a_2) + (1/a_1)(1/a_2)(1/a_3)+ \ldots$ Where $a_1\ge2$ and recurrence relation $a_{i+1}\ge a_i$ for $i \ge1$.

Also, $$A = a_0 + (1/a_1) + (1/(a_1-1)) (1/a_1) (1/a_2) \\+ (1/(a_1-1)) ((1/a_2-1)) (1/a_1)(1/a_2)(1/a_3) +\ldots$$

Briefly, explain where I am wrong in my example or not? Also, how to deduce one equation to other equations of A? A waiting your explanations and proofs.

Edit Is it applicable for any GIVEN real number, instead of any real number? If yes, how to proceed to complete the proof? can we deduce from one representation to other representations?

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Could you tell us where "in the literature" you found this? It might help us if we could see where this comes from. –  Old John Aug 28 '12 at 6:31
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If we are to take this at face value, real numbers $\le2$ don't exist. Curious. –  Harald Hanche-Olsen Aug 28 '12 at 6:52
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Yes, but which book does it come from? Who is the author? –  Old John Aug 28 '12 at 7:21
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As you have it there is no way of expressing zero (or any negative number, or indeed any real number less than 2.5) ... if you had $a_0$ the situation would be rather different, so you see it does matter what the statement is, and no-one will be able to answer unless you clarify the various matters noted in the comments. –  Mark Bennet Aug 28 '12 at 7:39
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@MarkBennet As a matter of fact, it is not so difficult to guess what would be a correct formulation of the first statement (and to prove it)... But I fully agree that the OP should first clarify (and maybe, o foolish hope, answer OldJohn's question about the source). –  Did Aug 28 '12 at 7:56

1 Answer 1

up vote 4 down vote accepted

I suspect that you are saying that real $A$ has a single representation as $$A = a_0 + (1/a_1) + (1/a_2) + (1/a_3) +\cdots$$ for integer $a_i$ where $a_1\ge2$ and the recurrence relation $a_{i+1}\ge a_i(a_i - 1) + 1$ for $i \ge 1$.

This is (at least for irrational $A$) often described as the Egyptian fraction representation: for example OEIS A001466 gives $$\pi = 3 + \frac{1}{8} + \frac{1}{61} + \frac{1}{5020} + \frac{1}{128541455}+\cdots .$$

To show it exists and is unique, consider the partial sum $A_i$ up to the $(1/a_i)$ term, where $a_0 = \lceil{A}\rceil - 1$ and $a_{i+1}=\left\lfloor{\frac{1}{A-A_i}}\right\rfloor +1 $.

A proof will use the following points:

  • $A_i \lt A \le A_{i-1} + \frac{1}{a_i - 1}$
  • $\frac{1}{n-1}-\frac{1}{n} = \frac{1}{n(n-1)}$
  • if $b_{j+1}= b_j(b_j - 1) + 1$ then $\sum_{j=k}^{\infty} \frac{1}{b_j} = \frac{1}{b_k -1}$

Your second expression can be handled in a similar way with suitable adjustments: for example the third bullet point becomes $\sum_{j=1}^{\infty} \left(\frac{1}{b_k}\right)^j = \frac{1}{b_k -1}$.

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HENRY Sir, you have added some good flavor to my post. Can you prove the my initial statement? I confused. please do not give hints. Prove the fist statement and then I will try for other two statements of A. –  vidyaojal Aug 28 '12 at 8:51
    
Sir! can you make proof of (i) by using your bullets? –  vidyaojal Aug 28 '12 at 10:03
    
Henry Sir, can you make proof of (i) by using your bullets? –  vidyaojal Aug 28 '12 at 10:03
    
Sir, kindly look at the proof of first part. Becoz, I applied your bullets. But those bullets are not enough to complete the proofs. Please look. I am waiting for your solution. –  vidyaojal Aug 28 '12 at 10:26
    
@vidyaojal You will understand better why this leads to a proof if you experiment with various values of $A$ and use $a_0 = \lceil{A}\rceil - 1$ and $a_{i+1}=\left\lfloor{\frac{1}{A-A_i}}\right\rfloor +1 $, keeping track of $A-A_i$ and $\frac{1}{A-A_i}$ –  Henry Aug 28 '12 at 12:50

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