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I've been looking into Egyptian fractions now, but information on certain topics seems scarce. Can you answer any of these questions that intrigue me:

1) What is known about the Egyptian fraction representation (by the greedy algorithm) of irrational numbers? Are Egyptian fractions known to be interesting in any similar sense as the continued fractions? I have not found any references to these expansions; OEIS lists a few terms for some numbers though.

2) What would be the "average" growth of the n:th term in the sequence? The slowest possible would be 2, 3, 7 etc. but for a random real, what is the expected size of the n:th term?

user102136
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  • You should give some thought toward making Q2 more precise. For instance, what do you mean by "random real"? Perhaps restricting to $(0,1)$ is worthwhile. Note that the growth rate of terms is quite rapid and the expected value of the first term is already infinite (assuming uniform distribution on $(0,1)$), so perhaps a more logarithmic measure of size is warranted. – Erick Wong Oct 20 '13 at 19:11
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    If I tried, I would most probably fail. I realize it's not accurate at all but I just want to see if someone has anything relevant to say about the growth of such sequences. Yes, but also there is a probability of 1/2 that the first term is 1/2, assuming uniform distribution on (0,1).

    I guess what I'd like to ask is, for a such a random real r, 0<r<1, which number is most likely to be the n:th term. This clearly has a solution.

     – user102136 Oct 20 '13 at 19:22

2 Answers2

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There's a little bit of info here:
H. E. Salzer, The Approximation of Numbers as Sums of Reciprocals, American Mathematical Monthly, Vol. 54, No. 3 (Mar., 1947), pp. 135-142. There are very few "naturally-occcurring" real numbers for which their expansion is known. One example where it is known is for (3-sqrt(5))/2, where the denominators are given by the (2^n)th Fibonacci number.

Jeffrey Shallit
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  • so using egyptian fraction ( greedy algorithm ) for irrational numbers means that we can write irrational numbers as sums of unit fraction (rational), but knowing that the sum of rationals is always rational leads to a contradiction , how can an irrational number be equal to a rational number(sum of rational numbers)? – Maths Survivor Oct 29 '17 at 20:35
  • For irrational numbers you get an infinite sum, not a finite one. – Jeffrey Shallit Oct 29 '17 at 21:14
  • but yet no matter if it is finite or infinite , the sum of rationals is still rational right? – Maths Survivor Oct 29 '17 at 21:20
  • No, of course not. Take any irrational number; then its base-10 representation expresses it as the infinite sum of rational numbers. – Jeffrey Shallit Oct 29 '17 at 22:07
  • Does this mean that irrational numbers are nothing but infinite sums of rationals? – Maths Survivor Oct 30 '17 at 08:02
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After trying to parse the following painstakingly-simple algorithm into plain English, I gave up all hope:

$$N = \underbrace{\underbrace{\underbrace{\underbrace{\underbrace{\underbrace{[N]}_{A_0}+\frac1{\left[N-A_0\right]}}_{A_1}+\frac1{\left[N-A_1\right]}}_{A_2}+\frac1{\left[N-A_2\right]}}_{A_3}+\frac1{\left[N-A_3\right]}}_{A_4}+\frac1{\left[N-A_4\right]}}_{A_5}+\ldots$$

If the terms are not allowed to oscillate ($\pm$) , then, with the exception of [N] , [...] should stand for upper-rounding. However, if fast convergence is what's desired, [...] should stand for rounding. In any case, continued fractions make for better convergents than the Egyptian ones under the same conditions/restrictions/limitations.

The famous Hungarian—Jewish mathematician Erdos devoted a significant part of his life to the study of Egyptian fractions. See: Greedy algorithm for Egyptian fractions , Odd greedy expansion , Erdos-Graham conjecture , and Erdos-Straus conjecture.

Lucian
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  • Your first sentence is much more suggestive of continued fractions, not Egyptian fractions. – Erick Wong Oct 20 '13 at 19:18
  • I'm not sure if you read my question at all. – user102136 Oct 20 '13 at 19:30
  • Take $\pi$. Calculate its integer part. It's 3. Subtract it, then calculate the inverse of its fractional part. It's 7. Now subtract $\tfrac17$ from $\pi-3$. Again, calculate the inverse of its fractional part. It's - 790. Now add $\tfrac1{790}$ to $\pi-3-\tfrac17$ , and calculate the inverse of its fractional part. It's 749896. Subtract $\tfrac1{749896}$ from $\pi-3-\tfrac17+\tfrac1{790}$ . Etc. – Lucian Oct 20 '13 at 19:30
  • @Lucian I agree with your calculation, but what you wrote (eliminate the integer part) does not really describe the action you perform (subtract $1/7$). The links you gave are relevant to the question, so I'm not accusing you of conflating Egyptian and continued fractions, but this doesn't contribute substantially towards an actual answer. – Erick Wong Oct 20 '13 at 20:01
  • I can read questions, but I can't read minds. Seven is the integer part of $\frac1{\pi-3}$. If the terms are not allowed to oscillate, then upper-rounding should replace the integer part. But oscillation makes for a better approximation. Either way, continuous fractions are better than both in this regard. – Lucian Oct 20 '13 at 20:11
  • @Lucian Yes, I understand that $7$ can be taken as the integer part of $1/(\pi-3)$. But I would have to be a mind-reader indeed in order to interpret "eliminate the integer part of $1/(\pi-3)$" (being $7$) to mean subtracting the reciprocal of $7$ from the reciprocal of $1/(\pi-3)$. On the other hand, the exact same sentence perfectly describes a continued fraction algorithm. – Erick Wong Oct 21 '13 at 03:43