Efficient Integer Representations of Irrationals - Page 3

Raman · 2009-10-28, 15:14

There is a power of 7, very early that is surprisingly very close to a power of 10.
7⁵¹⁰ ~ 10⁴³¹
This creates out a big number early in the continued fraction expansion for log 7.
How efficient is the approximation for log 7 as 431/510?

How good is the approximation for log 3 as 8519/17855?
Also, have a look up at the continued fraction expansions of the other logarithms too.
All these logarithms are to the base 10 only.
log 2, log 3, log 6, log 7, log 11...
log 5 = 1 - log 2: So that it creates up the same fractions within the continued fraction expansion essentially.

JonBarleycorn · 2009-11-22, 23:58

What are the most efficient integer representations of the irrational numbers π and e?

An efficient integer representation of an irrational number is defined as two integers p/q which, when expressed as a decimal, gives at least as many significant digits as the sum of the digits in p and in q. Thus, an efficient representation will be both mnemonic and accurate, requiring no fewer digits to write than the accuracy gained.

For example, 22/7 is an efficient integer representation of π (3.141592653589793238 …) since the sum of the integer digits (2 + 1) is 3 and 22/7 is accurate to just a bit better than 1 part in 791. Note that 22/7 – 1/791 = 355/113.

This representation, 355/113, is both efficient and memorable since it takes just 6 digits to get 7 digit accuracy (1 part in 3,748,629).

There was a nice discussion by ewmayer and CRGreathouse of a metric that gives a slight edge in efficiency to 22/7. But, 355/113 is reasonably memorable, and, it is the only representation to give a full digit of additional accuracy. Thus, this simple minded approach gives 355/113 the efficiency prize.

Finding a mnemonic representation for e (2.71828182845904523 …) is probably unnecessary because of the repeating ‘1828’ group. However, an efficient 3 digit representation is 19/7 which is good to about 1 part in 250. A 5 digit representation, 193/71, is good to 1 part in 35,675. An eight digit representation, 2721/1001, while accurate to 1 part in 9,076,277, is not efficient (8 digits needed for 7 digit accuracy). However, it is fairly easy to remember so it is being mentioned.

Others have pointed out that 49171/18089 = 2.718281828735. This is efficient and accurate to 1 part in 3,614,669,163. But, e itself, is just as easy to remember to 10 digits because of the repeating 1828 group so I cannot award 49171/18089 the prize.

Given that short term memory is limited to about 7 +/- 2 chunks of information, further exploration with larger fractions is probably fruitless.

JB

philmoore · 2009-11-23, 00:18

Thanks for the concise essay, JB. I remember reading somewhere of an Arabic mathematician of the middle ages questioning why 22/7 was being used for pi when it was "common knowledge" that 355/113 was a much better approximation!

biwema · 2009-11-23, 21:32

Quote:

Originally Posted by mart_r

A good one is 355/113.
There's this number 20776 at position 432 in the continued fraction expansion of pi, if I would calculate the fraction of the 431 previous terms I would also get an efficient integer representation, or whenever such a large number in the continued fraction appears.

The best known approximation is log(640320³+744)/sqrt(163) - it gives 31 significant digits!

Edit: but I don't know of any really good approximations for e.

There are efficient ways to write e.

(1 + 1/999!)^(999!)

gives more than 2500 digits of e.

Nevertheless i think it is not a valid solution because it is a general defintion of e:
(1+1/n)^n approximates e when n goes to infinity.

On the other hand the 31 digits approximation of pi is neat but neither a fraction as mentioined in this thread.

JonBarleycorn · 2009-11-24, 07:25

Forgive me if my calculator is not working right, but every time I tried log(640320 cubed + 744) / sqrt(163) I got 12.767145.. even when I used all the precision that Windows has to offer.

If I got it wrong then I owe mart_r an apology.

I'm sorry I wasn't more clear that I was looking for 2 integers p, q such that p/q was memorable and a good representation of pi and e. I hope that I was more clear in my statement of the problem the second time around.

That said, some of the excursions were really first class, and, I was really impressed by the caliber of discourse that my little puzzle elicited! Thanks to all who responded.

JB

retina · 2009-11-24, 07:42

Quote:

Originally Posted by JonBarleycorn

Forgive me if my calculator is not working right, but every time I tried log(640320 cubed + 744) / sqrt(163) I got 12.767145.. even when I used all the precision that Windows has to offer.

If I got it wrong then I owe mart_r an apology.

I'm sorry I wasn't more clear that I was looking for 2 integers p, q such that p/q was memorable and a good representation of pi and e. I hope that I was more clear in my statement of the problem the second time around.

That said, some of the excursions were really first class, and, I was really impressed by the caliber of discourse that my little puzzle elicited! Thanks to all who responded.

JB

For the "log" press the "ln" button.

But your answer above, 12.767145.., is simply sqrt(163). Perhaps your forgot to press the "=" button?

davieddy · 2009-11-24, 17:07

2^(7/12) = ~3/2

So 7 "equal" semitones make nearly a "perfect fifth"
and 5 nearly a "perfect fourth"

Johann Sebastian

Raman · 2012-03-24, 18:06

One of the ways to compute the efficiency of an approximation for irrationals
by using a given fraction
x = p/q
is to compute the error ratio for
(10^x)^q against (10^p) and then multiply it by using p.

Taking the value for x = log 2 ~ 3/10
for an example, the error ratio for
2¹⁰ against 10³ = 2.4% Multiplying it with the exponent 3, yields the ratio 7.2
for 6/20, 2²⁰ against 10⁶ = 4.8576% Multiplying it with the exponent 6 yields the ratio 29.1456
So, this ratio increases when the numerator, denominator together are being multiplied by using the multiplier. Lowest terms always give the best approximation factor ratios.

For log 2, the ratio for
28/93 = 27.014312
59/196 = 25.584038
146/485 = 15.190773
643/2136 = 10.474093
4004/13301 = 25.512806
30103/100000 = 3003.9994

For log 3, the ratio for
21/44 = 31.981105
73/153 = 7.52696
1074/2251 = 13.755307
8519/17855 = 5.923131

The fun is that
73/153, when both the numerator, denominator are being multiplied by using 2232, yields 162936/341496
but that the value for 162935/341496 is always being a better closer approximation to the value for log 3

For log 6, undoubtedly the best approximation =
463/595 with the ratio = 0.64197
against the fraction 7/9, with such ratio, factor = 5.43872

For log 7, undoubtedly the best approximation =
431/510 with the ratio = 0.040418

This is being the effect for the big term very early within the continued fraction expansion for log 7 =
1/1+ 1/5+ 1/2+ 1/5+ 1/6+ 1/1+ 1/4813+ ...

For log 6,
0 + 1/1+ 1/3+ 1/1+ 1/1+ 1/32+ 1/1+ 1/1+ 1/278+ ...

Comparing against the efficient approximations values for PI
22/7 = 14.1224069
333/106 = 683.2972314
355/113 = 2.46396731

Raman · 2012-03-24, 18:59

Taking the continued fraction expansion for (log b)
gives out the powers of b (within the denominator), digits (within the numerator), that are being increasingly closer to the powers of 10.

Question. Is there any way to give out the sequence for the powers of N that are being increasingly closer to the values for that following form, k.10ⁿ?

For this example, consider with the following terms that are being

2¹⁷¹ ≈ 3.10⁵¹
2¹³³⁷ ≈ 3.10⁴⁰²
2³⁰⁰⁷⁵ ≈ 3.10⁹⁰⁵³

2⁴⁶ ≈ 7.10¹³
2³³⁵ ≈ 7.10¹⁰⁰
2²⁴⁷¹ ≈ 7.10⁷⁴³
2¹⁵⁷⁷² ≈ 7.10⁴⁷⁴⁷
2¹⁵⁷³²⁶ ≈ 7.10⁴⁷³⁵⁹

2⁵³ ≈ 9.10¹⁵
2⁵³⁸ ≈ 9.10¹⁶¹
2²⁶⁷⁴ ≈ 9.10⁸⁰⁴
2⁴⁸¹⁰ ≈ 9.10¹⁴⁴⁷
2⁶⁰¹⁵⁰ ≈ 9.10¹⁸¹⁰⁶

Here are a few more good approximations
7.2²²³⁹ ≈ 71.10⁶⁷³
3^-14263 ≈ 66.10^-6807
3²¹⁹⁷³ ≈ 61.10¹⁰⁴⁸²

3⁶⁵⁶⁵ ≈ 2.10³¹³² ________ 3¹⁰⁴⁷⁵ ≈ 7.10⁴⁹⁹⁷ _______ 7.3¹⁷ ≈ 9.10⁸ _______ 3⁹⁴ ≈ 7.10⁴⁴
3¹³¹³⁰ ≈ 4.10⁶²⁶⁴ _______ 3⁴⁰⁹¹ ≈ 8.10¹⁹⁵¹ ________ 3³⁵ ≈ 5.10¹⁶ ________ 3¹³⁸ ≈ 7.10⁶⁵
3¹¹²⁹⁰ ≈ 5.10⁵³⁸⁶ _______ 3⁴⁷²⁵ ≈ 25.10²²⁵³ _______ 3⁴⁸ ≈ 8.10²² ________ 7.3¹⁰⁹ ≈ 71.10⁵¹
3⁶⁵⁶⁶ ≈ 6.10³¹³² ________ 3³⁵⁹² ≈ 66.10¹⁷¹² _______ 3⁸³ ≈ 4.10³⁹_________ 7.2¹⁰³ ≈ 71.10³⁰
3¹²⁷²⁶ ≈ 7.10⁶⁰⁷¹ _______ 3²¹⁴⁴⁷ ≈ 66.10¹⁰²³¹ _____ 3¹⁸⁸ ≈ 5.10⁸⁹
3¹⁸⁴⁰ ≈ 8.10⁸⁷⁷ _________ 3⁴¹¹⁸ ≈ 61.10¹⁹⁶³ _______ 3²²³ ≈ 25.10¹⁰⁵

Raman · 2012-03-30, 07:14

Quote:

Originally Posted by Raman

For this example, consider with the following terms that are being

27.2³⁰¹ ≈ 11.10⁹¹________267.2¹⁸ ≈ 7.10⁷
1249.2⁵⁶ ≈ 9.10¹⁹________266999.2²⁵ ≈ 8959.10⁹
9.5⁵³ ≈ 10³⁸_____________439.3¹³ ≈ 7.10⁸
2³¹⁷ ≈ 267.10⁹³__________7.3¹⁷ ≈ 904.10⁶
2⁵¹¹ ≈ 67.10¹⁵²__________7.3⁶ ≈ 51.10²
67.2²⁷ ≈ 9.10⁹___________7.3¹⁷ ≈ 9.10⁸

Taking the continued fraction expansion for (log b)
gives out the powers of b (within the denominator), digits (within the numerator), that are being increasingly closer to the powers of 10.

Question. Is there any way to give out the sequence for the powers of N that are being increasingly closer to the values for that following form, k.10ⁿ?

jasonp · 2012-03-30, 13:04

Yes, there is a standard estimate for the error term between a continued fraction and the decimal number it is supposed to represent, based only on the size of p and q (it's been ~15 years, I don't remember it). You can also prove that to get a better approximation of the error, you must use numbers larger than p and q.

2009-11-22, 23:58	#24
JonBarleycorn Oct 2009 5 Posts	Efficient Integer Representations of Irrationals What are the most efficient integer representations of the irrational numbers π and e? An efficient integer representation of an irrational number is defined as two integers p/q which, when expressed as a decimal, gives at least as many significant digits as the sum of the digits in p and in q. Thus, an efficient representation will be both mnemonic and accurate, requiring no fewer digits to write than the accuracy gained. For example, 22/7 is an efficient integer representation of π (3.141592653589793238 …) since the sum of the integer digits (2 + 1) is 3 and 22/7 is accurate to just a bit better than 1 part in 791. Note that 22/7 – 1/791 = 355/113. This representation, 355/113, is both efficient and memorable since it takes just 6 digits to get 7 digit accuracy (1 part in 3,748,629). There was a nice discussion by ewmayer and CRGreathouse of a metric that gives a slight edge in efficiency to 22/7. But, 355/113 is reasonably memorable, and, it is the only representation to give a full digit of additional accuracy. Thus, this simple minded approach gives 355/113 the efficiency prize. Finding a mnemonic representation for e (2.71828182845904523 …) is probably unnecessary because of the repeating ‘1828’ group. However, an efficient 3 digit representation is 19/7 which is good to about 1 part in 250. A 5 digit representation, 193/71, is good to 1 part in 35,675. An eight digit representation, 2721/1001, while accurate to 1 part in 9,076,277, is not efficient (8 digits needed for 7 digit accuracy). However, it is fairly easy to remember so it is being mentioned. Others have pointed out that 49171/18089 = 2.718281828735. This is efficient and accurate to 1 part in 3,614,669,163. But, e itself, is just as easy to remember to 10 digits because of the repeating 1828 group so I cannot award 49171/18089 the prize. Given that short term memory is limited to about 7 +/- 2 chunks of information, further exploration with larger fractions is probably fruitless. JB

2009-11-24, 17:07	#29
davieddy "Lucan" Dec 2006 England 194A₁₆ Posts	The Well-tempered Klavier 2^(7/12) = ~3/2 So 7 "equal" semitones make nearly a "perfect fifth" and 5 nearly a "perfect fourth" Johann Sebastian

2012-03-24, 18:06	#30
Raman Noodles "Mr. Tuch" Dec 2007 Chennai, India 3·419 Posts	One of the ways to compute the efficiency of an approximation for irrationals by using a given fraction x = p/q is to compute the error ratio for (10^x)^q against (10^p) and then multiply it by using p. Taking the value for x = log 2 ~ 3/10 for an example, the error ratio for 2¹⁰ against 10³ = 2.4% Multiplying it with the exponent 3, yields the ratio 7.2 for 6/20, 2²⁰ against 10⁶ = 4.8576% Multiplying it with the exponent 6 yields the ratio 29.1456 So, this ratio increases when the numerator, denominator together are being multiplied by using the multiplier. Lowest terms always give the best approximation factor ratios. For log 2, the ratio for 28/93 = 27.014312 59/196 = 25.584038 146/485 = 15.190773 643/2136 = 10.474093 4004/13301 = 25.512806 30103/100000 = 3003.9994 For log 3, the ratio for 21/44 = 31.981105 73/153 = 7.52696 1074/2251 = 13.755307 8519/17855 = 5.923131 The fun is that 73/153, when both the numerator, denominator are being multiplied by using 2232, yields 162936/341496 but that the value for 162935/341496 is always being a better closer approximation to the value for log 3 For log 6, undoubtedly the best approximation = 463/595 with the ratio = 0.64197 against the fraction 7/9, with such ratio, factor = 5.43872 For log 7, undoubtedly the best approximation = 431/510 with the ratio = 0.040418 This is being the effect for the big term very early within the continued fraction expansion for log 7 = 1/1+ 1/5+ 1/2+ 1/5+ 1/6+ 1/1+ 1/4813+ ... For log 6, 0 + 1/1+ 1/3+ 1/1+ 1/1+ 1/32+ 1/1+ 1/1+ 1/278+ ... Comparing against the efficient approximations values for PI 22/7 = 14.1224069 333/106 = 683.2972314 355/113 = 2.46396731 Last fiddled with by Raman on 2012-03-24 at 18:09

2012-03-24, 18:59	#31
Raman Noodles "Mr. Tuch" Dec 2007 Chennai, India 3·419 Posts	Question Taking the continued fraction expansion for (log b) gives out the powers of b (within the denominator), digits (within the numerator), that are being increasingly closer to the powers of 10. Question. Is there any way to give out the sequence for the powers of N that are being increasingly closer to the values for that following form, k.10ⁿ? For this example, consider with the following terms that are being 2¹⁷¹ ≈ 3.10⁵¹ 2¹³³⁷ ≈ 3.10⁴⁰² 2³⁰⁰⁷⁵ ≈ 3.10⁹⁰⁵³ 2⁴⁶ ≈ 7.10¹³ 2³³⁵ ≈ 7.10¹⁰⁰ 2²⁴⁷¹ ≈ 7.10⁷⁴³ 2¹⁵⁷⁷² ≈ 7.10⁴⁷⁴⁷ 2¹⁵⁷³²⁶ ≈ 7.10⁴⁷³⁵⁹ 2⁵³ ≈ 9.10¹⁵ 2⁵³⁸ ≈ 9.10¹⁶¹ 2²⁶⁷⁴ ≈ 9.10⁸⁰⁴ 2⁴⁸¹⁰ ≈ 9.10¹⁴⁴⁷ 2⁶⁰¹⁵⁰ ≈ 9.10¹⁸¹⁰⁶ Here are a few more good approximations 7.2²²³⁹ ≈ 71.10⁶⁷³ 3^-14263 ≈ 66.10^-6807 3²¹⁹⁷³ ≈ 61.10¹⁰⁴⁸² 3⁶⁵⁶⁵ ≈ 2.10³¹³² ________ 3¹⁰⁴⁷⁵ ≈ 7.10⁴⁹⁹⁷ _______ 7.3¹⁷ ≈ 9.10⁸ _______ 3⁹⁴ ≈ 7.10⁴⁴ 3¹³¹³⁰ ≈ 4.10⁶²⁶⁴ _______ 3⁴⁰⁹¹ ≈ 8.10¹⁹⁵¹ ________ 3³⁵ ≈ 5.10¹⁶ ________ 3¹³⁸ ≈ 7.10⁶⁵ 3¹¹²⁹⁰ ≈ 5.10⁵³⁸⁶ _______ 3⁴⁷²⁵ ≈ 25.10²²⁵³ _______ 3⁴⁸ ≈ 8.10²² ________ 7.3¹⁰⁹ ≈ 71.10⁵¹ 3⁶⁵⁶⁶ ≈ 6.10³¹³² ________ 3³⁵⁹² ≈ 66.10¹⁷¹² _______ 3⁸³ ≈ 4.10³⁹_________ 7.2¹⁰³ ≈ 71.10³⁰ 3¹²⁷²⁶ ≈ 7.10⁶⁰⁷¹ _______ 3²¹⁴⁴⁷ ≈ 66.10¹⁰²³¹ _____ 3¹⁸⁸ ≈ 5.10⁸⁹ 3¹⁸⁴⁰ ≈ 8.10⁸⁷⁷ _________ 3⁴¹¹⁸ ≈ 61.10¹⁹⁶³ _______ 3²²³ ≈ 25.10¹⁰⁵ Last fiddled with by Raman on 2012-03-24 at 19:43

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2009-10-28, 15:14	#23
Raman Noodles "Mr. Tuch" Dec 2007 Chennai, India 4E9₁₆ Posts	There is a power of 7, very early that is surprisingly very close to a power of 10. 7⁵¹⁰ ~ 10⁴³¹ This creates out a big number early in the continued fraction expansion for log 7. How efficient is the approximation for log 7 as 431/510? How good is the approximation for log 3 as 8519/17855? Also, have a look up at the continued fraction expansions of the other logarithms too. All these logarithms are to the base 10 only. log 2, log 3, log 6, log 7, log 11... log 5 = 1 - log 2: So that it creates up the same fractions within the continued fraction expansion essentially.

2009-11-23, 00:18	#25
philmoore "Phil" Sep 2002 Tracktown, U.S.A. 3×373 Posts	Thanks for the concise essay, JB. I remember reading somewhere of an Arabic mathematician of the middle ages questioning why 22/7 was being used for pi when it was "common knowledge" that 355/113 was a much better approximation!

2009-11-24, 07:25	#27
JonBarleycorn Oct 2009 5₁₀ Posts	Forgive me if my calculator is not working right, but every time I tried log(640320 cubed + 744) / sqrt(163) I got 12.767145.. even when I used all the precision that Windows has to offer. If I got it wrong then I owe mart_r an apology. I'm sorry I wasn't more clear that I was looking for 2 integers p, q such that p/q was memorable and a good representation of pi and e. I hope that I was more clear in my statement of the problem the second time around. That said, some of the excursions were really first class, and, I was really impressed by the caliber of discourse that my little puzzle elicited! Thanks to all who responded. JB

2012-03-30, 13:04	#33
jasonp Tribal Bullet Oct 2004 3·1,181 Posts	Yes, there is a standard estimate for the error term between a continued fraction and the decimal number it is supposed to represent, based only on the size of p and q (it's been ~15 years, I don't remember it). You can also prove that to get a better approximation of the error, you must use numbers larger than p and q.