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Approximation of pi based on sqrt(10) - Namir - 09-07-2017 02:44 PM
I was playing with my new NumWorks calculator (OK, OK, I bragging about having one in my possession :-) ) and decided to examine how close sqrt(10) is to pi and try to obtain a (reasonable) continuous fraction to calculate the difference between sqrt(10) and pi. I got the following
pi is approx = sqrt(10) - 1/(48 + 1/(2 + 1/(1 + 1/(9 + 1/2))))
let the RHS be p, I got
(p - pi)/pi * 1E9 = 110.26
Using the approximation of q=355/113 I get
(p - pi)/pi * 1E9 = 84.91
So my approximation, based on sqrt(10), is a bit less accurate than 355/113 but by not a whopping difference (ratio of errors is about 1.3)!
:-)
Namir
RE: Approximation of pi based on sqrt(10) - SlideRule - 09-07-2017 03:34 PM
Namir
Has this reference come to your attention?
[attachment=5160]
It contains numerous examples of approximation formulas/relationships, etc..
BEST!
SlideRule
RE: Approximation of pi based on sqrt(10) - Dieter - 09-07-2017 05:46 PM
(09-07-2017 02:44 PM)Namir Wrote: let the RHS be p, I got
(p - pi)/pi * 1E9 = 110.26
Using the approximation of q=355/113 I get
(p - pi)/pi * 1E9 = 84.91
I don't know what "RHS" is (assuming that you don't refer to the Retired Husband Syndrome), but I got different results for the relative error:
The √10-based approximation has a relative error of 9,11 E–9
while for 355/113 it's 8,49 E–8.
So the former yields seven correct decimal places while for the latter it's six digits: *)
p = 3,1415926822...
q = 3,1415929203...
In other words, your approximation actually is a bit more accurate than 355/113.
(09-07-2017 02:44 PM)Namir Wrote: So my approximation, based on sqrt(10), is a bit less accurate than 355/113 but by not a whopping difference (ratio of errors is about 1.3)!
I don't know how you get these results, but the √10-approximation is 9–10x more accurate than 355/113. That's why it yields one more correct digit.
BTW, the continued fraction part is just a complicated way of writing 61/2949. So pi ~ √10 – 61/2949. ;-)
Dieter
--
*) German is known as a language where composite nouns are quite common. For instance, the digits right of the decimal point/comma are simply "Nachkommastellen". Maybe a native speaker can tell me if there is a comparably compact term in English ?-)
RE: Approximation of pi based on sqrt(10) - Thomas Okken - 09-07-2017 07:10 PM
(09-07-2017 05:46 PM)Dieter Wrote: *) German is known as a language where composite nouns are quite common. For instance, the digits right of the decimal point/comma are simply "Nachkommastellen". Maybe a native speaker can tell me if there is a comparably compact term in English ?-)
Not a native speaker, but what you're looking for is "decimal places."
https://en.wiktionary.org/wiki/decimal_place
RE: Approximation of pi based on sqrt(10) - Namir - 09-07-2017 09:17 PM
(09-07-2017 03:34 PM)SlideRule Wrote: Namir
Has this reference come to your attention?
It contains numerous examples of approximation formulas/relationships, etc..
BEST!
SlideRule
I think I have an electronic version of that book.
Namir
RE: Approximation of pi based on sqrt(10) - Namir - 09-07-2017 09:18 PM
(09-07-2017 05:46 PM)Dieter Wrote:(09-07-2017 02:44 PM)Namir Wrote: let the RHS be p, I got
(p - pi)/pi * 1E9 = 110.26
Using the approximation of q=355/113 I get
(p - pi)/pi * 1E9 = 84.91
I don't know what "RHS" is (assuming that you don't refer to the Retired Husband Syndrome), but I got different results for the relative error:
The √10-based approximation has a relative error of 9,11 E–9
while for 355/113 it's 8,49 E–8.
So the former yields seven correct decimal places while for the latter it's six digits: *)
p = 3,1415926822...
q = 3,1415929203...
In other words, your approximation actually is a bit more accurate than 355/113.
(09-07-2017 02:44 PM)Namir Wrote: So my approximation, based on sqrt(10), is a bit less accurate than 355/113 but by not a whopping difference (ratio of errors is about 1.3)!
I don't know how you get these results, but the √10-approximation is 9–10x more accurate than 355/113. That's why it yields one more correct digit.
BTW, the continued fraction part is just a complicated way of writing 61/2949. So pi ~ √10 – 61/2949. ;-)
Dieter
--
*) German is known as a language where composite nouns are quite common. For instance, the digits right of the decimal point/comma are simply "Nachkommastellen". Maybe a native speaker can tell me if there is a comparably compact term in English ?-)
I was using the new NumWorks calculator. I am not surprised about the difference in the results you obtained.
Namir
RE: Approximation of pi based on sqrt(10) - Dieter - 09-08-2017 08:43 PM
(09-07-2017 09:18 PM)Namir Wrote: I was using the new NumWorks calculator. I am not surprised about the difference in the results you obtained.
The result for 355/113 is fine. But do you want to say that the relative error of the √10 approximation is returned as 1,1026 E–7 ?
You said:
Quote:let the RHS be p, I got
(p - pi)/pi * 1E9 = 110.26
Here p = √10 – 61/2949 = 3,141592682209749. Does the NumWorks calculator return a different value?
Pi = 3,141592653589793, and I assume that this is also true for the NumWorks calculator.
So (p – pi)/pi * 1E9 = 2,8619956E-8 / 3,141592653589793 * 1E9 = 9,1100149...
I cannot imagine that the NumWorks calculator really returns 110,26 for this.
On the other hand the relative error you mentioned (110,26 E–9) is returned if you assume that p = 3,141593 exactly instead of 3,14159268... May this be the reason for the result you got? Did you possibly do the calculation with p rounded to six decimal places?
Dieter
RE: Approximation of pi based on sqrt(10) - Dieter - 09-08-2017 08:58 PM
(09-07-2017 07:10 PM)Thomas Okken Wrote: Not a native speaker, but what you're looking for is "decimal places."
https://en.wiktionary.org/wiki/decimal_place
Thank you.
Then I read the wiktionary entry for "significant digit" and now I wonder if wiktionary really is a reliable source. According to the given definition ("A digit that is nonzero or, when zero, followed only by further zeros") 1000001 has just two significant digits. #-)
Dieter,
confused.
RE: Approximation of pi based on sqrt(10) - pier4r - 09-08-2017 09:09 PM
well wiktionary (or whatever wiki*) is as reliable as it gets. If someone as a better definition, be bold and put it in.
After enough iterations, it should be useful.
The drawback may be when "democracy" is applied to knowledge. In that case, well, the result could be not good enough. In those cases proper moderators/rules are needed. Wiki may miss moderators, but surely has enough rules.
RE: Approximation of pi based on sqrt(10) - Arno K - 09-08-2017 09:13 PM
(09-08-2017 08:58 PM)Dieter Wrote:No, reread your citation, followed by only Zeros.(09-07-2017 07:10 PM)Thomas Okken Wrote: Not a native speaker, but what you're looking for is "decimal places."
https://en.wiktionary.org/wiki/decimal_place
Thank you.
Then I read the wiktionary entry for "significant digit" and now I wonder if wiktionary really is a reliable source. According to the given definition ("A digit that is nonzero or, when zero, followed only by further zeros") 1000001 has just two significant digits. #-)
Arno
Dieter,
confused.
RE: Approximation of pi based on sqrt(10) - Dieter - 09-08-2017 09:27 PM
(09-08-2017 09:13 PM)Arno K Wrote: No, reread your citation, followed by only Zeros.
Arno
OK, here's the definition again:
"A digit that is nonzero" – this applies to the first and last digit of 1000001. This accounts for two significant digits.
"Or, when zero" (i.e. the zeroes in the middle of 1000001) "followed only by further zeros". This does not apply here since the five zeroes are followed by a 1. So according to the defintion these zeroes are considered non-significant digits.
Do I miss something here?
Dieter
RE: Approximation of pi based on sqrt(10) - Arno K - 09-08-2017 09:33 PM
I read that as 1.2340000 has 4 significant decimals, so your number 1000001 has 7 of them.
Arno
RE: Approximation of pi based on sqrt(10) - Tim Wessman - 09-08-2017 09:54 PM
I do find it a bit ridiculous they (NumWorks) specifically avoid mentioning the HP Prime in any of the comparisons they've published so far that do reference "old" calculators. They recognized that they'd come off looking bad in most of the "key points" (exception being the open-ness of course) if they included us...
RE: Approximation of pi based on sqrt(10) - brickviking - 09-08-2017 10:05 PM
Funnily enough, my first comparison of the NumWorks calculator was against the Prime, but then I thought of comparison against Casio's offerings too (the non-touchscreen ones). If this had programmability in, and could be prodded into RPN/RPL, it'd be even better. As it is, it's a "it'll do for kids who won't use it for anything else" graphing calculator. People seem to like it, the development's ongoing—it can only get better. The fact that it has a joypad that looks like an old Nintendo is offputting to me, but I'd live with it if the buttons were decent.
(Post 76)
RE: Approximation of pi based on sqrt(10) - Thomas Okken - 09-08-2017 10:38 PM
(09-08-2017 08:58 PM)Dieter Wrote:(09-07-2017 07:10 PM)Thomas Okken Wrote: Not a native speaker, but what you're looking for is "decimal places."
https://en.wiktionary.org/wiki/decimal_place
Thank you.
Then I read the wiktionary entry for "significant digit" and now I wonder if wiktionary really is a reliable source. According to the given definition ("A digit that is nonzero or, when zero, followed only by further zeros") 1000001 has just two significant digits. #-)
Dieter,
confused.
Language, being a tool of communication, is always going to be a matter of consensus, not absolute truth. I don't think there's anything controversial about the meaning of "decimal place," though, and I didn't post the wiktionary link as an appeal to authority; it's just a reference.
RE: Approximation of pi based on sqrt(10) - Claudio L. - 09-09-2017 02:41 AM
(09-08-2017 09:54 PM)Tim Wessman Wrote: I do find it a bit ridiculous they (NumWorks) specifically avoid mentioning the HP Prime in any of the comparisons they've published so far that do reference "old" calculators. They recognized that they'd come off looking bad in most of the "key points" (exception being the open-ness of course) if they included us...
I agree, they claim to have rejuvenated the calculator scene, when the Prime has been out for so many years, and their software is barely a wrapper on the C standard math library with some nice fonts and color graphics.
RE: Approximation of pi based on sqrt(10) - Namir - 09-09-2017 09:07 AM
(09-08-2017 09:54 PM)Tim Wessman Wrote: I do find it a bit ridiculous they (NumWorks) specifically avoid mentioning the HP Prime in any of the comparisons they've published so far that do reference "old" calculators. They recognized that they'd come off looking bad in most of the "key points" (exception being the open-ness of course) if they included us...
While the NumWorks calculator supports plotting functions, the software's decision for the initial range seems annoying to me. It uses a wide range by default. One has to tweak the range for the X (and sometimes the Y) values to get a better view of the graph. The calculator allows you to Zoom in on the graph. I think actually selecting the X (and also Y) range is a more direct approach. The lack of a regular manual (more than the single page on the company's web site) is annoying.
RE: Approximation of pi based on sqrt(10) - Gerson W. Barbosa - 09-09-2017 09:47 AM
(09-07-2017 05:46 PM)Dieter Wrote:(09-07-2017 02:44 PM)Namir Wrote: let the RHS be p, I got
(p - pi)/pi * 1E9 = 110.26
Using the approximation of q=355/113 I get
(p - pi)/pi * 1E9 = 84.91
I don't know what "RHS" is (assuming that you don't refer to the Retired Husband Syndrome),
Right-Hand Side (ONIK)
https://jakubmarian.com/right-side-vs-right-hand-side-in-english/
ONIK = only now I knew :-)
Gerson.
RE: Approximation of pi based on sqrt(10) - pier4r - 09-09-2017 11:34 AM
(09-09-2017 09:47 AM)Gerson W. Barbosa Wrote: ONIK = only now I knew :-)
TIL!
RE: Approximation of pi based on sqrt(10) - Gerson W. Barbosa - 09-09-2017 05:45 PM
(09-07-2017 09:18 PM)Namir Wrote:(09-07-2017 05:46 PM)Dieter Wrote: I don't know what "RHS" is (assuming that you don't refer to the Retired Husband Syndrome), but I got different results for the relative error:
The √10-based approximation has a relative error of 9,11 E–9
while for 355/113 it's 8,49 E–8.
So the former yields seven correct decimal places while for the latter it's six digits: *)
p = 3,1415926822...
q = 3,1415929203...
In other words, your approximation actually is a bit more accurate than 355/113.
I don't know how you get these results, but the √10-approximation is 9–10x more accurate than 355/113. That's why it yields one more correct digit.
BTW, the continued fraction part is just a complicated way of writing 61/2949. So pi ~ √10 – 61/2949. ;-)
Dieter
--
*) German is known as a language where composite nouns are quite common. For instance, the digits right of the decimal point/comma are simply "Nachkommastellen". Maybe a native speaker can tell me if there is a comparably compact term in English ?-)
I was using the new NumWorks calculator. I am not surprised about the difference in the results you obtained.
Namir
NumWorks appears to round the results to the number of digit in the display. SCI 5 and FIX 5 are the only available formats, but I may not have explored the online simulator enough yet. Anyway, thanks to this feature I was able to find yet another \(\pi\) approximation:
\[\frac{3141593-\frac{\sqrt{3 }}{5}}{1000000}\]
Please don't use NumWorks to see how good it is. Use the WP34S instead:-)
Gerson.
PS:
On NumWorks do
\(\pi\) EXE
Ans - \(\pi\) EXE
=> 3.464102E-7
The first significant digits of \(\sqrt{12}\) are easily recognizable, aren't they?