Is there any mathematical reason for this “digit-repetition-show”?












112












$begingroup$


The number $$sqrt{308642}$$ has a crazy decimal representation : $$555.5555777777773333333511111102222222719999970133335210666544640008cdots $$




Is there any mathematical reason for so many repetitions of the digits ?




A long block containing only a single digit would be easier to understand. This could mean that there are extremely good rational approximations. But here we have many long one-digit-blocks , some consecutive, some interrupted by a few digits. I did not calculate the probability of such a "digit-repitition-show", but I think it is extremely small.



Does anyone have an explanation ?






number-theory radicals decimal-expansion







share|cite|improve this question









$endgroup$








  • 23




    $begingroup$
    Hint: $308642=(5000^2+2)/9^2$.
    $endgroup$
    – Chen Wang
    Feb 8 '17 at 13:09






  • 5




    $begingroup$
    In interestingly the prime factorization of this number is $ 2 times 154321 $ I wonder if the 54321 has anything to do with it?
    $endgroup$
    – Riemann-bitcoin.
    Feb 8 '17 at 13:18








  • 4




    $begingroup$
    On a related note, see Schizophrenic number
    $endgroup$
    – PM 2Ring
    Feb 10 '17 at 10:29










  • $begingroup$
    Did this come up as an actual problem or just for fun?
    $endgroup$
    – Brian Risk
    Feb 10 '17 at 20:52








  • 1




    $begingroup$
    @BrianRisk Just for fun!
    $endgroup$
    – Peter
    Feb 11 '17 at 14:01
















112












$begingroup$


The number $$sqrt{308642}$$ has a crazy decimal representation : $$555.5555777777773333333511111102222222719999970133335210666544640008cdots $$




Is there any mathematical reason for so many repetitions of the digits ?




A long block containing only a single digit would be easier to understand. This could mean that there are extremely good rational approximations. But here we have many long one-digit-blocks , some consecutive, some interrupted by a few digits. I did not calculate the probability of such a "digit-repitition-show", but I think it is extremely small.



Does anyone have an explanation ?






number-theory radicals decimal-expansion







share|cite|improve this question









$endgroup$








  • 23




    $begingroup$
    Hint: $308642=(5000^2+2)/9^2$.
    $endgroup$
    – Chen Wang
    Feb 8 '17 at 13:09






  • 5




    $begingroup$
    In interestingly the prime factorization of this number is $ 2 times 154321 $ I wonder if the 54321 has anything to do with it?
    $endgroup$
    – Riemann-bitcoin.
    Feb 8 '17 at 13:18








  • 4




    $begingroup$
    On a related note, see Schizophrenic number
    $endgroup$
    – PM 2Ring
    Feb 10 '17 at 10:29










  • $begingroup$
    Did this come up as an actual problem or just for fun?
    $endgroup$
    – Brian Risk
    Feb 10 '17 at 20:52








  • 1




    $begingroup$
    @BrianRisk Just for fun!
    $endgroup$
    – Peter
    Feb 11 '17 at 14:01














112












112








112


36



$begingroup$


The number $$sqrt{308642}$$ has a crazy decimal representation : $$555.5555777777773333333511111102222222719999970133335210666544640008cdots $$




Is there any mathematical reason for so many repetitions of the digits ?




A long block containing only a single digit would be easier to understand. This could mean that there are extremely good rational approximations. But here we have many long one-digit-blocks , some consecutive, some interrupted by a few digits. I did not calculate the probability of such a "digit-repitition-show", but I think it is extremely small.



Does anyone have an explanation ?






number-theory radicals decimal-expansion







share|cite|improve this question









$endgroup$




The number $$sqrt{308642}$$ has a crazy decimal representation : $$555.5555777777773333333511111102222222719999970133335210666544640008cdots $$




Is there any mathematical reason for so many repetitions of the digits ?




A long block containing only a single digit would be easier to understand. This could mean that there are extremely good rational approximations. But here we have many long one-digit-blocks , some consecutive, some interrupted by a few digits. I did not calculate the probability of such a "digit-repitition-show", but I think it is extremely small.



Does anyone have an explanation ?






number-theory radicals decimal-expansion




number-theory radicals decimal-expansion






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Feb 8 '17 at 13:02









PeterPeter

47.7k1139131




47.7k1139131








  • 23




    $begingroup$
    Hint: $308642=(5000^2+2)/9^2$.
    $endgroup$
    – Chen Wang
    Feb 8 '17 at 13:09






  • 5




    $begingroup$
    In interestingly the prime factorization of this number is $ 2 times 154321 $ I wonder if the 54321 has anything to do with it?
    $endgroup$
    – Riemann-bitcoin.
    Feb 8 '17 at 13:18








  • 4




    $begingroup$
    On a related note, see Schizophrenic number
    $endgroup$
    – PM 2Ring
    Feb 10 '17 at 10:29










  • $begingroup$
    Did this come up as an actual problem or just for fun?
    $endgroup$
    – Brian Risk
    Feb 10 '17 at 20:52








  • 1




    $begingroup$
    @BrianRisk Just for fun!
    $endgroup$
    – Peter
    Feb 11 '17 at 14:01














  • 23




    $begingroup$
    Hint: $308642=(5000^2+2)/9^2$.
    $endgroup$
    – Chen Wang
    Feb 8 '17 at 13:09






  • 5




    $begingroup$
    In interestingly the prime factorization of this number is $ 2 times 154321 $ I wonder if the 54321 has anything to do with it?
    $endgroup$
    – Riemann-bitcoin.
    Feb 8 '17 at 13:18








  • 4




    $begingroup$
    On a related note, see Schizophrenic number
    $endgroup$
    – PM 2Ring
    Feb 10 '17 at 10:29










  • $begingroup$
    Did this come up as an actual problem or just for fun?
    $endgroup$
    – Brian Risk
    Feb 10 '17 at 20:52








  • 1




    $begingroup$
    @BrianRisk Just for fun!
    $endgroup$
    – Peter
    Feb 11 '17 at 14:01








23




23




$begingroup$
Hint: $308642=(5000^2+2)/9^2$.
$endgroup$
– Chen Wang
Feb 8 '17 at 13:09




$begingroup$
Hint: $308642=(5000^2+2)/9^2$.
$endgroup$
– Chen Wang
Feb 8 '17 at 13:09




5




5




$begingroup$
In interestingly the prime factorization of this number is $ 2 times 154321 $ I wonder if the 54321 has anything to do with it?
$endgroup$
– Riemann-bitcoin.
Feb 8 '17 at 13:18






$begingroup$
In interestingly the prime factorization of this number is $ 2 times 154321 $ I wonder if the 54321 has anything to do with it?
$endgroup$
– Riemann-bitcoin.
Feb 8 '17 at 13:18






4




4




$begingroup$
On a related note, see Schizophrenic number
$endgroup$
– PM 2Ring
Feb 10 '17 at 10:29




$begingroup$
On a related note, see Schizophrenic number
$endgroup$
– PM 2Ring
Feb 10 '17 at 10:29












$begingroup$
Did this come up as an actual problem or just for fun?
$endgroup$
– Brian Risk
Feb 10 '17 at 20:52






$begingroup$
Did this come up as an actual problem or just for fun?
$endgroup$
– Brian Risk
Feb 10 '17 at 20:52






1




1




$begingroup$
@BrianRisk Just for fun!
$endgroup$
– Peter
Feb 11 '17 at 14:01




$begingroup$
@BrianRisk Just for fun!
$endgroup$
– Peter
Feb 11 '17 at 14:01










3 Answers
3






active

oldest

votes


















129












$begingroup$

Repeated same numbers in a decimal representation can be converted to repeated zeros by multiplication with $9$. (try it out)



so if we multiply $9 sqrt{308642} = sqrt{308642 times 81} = sqrt{25 000 002}$ since this number is allmost $5000^2$ it has a lot of zeros in its decimal expansion






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Superb answer! (+1)
    $endgroup$
    – Peter
    Feb 8 '17 at 13:19






  • 27




    $begingroup$
    And the underlying reason here is the series expansion $$ sqrt{a^2+x} = a + frac{1}{2a}x - frac{1}{(2a)^3}x^2 + frac2{(2a)^5}x^3 - frac{5}{(2a)^7}x^4 + cdots $$ which can be derived from the generalized binomial theorem. When $2a$ is a large power of $10$, this gives a nice decimal representation of the square root.
    $endgroup$
    – Henning Makholm
    Feb 8 '17 at 14:04








  • 25




    $begingroup$
    To check that this is the "right" explanation I'd find it good to have other, similar examples. And here is another one: $sqrt{1975308642} = 44444.44444472222222222135416666667209201388884650336371564...$ which can be explained by noting that $1975308642 = (400000^2 + 2)/9^2$.
    $endgroup$
    – Michael Lugo
    Feb 8 '17 at 14:37






  • 2




    $begingroup$
    It is important to note that $25000002$ is not only almost $5000^2$, but also $>5000^2$. Otherwise the same argument would work for $$sqrt{30864color{red}1}approx 555.55467777708433223768894721... .$$ Does not look so very nice! It is because $81times 308641<5000^2$, but still close to $5000^2$.
    $endgroup$
    – M. Winter
    Jan 25 '18 at 9:57





















124












$begingroup$

The architect's answer, while explaining the absolutely crucial fact that $sqrt{308642}approx 5000/9=555.555ldots$, didn't quite make it clear why we get several runs of repeating decimals. I try to shed additional light to that using a different tool.



I want to emphasize the role of the binomial series. In particular the Taylor expansion
$$
sqrt{1+x}=1+frac x2-frac{x^2}8+frac{x^3}{16}-frac{5x^4}{128}+frac{7x^5}{256}-frac{21x^6}{1024}+cdots
$$
If we plug in $x=2/(5000)^2=8cdot10^{-8}$, we get
$$
M:=sqrt{1+8cdot10^{-8}}=1+4cdot10^{-8}-8cdot10^{-16}+32cdot10^{-24}-160cdot10^{-32}+cdots.
$$
Therefore
$$
begin{aligned}
sqrt{308462}&=frac{5000}9M=frac{5000}9+frac{20000}9cdot10^{-8}-frac{40000}9cdot10^{-16}+frac{160000}9cdot10^{-24}+cdots\
&=frac{5}9cdot10^3+frac29cdot10^{-4}-frac49cdot10^{-12}+frac{16}9cdot10^{-20}+cdots.
end{aligned}
$$
This explains both the runs, their starting points, as well as the origin and location of those extra digits not part of any run. For example, the run of $5+2=7$s begins when the first two terms of the above series are "active". When the third term joins in, we need to subtract a $4$ and a run of $3$s ensues et cetera.






share|cite|improve this answer











$endgroup$









  • 15




    $begingroup$
    @Peter It is quite common to unaccept an answer after a better answer appears. Doing so helps guide readers to the best answer, which is often not the highest voted one, due to many factors, e.g. earlier answers usually get more votes, and less technical answers usually get more votes from hot-list activity (as here). This is currently (by far) the best explanation you have.
    $endgroup$
    – Bill Dubuque
    Feb 8 '17 at 20:35








  • 11




    $begingroup$
    For the record: The reason I support Peter's decision to accept the architect's answer is that mine is building upon it. Without the observation that $5000/9$ is an extremely good approximation I most likely would not have bothered, and most certainly would not have come up with this refinement. IMHO Math.SE works at its best, when different users add different points of view refining earlier answers. The voters very clearly like both the answers. Sunshine and smiles to all!
    $endgroup$
    – Jyrki Lahtonen
    Feb 10 '17 at 7:36








  • 3




    $begingroup$
    @JyrkiLahtonen strong agreement. I love to see answers working in tandem, and the checkmark doesn't give all that many points. Best to have the answer that others build on be the one that people read first :)
    $endgroup$
    – hobbs
    Feb 10 '17 at 8:13






  • 1




    $begingroup$
    You may enjoy applying your skills to the Schizophrenic numbers. ;)
    $endgroup$
    – PM 2Ring
    Feb 10 '17 at 10:35






  • 1




    $begingroup$
    @JyrkiLahtonen's answer is the correct one. The repeating digits can be inferred from the Taylor expansion of the square root. the_architect's answer is merely an observation.
    $endgroup$
    – Klangen
    Mar 12 '18 at 10:37



















0












$begingroup$

I noticed another great thing.
Divide 308642 by sin(308642)
Approx 363943.9808
It cointains only digits 3,3×2,4,4×2,and 9
Perhaps 9 appears because of the architect's argument






share|cite|improve this answer









$endgroup$













  • $begingroup$
    You used degrees, right ? I do not understand what you mean with the last part of the answer.
    $endgroup$
    – Peter
    Jan 20 at 17:00











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3 Answers
3






active

oldest

votes








3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes









129












$begingroup$

Repeated same numbers in a decimal representation can be converted to repeated zeros by multiplication with $9$. (try it out)



so if we multiply $9 sqrt{308642} = sqrt{308642 times 81} = sqrt{25 000 002}$ since this number is allmost $5000^2$ it has a lot of zeros in its decimal expansion






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Superb answer! (+1)
    $endgroup$
    – Peter
    Feb 8 '17 at 13:19






  • 27




    $begingroup$
    And the underlying reason here is the series expansion $$ sqrt{a^2+x} = a + frac{1}{2a}x - frac{1}{(2a)^3}x^2 + frac2{(2a)^5}x^3 - frac{5}{(2a)^7}x^4 + cdots $$ which can be derived from the generalized binomial theorem. When $2a$ is a large power of $10$, this gives a nice decimal representation of the square root.
    $endgroup$
    – Henning Makholm
    Feb 8 '17 at 14:04








  • 25




    $begingroup$
    To check that this is the "right" explanation I'd find it good to have other, similar examples. And here is another one: $sqrt{1975308642} = 44444.44444472222222222135416666667209201388884650336371564...$ which can be explained by noting that $1975308642 = (400000^2 + 2)/9^2$.
    $endgroup$
    – Michael Lugo
    Feb 8 '17 at 14:37






  • 2




    $begingroup$
    It is important to note that $25000002$ is not only almost $5000^2$, but also $>5000^2$. Otherwise the same argument would work for $$sqrt{30864color{red}1}approx 555.55467777708433223768894721... .$$ Does not look so very nice! It is because $81times 308641<5000^2$, but still close to $5000^2$.
    $endgroup$
    – M. Winter
    Jan 25 '18 at 9:57


















129












$begingroup$

Repeated same numbers in a decimal representation can be converted to repeated zeros by multiplication with $9$. (try it out)



so if we multiply $9 sqrt{308642} = sqrt{308642 times 81} = sqrt{25 000 002}$ since this number is allmost $5000^2$ it has a lot of zeros in its decimal expansion






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Superb answer! (+1)
    $endgroup$
    – Peter
    Feb 8 '17 at 13:19






  • 27




    $begingroup$
    And the underlying reason here is the series expansion $$ sqrt{a^2+x} = a + frac{1}{2a}x - frac{1}{(2a)^3}x^2 + frac2{(2a)^5}x^3 - frac{5}{(2a)^7}x^4 + cdots $$ which can be derived from the generalized binomial theorem. When $2a$ is a large power of $10$, this gives a nice decimal representation of the square root.
    $endgroup$
    – Henning Makholm
    Feb 8 '17 at 14:04








  • 25




    $begingroup$
    To check that this is the "right" explanation I'd find it good to have other, similar examples. And here is another one: $sqrt{1975308642} = 44444.44444472222222222135416666667209201388884650336371564...$ which can be explained by noting that $1975308642 = (400000^2 + 2)/9^2$.
    $endgroup$
    – Michael Lugo
    Feb 8 '17 at 14:37






  • 2




    $begingroup$
    It is important to note that $25000002$ is not only almost $5000^2$, but also $>5000^2$. Otherwise the same argument would work for $$sqrt{30864color{red}1}approx 555.55467777708433223768894721... .$$ Does not look so very nice! It is because $81times 308641<5000^2$, but still close to $5000^2$.
    $endgroup$
    – M. Winter
    Jan 25 '18 at 9:57
















129












129








129





$begingroup$

Repeated same numbers in a decimal representation can be converted to repeated zeros by multiplication with $9$. (try it out)



so if we multiply $9 sqrt{308642} = sqrt{308642 times 81} = sqrt{25 000 002}$ since this number is allmost $5000^2$ it has a lot of zeros in its decimal expansion






share|cite|improve this answer











$endgroup$



Repeated same numbers in a decimal representation can be converted to repeated zeros by multiplication with $9$. (try it out)



so if we multiply $9 sqrt{308642} = sqrt{308642 times 81} = sqrt{25 000 002}$ since this number is allmost $5000^2$ it has a lot of zeros in its decimal expansion







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Feb 8 '17 at 13:18









Peter

47.7k1139131




47.7k1139131










answered Feb 8 '17 at 13:11









the_architectthe_architect

1,1041511




1,1041511








  • 1




    $begingroup$
    Superb answer! (+1)
    $endgroup$
    – Peter
    Feb 8 '17 at 13:19






  • 27




    $begingroup$
    And the underlying reason here is the series expansion $$ sqrt{a^2+x} = a + frac{1}{2a}x - frac{1}{(2a)^3}x^2 + frac2{(2a)^5}x^3 - frac{5}{(2a)^7}x^4 + cdots $$ which can be derived from the generalized binomial theorem. When $2a$ is a large power of $10$, this gives a nice decimal representation of the square root.
    $endgroup$
    – Henning Makholm
    Feb 8 '17 at 14:04








  • 25




    $begingroup$
    To check that this is the "right" explanation I'd find it good to have other, similar examples. And here is another one: $sqrt{1975308642} = 44444.44444472222222222135416666667209201388884650336371564...$ which can be explained by noting that $1975308642 = (400000^2 + 2)/9^2$.
    $endgroup$
    – Michael Lugo
    Feb 8 '17 at 14:37






  • 2




    $begingroup$
    It is important to note that $25000002$ is not only almost $5000^2$, but also $>5000^2$. Otherwise the same argument would work for $$sqrt{30864color{red}1}approx 555.55467777708433223768894721... .$$ Does not look so very nice! It is because $81times 308641<5000^2$, but still close to $5000^2$.
    $endgroup$
    – M. Winter
    Jan 25 '18 at 9:57
















  • 1




    $begingroup$
    Superb answer! (+1)
    $endgroup$
    – Peter
    Feb 8 '17 at 13:19






  • 27




    $begingroup$
    And the underlying reason here is the series expansion $$ sqrt{a^2+x} = a + frac{1}{2a}x - frac{1}{(2a)^3}x^2 + frac2{(2a)^5}x^3 - frac{5}{(2a)^7}x^4 + cdots $$ which can be derived from the generalized binomial theorem. When $2a$ is a large power of $10$, this gives a nice decimal representation of the square root.
    $endgroup$
    – Henning Makholm
    Feb 8 '17 at 14:04








  • 25




    $begingroup$
    To check that this is the "right" explanation I'd find it good to have other, similar examples. And here is another one: $sqrt{1975308642} = 44444.44444472222222222135416666667209201388884650336371564...$ which can be explained by noting that $1975308642 = (400000^2 + 2)/9^2$.
    $endgroup$
    – Michael Lugo
    Feb 8 '17 at 14:37






  • 2




    $begingroup$
    It is important to note that $25000002$ is not only almost $5000^2$, but also $>5000^2$. Otherwise the same argument would work for $$sqrt{30864color{red}1}approx 555.55467777708433223768894721... .$$ Does not look so very nice! It is because $81times 308641<5000^2$, but still close to $5000^2$.
    $endgroup$
    – M. Winter
    Jan 25 '18 at 9:57










1




1




$begingroup$
Superb answer! (+1)
$endgroup$
– Peter
Feb 8 '17 at 13:19




$begingroup$
Superb answer! (+1)
$endgroup$
– Peter
Feb 8 '17 at 13:19




27




27




$begingroup$
And the underlying reason here is the series expansion $$ sqrt{a^2+x} = a + frac{1}{2a}x - frac{1}{(2a)^3}x^2 + frac2{(2a)^5}x^3 - frac{5}{(2a)^7}x^4 + cdots $$ which can be derived from the generalized binomial theorem. When $2a$ is a large power of $10$, this gives a nice decimal representation of the square root.
$endgroup$
– Henning Makholm
Feb 8 '17 at 14:04






$begingroup$
And the underlying reason here is the series expansion $$ sqrt{a^2+x} = a + frac{1}{2a}x - frac{1}{(2a)^3}x^2 + frac2{(2a)^5}x^3 - frac{5}{(2a)^7}x^4 + cdots $$ which can be derived from the generalized binomial theorem. When $2a$ is a large power of $10$, this gives a nice decimal representation of the square root.
$endgroup$
– Henning Makholm
Feb 8 '17 at 14:04






25




25




$begingroup$
To check that this is the "right" explanation I'd find it good to have other, similar examples. And here is another one: $sqrt{1975308642} = 44444.44444472222222222135416666667209201388884650336371564...$ which can be explained by noting that $1975308642 = (400000^2 + 2)/9^2$.
$endgroup$
– Michael Lugo
Feb 8 '17 at 14:37




$begingroup$
To check that this is the "right" explanation I'd find it good to have other, similar examples. And here is another one: $sqrt{1975308642} = 44444.44444472222222222135416666667209201388884650336371564...$ which can be explained by noting that $1975308642 = (400000^2 + 2)/9^2$.
$endgroup$
– Michael Lugo
Feb 8 '17 at 14:37




2




2




$begingroup$
It is important to note that $25000002$ is not only almost $5000^2$, but also $>5000^2$. Otherwise the same argument would work for $$sqrt{30864color{red}1}approx 555.55467777708433223768894721... .$$ Does not look so very nice! It is because $81times 308641<5000^2$, but still close to $5000^2$.
$endgroup$
– M. Winter
Jan 25 '18 at 9:57






$begingroup$
It is important to note that $25000002$ is not only almost $5000^2$, but also $>5000^2$. Otherwise the same argument would work for $$sqrt{30864color{red}1}approx 555.55467777708433223768894721... .$$ Does not look so very nice! It is because $81times 308641<5000^2$, but still close to $5000^2$.
$endgroup$
– M. Winter
Jan 25 '18 at 9:57













124












$begingroup$

The architect's answer, while explaining the absolutely crucial fact that $sqrt{308642}approx 5000/9=555.555ldots$, didn't quite make it clear why we get several runs of repeating decimals. I try to shed additional light to that using a different tool.



I want to emphasize the role of the binomial series. In particular the Taylor expansion
$$
sqrt{1+x}=1+frac x2-frac{x^2}8+frac{x^3}{16}-frac{5x^4}{128}+frac{7x^5}{256}-frac{21x^6}{1024}+cdots
$$
If we plug in $x=2/(5000)^2=8cdot10^{-8}$, we get
$$
M:=sqrt{1+8cdot10^{-8}}=1+4cdot10^{-8}-8cdot10^{-16}+32cdot10^{-24}-160cdot10^{-32}+cdots.
$$
Therefore
$$
begin{aligned}
sqrt{308462}&=frac{5000}9M=frac{5000}9+frac{20000}9cdot10^{-8}-frac{40000}9cdot10^{-16}+frac{160000}9cdot10^{-24}+cdots\
&=frac{5}9cdot10^3+frac29cdot10^{-4}-frac49cdot10^{-12}+frac{16}9cdot10^{-20}+cdots.
end{aligned}
$$
This explains both the runs, their starting points, as well as the origin and location of those extra digits not part of any run. For example, the run of $5+2=7$s begins when the first two terms of the above series are "active". When the third term joins in, we need to subtract a $4$ and a run of $3$s ensues et cetera.






share|cite|improve this answer











$endgroup$









  • 15




    $begingroup$
    @Peter It is quite common to unaccept an answer after a better answer appears. Doing so helps guide readers to the best answer, which is often not the highest voted one, due to many factors, e.g. earlier answers usually get more votes, and less technical answers usually get more votes from hot-list activity (as here). This is currently (by far) the best explanation you have.
    $endgroup$
    – Bill Dubuque
    Feb 8 '17 at 20:35








  • 11




    $begingroup$
    For the record: The reason I support Peter's decision to accept the architect's answer is that mine is building upon it. Without the observation that $5000/9$ is an extremely good approximation I most likely would not have bothered, and most certainly would not have come up with this refinement. IMHO Math.SE works at its best, when different users add different points of view refining earlier answers. The voters very clearly like both the answers. Sunshine and smiles to all!
    $endgroup$
    – Jyrki Lahtonen
    Feb 10 '17 at 7:36








  • 3




    $begingroup$
    @JyrkiLahtonen strong agreement. I love to see answers working in tandem, and the checkmark doesn't give all that many points. Best to have the answer that others build on be the one that people read first :)
    $endgroup$
    – hobbs
    Feb 10 '17 at 8:13






  • 1




    $begingroup$
    You may enjoy applying your skills to the Schizophrenic numbers. ;)
    $endgroup$
    – PM 2Ring
    Feb 10 '17 at 10:35






  • 1




    $begingroup$
    @JyrkiLahtonen's answer is the correct one. The repeating digits can be inferred from the Taylor expansion of the square root. the_architect's answer is merely an observation.
    $endgroup$
    – Klangen
    Mar 12 '18 at 10:37
















124












$begingroup$

The architect's answer, while explaining the absolutely crucial fact that $sqrt{308642}approx 5000/9=555.555ldots$, didn't quite make it clear why we get several runs of repeating decimals. I try to shed additional light to that using a different tool.



I want to emphasize the role of the binomial series. In particular the Taylor expansion
$$
sqrt{1+x}=1+frac x2-frac{x^2}8+frac{x^3}{16}-frac{5x^4}{128}+frac{7x^5}{256}-frac{21x^6}{1024}+cdots
$$
If we plug in $x=2/(5000)^2=8cdot10^{-8}$, we get
$$
M:=sqrt{1+8cdot10^{-8}}=1+4cdot10^{-8}-8cdot10^{-16}+32cdot10^{-24}-160cdot10^{-32}+cdots.
$$
Therefore
$$
begin{aligned}
sqrt{308462}&=frac{5000}9M=frac{5000}9+frac{20000}9cdot10^{-8}-frac{40000}9cdot10^{-16}+frac{160000}9cdot10^{-24}+cdots\
&=frac{5}9cdot10^3+frac29cdot10^{-4}-frac49cdot10^{-12}+frac{16}9cdot10^{-20}+cdots.
end{aligned}
$$
This explains both the runs, their starting points, as well as the origin and location of those extra digits not part of any run. For example, the run of $5+2=7$s begins when the first two terms of the above series are "active". When the third term joins in, we need to subtract a $4$ and a run of $3$s ensues et cetera.






share|cite|improve this answer











$endgroup$









  • 15




    $begingroup$
    @Peter It is quite common to unaccept an answer after a better answer appears. Doing so helps guide readers to the best answer, which is often not the highest voted one, due to many factors, e.g. earlier answers usually get more votes, and less technical answers usually get more votes from hot-list activity (as here). This is currently (by far) the best explanation you have.
    $endgroup$
    – Bill Dubuque
    Feb 8 '17 at 20:35








  • 11




    $begingroup$
    For the record: The reason I support Peter's decision to accept the architect's answer is that mine is building upon it. Without the observation that $5000/9$ is an extremely good approximation I most likely would not have bothered, and most certainly would not have come up with this refinement. IMHO Math.SE works at its best, when different users add different points of view refining earlier answers. The voters very clearly like both the answers. Sunshine and smiles to all!
    $endgroup$
    – Jyrki Lahtonen
    Feb 10 '17 at 7:36








  • 3




    $begingroup$
    @JyrkiLahtonen strong agreement. I love to see answers working in tandem, and the checkmark doesn't give all that many points. Best to have the answer that others build on be the one that people read first :)
    $endgroup$
    – hobbs
    Feb 10 '17 at 8:13






  • 1




    $begingroup$
    You may enjoy applying your skills to the Schizophrenic numbers. ;)
    $endgroup$
    – PM 2Ring
    Feb 10 '17 at 10:35






  • 1




    $begingroup$
    @JyrkiLahtonen's answer is the correct one. The repeating digits can be inferred from the Taylor expansion of the square root. the_architect's answer is merely an observation.
    $endgroup$
    – Klangen
    Mar 12 '18 at 10:37














124












124








124





$begingroup$

The architect's answer, while explaining the absolutely crucial fact that $sqrt{308642}approx 5000/9=555.555ldots$, didn't quite make it clear why we get several runs of repeating decimals. I try to shed additional light to that using a different tool.



I want to emphasize the role of the binomial series. In particular the Taylor expansion
$$
sqrt{1+x}=1+frac x2-frac{x^2}8+frac{x^3}{16}-frac{5x^4}{128}+frac{7x^5}{256}-frac{21x^6}{1024}+cdots
$$
If we plug in $x=2/(5000)^2=8cdot10^{-8}$, we get
$$
M:=sqrt{1+8cdot10^{-8}}=1+4cdot10^{-8}-8cdot10^{-16}+32cdot10^{-24}-160cdot10^{-32}+cdots.
$$
Therefore
$$
begin{aligned}
sqrt{308462}&=frac{5000}9M=frac{5000}9+frac{20000}9cdot10^{-8}-frac{40000}9cdot10^{-16}+frac{160000}9cdot10^{-24}+cdots\
&=frac{5}9cdot10^3+frac29cdot10^{-4}-frac49cdot10^{-12}+frac{16}9cdot10^{-20}+cdots.
end{aligned}
$$
This explains both the runs, their starting points, as well as the origin and location of those extra digits not part of any run. For example, the run of $5+2=7$s begins when the first two terms of the above series are "active". When the third term joins in, we need to subtract a $4$ and a run of $3$s ensues et cetera.






share|cite|improve this answer











$endgroup$



The architect's answer, while explaining the absolutely crucial fact that $sqrt{308642}approx 5000/9=555.555ldots$, didn't quite make it clear why we get several runs of repeating decimals. I try to shed additional light to that using a different tool.



I want to emphasize the role of the binomial series. In particular the Taylor expansion
$$
sqrt{1+x}=1+frac x2-frac{x^2}8+frac{x^3}{16}-frac{5x^4}{128}+frac{7x^5}{256}-frac{21x^6}{1024}+cdots
$$
If we plug in $x=2/(5000)^2=8cdot10^{-8}$, we get
$$
M:=sqrt{1+8cdot10^{-8}}=1+4cdot10^{-8}-8cdot10^{-16}+32cdot10^{-24}-160cdot10^{-32}+cdots.
$$
Therefore
$$
begin{aligned}
sqrt{308462}&=frac{5000}9M=frac{5000}9+frac{20000}9cdot10^{-8}-frac{40000}9cdot10^{-16}+frac{160000}9cdot10^{-24}+cdots\
&=frac{5}9cdot10^3+frac29cdot10^{-4}-frac49cdot10^{-12}+frac{16}9cdot10^{-20}+cdots.
end{aligned}
$$
This explains both the runs, their starting points, as well as the origin and location of those extra digits not part of any run. For example, the run of $5+2=7$s begins when the first two terms of the above series are "active". When the third term joins in, we need to subtract a $4$ and a run of $3$s ensues et cetera.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Feb 9 '17 at 21:34

























answered Feb 8 '17 at 14:00









Jyrki LahtonenJyrki Lahtonen

109k13169374




109k13169374








  • 15




    $begingroup$
    @Peter It is quite common to unaccept an answer after a better answer appears. Doing so helps guide readers to the best answer, which is often not the highest voted one, due to many factors, e.g. earlier answers usually get more votes, and less technical answers usually get more votes from hot-list activity (as here). This is currently (by far) the best explanation you have.
    $endgroup$
    – Bill Dubuque
    Feb 8 '17 at 20:35








  • 11




    $begingroup$
    For the record: The reason I support Peter's decision to accept the architect's answer is that mine is building upon it. Without the observation that $5000/9$ is an extremely good approximation I most likely would not have bothered, and most certainly would not have come up with this refinement. IMHO Math.SE works at its best, when different users add different points of view refining earlier answers. The voters very clearly like both the answers. Sunshine and smiles to all!
    $endgroup$
    – Jyrki Lahtonen
    Feb 10 '17 at 7:36








  • 3




    $begingroup$
    @JyrkiLahtonen strong agreement. I love to see answers working in tandem, and the checkmark doesn't give all that many points. Best to have the answer that others build on be the one that people read first :)
    $endgroup$
    – hobbs
    Feb 10 '17 at 8:13






  • 1




    $begingroup$
    You may enjoy applying your skills to the Schizophrenic numbers. ;)
    $endgroup$
    – PM 2Ring
    Feb 10 '17 at 10:35






  • 1




    $begingroup$
    @JyrkiLahtonen's answer is the correct one. The repeating digits can be inferred from the Taylor expansion of the square root. the_architect's answer is merely an observation.
    $endgroup$
    – Klangen
    Mar 12 '18 at 10:37














  • 15




    $begingroup$
    @Peter It is quite common to unaccept an answer after a better answer appears. Doing so helps guide readers to the best answer, which is often not the highest voted one, due to many factors, e.g. earlier answers usually get more votes, and less technical answers usually get more votes from hot-list activity (as here). This is currently (by far) the best explanation you have.
    $endgroup$
    – Bill Dubuque
    Feb 8 '17 at 20:35








  • 11




    $begingroup$
    For the record: The reason I support Peter's decision to accept the architect's answer is that mine is building upon it. Without the observation that $5000/9$ is an extremely good approximation I most likely would not have bothered, and most certainly would not have come up with this refinement. IMHO Math.SE works at its best, when different users add different points of view refining earlier answers. The voters very clearly like both the answers. Sunshine and smiles to all!
    $endgroup$
    – Jyrki Lahtonen
    Feb 10 '17 at 7:36








  • 3




    $begingroup$
    @JyrkiLahtonen strong agreement. I love to see answers working in tandem, and the checkmark doesn't give all that many points. Best to have the answer that others build on be the one that people read first :)
    $endgroup$
    – hobbs
    Feb 10 '17 at 8:13






  • 1




    $begingroup$
    You may enjoy applying your skills to the Schizophrenic numbers. ;)
    $endgroup$
    – PM 2Ring
    Feb 10 '17 at 10:35






  • 1




    $begingroup$
    @JyrkiLahtonen's answer is the correct one. The repeating digits can be inferred from the Taylor expansion of the square root. the_architect's answer is merely an observation.
    $endgroup$
    – Klangen
    Mar 12 '18 at 10:37








15




15




$begingroup$
@Peter It is quite common to unaccept an answer after a better answer appears. Doing so helps guide readers to the best answer, which is often not the highest voted one, due to many factors, e.g. earlier answers usually get more votes, and less technical answers usually get more votes from hot-list activity (as here). This is currently (by far) the best explanation you have.
$endgroup$
– Bill Dubuque
Feb 8 '17 at 20:35






$begingroup$
@Peter It is quite common to unaccept an answer after a better answer appears. Doing so helps guide readers to the best answer, which is often not the highest voted one, due to many factors, e.g. earlier answers usually get more votes, and less technical answers usually get more votes from hot-list activity (as here). This is currently (by far) the best explanation you have.
$endgroup$
– Bill Dubuque
Feb 8 '17 at 20:35






11




11




$begingroup$
For the record: The reason I support Peter's decision to accept the architect's answer is that mine is building upon it. Without the observation that $5000/9$ is an extremely good approximation I most likely would not have bothered, and most certainly would not have come up with this refinement. IMHO Math.SE works at its best, when different users add different points of view refining earlier answers. The voters very clearly like both the answers. Sunshine and smiles to all!
$endgroup$
– Jyrki Lahtonen
Feb 10 '17 at 7:36






$begingroup$
For the record: The reason I support Peter's decision to accept the architect's answer is that mine is building upon it. Without the observation that $5000/9$ is an extremely good approximation I most likely would not have bothered, and most certainly would not have come up with this refinement. IMHO Math.SE works at its best, when different users add different points of view refining earlier answers. The voters very clearly like both the answers. Sunshine and smiles to all!
$endgroup$
– Jyrki Lahtonen
Feb 10 '17 at 7:36






3




3




$begingroup$
@JyrkiLahtonen strong agreement. I love to see answers working in tandem, and the checkmark doesn't give all that many points. Best to have the answer that others build on be the one that people read first :)
$endgroup$
– hobbs
Feb 10 '17 at 8:13




$begingroup$
@JyrkiLahtonen strong agreement. I love to see answers working in tandem, and the checkmark doesn't give all that many points. Best to have the answer that others build on be the one that people read first :)
$endgroup$
– hobbs
Feb 10 '17 at 8:13




1




1




$begingroup$
You may enjoy applying your skills to the Schizophrenic numbers. ;)
$endgroup$
– PM 2Ring
Feb 10 '17 at 10:35




$begingroup$
You may enjoy applying your skills to the Schizophrenic numbers. ;)
$endgroup$
– PM 2Ring
Feb 10 '17 at 10:35




1




1




$begingroup$
@JyrkiLahtonen's answer is the correct one. The repeating digits can be inferred from the Taylor expansion of the square root. the_architect's answer is merely an observation.
$endgroup$
– Klangen
Mar 12 '18 at 10:37




$begingroup$
@JyrkiLahtonen's answer is the correct one. The repeating digits can be inferred from the Taylor expansion of the square root. the_architect's answer is merely an observation.
$endgroup$
– Klangen
Mar 12 '18 at 10:37











0












$begingroup$

I noticed another great thing.
Divide 308642 by sin(308642)
Approx 363943.9808
It cointains only digits 3,3×2,4,4×2,and 9
Perhaps 9 appears because of the architect's argument






share|cite|improve this answer









$endgroup$













  • $begingroup$
    You used degrees, right ? I do not understand what you mean with the last part of the answer.
    $endgroup$
    – Peter
    Jan 20 at 17:00
















0












$begingroup$

I noticed another great thing.
Divide 308642 by sin(308642)
Approx 363943.9808
It cointains only digits 3,3×2,4,4×2,and 9
Perhaps 9 appears because of the architect's argument






share|cite|improve this answer









$endgroup$













  • $begingroup$
    You used degrees, right ? I do not understand what you mean with the last part of the answer.
    $endgroup$
    – Peter
    Jan 20 at 17:00














0












0








0





$begingroup$

I noticed another great thing.
Divide 308642 by sin(308642)
Approx 363943.9808
It cointains only digits 3,3×2,4,4×2,and 9
Perhaps 9 appears because of the architect's argument






share|cite|improve this answer









$endgroup$



I noticed another great thing.
Divide 308642 by sin(308642)
Approx 363943.9808
It cointains only digits 3,3×2,4,4×2,and 9
Perhaps 9 appears because of the architect's argument







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 20 at 16:25







user636268



















  • $begingroup$
    You used degrees, right ? I do not understand what you mean with the last part of the answer.
    $endgroup$
    – Peter
    Jan 20 at 17:00


















  • $begingroup$
    You used degrees, right ? I do not understand what you mean with the last part of the answer.
    $endgroup$
    – Peter
    Jan 20 at 17:00
















$begingroup$
You used degrees, right ? I do not understand what you mean with the last part of the answer.
$endgroup$
– Peter
Jan 20 at 17:00




$begingroup$
You used degrees, right ? I do not understand what you mean with the last part of the answer.
$endgroup$
– Peter
Jan 20 at 17:00


















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