How many terms are required for harmonic series of degrees to “cover” a full $360^text{o}$ circle?
$begingroup$
This question accidentally came to my mind when reading about harmonic series. I've never been able to find an answer on the Internet. Consider $H_n$ which is the $n$-th harmonic number:
$$
H_n = 1 + {1over 2} + {1over 3} + cdots + {1over n}
$$
Not sure whether it's valid but suppose we have a circle which's full rotation is known to be $360^text{o}$. Lets define the following series:
$$
S_n^text{o} = (1)^text{o} + left({1over 2}right)^text{o} + left({1over 3}right)^text{o} + cdots
$$
Since harmonic series is divergent and its sum tends to infinity, then at some point we should have "covered" the whole circumference. If we now define an "$n$-th harmonic degree number" as $H_n^text{o}$. Then we have to solve the following inequality for $n$:
$$
H_n^text{o} ge 360^text{o}
$$
Please note that i'm not very familiar with series and only have basic calculus knowledge like limits, derivatives and Taylor expansion. Also I may have misused a lot of terms in this question, so please comment in order for me to improve it. Apart from that I'm basically interested in two things:
- Is it valid to consider the harmonic sum of degrees rather than the sum of rationals?
- If so what would be the way to find the index of $H_n^text{o}$ such that the whole circle is "covered"?
calculus sequences-and-series summation harmonic-numbers
edited Jan 22 at 14:50
José Carlos Santos
164k22131234
asked Jan 22 at 14:35
romanroman
2,30921224
$endgroup$
add a comment |
$begingroup$
This question accidentally came to my mind when reading about harmonic series. I've never been able to find an answer on the Internet. Consider $H_n$ which is the $n$-th harmonic number:
$$
H_n = 1 + {1over 2} + {1over 3} + cdots + {1over n}
$$
Not sure whether it's valid but suppose we have a circle which's full rotation is known to be $360^text{o}$. Lets define the following series:
$$
S_n^text{o} = (1)^text{o} + left({1over 2}right)^text{o} + left({1over 3}right)^text{o} + cdots
$$
Since harmonic series is divergent and its sum tends to infinity, then at some point we should have "covered" the whole circumference. If we now define an "$n$-th harmonic degree number" as $H_n^text{o}$. Then we have to solve the following inequality for $n$:
$$
H_n^text{o} ge 360^text{o}
$$
Please note that i'm not very familiar with series and only have basic calculus knowledge like limits, derivatives and Taylor expansion. Also I may have misused a lot of terms in this question, so please comment in order for me to improve it. Apart from that I'm basically interested in two things:
- Is it valid to consider the harmonic sum of degrees rather than the sum of rationals?
- If so what would be the way to find the index of $H_n^text{o}$ such that the whole circle is "covered"?
calculus sequences-and-series summation harmonic-numbers
edited Jan 22 at 14:50
José Carlos Santos
164k22131234
asked Jan 22 at 14:35
romanroman
2,30921224
$endgroup$
add a comment |
$begingroup$
This question accidentally came to my mind when reading about harmonic series. I've never been able to find an answer on the Internet. Consider $H_n$ which is the $n$-th harmonic number:
$$
H_n = 1 + {1over 2} + {1over 3} + cdots + {1over n}
$$
Not sure whether it's valid but suppose we have a circle which's full rotation is known to be $360^text{o}$. Lets define the following series:
$$
S_n^text{o} = (1)^text{o} + left({1over 2}right)^text{o} + left({1over 3}right)^text{o} + cdots
$$
Since harmonic series is divergent and its sum tends to infinity, then at some point we should have "covered" the whole circumference. If we now define an "$n$-th harmonic degree number" as $H_n^text{o}$. Then we have to solve the following inequality for $n$:
$$
H_n^text{o} ge 360^text{o}
$$
Please note that i'm not very familiar with series and only have basic calculus knowledge like limits, derivatives and Taylor expansion. Also I may have misused a lot of terms in this question, so please comment in order for me to improve it. Apart from that I'm basically interested in two things:
- Is it valid to consider the harmonic sum of degrees rather than the sum of rationals?
- If so what would be the way to find the index of $H_n^text{o}$ such that the whole circle is "covered"?
calculus sequences-and-series summation harmonic-numbers
edited Jan 22 at 14:50
José Carlos Santos
164k22131234
asked Jan 22 at 14:35
romanroman
2,30921224
$endgroup$
This question accidentally came to my mind when reading about harmonic series. I've never been able to find an answer on the Internet. Consider $H_n$ which is the $n$-th harmonic number:
$$
H_n = 1 + {1over 2} + {1over 3} + cdots + {1over n}
$$
Not sure whether it's valid but suppose we have a circle which's full rotation is known to be $360^text{o}$. Lets define the following series:
$$
S_n^text{o} = (1)^text{o} + left({1over 2}right)^text{o} + left({1over 3}right)^text{o} + cdots
$$
Since harmonic series is divergent and its sum tends to infinity, then at some point we should have "covered" the whole circumference. If we now define an "$n$-th harmonic degree number" as $H_n^text{o}$. Then we have to solve the following inequality for $n$:
$$
H_n^text{o} ge 360^text{o}
$$
Please note that i'm not very familiar with series and only have basic calculus knowledge like limits, derivatives and Taylor expansion. Also I may have misused a lot of terms in this question, so please comment in order for me to improve it. Apart from that I'm basically interested in two things:
- Is it valid to consider the harmonic sum of degrees rather than the sum of rationals?
- If so what would be the way to find the index of $H_n^text{o}$ such that the whole circle is "covered"?
calculus sequences-and-series summation harmonic-numbers
calculus sequences-and-series summation harmonic-numbers
edited Jan 22 at 14:50
José Carlos Santos
164k22131234
asked Jan 22 at 14:35
romanroman
2,30921224
edited Jan 22 at 14:50
José Carlos Santos
164k22131234
asked Jan 22 at 14:35
romanroman
2,30921224
edited Jan 22 at 14:50
José Carlos Santos
164k22131234
edited Jan 22 at 14:50
José Carlos Santos
164k22131234
edited Jan 22 at 14:50
José Carlos Santos
164k22131234
164k22131234
asked Jan 22 at 14:35
romanroman
2,30921224
asked Jan 22 at 14:35
romanroman
2,30921224
asked Jan 22 at 14:35
romanroman
2,30921224
2,30921224
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Since $H_n$ is close to $log n$ if $n$ is large enough, you will have to take $n$ about $e^{360-gamma}$, which is about $1.25times10^{156}$. Here, $gamma$ is the Euler-Mascheroni constant.
answered Jan 22 at 14:44
José Carlos SantosJosé Carlos Santos
164k22131234
$endgroup$
$begingroup$
Indeed. Do you mind if I edit my answer, adding to it your suggestion?
$endgroup$
– José Carlos Santos
Jan 22 at 14:49
$begingroup$
whoa, that's a long way round. Thanks!
$endgroup$
– roman
Jan 22 at 14:53
add a comment |
$begingroup$
Note that for large $n$, $H_nsim ln n$, so $ln napprox 360$ means that $napprox 2.2times 10^{156}$.
answered Jan 22 at 14:44
rogerlrogerl
18k22747
$endgroup$
1
$begingroup$
It is misleading to give so many significant figures! The approximation $H_nsim ln n$ is not nearly so precise.
$endgroup$
– TonyK
Jan 22 at 14:51
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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votes
$begingroup$
Since $H_n$ is close to $log n$ if $n$ is large enough, you will have to take $n$ about $e^{360-gamma}$, which is about $1.25times10^{156}$. Here, $gamma$ is the Euler-Mascheroni constant.
answered Jan 22 at 14:44
José Carlos SantosJosé Carlos Santos
164k22131234
$endgroup$
$begingroup$
Indeed. Do you mind if I edit my answer, adding to it your suggestion?
$endgroup$
– José Carlos Santos
Jan 22 at 14:49
$begingroup$
whoa, that's a long way round. Thanks!
$endgroup$
– roman
Jan 22 at 14:53
add a comment |
$begingroup$
Since $H_n$ is close to $log n$ if $n$ is large enough, you will have to take $n$ about $e^{360-gamma}$, which is about $1.25times10^{156}$. Here, $gamma$ is the Euler-Mascheroni constant.
answered Jan 22 at 14:44
José Carlos SantosJosé Carlos Santos
164k22131234
$endgroup$
$begingroup$
Indeed. Do you mind if I edit my answer, adding to it your suggestion?
$endgroup$
– José Carlos Santos
Jan 22 at 14:49
$begingroup$
whoa, that's a long way round. Thanks!
$endgroup$
– roman
Jan 22 at 14:53
add a comment |
$begingroup$
Since $H_n$ is close to $log n$ if $n$ is large enough, you will have to take $n$ about $e^{360-gamma}$, which is about $1.25times10^{156}$. Here, $gamma$ is the Euler-Mascheroni constant.
answered Jan 22 at 14:44
José Carlos SantosJosé Carlos Santos
164k22131234
$endgroup$
Since $H_n$ is close to $log n$ if $n$ is large enough, you will have to take $n$ about $e^{360-gamma}$, which is about $1.25times10^{156}$. Here, $gamma$ is the Euler-Mascheroni constant.
answered Jan 22 at 14:44
José Carlos SantosJosé Carlos Santos
164k22131234
edited Jan 22 at 14:53
answered Jan 22 at 14:44
José Carlos SantosJosé Carlos Santos
164k22131234
answered Jan 22 at 14:44
José Carlos SantosJosé Carlos Santos
164k22131234
answered Jan 22 at 14:44
José Carlos SantosJosé Carlos Santos
164k22131234
164k22131234
$begingroup$
Indeed. Do you mind if I edit my answer, adding to it your suggestion?
$endgroup$
– José Carlos Santos
Jan 22 at 14:49
$begingroup$
whoa, that's a long way round. Thanks!
$endgroup$
– roman
Jan 22 at 14:53
add a comment |
$begingroup$
Indeed. Do you mind if I edit my answer, adding to it your suggestion?
$endgroup$
– José Carlos Santos
Jan 22 at 14:49
$begingroup$
whoa, that's a long way round. Thanks!
$endgroup$
– roman
Jan 22 at 14:53
$begingroup$
Indeed. Do you mind if I edit my answer, adding to it your suggestion?
$endgroup$
– José Carlos Santos
Jan 22 at 14:49
$begingroup$
Indeed. Do you mind if I edit my answer, adding to it your suggestion?
$endgroup$
– José Carlos Santos
Jan 22 at 14:49
$begingroup$
whoa, that's a long way round. Thanks!
$endgroup$
– roman
Jan 22 at 14:53
$begingroup$
whoa, that's a long way round. Thanks!
$endgroup$
– roman
Jan 22 at 14:53
add a comment |
$begingroup$
Note that for large $n$, $H_nsim ln n$, so $ln napprox 360$ means that $napprox 2.2times 10^{156}$.
answered Jan 22 at 14:44
rogerlrogerl
18k22747
$endgroup$
1
$begingroup$
It is misleading to give so many significant figures! The approximation $H_nsim ln n$ is not nearly so precise.
$endgroup$
– TonyK
Jan 22 at 14:51
add a comment |
$begingroup$
Note that for large $n$, $H_nsim ln n$, so $ln napprox 360$ means that $napprox 2.2times 10^{156}$.
answered Jan 22 at 14:44
rogerlrogerl
18k22747
$endgroup$
1
$begingroup$
It is misleading to give so many significant figures! The approximation $H_nsim ln n$ is not nearly so precise.
$endgroup$
– TonyK
Jan 22 at 14:51
add a comment |
$begingroup$
Note that for large $n$, $H_nsim ln n$, so $ln napprox 360$ means that $napprox 2.2times 10^{156}$.
answered Jan 22 at 14:44
rogerlrogerl
18k22747
$endgroup$
Note that for large $n$, $H_nsim ln n$, so $ln napprox 360$ means that $napprox 2.2times 10^{156}$.
answered Jan 22 at 14:44
rogerlrogerl
18k22747
edited Jan 22 at 14:56
answered Jan 22 at 14:44
rogerlrogerl
18k22747
answered Jan 22 at 14:44
rogerlrogerl
18k22747
answered Jan 22 at 14:44
rogerlrogerl
18k22747
18k22747
1
$begingroup$
It is misleading to give so many significant figures! The approximation $H_nsim ln n$ is not nearly so precise.
$endgroup$
– TonyK
Jan 22 at 14:51
add a comment |
1
$begingroup$
It is misleading to give so many significant figures! The approximation $H_nsim ln n$ is not nearly so precise.
$endgroup$
– TonyK
Jan 22 at 14:51
1
1
$begingroup$
It is misleading to give so many significant figures! The approximation $H_nsim ln n$ is not nearly so precise.
$endgroup$
– TonyK
Jan 22 at 14:51
$begingroup$
It is misleading to give so many significant figures! The approximation $H_nsim ln n$ is not nearly so precise.
$endgroup$
– TonyK
Jan 22 at 14:51
add a comment |
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