Explanation of summation equation












1












$begingroup$


Could somebody please explain the following equation to me?
enter image description here



I have no clue what H represents, nor how theta(ln(n)) - theta(ln(k)) results in theta(ln (n/k))



Any explanation would be appreciated.



Thanks!






summation algorithms logarithms







share|cite|improve this question









$endgroup$












  • $begingroup$
    H indicates the harmonic numbers
    $endgroup$
    – G Cab
    Dec 7 '18 at 0:56
















1












$begingroup$


Could somebody please explain the following equation to me?
enter image description here



I have no clue what H represents, nor how theta(ln(n)) - theta(ln(k)) results in theta(ln (n/k))



Any explanation would be appreciated.



Thanks!






summation algorithms logarithms







share|cite|improve this question









$endgroup$












  • $begingroup$
    H indicates the harmonic numbers
    $endgroup$
    – G Cab
    Dec 7 '18 at 0:56














1












1








1





$begingroup$


Could somebody please explain the following equation to me?
enter image description here



I have no clue what H represents, nor how theta(ln(n)) - theta(ln(k)) results in theta(ln (n/k))



Any explanation would be appreciated.



Thanks!






summation algorithms logarithms







share|cite|improve this question









$endgroup$




Could somebody please explain the following equation to me?
enter image description here



I have no clue what H represents, nor how theta(ln(n)) - theta(ln(k)) results in theta(ln (n/k))



Any explanation would be appreciated.



Thanks!






summation algorithms logarithms




summation algorithms logarithms






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share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 7 '18 at 0:50









Jerry M.Jerry M.

1084




1084












  • $begingroup$
    H indicates the harmonic numbers
    $endgroup$
    – G Cab
    Dec 7 '18 at 0:56


















  • $begingroup$
    H indicates the harmonic numbers
    $endgroup$
    – G Cab
    Dec 7 '18 at 0:56
















$begingroup$
H indicates the harmonic numbers
$endgroup$
– G Cab
Dec 7 '18 at 0:56




$begingroup$
H indicates the harmonic numbers
$endgroup$
– G Cab
Dec 7 '18 at 0:56










1 Answer
1






active

oldest

votes


















1












$begingroup$

The $H$ notation used is more a shorthand for the harmonic series' partial sums. More explicitly,



$$H(n) = text{(the n-th partial sum of the harmonic series)} = sum_{k=1}^n frac{1}{k}$$



Edit: You could also say that $H(n)$ is the $n^{th}$ harmonic number, as the definition is the same.



Going from there, I'm just guessing on what $Theta$ represents from what I could Google up. My guess is that it is a function that means "on the order of" or something of the sort. For that, it might be important to note that we can represent the harmonic series in a sort of "closed" form:



$$sum_{k=1}^n frac{1}{k} = ln(n) + gamma + epsilon_n approx ln(n)$$



In this context, $gamma$ is the Euler-Mascheroni constant, and $epsilon_n$ is a sort of "error constant" that establishes the equality. It is proportional to $1/2n$ and thus $epsilon_n to 0$ as $n to infty$.



And then I guess because $ln(n) - ln(k) = ln(n/k)$ per a property of logarithms, that might explain your other question.



But I want to emphasize that I'm kinda just guessing here since I'm not particularly familiar with the big-theta notation or how to utilize it. I think that's what's at play here, but it's just a guess.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Yes, $Theta$ basically means "on the order of", i.e., bounded both above and below up to a constant. See Big-O-notation
    $endgroup$
    – Jair Taylor
    Dec 7 '18 at 1:05










  • $begingroup$
    This was an absolutely amazing explanation. Thank you so much for taking the time to explain this to me.
    $endgroup$
    – Jerry M.
    Dec 7 '18 at 1:37











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1 Answer
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active

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

The $H$ notation used is more a shorthand for the harmonic series' partial sums. More explicitly,



$$H(n) = text{(the n-th partial sum of the harmonic series)} = sum_{k=1}^n frac{1}{k}$$



Edit: You could also say that $H(n)$ is the $n^{th}$ harmonic number, as the definition is the same.



Going from there, I'm just guessing on what $Theta$ represents from what I could Google up. My guess is that it is a function that means "on the order of" or something of the sort. For that, it might be important to note that we can represent the harmonic series in a sort of "closed" form:



$$sum_{k=1}^n frac{1}{k} = ln(n) + gamma + epsilon_n approx ln(n)$$



In this context, $gamma$ is the Euler-Mascheroni constant, and $epsilon_n$ is a sort of "error constant" that establishes the equality. It is proportional to $1/2n$ and thus $epsilon_n to 0$ as $n to infty$.



And then I guess because $ln(n) - ln(k) = ln(n/k)$ per a property of logarithms, that might explain your other question.



But I want to emphasize that I'm kinda just guessing here since I'm not particularly familiar with the big-theta notation or how to utilize it. I think that's what's at play here, but it's just a guess.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Yes, $Theta$ basically means "on the order of", i.e., bounded both above and below up to a constant. See Big-O-notation
    $endgroup$
    – Jair Taylor
    Dec 7 '18 at 1:05










  • $begingroup$
    This was an absolutely amazing explanation. Thank you so much for taking the time to explain this to me.
    $endgroup$
    – Jerry M.
    Dec 7 '18 at 1:37
















1












$begingroup$

The $H$ notation used is more a shorthand for the harmonic series' partial sums. More explicitly,



$$H(n) = text{(the n-th partial sum of the harmonic series)} = sum_{k=1}^n frac{1}{k}$$



Edit: You could also say that $H(n)$ is the $n^{th}$ harmonic number, as the definition is the same.



Going from there, I'm just guessing on what $Theta$ represents from what I could Google up. My guess is that it is a function that means "on the order of" or something of the sort. For that, it might be important to note that we can represent the harmonic series in a sort of "closed" form:



$$sum_{k=1}^n frac{1}{k} = ln(n) + gamma + epsilon_n approx ln(n)$$



In this context, $gamma$ is the Euler-Mascheroni constant, and $epsilon_n$ is a sort of "error constant" that establishes the equality. It is proportional to $1/2n$ and thus $epsilon_n to 0$ as $n to infty$.



And then I guess because $ln(n) - ln(k) = ln(n/k)$ per a property of logarithms, that might explain your other question.



But I want to emphasize that I'm kinda just guessing here since I'm not particularly familiar with the big-theta notation or how to utilize it. I think that's what's at play here, but it's just a guess.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Yes, $Theta$ basically means "on the order of", i.e., bounded both above and below up to a constant. See Big-O-notation
    $endgroup$
    – Jair Taylor
    Dec 7 '18 at 1:05










  • $begingroup$
    This was an absolutely amazing explanation. Thank you so much for taking the time to explain this to me.
    $endgroup$
    – Jerry M.
    Dec 7 '18 at 1:37














1












1








1





$begingroup$

The $H$ notation used is more a shorthand for the harmonic series' partial sums. More explicitly,



$$H(n) = text{(the n-th partial sum of the harmonic series)} = sum_{k=1}^n frac{1}{k}$$



Edit: You could also say that $H(n)$ is the $n^{th}$ harmonic number, as the definition is the same.



Going from there, I'm just guessing on what $Theta$ represents from what I could Google up. My guess is that it is a function that means "on the order of" or something of the sort. For that, it might be important to note that we can represent the harmonic series in a sort of "closed" form:



$$sum_{k=1}^n frac{1}{k} = ln(n) + gamma + epsilon_n approx ln(n)$$



In this context, $gamma$ is the Euler-Mascheroni constant, and $epsilon_n$ is a sort of "error constant" that establishes the equality. It is proportional to $1/2n$ and thus $epsilon_n to 0$ as $n to infty$.



And then I guess because $ln(n) - ln(k) = ln(n/k)$ per a property of logarithms, that might explain your other question.



But I want to emphasize that I'm kinda just guessing here since I'm not particularly familiar with the big-theta notation or how to utilize it. I think that's what's at play here, but it's just a guess.






share|cite|improve this answer









$endgroup$



The $H$ notation used is more a shorthand for the harmonic series' partial sums. More explicitly,



$$H(n) = text{(the n-th partial sum of the harmonic series)} = sum_{k=1}^n frac{1}{k}$$



Edit: You could also say that $H(n)$ is the $n^{th}$ harmonic number, as the definition is the same.



Going from there, I'm just guessing on what $Theta$ represents from what I could Google up. My guess is that it is a function that means "on the order of" or something of the sort. For that, it might be important to note that we can represent the harmonic series in a sort of "closed" form:



$$sum_{k=1}^n frac{1}{k} = ln(n) + gamma + epsilon_n approx ln(n)$$



In this context, $gamma$ is the Euler-Mascheroni constant, and $epsilon_n$ is a sort of "error constant" that establishes the equality. It is proportional to $1/2n$ and thus $epsilon_n to 0$ as $n to infty$.



And then I guess because $ln(n) - ln(k) = ln(n/k)$ per a property of logarithms, that might explain your other question.



But I want to emphasize that I'm kinda just guessing here since I'm not particularly familiar with the big-theta notation or how to utilize it. I think that's what's at play here, but it's just a guess.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 7 '18 at 1:01









Eevee TrainerEevee Trainer

5,7961936




5,7961936












  • $begingroup$
    Yes, $Theta$ basically means "on the order of", i.e., bounded both above and below up to a constant. See Big-O-notation
    $endgroup$
    – Jair Taylor
    Dec 7 '18 at 1:05










  • $begingroup$
    This was an absolutely amazing explanation. Thank you so much for taking the time to explain this to me.
    $endgroup$
    – Jerry M.
    Dec 7 '18 at 1:37


















  • $begingroup$
    Yes, $Theta$ basically means "on the order of", i.e., bounded both above and below up to a constant. See Big-O-notation
    $endgroup$
    – Jair Taylor
    Dec 7 '18 at 1:05










  • $begingroup$
    This was an absolutely amazing explanation. Thank you so much for taking the time to explain this to me.
    $endgroup$
    – Jerry M.
    Dec 7 '18 at 1:37
















$begingroup$
Yes, $Theta$ basically means "on the order of", i.e., bounded both above and below up to a constant. See Big-O-notation
$endgroup$
– Jair Taylor
Dec 7 '18 at 1:05




$begingroup$
Yes, $Theta$ basically means "on the order of", i.e., bounded both above and below up to a constant. See Big-O-notation
$endgroup$
– Jair Taylor
Dec 7 '18 at 1:05












$begingroup$
This was an absolutely amazing explanation. Thank you so much for taking the time to explain this to me.
$endgroup$
– Jerry M.
Dec 7 '18 at 1:37




$begingroup$
This was an absolutely amazing explanation. Thank you so much for taking the time to explain this to me.
$endgroup$
– Jerry M.
Dec 7 '18 at 1:37


















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