Closed Form of the Real Portion of the Product of Exponentials
$begingroup$
I am wondering if it is possible to express an equation in closed form. I currently have:
$f(n) = prod_{m=2}^{n-1} e^frac{pi i n}{m}$
Where $i$ is the $sqrt{-1}$, which I know it commonly represents but due to the finite product I figured I would add this note for clarity.
Context: I am trying to work towards a solution for Simplify Product of sines, yet so far I have only arrived here.
Some options I have been considering to try and find a closed form include cases such as I only actually care when $n$ is an odd number so discarding even cases is fine if that simplifies it. I am also fine with having the series start a $m=1$ instead of $m=2$ if that somehow simplifies the answer. The other potential saving grace is that I may only care about the real portion of the answer and not the imaginary portion if that simplifies things.
Sorry I am a computer scientist not a mathematician so please let me know if I should adjust the title or the wording of the question for clarity. Any leads or help of any form would be appreciated as I am currently lost.
exponential-function closed-form products
asked 1 hour ago
cytinuscytinus
1164
New contributor
cytinus is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I am wondering if it is possible to express an equation in closed form. I currently have:
$f(n) = prod_{m=2}^{n-1} e^frac{pi i n}{m}$
Where $i$ is the $sqrt{-1}$, which I know it commonly represents but due to the finite product I figured I would add this note for clarity.
Context: I am trying to work towards a solution for Simplify Product of sines, yet so far I have only arrived here.
Some options I have been considering to try and find a closed form include cases such as I only actually care when $n$ is an odd number so discarding even cases is fine if that simplifies it. I am also fine with having the series start a $m=1$ instead of $m=2$ if that somehow simplifies the answer. The other potential saving grace is that I may only care about the real portion of the answer and not the imaginary portion if that simplifies things.
Sorry I am a computer scientist not a mathematician so please let me know if I should adjust the title or the wording of the question for clarity. Any leads or help of any form would be appreciated as I am currently lost.
exponential-function closed-form products
asked 1 hour ago
cytinuscytinus
1164
New contributor
cytinus is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Please give more context. Providing context not only assures that this is not simply copied from a homework assignment, but also allows answers to be better directed at where the problem lies and to be within the proper scope. Please avoid "I have no clue" questions. Defining keywords and trying a simpler, similar problem often helps.
$endgroup$
– robjohn♦
1 hour ago
$begingroup$
Hi, I guess the context is I was trying to work a little more on this, math.stackexchange.com/questions/1689831/…. I'm not in school and this is not homework, I am just a computer scientist who likes to dabble in some of my free time. I was trying to work towards a closed form of the equation linked and so I was breaking it into parts and this is where I got stuck. Its something I look at every few months so even a nudge in any direction would be helpful.
$endgroup$
– cytinus
1 hour ago
$begingroup$
Hint: $prodlimits_{m=2}^{n-1}e^{a_m}=e^{sumlimits_{m=2}^{n-1}a_m}$
$endgroup$
– robjohn♦
58 mins ago
1
$begingroup$
It would be helpful to add that context to the question itself, rather than in a comment. I have undeleted my answer.
$endgroup$
– robjohn♦
55 mins ago
add a comment |
$begingroup$
I am wondering if it is possible to express an equation in closed form. I currently have:
$f(n) = prod_{m=2}^{n-1} e^frac{pi i n}{m}$
Where $i$ is the $sqrt{-1}$, which I know it commonly represents but due to the finite product I figured I would add this note for clarity.
Context: I am trying to work towards a solution for Simplify Product of sines, yet so far I have only arrived here.
Some options I have been considering to try and find a closed form include cases such as I only actually care when $n$ is an odd number so discarding even cases is fine if that simplifies it. I am also fine with having the series start a $m=1$ instead of $m=2$ if that somehow simplifies the answer. The other potential saving grace is that I may only care about the real portion of the answer and not the imaginary portion if that simplifies things.
Sorry I am a computer scientist not a mathematician so please let me know if I should adjust the title or the wording of the question for clarity. Any leads or help of any form would be appreciated as I am currently lost.
exponential-function closed-form products
asked 1 hour ago
cytinuscytinus
1164
New contributor
cytinus is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I am wondering if it is possible to express an equation in closed form. I currently have:
$f(n) = prod_{m=2}^{n-1} e^frac{pi i n}{m}$
Where $i$ is the $sqrt{-1}$, which I know it commonly represents but due to the finite product I figured I would add this note for clarity.
Context: I am trying to work towards a solution for Simplify Product of sines, yet so far I have only arrived here.
Some options I have been considering to try and find a closed form include cases such as I only actually care when $n$ is an odd number so discarding even cases is fine if that simplifies it. I am also fine with having the series start a $m=1$ instead of $m=2$ if that somehow simplifies the answer. The other potential saving grace is that I may only care about the real portion of the answer and not the imaginary portion if that simplifies things.
Sorry I am a computer scientist not a mathematician so please let me know if I should adjust the title or the wording of the question for clarity. Any leads or help of any form would be appreciated as I am currently lost.
exponential-function closed-form products
exponential-function closed-form products
asked 1 hour ago
cytinuscytinus
1164
New contributor
cytinus is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
cytinuscytinus
1164
New contributor
cytinus is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 52 mins ago
cytinus
asked 1 hour ago
cytinuscytinus
1164
New contributor
cytinus is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
cytinuscytinus
1164
asked 1 hour ago
cytinuscytinus
1164
1164
New contributor
cytinus is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
cytinus is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
cytinus is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
Please give more context. Providing context not only assures that this is not simply copied from a homework assignment, but also allows answers to be better directed at where the problem lies and to be within the proper scope. Please avoid "I have no clue" questions. Defining keywords and trying a simpler, similar problem often helps.
$endgroup$
– robjohn♦
1 hour ago
$begingroup$
Hi, I guess the context is I was trying to work a little more on this, math.stackexchange.com/questions/1689831/…. I'm not in school and this is not homework, I am just a computer scientist who likes to dabble in some of my free time. I was trying to work towards a closed form of the equation linked and so I was breaking it into parts and this is where I got stuck. Its something I look at every few months so even a nudge in any direction would be helpful.
$endgroup$
– cytinus
1 hour ago
$begingroup$
Hint: $prodlimits_{m=2}^{n-1}e^{a_m}=e^{sumlimits_{m=2}^{n-1}a_m}$
$endgroup$
– robjohn♦
58 mins ago
1
$begingroup$
It would be helpful to add that context to the question itself, rather than in a comment. I have undeleted my answer.
$endgroup$
– robjohn♦
55 mins ago
add a comment |
$begingroup$
Please give more context. Providing context not only assures that this is not simply copied from a homework assignment, but also allows answers to be better directed at where the problem lies and to be within the proper scope. Please avoid "I have no clue" questions. Defining keywords and trying a simpler, similar problem often helps.
$endgroup$
– robjohn♦
1 hour ago
$begingroup$
Hi, I guess the context is I was trying to work a little more on this, math.stackexchange.com/questions/1689831/…. I'm not in school and this is not homework, I am just a computer scientist who likes to dabble in some of my free time. I was trying to work towards a closed form of the equation linked and so I was breaking it into parts and this is where I got stuck. Its something I look at every few months so even a nudge in any direction would be helpful.
$endgroup$
– cytinus
1 hour ago
$begingroup$
Hint: $prodlimits_{m=2}^{n-1}e^{a_m}=e^{sumlimits_{m=2}^{n-1}a_m}$
$endgroup$
– robjohn♦
58 mins ago
1
$begingroup$
It would be helpful to add that context to the question itself, rather than in a comment. I have undeleted my answer.
$endgroup$
– robjohn♦
55 mins ago
$begingroup$
Please give more context. Providing context not only assures that this is not simply copied from a homework assignment, but also allows answers to be better directed at where the problem lies and to be within the proper scope. Please avoid "I have no clue" questions. Defining keywords and trying a simpler, similar problem often helps.
$endgroup$
– robjohn♦
1 hour ago
$begingroup$
Please give more context. Providing context not only assures that this is not simply copied from a homework assignment, but also allows answers to be better directed at where the problem lies and to be within the proper scope. Please avoid "I have no clue" questions. Defining keywords and trying a simpler, similar problem often helps.
$endgroup$
– robjohn♦
1 hour ago
$begingroup$
Hi, I guess the context is I was trying to work a little more on this, math.stackexchange.com/questions/1689831/…. I'm not in school and this is not homework, I am just a computer scientist who likes to dabble in some of my free time. I was trying to work towards a closed form of the equation linked and so I was breaking it into parts and this is where I got stuck. Its something I look at every few months so even a nudge in any direction would be helpful.
$endgroup$
– cytinus
1 hour ago
$begingroup$
Hi, I guess the context is I was trying to work a little more on this, math.stackexchange.com/questions/1689831/…. I'm not in school and this is not homework, I am just a computer scientist who likes to dabble in some of my free time. I was trying to work towards a closed form of the equation linked and so I was breaking it into parts and this is where I got stuck. Its something I look at every few months so even a nudge in any direction would be helpful.
$endgroup$
– cytinus
1 hour ago
$begingroup$
Hint: $prodlimits_{m=2}^{n-1}e^{a_m}=e^{sumlimits_{m=2}^{n-1}a_m}$
$endgroup$
– robjohn♦
58 mins ago
$begingroup$
Hint: $prodlimits_{m=2}^{n-1}e^{a_m}=e^{sumlimits_{m=2}^{n-1}a_m}$
$endgroup$
– robjohn♦
58 mins ago
1
1
$begingroup$
It would be helpful to add that context to the question itself, rather than in a comment. I have undeleted my answer.
$endgroup$
– robjohn♦
55 mins ago
$begingroup$
It would be helpful to add that context to the question itself, rather than in a comment. I have undeleted my answer.
$endgroup$
– robjohn♦
55 mins ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$$begin{align}
f(n)=&prod_{m=2}^{n-1}expfrac{ipi n}{m}\
&=expleft[sum_{m=2}^{n-1}frac{ipi n}{m}right]\
&=expleft[ipi nsum_{m=2}^{n-1}frac{1}{m}right]\
end{align}$$
Recalling the definition of the harmonic numbers:
$$H_n=sum_{m=1}^nfrac1m$$
We have that
$$f(n)=expleft[ipi n(H_{n-1}-1)right]$$
Then using $e^{itheta}=costheta+isintheta$,
$$f(n)=cosleft[pi n(H_{n-1}-1)right]+isinleft[pi n(H_{n-1}-1)right]$$
So
$$text{Re}f(n)=cosleft[pi n(H_{n-1}-1)right]$$
answered 1 hour ago
clathratusclathratus
3,606332
$endgroup$
1
$begingroup$
Why is $expleft[ipi n(H_{n-1}-1)right]=exp(ipi n)exp(H_{n-1}-1)$?
$endgroup$
– greelious
54 mins ago
$begingroup$
@greelious because I messed up... Thanks! :)
$endgroup$
– clathratus
53 mins ago
$begingroup$
I think it is still not correct: $expleft[ipi n(H_{n-1}-1)right]=exp(i pi n H_{n-1})exp(-i pi n)=exp(i pi n H_{n-1})(-1)^n$
$endgroup$
– greelious
46 mins ago
$begingroup$
@greelious okay look now
$endgroup$
– clathratus
43 mins ago
$begingroup$
Looks good now.
$endgroup$
– greelious
39 mins ago
add a comment |
$begingroup$
$$
begin{align}
prod_{m=2}^{n-1}e^{frac{pi in}m}
&=e^{pi in(H_{n-1}-1)}\
&=(-1)^ne^{pi inH_{n-1}}\[9pt]
&=(-1)^{n-1}e^{pi inH_n}tag1
end{align}
$$
where $H_n$ is the $n^text{th}$ Harmonic Number.
$$
H_nsimlog(n)+gamma+frac1{2n}-frac1{12n^2}+frac1{120n^4}-frac1{252n^6}+frac1{240n^8}-frac1{132n^{10}}tag2
$$
and $gamma$ is the Euler-Mascheroni constant.
The real portion of $(1)$ is
$$
operatorname{Re}left(prod_{m=2}^{n-1}e^{frac{pi in}m}right)=(-1)^{n-1}cosleft(pi nH_nright)tag3
$$
answered 1 hour ago
robjohn♦robjohn
265k27304626
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$$begin{align}
f(n)=&prod_{m=2}^{n-1}expfrac{ipi n}{m}\
&=expleft[sum_{m=2}^{n-1}frac{ipi n}{m}right]\
&=expleft[ipi nsum_{m=2}^{n-1}frac{1}{m}right]\
end{align}$$
Recalling the definition of the harmonic numbers:
$$H_n=sum_{m=1}^nfrac1m$$
We have that
$$f(n)=expleft[ipi n(H_{n-1}-1)right]$$
Then using $e^{itheta}=costheta+isintheta$,
$$f(n)=cosleft[pi n(H_{n-1}-1)right]+isinleft[pi n(H_{n-1}-1)right]$$
So
$$text{Re}f(n)=cosleft[pi n(H_{n-1}-1)right]$$
answered 1 hour ago
clathratusclathratus
3,606332
$endgroup$
1
$begingroup$
Why is $expleft[ipi n(H_{n-1}-1)right]=exp(ipi n)exp(H_{n-1}-1)$?
$endgroup$
– greelious
54 mins ago
$begingroup$
@greelious because I messed up... Thanks! :)
$endgroup$
– clathratus
53 mins ago
$begingroup$
I think it is still not correct: $expleft[ipi n(H_{n-1}-1)right]=exp(i pi n H_{n-1})exp(-i pi n)=exp(i pi n H_{n-1})(-1)^n$
$endgroup$
– greelious
46 mins ago
$begingroup$
@greelious okay look now
$endgroup$
– clathratus
43 mins ago
$begingroup$
Looks good now.
$endgroup$
– greelious
39 mins ago
add a comment |
$begingroup$
$$begin{align}
f(n)=&prod_{m=2}^{n-1}expfrac{ipi n}{m}\
&=expleft[sum_{m=2}^{n-1}frac{ipi n}{m}right]\
&=expleft[ipi nsum_{m=2}^{n-1}frac{1}{m}right]\
end{align}$$
Recalling the definition of the harmonic numbers:
$$H_n=sum_{m=1}^nfrac1m$$
We have that
$$f(n)=expleft[ipi n(H_{n-1}-1)right]$$
Then using $e^{itheta}=costheta+isintheta$,
$$f(n)=cosleft[pi n(H_{n-1}-1)right]+isinleft[pi n(H_{n-1}-1)right]$$
So
$$text{Re}f(n)=cosleft[pi n(H_{n-1}-1)right]$$
answered 1 hour ago
clathratusclathratus
3,606332
$endgroup$
1
$begingroup$
Why is $expleft[ipi n(H_{n-1}-1)right]=exp(ipi n)exp(H_{n-1}-1)$?
$endgroup$
– greelious
54 mins ago
$begingroup$
@greelious because I messed up... Thanks! :)
$endgroup$
– clathratus
53 mins ago
$begingroup$
I think it is still not correct: $expleft[ipi n(H_{n-1}-1)right]=exp(i pi n H_{n-1})exp(-i pi n)=exp(i pi n H_{n-1})(-1)^n$
$endgroup$
– greelious
46 mins ago
$begingroup$
@greelious okay look now
$endgroup$
– clathratus
43 mins ago
$begingroup$
Looks good now.
$endgroup$
– greelious
39 mins ago
add a comment |
$begingroup$
$$begin{align}
f(n)=&prod_{m=2}^{n-1}expfrac{ipi n}{m}\
&=expleft[sum_{m=2}^{n-1}frac{ipi n}{m}right]\
&=expleft[ipi nsum_{m=2}^{n-1}frac{1}{m}right]\
end{align}$$
Recalling the definition of the harmonic numbers:
$$H_n=sum_{m=1}^nfrac1m$$
We have that
$$f(n)=expleft[ipi n(H_{n-1}-1)right]$$
Then using $e^{itheta}=costheta+isintheta$,
$$f(n)=cosleft[pi n(H_{n-1}-1)right]+isinleft[pi n(H_{n-1}-1)right]$$
So
$$text{Re}f(n)=cosleft[pi n(H_{n-1}-1)right]$$
answered 1 hour ago
clathratusclathratus
3,606332
$endgroup$
$$begin{align}
f(n)=&prod_{m=2}^{n-1}expfrac{ipi n}{m}\
&=expleft[sum_{m=2}^{n-1}frac{ipi n}{m}right]\
&=expleft[ipi nsum_{m=2}^{n-1}frac{1}{m}right]\
end{align}$$
Recalling the definition of the harmonic numbers:
$$H_n=sum_{m=1}^nfrac1m$$
We have that
$$f(n)=expleft[ipi n(H_{n-1}-1)right]$$
Then using $e^{itheta}=costheta+isintheta$,
$$f(n)=cosleft[pi n(H_{n-1}-1)right]+isinleft[pi n(H_{n-1}-1)right]$$
So
$$text{Re}f(n)=cosleft[pi n(H_{n-1}-1)right]$$
answered 1 hour ago
clathratusclathratus
3,606332
edited 43 mins ago
answered 1 hour ago
clathratusclathratus
3,606332
answered 1 hour ago
clathratusclathratus
3,606332
answered 1 hour ago
clathratusclathratus
3,606332
3,606332
1
$begingroup$
Why is $expleft[ipi n(H_{n-1}-1)right]=exp(ipi n)exp(H_{n-1}-1)$?
$endgroup$
– greelious
54 mins ago
$begingroup$
@greelious because I messed up... Thanks! :)
$endgroup$
– clathratus
53 mins ago
$begingroup$
I think it is still not correct: $expleft[ipi n(H_{n-1}-1)right]=exp(i pi n H_{n-1})exp(-i pi n)=exp(i pi n H_{n-1})(-1)^n$
$endgroup$
– greelious
46 mins ago
$begingroup$
@greelious okay look now
$endgroup$
– clathratus
43 mins ago
$begingroup$
Looks good now.
$endgroup$
– greelious
39 mins ago
add a comment |
1
$begingroup$
Why is $expleft[ipi n(H_{n-1}-1)right]=exp(ipi n)exp(H_{n-1}-1)$?
$endgroup$
– greelious
54 mins ago
$begingroup$
@greelious because I messed up... Thanks! :)
$endgroup$
– clathratus
53 mins ago
$begingroup$
I think it is still not correct: $expleft[ipi n(H_{n-1}-1)right]=exp(i pi n H_{n-1})exp(-i pi n)=exp(i pi n H_{n-1})(-1)^n$
$endgroup$
– greelious
46 mins ago
$begingroup$
@greelious okay look now
$endgroup$
– clathratus
43 mins ago
$begingroup$
Looks good now.
$endgroup$
– greelious
39 mins ago
1
1
$begingroup$
Why is $expleft[ipi n(H_{n-1}-1)right]=exp(ipi n)exp(H_{n-1}-1)$?
$endgroup$
– greelious
54 mins ago
$begingroup$
Why is $expleft[ipi n(H_{n-1}-1)right]=exp(ipi n)exp(H_{n-1}-1)$?
$endgroup$
– greelious
54 mins ago
$begingroup$
@greelious because I messed up... Thanks! :)
$endgroup$
– clathratus
53 mins ago
$begingroup$
@greelious because I messed up... Thanks! :)
$endgroup$
– clathratus
53 mins ago
$begingroup$
I think it is still not correct: $expleft[ipi n(H_{n-1}-1)right]=exp(i pi n H_{n-1})exp(-i pi n)=exp(i pi n H_{n-1})(-1)^n$
$endgroup$
– greelious
46 mins ago
$begingroup$
I think it is still not correct: $expleft[ipi n(H_{n-1}-1)right]=exp(i pi n H_{n-1})exp(-i pi n)=exp(i pi n H_{n-1})(-1)^n$
$endgroup$
– greelious
46 mins ago
$begingroup$
@greelious okay look now
$endgroup$
– clathratus
43 mins ago
$begingroup$
@greelious okay look now
$endgroup$
– clathratus
43 mins ago
$begingroup$
Looks good now.
$endgroup$
– greelious
39 mins ago
$begingroup$
Looks good now.
$endgroup$
– greelious
39 mins ago
add a comment |
$begingroup$
$$
begin{align}
prod_{m=2}^{n-1}e^{frac{pi in}m}
&=e^{pi in(H_{n-1}-1)}\
&=(-1)^ne^{pi inH_{n-1}}\[9pt]
&=(-1)^{n-1}e^{pi inH_n}tag1
end{align}
$$
where $H_n$ is the $n^text{th}$ Harmonic Number.
$$
H_nsimlog(n)+gamma+frac1{2n}-frac1{12n^2}+frac1{120n^4}-frac1{252n^6}+frac1{240n^8}-frac1{132n^{10}}tag2
$$
and $gamma$ is the Euler-Mascheroni constant.
The real portion of $(1)$ is
$$
operatorname{Re}left(prod_{m=2}^{n-1}e^{frac{pi in}m}right)=(-1)^{n-1}cosleft(pi nH_nright)tag3
$$
answered 1 hour ago
robjohn♦robjohn
265k27304626
$endgroup$
add a comment |
$begingroup$
$$
begin{align}
prod_{m=2}^{n-1}e^{frac{pi in}m}
&=e^{pi in(H_{n-1}-1)}\
&=(-1)^ne^{pi inH_{n-1}}\[9pt]
&=(-1)^{n-1}e^{pi inH_n}tag1
end{align}
$$
where $H_n$ is the $n^text{th}$ Harmonic Number.
$$
H_nsimlog(n)+gamma+frac1{2n}-frac1{12n^2}+frac1{120n^4}-frac1{252n^6}+frac1{240n^8}-frac1{132n^{10}}tag2
$$
and $gamma$ is the Euler-Mascheroni constant.
The real portion of $(1)$ is
$$
operatorname{Re}left(prod_{m=2}^{n-1}e^{frac{pi in}m}right)=(-1)^{n-1}cosleft(pi nH_nright)tag3
$$
answered 1 hour ago
robjohn♦robjohn
265k27304626
$endgroup$
add a comment |
$begingroup$
$$
begin{align}
prod_{m=2}^{n-1}e^{frac{pi in}m}
&=e^{pi in(H_{n-1}-1)}\
&=(-1)^ne^{pi inH_{n-1}}\[9pt]
&=(-1)^{n-1}e^{pi inH_n}tag1
end{align}
$$
where $H_n$ is the $n^text{th}$ Harmonic Number.
$$
H_nsimlog(n)+gamma+frac1{2n}-frac1{12n^2}+frac1{120n^4}-frac1{252n^6}+frac1{240n^8}-frac1{132n^{10}}tag2
$$
and $gamma$ is the Euler-Mascheroni constant.
The real portion of $(1)$ is
$$
operatorname{Re}left(prod_{m=2}^{n-1}e^{frac{pi in}m}right)=(-1)^{n-1}cosleft(pi nH_nright)tag3
$$
answered 1 hour ago
robjohn♦robjohn
265k27304626
$endgroup$
$$
begin{align}
prod_{m=2}^{n-1}e^{frac{pi in}m}
&=e^{pi in(H_{n-1}-1)}\
&=(-1)^ne^{pi inH_{n-1}}\[9pt]
&=(-1)^{n-1}e^{pi inH_n}tag1
end{align}
$$
where $H_n$ is the $n^text{th}$ Harmonic Number.
$$
H_nsimlog(n)+gamma+frac1{2n}-frac1{12n^2}+frac1{120n^4}-frac1{252n^6}+frac1{240n^8}-frac1{132n^{10}}tag2
$$
and $gamma$ is the Euler-Mascheroni constant.
The real portion of $(1)$ is
$$
operatorname{Re}left(prod_{m=2}^{n-1}e^{frac{pi in}m}right)=(-1)^{n-1}cosleft(pi nH_nright)tag3
$$
answered 1 hour ago
robjohn♦robjohn
265k27304626
edited 57 mins ago
answered 1 hour ago
robjohn♦robjohn
265k27304626
answered 1 hour ago
robjohn♦robjohn
265k27304626
answered 1 hour ago
robjohn♦robjohn
265k27304626
265k27304626
add a comment |
add a comment |
cytinus is a new contributor. Be nice, and check out our Code of Conduct.
cytinus is a new contributor. Be nice, and check out our Code of Conduct.
cytinus is a new contributor. Be nice, and check out our Code of Conduct.
cytinus is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
Please give more context. Providing context not only assures that this is not simply copied from a homework assignment, but also allows answers to be better directed at where the problem lies and to be within the proper scope. Please avoid "I have no clue" questions. Defining keywords and trying a simpler, similar problem often helps.
$endgroup$
– robjohn♦
1 hour ago
$begingroup$
Hi, I guess the context is I was trying to work a little more on this, math.stackexchange.com/questions/1689831/…. I'm not in school and this is not homework, I am just a computer scientist who likes to dabble in some of my free time. I was trying to work towards a closed form of the equation linked and so I was breaking it into parts and this is where I got stuck. Its something I look at every few months so even a nudge in any direction would be helpful.
$endgroup$
– cytinus
1 hour ago
$begingroup$
Hint: $prodlimits_{m=2}^{n-1}e^{a_m}=e^{sumlimits_{m=2}^{n-1}a_m}$
$endgroup$
– robjohn♦
58 mins ago
1
$begingroup$
It would be helpful to add that context to the question itself, rather than in a comment. I have undeleted my answer.
$endgroup$
– robjohn♦
55 mins ago