Mixing
Closed Form of the Real Portion of $f(n) = prod_{m=2}^{n-1} e^{pi i n/m}$
$begingroup$
I am wondering if it is possible to express an equation in closed form. I currently have:
$$f(n) = prod_{m=2}^{n-1} e^{pi i n/m}$$
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
edited Jan 17 at 8:26
Asaf Karagila♦
305k33435766
asked Jan 17 at 2:44
cytinuscytinus
1164
$endgroup$
add a comment |
$begingroup$
I am wondering if it is possible to express an equation in closed form. I currently have:
$$f(n) = prod_{m=2}^{n-1} e^{pi i n/m}$$
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
edited Jan 17 at 8:26
Asaf Karagila♦
305k33435766
asked Jan 17 at 2:44
cytinuscytinus
1164
$endgroup$
$begingroup$
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♦
Jan 17 at 3:15
1
$begingroup$
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
Jan 17 at 3:22
$begingroup$
Hint: $prodlimits_{m=2}^{n-1}e^{a_m}=e^{sumlimits_{m=2}^{n-1}a_m}$
$endgroup$
– robjohn♦
Jan 17 at 3:25
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♦
Jan 17 at 3:29
add a comment |
$begingroup$
I am wondering if it is possible to express an equation in closed form. I currently have:
$$f(n) = prod_{m=2}^{n-1} e^{pi i n/m}$$
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
edited Jan 17 at 8:26
Asaf Karagila♦
305k33435766
asked Jan 17 at 2:44
cytinuscytinus
1164
$endgroup$
I am wondering if it is possible to express an equation in closed form. I currently have:
$$f(n) = prod_{m=2}^{n-1} e^{pi i n/m}$$
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
edited Jan 17 at 8:26
Asaf Karagila♦
305k33435766
asked Jan 17 at 2:44
cytinuscytinus
1164
edited Jan 17 at 8:26
Asaf Karagila♦
305k33435766
asked Jan 17 at 2:44
cytinuscytinus
1164
edited Jan 17 at 8:26
Asaf Karagila♦
305k33435766
edited Jan 17 at 8:26
Asaf Karagila♦
305k33435766
edited Jan 17 at 8:26
Asaf Karagila♦
305k33435766
305k33435766
asked Jan 17 at 2:44
cytinuscytinus
1164
asked Jan 17 at 2:44
cytinuscytinus
1164
asked Jan 17 at 2:44
cytinuscytinus
1164
1164
$begingroup$
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♦
Jan 17 at 3:15
1
$begingroup$
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
Jan 17 at 3:22
$begingroup$
Hint: $prodlimits_{m=2}^{n-1}e^{a_m}=e^{sumlimits_{m=2}^{n-1}a_m}$
$endgroup$
– robjohn♦
Jan 17 at 3:25
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♦
Jan 17 at 3:29
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♦
Jan 17 at 3:15
1
$begingroup$
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
Jan 17 at 3:22
$begingroup$
Hint: $prodlimits_{m=2}^{n-1}e^{a_m}=e^{sumlimits_{m=2}^{n-1}a_m}$
$endgroup$
– robjohn♦
Jan 17 at 3:25
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♦
Jan 17 at 3:29
$begingroup$
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♦
Jan 17 at 3:15
$begingroup$
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♦
Jan 17 at 3:15
1
1
$begingroup$
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
Jan 17 at 3:22
$begingroup$
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
Jan 17 at 3:22
$begingroup$
Hint: $prodlimits_{m=2}^{n-1}e^{a_m}=e^{sumlimits_{m=2}^{n-1}a_m}$
$endgroup$
– robjohn♦
Jan 17 at 3:25
$begingroup$
Hint: $prodlimits_{m=2}^{n-1}e^{a_m}=e^{sumlimits_{m=2}^{n-1}a_m}$
$endgroup$
– robjohn♦
Jan 17 at 3:25
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♦
Jan 17 at 3:29
$begingroup$
It would be helpful to add that context to the question itself, rather than in a comment. I have undeleted my answer.
$endgroup$
– robjohn♦
Jan 17 at 3:29
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$$
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
$$
answered Jan 17 at 2:58
robjohn♦robjohn
268k27308633
$endgroup$
add a comment |
$begingroup$
$$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]$$
answered Jan 17 at 3:21
clathratusclathratus
4,640337
$endgroup$
1
$begingroup$
Why is $expleft[ipi n(H_{n-1}-1)right]=exp(ipi n)exp(H_{n-1}-1)$?
$endgroup$
– greelious
Jan 17 at 3:29
$begingroup$
@greelious because I messed up... Thanks! :)
$endgroup$
– clathratus
Jan 17 at 3:30
$begingroup$
I think it is still not correct: $expleft[ipi n(H_{n-1}-1)right]=exp(i pi n H_{n-1})exp(-i pi n)=exp(i pi n H_{n-1})(-1)^n$
$endgroup$
– greelious
Jan 17 at 3:37
$begingroup$
@greelious okay look now
$endgroup$
– clathratus
Jan 17 at 3:41
$begingroup$
Looks good now.
$endgroup$
– greelious
Jan 17 at 3:45
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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active
oldest
votes
$begingroup$
$$
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
$$
answered Jan 17 at 2:58
robjohn♦robjohn
268k27308633
$endgroup$
add a comment |
$begingroup$
$$
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
$$
answered Jan 17 at 2:58
robjohn♦robjohn
268k27308633
$endgroup$
add a comment |
$begingroup$
$$
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
$$
answered Jan 17 at 2:58
robjohn♦robjohn
268k27308633
$endgroup$
$$
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
$$
answered Jan 17 at 2:58
robjohn♦robjohn
268k27308633
edited Jan 17 at 3:27
answered Jan 17 at 2:58
robjohn♦robjohn
268k27308633
answered Jan 17 at 2:58
robjohn♦robjohn
268k27308633
answered Jan 17 at 2:58
robjohn♦robjohn
268k27308633
268k27308633
add a comment |
add a comment |
$begingroup$
$$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]$$
answered Jan 17 at 3:21
clathratusclathratus
4,640337
$endgroup$
1
$begingroup$
Why is $expleft[ipi n(H_{n-1}-1)right]=exp(ipi n)exp(H_{n-1}-1)$?
$endgroup$
– greelious
Jan 17 at 3:29
$begingroup$
@greelious because I messed up... Thanks! :)
$endgroup$
– clathratus
Jan 17 at 3:30
$begingroup$
I think it is still not correct: $expleft[ipi n(H_{n-1}-1)right]=exp(i pi n H_{n-1})exp(-i pi n)=exp(i pi n H_{n-1})(-1)^n$
$endgroup$
– greelious
Jan 17 at 3:37
$begingroup$
@greelious okay look now
$endgroup$
– clathratus
Jan 17 at 3:41
$begingroup$
Looks good now.
$endgroup$
– greelious
Jan 17 at 3:45
add a comment |
$begingroup$
$$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]$$
answered Jan 17 at 3:21
clathratusclathratus
4,640337
$endgroup$
1
$begingroup$
Why is $expleft[ipi n(H_{n-1}-1)right]=exp(ipi n)exp(H_{n-1}-1)$?
$endgroup$
– greelious
Jan 17 at 3:29
$begingroup$
@greelious because I messed up... Thanks! :)
$endgroup$
– clathratus
Jan 17 at 3:30
$begingroup$
I think it is still not correct: $expleft[ipi n(H_{n-1}-1)right]=exp(i pi n H_{n-1})exp(-i pi n)=exp(i pi n H_{n-1})(-1)^n$
$endgroup$
– greelious
Jan 17 at 3:37
$begingroup$
@greelious okay look now
$endgroup$
– clathratus
Jan 17 at 3:41
$begingroup$
Looks good now.
$endgroup$
– greelious
Jan 17 at 3:45
add a comment |
$begingroup$
$$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]$$
answered Jan 17 at 3:21
clathratusclathratus
4,640337
$endgroup$
$$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 Jan 17 at 3:21
clathratusclathratus
4,640337
edited Jan 17 at 3:41
answered Jan 17 at 3:21
clathratusclathratus
4,640337
answered Jan 17 at 3:21
clathratusclathratus
4,640337
answered Jan 17 at 3:21
clathratusclathratus
4,640337
4,640337
1
$begingroup$
Why is $expleft[ipi n(H_{n-1}-1)right]=exp(ipi n)exp(H_{n-1}-1)$?
$endgroup$
– greelious
Jan 17 at 3:29
$begingroup$
@greelious because I messed up... Thanks! :)
$endgroup$
– clathratus
Jan 17 at 3:30
$begingroup$
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
Jan 17 at 3:37
$begingroup$
@greelious okay look now
$endgroup$
– clathratus
Jan 17 at 3:41
$begingroup$
Looks good now.
$endgroup$
– greelious
Jan 17 at 3:45
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
Jan 17 at 3:29
$begingroup$
@greelious because I messed up... Thanks! :)
$endgroup$
– clathratus
Jan 17 at 3:30
$begingroup$
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
Jan 17 at 3:37
$begingroup$
@greelious okay look now
$endgroup$
– clathratus
Jan 17 at 3:41
$begingroup$
Looks good now.
$endgroup$
– greelious
Jan 17 at 3:45
1
1
$begingroup$
Why is $expleft[ipi n(H_{n-1}-1)right]=exp(ipi n)exp(H_{n-1}-1)$?
$endgroup$
– greelious
Jan 17 at 3:29
$begingroup$
Why is $expleft[ipi n(H_{n-1}-1)right]=exp(ipi n)exp(H_{n-1}-1)$?
$endgroup$
– greelious
Jan 17 at 3:29
$begingroup$
@greelious because I messed up... Thanks! :)
$endgroup$
– clathratus
Jan 17 at 3:30
$begingroup$
@greelious because I messed up... Thanks! :)
$endgroup$
– clathratus
Jan 17 at 3:30
$begingroup$
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
Jan 17 at 3:37
$begingroup$
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
Jan 17 at 3:37
$begingroup$
@greelious okay look now
$endgroup$
– clathratus
Jan 17 at 3:41
$begingroup$
@greelious okay look now
$endgroup$
– clathratus
Jan 17 at 3:41
$begingroup$
Looks good now.
$endgroup$
– greelious
Jan 17 at 3:45
$begingroup$
Looks good now.
$endgroup$
– greelious
Jan 17 at 3:45
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
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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♦
Jan 17 at 3:15
1
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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.
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– cytinus
Jan 17 at 3:22
$begingroup$
Hint: $prodlimits_{m=2}^{n-1}e^{a_m}=e^{sumlimits_{m=2}^{n-1}a_m}$
$endgroup$
– robjohn♦
Jan 17 at 3:25
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♦
Jan 17 at 3:29