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What is the value of: $lim_{nrightarrow infty}sum_{r=1}^{n-1}frac{cot^2(rpi/n)}{n^2}$
$begingroup$
$$lim_{nrightarrow infty}sum_{r=1}^{n-1}frac{cot^2(rpi/n)}{n^2}$$
How can I calculate the value of this trigonometric function where limits tends to infinity? I have thought and tried various of methods like using:
- Tan(x)/x property
- Converting lim to
Integration - Sandwich theorem
- Trigonometric series
But none worked out for me.
So please tell me a good and easy approach for this question :)
sequences-and-series limits trigonometry summation
edited Jan 8 at 17:59
Mark Viola
131k1275171
asked Jan 8 at 16:20
Shubham GawriShubham Gawri
286
$endgroup$
add a comment |
$begingroup$
$$lim_{nrightarrow infty}sum_{r=1}^{n-1}frac{cot^2(rpi/n)}{n^2}$$
How can I calculate the value of this trigonometric function where limits tends to infinity? I have thought and tried various of methods like using:
- Tan(x)/x property
- Converting lim to
Integration - Sandwich theorem
- Trigonometric series
But none worked out for me.
So please tell me a good and easy approach for this question :)
sequences-and-series limits trigonometry summation
edited Jan 8 at 17:59
Mark Viola
131k1275171
asked Jan 8 at 16:20
Shubham GawriShubham Gawri
286
$endgroup$
$begingroup$
But where exactly are you stuck had you really tried any of these methods? It's way too easy to simply name a few standard methods.
$endgroup$
– Saad
Jan 8 at 16:23
$begingroup$
I m stuck because I cannot use the summation of (1/r^2) which equals to π^2/6 because this is from my exam and therefore I need another approach
$endgroup$
– Shubham Gawri
Jan 8 at 16:25
1
$begingroup$
Some useful hints here and here... and it looks like the limit is $frac{1}{3}$
$endgroup$
– rtybase
Jan 8 at 18:30
add a comment |
$begingroup$
$$lim_{nrightarrow infty}sum_{r=1}^{n-1}frac{cot^2(rpi/n)}{n^2}$$
How can I calculate the value of this trigonometric function where limits tends to infinity? I have thought and tried various of methods like using:
- Tan(x)/x property
- Converting lim to
Integration - Sandwich theorem
- Trigonometric series
But none worked out for me.
So please tell me a good and easy approach for this question :)
sequences-and-series limits trigonometry summation
edited Jan 8 at 17:59
Mark Viola
131k1275171
asked Jan 8 at 16:20
Shubham GawriShubham Gawri
286
$endgroup$
$$lim_{nrightarrow infty}sum_{r=1}^{n-1}frac{cot^2(rpi/n)}{n^2}$$
How can I calculate the value of this trigonometric function where limits tends to infinity? I have thought and tried various of methods like using:
- Tan(x)/x property
- Converting lim to
Integration - Sandwich theorem
- Trigonometric series
But none worked out for me.
So please tell me a good and easy approach for this question :)
sequences-and-series limits trigonometry summation
sequences-and-series limits trigonometry summation
edited Jan 8 at 17:59
Mark Viola
131k1275171
asked Jan 8 at 16:20
Shubham GawriShubham Gawri
286
edited Jan 8 at 17:59
Mark Viola
131k1275171
asked Jan 8 at 16:20
Shubham GawriShubham Gawri
286
edited Jan 8 at 17:59
Mark Viola
131k1275171
edited Jan 8 at 17:59
Mark Viola
131k1275171
edited Jan 8 at 17:59
Mark Viola
131k1275171
131k1275171
asked Jan 8 at 16:20
Shubham GawriShubham Gawri
286
asked Jan 8 at 16:20
Shubham GawriShubham Gawri
286
asked Jan 8 at 16:20
Shubham GawriShubham Gawri
286
286
$begingroup$
But where exactly are you stuck had you really tried any of these methods? It's way too easy to simply name a few standard methods.
$endgroup$
– Saad
Jan 8 at 16:23
$begingroup$
I m stuck because I cannot use the summation of (1/r^2) which equals to π^2/6 because this is from my exam and therefore I need another approach
$endgroup$
– Shubham Gawri
Jan 8 at 16:25
1
$begingroup$
Some useful hints here and here... and it looks like the limit is $frac{1}{3}$
$endgroup$
– rtybase
Jan 8 at 18:30
add a comment |
$begingroup$
But where exactly are you stuck had you really tried any of these methods? It's way too easy to simply name a few standard methods.
$endgroup$
– Saad
Jan 8 at 16:23
$begingroup$
I m stuck because I cannot use the summation of (1/r^2) which equals to π^2/6 because this is from my exam and therefore I need another approach
$endgroup$
– Shubham Gawri
Jan 8 at 16:25
1
$begingroup$
Some useful hints here and here... and it looks like the limit is $frac{1}{3}$
$endgroup$
– rtybase
Jan 8 at 18:30
$begingroup$
But where exactly are you stuck had you really tried any of these methods? It's way too easy to simply name a few standard methods.
$endgroup$
– Saad
Jan 8 at 16:23
$begingroup$
But where exactly are you stuck had you really tried any of these methods? It's way too easy to simply name a few standard methods.
$endgroup$
– Saad
Jan 8 at 16:23
$begingroup$
I m stuck because I cannot use the summation of (1/r^2) which equals to π^2/6 because this is from my exam and therefore I need another approach
$endgroup$
– Shubham Gawri
Jan 8 at 16:25
$begingroup$
I m stuck because I cannot use the summation of (1/r^2) which equals to π^2/6 because this is from my exam and therefore I need another approach
$endgroup$
– Shubham Gawri
Jan 8 at 16:25
1
1
$begingroup$
Some useful hints here and here... and it looks like the limit is $frac{1}{3}$
$endgroup$
– rtybase
Jan 8 at 18:30
$begingroup$
Some useful hints here and here... and it looks like the limit is $frac{1}{3}$
$endgroup$
– rtybase
Jan 8 at 18:30
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
It would be naïve and incorrect to proceed as follows
$$begin{align}
sum_{k=1}^{n-1}frac{cot^2(pi k/n)}{n^2}&underbrace{approx}_{text{WRONG!}} frac1nint_{1/n}^{1-1/n}cot^2(pi x),dx\\
&=frac1nleft.left(-x-frac1pi cot(pi x)right)right|_{1/n}^{1-1/n}\\
&=frac2{n^2}-frac1n +frac2{npi}cot(pi/n)\\
&to frac2{pi^2}
end{align}$$
Instead, we use $cot^2(x)=csc^2(x)-1$ to write
$$begin{align}
sum_{k=1}^{n-1}frac{cot^2(pi k/n)}{n^2}&=frac1n-frac1{n^2}+sum_{k=1}^{n-1}frac{1}{n^2,sin^2(pi k/n)}\\
&=frac1n-frac1{n^2}+2sum_{k=1}^{lfloor n/2rfloor-1}frac{1}{n^2,sin^2(pi k/n)}
end{align}$$
Next, we note that for $pi/2>x>0$, $left(x-frac16 x^3right)^2le sin^2(x)le x^2$. Hence, we have
$$begin{align}
frac2{pi^2}sum_{k=1}^{lfloor n/2rfloor-1}frac1{k^2} le frac2{n^2}sum_{k=1}^{lfloor n/2rfloor-1}frac{1}{n^2,sin^2(pi k/n)}&le frac2{pi^2}sum_{k=1}^{lfloor n/2rfloor-1}frac1{k^2left(1-frac16frac{pi^2k^2}{n^2}right)^2}\&=frac2{pi^2}sum_{k=1}^{lfloor n/2rfloor-1}frac1{k^2}+Oleft(frac1nright)
end{align}$$
whence letting $nto infty$ and applying the squeeze theorem yields the coveted limit
$$lim_{ntoinfty}sum_{k=1}^{n-1}frac{cot^2(kpi/n)}{n^2}=frac13$$
answered Jan 8 at 19:02
Mark ViolaMark Viola
131k1275171
$endgroup$
$begingroup$
+1. Nice job.${}$
$endgroup$
– Felix Marin
Jan 8 at 20:41
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@FelixMarin Hi Felix! Happy New Year. And thank you. Much appreciated.
$endgroup$
– Mark Viola
Jan 8 at 20:58
$begingroup$
Happy New Year, Mark Viola.
$endgroup$
– Felix Marin
Jan 8 at 21:07
$begingroup$
@MarkViola Then what is the mistake if I write cot(x) in terms of tan(x), use tan(x)/x property with basel problem where sum(1/k^2)= π^2/6 because that eventually results in with a wrong answer :1/6 though the answer is 1/3.What is my mistake here?
$endgroup$
– Shubham Gawri
Jan 9 at 19:26
1
$begingroup$
So, you have $$frac{cot^2(kpi/n)}{n^2}=frac{1}{n^2tan^2(kpi /n)}=frac{1/(kpi/n)^2}{n^2 frac{tan^2(kpi/n)}{(kpi/n)^2}}=frac{1}{pi^2k^2}times frac{1}{frac{tan^2(kpi/n)}{(kpi/n)^2}}$$Now what? While $tan(x)/xto 1$ as $xto 0$, there are many terms in the series such that $kpi/n$ is not close to zero. Keep in mind that $1le kle n-1$.
$endgroup$
– Mark Viola
Jan 9 at 22:22
|
show 2 more comments
$begingroup$
By applying Vieta's formulas to Chebyshev polynomials of the second kind we have
$$ sum_{r=1}^{n-1}cot^2left(frac{pi r}{n}right)=frac{(n-1)(n-2)}{3} $$
(compare Cauchy's proof of $zeta(2)=frac{pi^2}{6}$ in his Cours d'Analyse) hence the wanted limit is clearly $frac{1}{3}$.
answered Jan 8 at 18:45
Jack D'AurizioJack D'Aurizio
1
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It's quite short your answer. We were expecting a little more details.
$endgroup$
– Felix Marin
Jan 10 at 23:41
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@FelixMarin: details can be found in my notes. After a few years on MSE, I am growing tired/lazy about re-writing the exact same things over and over.
$endgroup$
– Jack D'Aurizio
Jan 10 at 23:47
$begingroup$
Fine. I'll see your notes. Happy New Year 2019. Thanks.
$endgroup$
– Felix Marin
Jan 11 at 0:14
add a comment |
$begingroup$
Here I came up with a solution involving only elementary calculation, I hope it's appreciable...
(Sorry as it's in image form)
answered Jan 9 at 7:30
Shubham GawriShubham Gawri
286
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It would be naïve and incorrect to proceed as follows
$$begin{align}
sum_{k=1}^{n-1}frac{cot^2(pi k/n)}{n^2}&underbrace{approx}_{text{WRONG!}} frac1nint_{1/n}^{1-1/n}cot^2(pi x),dx\\
&=frac1nleft.left(-x-frac1pi cot(pi x)right)right|_{1/n}^{1-1/n}\\
&=frac2{n^2}-frac1n +frac2{npi}cot(pi/n)\\
&to frac2{pi^2}
end{align}$$
Instead, we use $cot^2(x)=csc^2(x)-1$ to write
$$begin{align}
sum_{k=1}^{n-1}frac{cot^2(pi k/n)}{n^2}&=frac1n-frac1{n^2}+sum_{k=1}^{n-1}frac{1}{n^2,sin^2(pi k/n)}\\
&=frac1n-frac1{n^2}+2sum_{k=1}^{lfloor n/2rfloor-1}frac{1}{n^2,sin^2(pi k/n)}
end{align}$$
Next, we note that for $pi/2>x>0$, $left(x-frac16 x^3right)^2le sin^2(x)le x^2$. Hence, we have
$$begin{align}
frac2{pi^2}sum_{k=1}^{lfloor n/2rfloor-1}frac1{k^2} le frac2{n^2}sum_{k=1}^{lfloor n/2rfloor-1}frac{1}{n^2,sin^2(pi k/n)}&le frac2{pi^2}sum_{k=1}^{lfloor n/2rfloor-1}frac1{k^2left(1-frac16frac{pi^2k^2}{n^2}right)^2}\&=frac2{pi^2}sum_{k=1}^{lfloor n/2rfloor-1}frac1{k^2}+Oleft(frac1nright)
end{align}$$
whence letting $nto infty$ and applying the squeeze theorem yields the coveted limit
$$lim_{ntoinfty}sum_{k=1}^{n-1}frac{cot^2(kpi/n)}{n^2}=frac13$$
answered Jan 8 at 19:02
Mark ViolaMark Viola
131k1275171
$endgroup$
$begingroup$
+1. Nice job.${}$
$endgroup$
– Felix Marin
Jan 8 at 20:41
$begingroup$
@FelixMarin Hi Felix! Happy New Year. And thank you. Much appreciated.
$endgroup$
– Mark Viola
Jan 8 at 20:58
$begingroup$
Happy New Year, Mark Viola.
$endgroup$
– Felix Marin
Jan 8 at 21:07
$begingroup$
@MarkViola Then what is the mistake if I write cot(x) in terms of tan(x), use tan(x)/x property with basel problem where sum(1/k^2)= π^2/6 because that eventually results in with a wrong answer :1/6 though the answer is 1/3.What is my mistake here?
$endgroup$
– Shubham Gawri
Jan 9 at 19:26
1
$begingroup$
So, you have $$frac{cot^2(kpi/n)}{n^2}=frac{1}{n^2tan^2(kpi /n)}=frac{1/(kpi/n)^2}{n^2 frac{tan^2(kpi/n)}{(kpi/n)^2}}=frac{1}{pi^2k^2}times frac{1}{frac{tan^2(kpi/n)}{(kpi/n)^2}}$$Now what? While $tan(x)/xto 1$ as $xto 0$, there are many terms in the series such that $kpi/n$ is not close to zero. Keep in mind that $1le kle n-1$.
$endgroup$
– Mark Viola
Jan 9 at 22:22
|
show 2 more comments
$begingroup$
It would be naïve and incorrect to proceed as follows
$$begin{align}
sum_{k=1}^{n-1}frac{cot^2(pi k/n)}{n^2}&underbrace{approx}_{text{WRONG!}} frac1nint_{1/n}^{1-1/n}cot^2(pi x),dx\\
&=frac1nleft.left(-x-frac1pi cot(pi x)right)right|_{1/n}^{1-1/n}\\
&=frac2{n^2}-frac1n +frac2{npi}cot(pi/n)\\
&to frac2{pi^2}
end{align}$$
Instead, we use $cot^2(x)=csc^2(x)-1$ to write
$$begin{align}
sum_{k=1}^{n-1}frac{cot^2(pi k/n)}{n^2}&=frac1n-frac1{n^2}+sum_{k=1}^{n-1}frac{1}{n^2,sin^2(pi k/n)}\\
&=frac1n-frac1{n^2}+2sum_{k=1}^{lfloor n/2rfloor-1}frac{1}{n^2,sin^2(pi k/n)}
end{align}$$
Next, we note that for $pi/2>x>0$, $left(x-frac16 x^3right)^2le sin^2(x)le x^2$. Hence, we have
$$begin{align}
frac2{pi^2}sum_{k=1}^{lfloor n/2rfloor-1}frac1{k^2} le frac2{n^2}sum_{k=1}^{lfloor n/2rfloor-1}frac{1}{n^2,sin^2(pi k/n)}&le frac2{pi^2}sum_{k=1}^{lfloor n/2rfloor-1}frac1{k^2left(1-frac16frac{pi^2k^2}{n^2}right)^2}\&=frac2{pi^2}sum_{k=1}^{lfloor n/2rfloor-1}frac1{k^2}+Oleft(frac1nright)
end{align}$$
whence letting $nto infty$ and applying the squeeze theorem yields the coveted limit
$$lim_{ntoinfty}sum_{k=1}^{n-1}frac{cot^2(kpi/n)}{n^2}=frac13$$
answered Jan 8 at 19:02
Mark ViolaMark Viola
131k1275171
$endgroup$
$begingroup$
+1. Nice job.${}$
$endgroup$
– Felix Marin
Jan 8 at 20:41
$begingroup$
@FelixMarin Hi Felix! Happy New Year. And thank you. Much appreciated.
$endgroup$
– Mark Viola
Jan 8 at 20:58
$begingroup$
Happy New Year, Mark Viola.
$endgroup$
– Felix Marin
Jan 8 at 21:07
$begingroup$
@MarkViola Then what is the mistake if I write cot(x) in terms of tan(x), use tan(x)/x property with basel problem where sum(1/k^2)= π^2/6 because that eventually results in with a wrong answer :1/6 though the answer is 1/3.What is my mistake here?
$endgroup$
– Shubham Gawri
Jan 9 at 19:26
1
$begingroup$
So, you have $$frac{cot^2(kpi/n)}{n^2}=frac{1}{n^2tan^2(kpi /n)}=frac{1/(kpi/n)^2}{n^2 frac{tan^2(kpi/n)}{(kpi/n)^2}}=frac{1}{pi^2k^2}times frac{1}{frac{tan^2(kpi/n)}{(kpi/n)^2}}$$Now what? While $tan(x)/xto 1$ as $xto 0$, there are many terms in the series such that $kpi/n$ is not close to zero. Keep in mind that $1le kle n-1$.
$endgroup$
– Mark Viola
Jan 9 at 22:22
|
show 2 more comments
$begingroup$
It would be naïve and incorrect to proceed as follows
$$begin{align}
sum_{k=1}^{n-1}frac{cot^2(pi k/n)}{n^2}&underbrace{approx}_{text{WRONG!}} frac1nint_{1/n}^{1-1/n}cot^2(pi x),dx\\
&=frac1nleft.left(-x-frac1pi cot(pi x)right)right|_{1/n}^{1-1/n}\\
&=frac2{n^2}-frac1n +frac2{npi}cot(pi/n)\\
&to frac2{pi^2}
end{align}$$
Instead, we use $cot^2(x)=csc^2(x)-1$ to write
$$begin{align}
sum_{k=1}^{n-1}frac{cot^2(pi k/n)}{n^2}&=frac1n-frac1{n^2}+sum_{k=1}^{n-1}frac{1}{n^2,sin^2(pi k/n)}\\
&=frac1n-frac1{n^2}+2sum_{k=1}^{lfloor n/2rfloor-1}frac{1}{n^2,sin^2(pi k/n)}
end{align}$$
Next, we note that for $pi/2>x>0$, $left(x-frac16 x^3right)^2le sin^2(x)le x^2$. Hence, we have
$$begin{align}
frac2{pi^2}sum_{k=1}^{lfloor n/2rfloor-1}frac1{k^2} le frac2{n^2}sum_{k=1}^{lfloor n/2rfloor-1}frac{1}{n^2,sin^2(pi k/n)}&le frac2{pi^2}sum_{k=1}^{lfloor n/2rfloor-1}frac1{k^2left(1-frac16frac{pi^2k^2}{n^2}right)^2}\&=frac2{pi^2}sum_{k=1}^{lfloor n/2rfloor-1}frac1{k^2}+Oleft(frac1nright)
end{align}$$
whence letting $nto infty$ and applying the squeeze theorem yields the coveted limit
$$lim_{ntoinfty}sum_{k=1}^{n-1}frac{cot^2(kpi/n)}{n^2}=frac13$$
answered Jan 8 at 19:02
Mark ViolaMark Viola
131k1275171
$endgroup$
It would be naïve and incorrect to proceed as follows
$$begin{align}
sum_{k=1}^{n-1}frac{cot^2(pi k/n)}{n^2}&underbrace{approx}_{text{WRONG!}} frac1nint_{1/n}^{1-1/n}cot^2(pi x),dx\\
&=frac1nleft.left(-x-frac1pi cot(pi x)right)right|_{1/n}^{1-1/n}\\
&=frac2{n^2}-frac1n +frac2{npi}cot(pi/n)\\
&to frac2{pi^2}
end{align}$$
Instead, we use $cot^2(x)=csc^2(x)-1$ to write
$$begin{align}
sum_{k=1}^{n-1}frac{cot^2(pi k/n)}{n^2}&=frac1n-frac1{n^2}+sum_{k=1}^{n-1}frac{1}{n^2,sin^2(pi k/n)}\\
&=frac1n-frac1{n^2}+2sum_{k=1}^{lfloor n/2rfloor-1}frac{1}{n^2,sin^2(pi k/n)}
end{align}$$
Next, we note that for $pi/2>x>0$, $left(x-frac16 x^3right)^2le sin^2(x)le x^2$. Hence, we have
$$begin{align}
frac2{pi^2}sum_{k=1}^{lfloor n/2rfloor-1}frac1{k^2} le frac2{n^2}sum_{k=1}^{lfloor n/2rfloor-1}frac{1}{n^2,sin^2(pi k/n)}&le frac2{pi^2}sum_{k=1}^{lfloor n/2rfloor-1}frac1{k^2left(1-frac16frac{pi^2k^2}{n^2}right)^2}\&=frac2{pi^2}sum_{k=1}^{lfloor n/2rfloor-1}frac1{k^2}+Oleft(frac1nright)
end{align}$$
whence letting $nto infty$ and applying the squeeze theorem yields the coveted limit
$$lim_{ntoinfty}sum_{k=1}^{n-1}frac{cot^2(kpi/n)}{n^2}=frac13$$
answered Jan 8 at 19:02
Mark ViolaMark Viola
131k1275171
edited Jan 8 at 19:08
answered Jan 8 at 19:02
Mark ViolaMark Viola
131k1275171
answered Jan 8 at 19:02
Mark ViolaMark Viola
131k1275171
answered Jan 8 at 19:02
Mark ViolaMark Viola
131k1275171
131k1275171
$begingroup$
+1. Nice job.${}$
$endgroup$
– Felix Marin
Jan 8 at 20:41
$begingroup$
@FelixMarin Hi Felix! Happy New Year. And thank you. Much appreciated.
$endgroup$
– Mark Viola
Jan 8 at 20:58
$begingroup$
Happy New Year, Mark Viola.
$endgroup$
– Felix Marin
Jan 8 at 21:07
$begingroup$
@MarkViola Then what is the mistake if I write cot(x) in terms of tan(x), use tan(x)/x property with basel problem where sum(1/k^2)= π^2/6 because that eventually results in with a wrong answer :1/6 though the answer is 1/3.What is my mistake here?
$endgroup$
– Shubham Gawri
Jan 9 at 19:26
1
$begingroup$
So, you have $$frac{cot^2(kpi/n)}{n^2}=frac{1}{n^2tan^2(kpi /n)}=frac{1/(kpi/n)^2}{n^2 frac{tan^2(kpi/n)}{(kpi/n)^2}}=frac{1}{pi^2k^2}times frac{1}{frac{tan^2(kpi/n)}{(kpi/n)^2}}$$Now what? While $tan(x)/xto 1$ as $xto 0$, there are many terms in the series such that $kpi/n$ is not close to zero. Keep in mind that $1le kle n-1$.
$endgroup$
– Mark Viola
Jan 9 at 22:22
|
show 2 more comments
$begingroup$
+1. Nice job.${}$
$endgroup$
– Felix Marin
Jan 8 at 20:41
$begingroup$
@FelixMarin Hi Felix! Happy New Year. And thank you. Much appreciated.
$endgroup$
– Mark Viola
Jan 8 at 20:58
$begingroup$
Happy New Year, Mark Viola.
$endgroup$
– Felix Marin
Jan 8 at 21:07
$begingroup$
@MarkViola Then what is the mistake if I write cot(x) in terms of tan(x), use tan(x)/x property with basel problem where sum(1/k^2)= π^2/6 because that eventually results in with a wrong answer :1/6 though the answer is 1/3.What is my mistake here?
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– Shubham Gawri
Jan 9 at 19:26
1
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So, you have $$frac{cot^2(kpi/n)}{n^2}=frac{1}{n^2tan^2(kpi /n)}=frac{1/(kpi/n)^2}{n^2 frac{tan^2(kpi/n)}{(kpi/n)^2}}=frac{1}{pi^2k^2}times frac{1}{frac{tan^2(kpi/n)}{(kpi/n)^2}}$$Now what? While $tan(x)/xto 1$ as $xto 0$, there are many terms in the series such that $kpi/n$ is not close to zero. Keep in mind that $1le kle n-1$.
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– Mark Viola
Jan 9 at 22:22
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+1. Nice job.${}$
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– Felix Marin
Jan 8 at 20:41
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+1. Nice job.${}$
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– Felix Marin
Jan 8 at 20:41
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@FelixMarin Hi Felix! Happy New Year. And thank you. Much appreciated.
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– Mark Viola
Jan 8 at 20:58
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@FelixMarin Hi Felix! Happy New Year. And thank you. Much appreciated.
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– Mark Viola
Jan 8 at 20:58
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Happy New Year, Mark Viola.
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– Felix Marin
Jan 8 at 21:07
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Happy New Year, Mark Viola.
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– Felix Marin
Jan 8 at 21:07
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@MarkViola Then what is the mistake if I write cot(x) in terms of tan(x), use tan(x)/x property with basel problem where sum(1/k^2)= π^2/6 because that eventually results in with a wrong answer :1/6 though the answer is 1/3.What is my mistake here?
$endgroup$
– Shubham Gawri
Jan 9 at 19:26
$begingroup$
@MarkViola Then what is the mistake if I write cot(x) in terms of tan(x), use tan(x)/x property with basel problem where sum(1/k^2)= π^2/6 because that eventually results in with a wrong answer :1/6 though the answer is 1/3.What is my mistake here?
$endgroup$
– Shubham Gawri
Jan 9 at 19:26
1
1
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So, you have $$frac{cot^2(kpi/n)}{n^2}=frac{1}{n^2tan^2(kpi /n)}=frac{1/(kpi/n)^2}{n^2 frac{tan^2(kpi/n)}{(kpi/n)^2}}=frac{1}{pi^2k^2}times frac{1}{frac{tan^2(kpi/n)}{(kpi/n)^2}}$$Now what? While $tan(x)/xto 1$ as $xto 0$, there are many terms in the series such that $kpi/n$ is not close to zero. Keep in mind that $1le kle n-1$.
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– Mark Viola
Jan 9 at 22:22
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So, you have $$frac{cot^2(kpi/n)}{n^2}=frac{1}{n^2tan^2(kpi /n)}=frac{1/(kpi/n)^2}{n^2 frac{tan^2(kpi/n)}{(kpi/n)^2}}=frac{1}{pi^2k^2}times frac{1}{frac{tan^2(kpi/n)}{(kpi/n)^2}}$$Now what? While $tan(x)/xto 1$ as $xto 0$, there are many terms in the series such that $kpi/n$ is not close to zero. Keep in mind that $1le kle n-1$.
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– Mark Viola
Jan 9 at 22:22
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show 2 more comments
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By applying Vieta's formulas to Chebyshev polynomials of the second kind we have
$$ sum_{r=1}^{n-1}cot^2left(frac{pi r}{n}right)=frac{(n-1)(n-2)}{3} $$
(compare Cauchy's proof of $zeta(2)=frac{pi^2}{6}$ in his Cours d'Analyse) hence the wanted limit is clearly $frac{1}{3}$.
answered Jan 8 at 18:45
Jack D'AurizioJack D'Aurizio
1
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It's quite short your answer. We were expecting a little more details.
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– Felix Marin
Jan 10 at 23:41
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@FelixMarin: details can be found in my notes. After a few years on MSE, I am growing tired/lazy about re-writing the exact same things over and over.
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– Jack D'Aurizio
Jan 10 at 23:47
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Fine. I'll see your notes. Happy New Year 2019. Thanks.
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– Felix Marin
Jan 11 at 0:14
add a comment |
$begingroup$
By applying Vieta's formulas to Chebyshev polynomials of the second kind we have
$$ sum_{r=1}^{n-1}cot^2left(frac{pi r}{n}right)=frac{(n-1)(n-2)}{3} $$
(compare Cauchy's proof of $zeta(2)=frac{pi^2}{6}$ in his Cours d'Analyse) hence the wanted limit is clearly $frac{1}{3}$.
answered Jan 8 at 18:45
Jack D'AurizioJack D'Aurizio
1
$endgroup$
$begingroup$
It's quite short your answer. We were expecting a little more details.
$endgroup$
– Felix Marin
Jan 10 at 23:41
$begingroup$
@FelixMarin: details can be found in my notes. After a few years on MSE, I am growing tired/lazy about re-writing the exact same things over and over.
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– Jack D'Aurizio
Jan 10 at 23:47
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Fine. I'll see your notes. Happy New Year 2019. Thanks.
$endgroup$
– Felix Marin
Jan 11 at 0:14
add a comment |
$begingroup$
By applying Vieta's formulas to Chebyshev polynomials of the second kind we have
$$ sum_{r=1}^{n-1}cot^2left(frac{pi r}{n}right)=frac{(n-1)(n-2)}{3} $$
(compare Cauchy's proof of $zeta(2)=frac{pi^2}{6}$ in his Cours d'Analyse) hence the wanted limit is clearly $frac{1}{3}$.
answered Jan 8 at 18:45
Jack D'AurizioJack D'Aurizio
1
$endgroup$
By applying Vieta's formulas to Chebyshev polynomials of the second kind we have
$$ sum_{r=1}^{n-1}cot^2left(frac{pi r}{n}right)=frac{(n-1)(n-2)}{3} $$
(compare Cauchy's proof of $zeta(2)=frac{pi^2}{6}$ in his Cours d'Analyse) hence the wanted limit is clearly $frac{1}{3}$.
answered Jan 8 at 18:45
Jack D'AurizioJack D'Aurizio
1
answered Jan 8 at 18:45
Jack D'AurizioJack D'Aurizio
1
answered Jan 8 at 18:45
Jack D'AurizioJack D'Aurizio
1
answered Jan 8 at 18:45
Jack D'AurizioJack D'Aurizio
1
1
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It's quite short your answer. We were expecting a little more details.
$endgroup$
– Felix Marin
Jan 10 at 23:41
$begingroup$
@FelixMarin: details can be found in my notes. After a few years on MSE, I am growing tired/lazy about re-writing the exact same things over and over.
$endgroup$
– Jack D'Aurizio
Jan 10 at 23:47
$begingroup$
Fine. I'll see your notes. Happy New Year 2019. Thanks.
$endgroup$
– Felix Marin
Jan 11 at 0:14
add a comment |
$begingroup$
It's quite short your answer. We were expecting a little more details.
$endgroup$
– Felix Marin
Jan 10 at 23:41
$begingroup$
@FelixMarin: details can be found in my notes. After a few years on MSE, I am growing tired/lazy about re-writing the exact same things over and over.
$endgroup$
– Jack D'Aurizio
Jan 10 at 23:47
$begingroup$
Fine. I'll see your notes. Happy New Year 2019. Thanks.
$endgroup$
– Felix Marin
Jan 11 at 0:14
$begingroup$
It's quite short your answer. We were expecting a little more details.
$endgroup$
– Felix Marin
Jan 10 at 23:41
$begingroup$
It's quite short your answer. We were expecting a little more details.
$endgroup$
– Felix Marin
Jan 10 at 23:41
$begingroup$
@FelixMarin: details can be found in my notes. After a few years on MSE, I am growing tired/lazy about re-writing the exact same things over and over.
$endgroup$
– Jack D'Aurizio
Jan 10 at 23:47
$begingroup$
@FelixMarin: details can be found in my notes. After a few years on MSE, I am growing tired/lazy about re-writing the exact same things over and over.
$endgroup$
– Jack D'Aurizio
Jan 10 at 23:47
$begingroup$
Fine. I'll see your notes. Happy New Year 2019. Thanks.
$endgroup$
– Felix Marin
Jan 11 at 0:14
$begingroup$
Fine. I'll see your notes. Happy New Year 2019. Thanks.
$endgroup$
– Felix Marin
Jan 11 at 0:14
add a comment |
$begingroup$
Here I came up with a solution involving only elementary calculation, I hope it's appreciable...
(Sorry as it's in image form)
answered Jan 9 at 7:30
Shubham GawriShubham Gawri
286
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add a comment |
$begingroup$
Here I came up with a solution involving only elementary calculation, I hope it's appreciable...
(Sorry as it's in image form)
answered Jan 9 at 7:30
Shubham GawriShubham Gawri
286
$endgroup$
add a comment |
$begingroup$
Here I came up with a solution involving only elementary calculation, I hope it's appreciable...
(Sorry as it's in image form)
answered Jan 9 at 7:30
Shubham GawriShubham Gawri
286
$endgroup$
Here I came up with a solution involving only elementary calculation, I hope it's appreciable...
(Sorry as it's in image form)
answered Jan 9 at 7:30
Shubham GawriShubham Gawri
286
answered Jan 9 at 7:30
Shubham GawriShubham Gawri
286
answered Jan 9 at 7:30
Shubham GawriShubham Gawri
286
answered Jan 9 at 7:30
Shubham GawriShubham Gawri
286
286
add a comment |
add a comment |
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But where exactly are you stuck had you really tried any of these methods? It's way too easy to simply name a few standard methods.
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– Saad
Jan 8 at 16:23
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I m stuck because I cannot use the summation of (1/r^2) which equals to π^2/6 because this is from my exam and therefore I need another approach
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– Shubham Gawri
Jan 8 at 16:25
1
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Some useful hints here and here... and it looks like the limit is $frac{1}{3}$
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– rtybase
Jan 8 at 18:30