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Limit of a sum using complex analysis.
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I'm trying to find the limit of this sum: $$S_n =frac1nleft(frac12+sum_k=1^ncos(kx)right)$$
I tried to find a formula for the inner sum first and I ended up getting zero as an answer.
The sum is supposed to converge to $cot(xover 2)$ and that appears in my last expression but it goes to zero.
Here is what I got with a rather long and clumsy reasoning:
$$S_n = frac12nleft(cotleft(fracx2right)sin(nx)+cos(nx)right)$$
Thanks in advance.
complex-analysis limits complex-numbers trigonometric-series
asked 1 hour ago
FuzzyPixelz
191111
add a comment |Â
up vote
2
down vote
favorite
I'm trying to find the limit of this sum: $$S_n =frac1nleft(frac12+sum_k=1^ncos(kx)right)$$
I tried to find a formula for the inner sum first and I ended up getting zero as an answer.
The sum is supposed to converge to $cot(xover 2)$ and that appears in my last expression but it goes to zero.
Here is what I got with a rather long and clumsy reasoning:
$$S_n = frac12nleft(cotleft(fracx2right)sin(nx)+cos(nx)right)$$
Thanks in advance.
complex-analysis limits complex-numbers trigonometric-series
asked 1 hour ago
FuzzyPixelz
191111
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I'm trying to find the limit of this sum: $$S_n =frac1nleft(frac12+sum_k=1^ncos(kx)right)$$
I tried to find a formula for the inner sum first and I ended up getting zero as an answer.
The sum is supposed to converge to $cot(xover 2)$ and that appears in my last expression but it goes to zero.
Here is what I got with a rather long and clumsy reasoning:
$$S_n = frac12nleft(cotleft(fracx2right)sin(nx)+cos(nx)right)$$
Thanks in advance.
complex-analysis limits complex-numbers trigonometric-series
asked 1 hour ago
FuzzyPixelz
191111
I'm trying to find the limit of this sum: $$S_n =frac1nleft(frac12+sum_k=1^ncos(kx)right)$$
I tried to find a formula for the inner sum first and I ended up getting zero as an answer.
The sum is supposed to converge to $cot(xover 2)$ and that appears in my last expression but it goes to zero.
Here is what I got with a rather long and clumsy reasoning:
$$S_n = frac12nleft(cotleft(fracx2right)sin(nx)+cos(nx)right)$$
Thanks in advance.
complex-analysis limits complex-numbers trigonometric-series
complex-analysis limits complex-numbers trigonometric-series
asked 1 hour ago
FuzzyPixelz
191111
asked 1 hour ago
FuzzyPixelz
191111
asked 1 hour ago
FuzzyPixelz
191111
asked 1 hour ago
FuzzyPixelz
191111
asked 1 hour ago
FuzzyPixelz
191111
191111
add a comment |Â
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
3
down vote
accepted
You have, for all $x neq 2kpi$,
$$sum_k=1^n cos(kx) = mathrmReleft( sum_k=1^n e^ikxright) = mathrmReleft( e^ix frac1-e^inx1-e^ix right)$$
Moreover you have
$$left|e^ix frac1-e^inx1-e^ixright| leq frac2$$
So $$left| sum_k=1^n cos(kx) right| leq frac2$$
And therefore you get $S_n rightarrow 0$.
If $x = 2kpi$ for $k in mathbbZ$, you easily have $S_n rightarrow 1$.
answered 56 mins ago
TheSilverDoe
1,10511
Remember that $|e^iz|=1$ for all $z in mathbbR$. So by the triangle inequality, $|1-e^inx| leq 2$.
– TheSilverDoe
50 mins ago
Thank you very much, this was very helpful.
– FuzzyPixelz
48 mins ago
add a comment |Â
up vote
2
down vote
Just write $cos (kx)$ as the real part of $e^ikx$.
answered 1 hour ago
AlexL
992
New contributor
AlexL is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I already did that but it didn't work.
– FuzzyPixelz
1 hour ago
1
Then the limit is 0. Why do you think the answer is a cotan ?
– AlexL
57 mins ago
I don't know either way, but it can't be zero because the question asks for an answer in terms of $x$
– FuzzyPixelz
56 mins ago
1
You must have forgotten something.
– AlexL
54 mins ago
TheSilverDoe's proof convinced me and it has two different answers depending on the value of $x$ which solves my troubles, so yes not $cot$, my bad.
– FuzzyPixelz
45 mins ago
add a comment |Â
up vote
0
down vote
Another view of the question.
Instead of taking the real part of $e^ikx$ as given by the answers from TheSilverDoe and AlexL, the complex exponential can be more symmetric:
$$beginalign*
cos kx &= frace^-ikx+e^ikx2\
frac12+sum_k=1^ncos kx &= frac12sum_k=-n^ne^ikx\
&= frac12 cdot e^-inxfrace^i(2n+1)x-1e^ix-1\
&= frac12 cdot frace^i(n+1)x-e^-inxe^ix-1\
&= frac12 cdot frace^ileft(n+frac12right)x-e^-ileft(n+frac12right)xe^ifrac12x-e^-ifrac12x\
&= frac12 cdot fracsinleft(n+frac12right)xsinfrac12x\
S_n&= fracsinleft(n+frac12right)x2nsinfrac12x\
endalign*$$
When $e^ix ne 1$, i.e. $xne 2pi m$ for integer $m$, $S_n to 0$.
For other $x$'s, i.e. when $x=2pi m$,
$$beginalign*
cos kx = cos 2pi km &= 1\
sum_k=1^ncos kx &= n\
frac12 + sum_k=1^ncos kx &= frac12+ n\
S_n &= frac1nleft(frac12 + nright)\
&= frac12n + 1\
&to 1
endalign*$$
answered 18 mins ago
peterwhy
11.4k21127
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
You have, for all $x neq 2kpi$,
$$sum_k=1^n cos(kx) = mathrmReleft( sum_k=1^n e^ikxright) = mathrmReleft( e^ix frac1-e^inx1-e^ix right)$$
Moreover you have
$$left|e^ix frac1-e^inx1-e^ixright| leq frac2$$
So $$left| sum_k=1^n cos(kx) right| leq frac2$$
And therefore you get $S_n rightarrow 0$.
If $x = 2kpi$ for $k in mathbbZ$, you easily have $S_n rightarrow 1$.
answered 56 mins ago
TheSilverDoe
1,10511
Remember that $|e^iz|=1$ for all $z in mathbbR$. So by the triangle inequality, $|1-e^inx| leq 2$.
– TheSilverDoe
50 mins ago
Thank you very much, this was very helpful.
– FuzzyPixelz
48 mins ago
add a comment |Â
up vote
3
down vote
accepted
You have, for all $x neq 2kpi$,
$$sum_k=1^n cos(kx) = mathrmReleft( sum_k=1^n e^ikxright) = mathrmReleft( e^ix frac1-e^inx1-e^ix right)$$
Moreover you have
$$left|e^ix frac1-e^inx1-e^ixright| leq frac2$$
So $$left| sum_k=1^n cos(kx) right| leq frac2$$
And therefore you get $S_n rightarrow 0$.
If $x = 2kpi$ for $k in mathbbZ$, you easily have $S_n rightarrow 1$.
answered 56 mins ago
TheSilverDoe
1,10511
Remember that $|e^iz|=1$ for all $z in mathbbR$. So by the triangle inequality, $|1-e^inx| leq 2$.
– TheSilverDoe
50 mins ago
Thank you very much, this was very helpful.
– FuzzyPixelz
48 mins ago
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
You have, for all $x neq 2kpi$,
$$sum_k=1^n cos(kx) = mathrmReleft( sum_k=1^n e^ikxright) = mathrmReleft( e^ix frac1-e^inx1-e^ix right)$$
Moreover you have
$$left|e^ix frac1-e^inx1-e^ixright| leq frac2$$
So $$left| sum_k=1^n cos(kx) right| leq frac2$$
And therefore you get $S_n rightarrow 0$.
If $x = 2kpi$ for $k in mathbbZ$, you easily have $S_n rightarrow 1$.
answered 56 mins ago
TheSilverDoe
1,10511
You have, for all $x neq 2kpi$,
$$sum_k=1^n cos(kx) = mathrmReleft( sum_k=1^n e^ikxright) = mathrmReleft( e^ix frac1-e^inx1-e^ix right)$$
Moreover you have
$$left|e^ix frac1-e^inx1-e^ixright| leq frac2$$
So $$left| sum_k=1^n cos(kx) right| leq frac2$$
And therefore you get $S_n rightarrow 0$.
If $x = 2kpi$ for $k in mathbbZ$, you easily have $S_n rightarrow 1$.
answered 56 mins ago
TheSilverDoe
1,10511
answered 56 mins ago
TheSilverDoe
1,10511
answered 56 mins ago
TheSilverDoe
1,10511
answered 56 mins ago
TheSilverDoe
1,10511
1,10511
Remember that $|e^iz|=1$ for all $z in mathbbR$. So by the triangle inequality, $|1-e^inx| leq 2$.
– TheSilverDoe
50 mins ago
Thank you very much, this was very helpful.
– FuzzyPixelz
48 mins ago
add a comment |Â
Remember that $|e^iz|=1$ for all $z in mathbbR$. So by the triangle inequality, $|1-e^inx| leq 2$.
– TheSilverDoe
50 mins ago
Thank you very much, this was very helpful.
– FuzzyPixelz
48 mins ago
Remember that $|e^iz|=1$ for all $z in mathbbR$. So by the triangle inequality, $|1-e^inx| leq 2$.
– TheSilverDoe
50 mins ago
Remember that $|e^iz|=1$ for all $z in mathbbR$. So by the triangle inequality, $|1-e^inx| leq 2$.
– TheSilverDoe
50 mins ago
Thank you very much, this was very helpful.
– FuzzyPixelz
48 mins ago
Thank you very much, this was very helpful.
– FuzzyPixelz
48 mins ago
add a comment |Â
up vote
2
down vote
Just write $cos (kx)$ as the real part of $e^ikx$.
answered 1 hour ago
AlexL
992
New contributor
AlexL is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I already did that but it didn't work.
– FuzzyPixelz
1 hour ago
1
Then the limit is 0. Why do you think the answer is a cotan ?
– AlexL
57 mins ago
I don't know either way, but it can't be zero because the question asks for an answer in terms of $x$
– FuzzyPixelz
56 mins ago
1
You must have forgotten something.
– AlexL
54 mins ago
TheSilverDoe's proof convinced me and it has two different answers depending on the value of $x$ which solves my troubles, so yes not $cot$, my bad.
– FuzzyPixelz
45 mins ago
add a comment |Â
up vote
2
down vote
Just write $cos (kx)$ as the real part of $e^ikx$.
answered 1 hour ago
AlexL
992
New contributor
AlexL is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I already did that but it didn't work.
– FuzzyPixelz
1 hour ago
1
Then the limit is 0. Why do you think the answer is a cotan ?
– AlexL
57 mins ago
I don't know either way, but it can't be zero because the question asks for an answer in terms of $x$
– FuzzyPixelz
56 mins ago
1
You must have forgotten something.
– AlexL
54 mins ago
TheSilverDoe's proof convinced me and it has two different answers depending on the value of $x$ which solves my troubles, so yes not $cot$, my bad.
– FuzzyPixelz
45 mins ago
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Just write $cos (kx)$ as the real part of $e^ikx$.
answered 1 hour ago
AlexL
992
New contributor
AlexL is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Just write $cos (kx)$ as the real part of $e^ikx$.
answered 1 hour ago
AlexL
992
New contributor
AlexL is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 1 hour ago
AlexL
992
New contributor
AlexL is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 1 hour ago
AlexL
992
answered 1 hour ago
AlexL
992
992
New contributor
AlexL is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
AlexL is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
AlexL is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I already did that but it didn't work.
– FuzzyPixelz
1 hour ago
1
Then the limit is 0. Why do you think the answer is a cotan ?
– AlexL
57 mins ago
I don't know either way, but it can't be zero because the question asks for an answer in terms of $x$
– FuzzyPixelz
56 mins ago
1
You must have forgotten something.
– AlexL
54 mins ago
TheSilverDoe's proof convinced me and it has two different answers depending on the value of $x$ which solves my troubles, so yes not $cot$, my bad.
– FuzzyPixelz
45 mins ago
add a comment |Â
I already did that but it didn't work.
– FuzzyPixelz
1 hour ago
1
Then the limit is 0. Why do you think the answer is a cotan ?
– AlexL
57 mins ago
I don't know either way, but it can't be zero because the question asks for an answer in terms of $x$
– FuzzyPixelz
56 mins ago
1
You must have forgotten something.
– AlexL
54 mins ago
TheSilverDoe's proof convinced me and it has two different answers depending on the value of $x$ which solves my troubles, so yes not $cot$, my bad.
– FuzzyPixelz
45 mins ago
I already did that but it didn't work.
– FuzzyPixelz
1 hour ago
I already did that but it didn't work.
– FuzzyPixelz
1 hour ago
1
1
Then the limit is 0. Why do you think the answer is a cotan ?
– AlexL
57 mins ago
Then the limit is 0. Why do you think the answer is a cotan ?
– AlexL
57 mins ago
I don't know either way, but it can't be zero because the question asks for an answer in terms of $x$
– FuzzyPixelz
56 mins ago
I don't know either way, but it can't be zero because the question asks for an answer in terms of $x$
– FuzzyPixelz
56 mins ago
1
1
You must have forgotten something.
– AlexL
54 mins ago
You must have forgotten something.
– AlexL
54 mins ago
TheSilverDoe's proof convinced me and it has two different answers depending on the value of $x$ which solves my troubles, so yes not $cot$, my bad.
– FuzzyPixelz
45 mins ago
TheSilverDoe's proof convinced me and it has two different answers depending on the value of $x$ which solves my troubles, so yes not $cot$, my bad.
– FuzzyPixelz
45 mins ago
add a comment |Â
up vote
0
down vote
Another view of the question.
Instead of taking the real part of $e^ikx$ as given by the answers from TheSilverDoe and AlexL, the complex exponential can be more symmetric:
$$beginalign*
cos kx &= frace^-ikx+e^ikx2\
frac12+sum_k=1^ncos kx &= frac12sum_k=-n^ne^ikx\
&= frac12 cdot e^-inxfrace^i(2n+1)x-1e^ix-1\
&= frac12 cdot frace^i(n+1)x-e^-inxe^ix-1\
&= frac12 cdot frace^ileft(n+frac12right)x-e^-ileft(n+frac12right)xe^ifrac12x-e^-ifrac12x\
&= frac12 cdot fracsinleft(n+frac12right)xsinfrac12x\
S_n&= fracsinleft(n+frac12right)x2nsinfrac12x\
endalign*$$
When $e^ix ne 1$, i.e. $xne 2pi m$ for integer $m$, $S_n to 0$.
For other $x$'s, i.e. when $x=2pi m$,
$$beginalign*
cos kx = cos 2pi km &= 1\
sum_k=1^ncos kx &= n\
frac12 + sum_k=1^ncos kx &= frac12+ n\
S_n &= frac1nleft(frac12 + nright)\
&= frac12n + 1\
&to 1
endalign*$$
answered 18 mins ago
peterwhy
11.4k21127
add a comment |Â
up vote
0
down vote
Another view of the question.
Instead of taking the real part of $e^ikx$ as given by the answers from TheSilverDoe and AlexL, the complex exponential can be more symmetric:
$$beginalign*
cos kx &= frace^-ikx+e^ikx2\
frac12+sum_k=1^ncos kx &= frac12sum_k=-n^ne^ikx\
&= frac12 cdot e^-inxfrace^i(2n+1)x-1e^ix-1\
&= frac12 cdot frace^i(n+1)x-e^-inxe^ix-1\
&= frac12 cdot frace^ileft(n+frac12right)x-e^-ileft(n+frac12right)xe^ifrac12x-e^-ifrac12x\
&= frac12 cdot fracsinleft(n+frac12right)xsinfrac12x\
S_n&= fracsinleft(n+frac12right)x2nsinfrac12x\
endalign*$$
When $e^ix ne 1$, i.e. $xne 2pi m$ for integer $m$, $S_n to 0$.
For other $x$'s, i.e. when $x=2pi m$,
$$beginalign*
cos kx = cos 2pi km &= 1\
sum_k=1^ncos kx &= n\
frac12 + sum_k=1^ncos kx &= frac12+ n\
S_n &= frac1nleft(frac12 + nright)\
&= frac12n + 1\
&to 1
endalign*$$
answered 18 mins ago
peterwhy
11.4k21127
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Another view of the question.
Instead of taking the real part of $e^ikx$ as given by the answers from TheSilverDoe and AlexL, the complex exponential can be more symmetric:
$$beginalign*
cos kx &= frace^-ikx+e^ikx2\
frac12+sum_k=1^ncos kx &= frac12sum_k=-n^ne^ikx\
&= frac12 cdot e^-inxfrace^i(2n+1)x-1e^ix-1\
&= frac12 cdot frace^i(n+1)x-e^-inxe^ix-1\
&= frac12 cdot frace^ileft(n+frac12right)x-e^-ileft(n+frac12right)xe^ifrac12x-e^-ifrac12x\
&= frac12 cdot fracsinleft(n+frac12right)xsinfrac12x\
S_n&= fracsinleft(n+frac12right)x2nsinfrac12x\
endalign*$$
When $e^ix ne 1$, i.e. $xne 2pi m$ for integer $m$, $S_n to 0$.
For other $x$'s, i.e. when $x=2pi m$,
$$beginalign*
cos kx = cos 2pi km &= 1\
sum_k=1^ncos kx &= n\
frac12 + sum_k=1^ncos kx &= frac12+ n\
S_n &= frac1nleft(frac12 + nright)\
&= frac12n + 1\
&to 1
endalign*$$
answered 18 mins ago
peterwhy
11.4k21127
Another view of the question.
Instead of taking the real part of $e^ikx$ as given by the answers from TheSilverDoe and AlexL, the complex exponential can be more symmetric:
$$beginalign*
cos kx &= frace^-ikx+e^ikx2\
frac12+sum_k=1^ncos kx &= frac12sum_k=-n^ne^ikx\
&= frac12 cdot e^-inxfrace^i(2n+1)x-1e^ix-1\
&= frac12 cdot frace^i(n+1)x-e^-inxe^ix-1\
&= frac12 cdot frace^ileft(n+frac12right)x-e^-ileft(n+frac12right)xe^ifrac12x-e^-ifrac12x\
&= frac12 cdot fracsinleft(n+frac12right)xsinfrac12x\
S_n&= fracsinleft(n+frac12right)x2nsinfrac12x\
endalign*$$
When $e^ix ne 1$, i.e. $xne 2pi m$ for integer $m$, $S_n to 0$.
For other $x$'s, i.e. when $x=2pi m$,
$$beginalign*
cos kx = cos 2pi km &= 1\
sum_k=1^ncos kx &= n\
frac12 + sum_k=1^ncos kx &= frac12+ n\
S_n &= frac1nleft(frac12 + nright)\
&= frac12n + 1\
&to 1
endalign*$$
answered 18 mins ago
peterwhy
11.4k21127
edited 11 mins ago
answered 18 mins ago
peterwhy
11.4k21127
answered 18 mins ago
peterwhy
11.4k21127
answered 18 mins ago
peterwhy
11.4k21127
11.4k21127
add a comment |Â
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