Mixing
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The sum of the series $ cos(x)-cos(2x)+cos(3x)-…$
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In the book "The spirit of mathematical analysis" of Martin Ohm, the author gives an example of differentiating an infinite series and obtaining an absurd result (page 2)
From the series
$frac{x}{2}=sin(x)-frac{1}{2}sin2x+frac{1}{3}sin(3x)-...$ (1)
If one differentiate terms by terms, one obtains this series
$frac{1}{2}=cos(x)-cos(2x)+cos(3x)-...$ (2)
The author said the last series is divergent, therefore this result is nonsensical.
My question is how does one obtain the first series? What transformation do you perform to obtain $frac{x}{2}$ on the left hand side.
My second question how can we prove that the second series is divergent?
calculus
asked Feb 2 at 20:17
James WarthingtonJames Warthington
48929
$endgroup$
|
show 9 more comments
$begingroup$
In the book "The spirit of mathematical analysis" of Martin Ohm, the author gives an example of differentiating an infinite series and obtaining an absurd result (page 2)
From the series
$frac{x}{2}=sin(x)-frac{1}{2}sin2x+frac{1}{3}sin(3x)-...$ (1)
If one differentiate terms by terms, one obtains this series
$frac{1}{2}=cos(x)-cos(2x)+cos(3x)-...$ (2)
The author said the last series is divergent, therefore this result is nonsensical.
My question is how does one obtain the first series? What transformation do you perform to obtain $frac{x}{2}$ on the left hand side.
My second question how can we prove that the second series is divergent?
calculus
asked Feb 2 at 20:17
James WarthingtonJames Warthington
48929
$endgroup$
1
$begingroup$
Do you know hay Fourier series are.?
$endgroup$
– Julián Aguirre
Feb 2 at 20:22
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Your second question is easy enough: $cos(nx)$ does not tend to $0$ as $n to infty$, which means the series fails the divergence test.
$endgroup$
– Theo Bendit
Feb 2 at 20:24
$begingroup$
I haven't learnt Fourier series yet. But I am eager to learn, sir.
$endgroup$
– James Warthington
Feb 2 at 20:24
$begingroup$
@Theo Bendit Which test should you perform?
$endgroup$
– James Warthington
Feb 2 at 20:25
$begingroup$
@James Warthington, it is just a basic theorem about series. If $sum a_n$ converges then $a_nto 0$.
$endgroup$
– Mark
Feb 2 at 20:26
|
show 9 more comments
$begingroup$
In the book "The spirit of mathematical analysis" of Martin Ohm, the author gives an example of differentiating an infinite series and obtaining an absurd result (page 2)
From the series
$frac{x}{2}=sin(x)-frac{1}{2}sin2x+frac{1}{3}sin(3x)-...$ (1)
If one differentiate terms by terms, one obtains this series
$frac{1}{2}=cos(x)-cos(2x)+cos(3x)-...$ (2)
The author said the last series is divergent, therefore this result is nonsensical.
My question is how does one obtain the first series? What transformation do you perform to obtain $frac{x}{2}$ on the left hand side.
My second question how can we prove that the second series is divergent?
calculus
asked Feb 2 at 20:17
James WarthingtonJames Warthington
48929
$endgroup$
In the book "The spirit of mathematical analysis" of Martin Ohm, the author gives an example of differentiating an infinite series and obtaining an absurd result (page 2)
From the series
$frac{x}{2}=sin(x)-frac{1}{2}sin2x+frac{1}{3}sin(3x)-...$ (1)
If one differentiate terms by terms, one obtains this series
$frac{1}{2}=cos(x)-cos(2x)+cos(3x)-...$ (2)
The author said the last series is divergent, therefore this result is nonsensical.
My question is how does one obtain the first series? What transformation do you perform to obtain $frac{x}{2}$ on the left hand side.
My second question how can we prove that the second series is divergent?
calculus
calculus
asked Feb 2 at 20:17
James WarthingtonJames Warthington
48929
asked Feb 2 at 20:17
James WarthingtonJames Warthington
48929
edited Feb 3 at 1:39
James Warthington
asked Feb 2 at 20:17
James WarthingtonJames Warthington
48929
asked Feb 2 at 20:17
James WarthingtonJames Warthington
48929
asked Feb 2 at 20:17
James WarthingtonJames Warthington
48929
48929
1
$begingroup$
Do you know hay Fourier series are.?
$endgroup$
– Julián Aguirre
Feb 2 at 20:22
$begingroup$
Your second question is easy enough: $cos(nx)$ does not tend to $0$ as $n to infty$, which means the series fails the divergence test.
$endgroup$
– Theo Bendit
Feb 2 at 20:24
$begingroup$
I haven't learnt Fourier series yet. But I am eager to learn, sir.
$endgroup$
– James Warthington
Feb 2 at 20:24
$begingroup$
@Theo Bendit Which test should you perform?
$endgroup$
– James Warthington
Feb 2 at 20:25
$begingroup$
@James Warthington, it is just a basic theorem about series. If $sum a_n$ converges then $a_nto 0$.
$endgroup$
– Mark
Feb 2 at 20:26
|
show 9 more comments
1
$begingroup$
Do you know hay Fourier series are.?
$endgroup$
– Julián Aguirre
Feb 2 at 20:22
$begingroup$
Your second question is easy enough: $cos(nx)$ does not tend to $0$ as $n to infty$, which means the series fails the divergence test.
$endgroup$
– Theo Bendit
Feb 2 at 20:24
$begingroup$
I haven't learnt Fourier series yet. But I am eager to learn, sir.
$endgroup$
– James Warthington
Feb 2 at 20:24
$begingroup$
@Theo Bendit Which test should you perform?
$endgroup$
– James Warthington
Feb 2 at 20:25
$begingroup$
@James Warthington, it is just a basic theorem about series. If $sum a_n$ converges then $a_nto 0$.
$endgroup$
– Mark
Feb 2 at 20:26
1
1
$begingroup$
Do you know hay Fourier series are.?
$endgroup$
– Julián Aguirre
Feb 2 at 20:22
$begingroup$
Do you know hay Fourier series are.?
$endgroup$
– Julián Aguirre
Feb 2 at 20:22
$begingroup$
Your second question is easy enough: $cos(nx)$ does not tend to $0$ as $n to infty$, which means the series fails the divergence test.
$endgroup$
– Theo Bendit
Feb 2 at 20:24
$begingroup$
Your second question is easy enough: $cos(nx)$ does not tend to $0$ as $n to infty$, which means the series fails the divergence test.
$endgroup$
– Theo Bendit
Feb 2 at 20:24
$begingroup$
I haven't learnt Fourier series yet. But I am eager to learn, sir.
$endgroup$
– James Warthington
Feb 2 at 20:24
$begingroup$
I haven't learnt Fourier series yet. But I am eager to learn, sir.
$endgroup$
– James Warthington
Feb 2 at 20:24
$begingroup$
@Theo Bendit Which test should you perform?
$endgroup$
– James Warthington
Feb 2 at 20:25
$begingroup$
@Theo Bendit Which test should you perform?
$endgroup$
– James Warthington
Feb 2 at 20:25
$begingroup$
@James Warthington, it is just a basic theorem about series. If $sum a_n$ converges then $a_nto 0$.
$endgroup$
– Mark
Feb 2 at 20:26
$begingroup$
@James Warthington, it is just a basic theorem about series. If $sum a_n$ converges then $a_nto 0$.
$endgroup$
– Mark
Feb 2 at 20:26
|
show 9 more comments
4 Answers
4
active
oldest
votes
$begingroup$
It might to better to first study the well known convergent series
$$ln cos left(frac{x}{2}right)=-ln 2 + sum_{k=1}^infty frac{ (-1)^{k-1} cos(kx)}{k}tag{1}$$
You can then naively differentiate this to give the divergent series
$$frac{1}{2}tanleft(frac{x}{2}right)=sum_{k=1}^infty (-1)^{k-1}sin(kx)$$
You can examine the nature of the divergences through looking at the partial sums. They appear to be correlated plus and minus going divergences. If you want to claim that this function can be integrated with rigour (which I've read somewhere is possible [possibly in a pamphlet by Hardy on Fourier Series] ) I think you must prove that these "correlated" plus and minus going divergences somehow cancel each other out in the integration process.
At $x=pi/2$ you have the series.
$$frac{1}{2}=1-1+1-1+...$$
A proof for series (1) is here and you should be able to prove your series in a similar fashion by changing to the exponential form.
answered Feb 2 at 21:35
James ArathoonJames Arathoon
1,608423
$endgroup$
$begingroup$
I have found these series in chapter 3 of Fourier's Analytical Theory of Heat:books.google.com/…
$endgroup$
– James Warthington
Feb 2 at 21:46
add a comment |
$begingroup$
The first series is nothing more than the Fourier series of the $2pi$-periodic odd function
$$
y=frac{x}{2},quadpi<x<pi.
$$
Indeed the coefficients of the expansion $frac{x}{2}=sumlimits_{n=1}^infty b_nsin nx$ are:
$$
b_n=frac{1}{pi}int_{-pi}^pifrac{x}{2}sin nx;dx=frac{1}{pi}left[-frac{x}{2n}cos nxright]_{-pi}^pi=-frac{1}{npi}frac{pi(-1)^n+pi(-1)^n}{2}=frac{(-1)^{n+1}}{n}.
$$
The answer to the second question is most easily obtained through exponential representation of cosine and well-known expression for the partial sums of geometrical series:
$$begin {array}{}
S_N(x)&=sum_{n=1}^N (-1)^{n+1}cos nx\
&=sum_{n=1}^N(-1)^{n+1}frac {e^{inx}+e^{-inx}}{2}\
&=frac12left [frac {e^{ix}-(-1)^{N+2}e^{(N+1)ix}}{1+e^{ix}}+
frac {e^{-ix}-(-1)^{N+2}e^{-(N+1)ix}}{1+e^{-ix}}right]\
&=frac12left [frac {e^{ix/2}-(-1)^{N+2}e^{(N+1/2)ix}}{e^{-ix/2}+e^{ix/2}}+
frac {e^{-ix/2}-(-1)^{N+2}e^{-(N+1/2)ix}}{e^{ix/2}+e^{-ix/2}}right]\
&=frac12left [1-(-1)^{N}frac{cos(N+1/2)x}{cos (x/2)}right].
end {array}
$$
From this one can conclude that $S_N (x) $ does not converge for any value of $x $.
answered Feb 2 at 22:43
useruser
6,56011031
$endgroup$
add a comment |
$begingroup$
For the first series, consider that you look for the imaginary part of
$$sum_{n=1}^infty (-1)^{n-1} frac {e^{i n x}} n=sum_{n=1}^infty (-1)^{n-1} frac {left( e^{i x}right)^n} n=log left(1+e^{i x}right)$$So
$$sum_{n=1}^infty (-1)^{n-1} frac {sin(n x)} n=frac{1}{2} i left(log left(1+e^{-i x}right)-log left(1+e^{i x}right)right)=frac{1}{2} left(arg left(1+e^{i x}right)- arg left(1+e^{-i x}right)right)=frac x 2$$ provided $-pi leq x leq pi$ .
answered Feb 3 at 5:26
Claude LeiboviciClaude Leibovici
126k1158135
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Do you know anyone to obtain this result without recoursing to complex analysis? I haven't studied complex analysis yet so I don't understand your answer.
$endgroup$
– James Warthington
Feb 27 at 1:30
add a comment |
$begingroup$
The first equation's right-hand side is $Imln(1+exp ix)$. The second equation's right-hand side can only have convergent partial sums if $lim_{ntoinfty}cos nx=0$, but this clearly fails for $x/pi$ rational. A famous but less obvious result is that other real $x$ also do not obtain such a limit.
answered Feb 2 at 20:44
J.G.J.G.
33.5k23252
$endgroup$
$begingroup$
I have never seen $ln(1+expix)$ before, what is it?
$endgroup$
– James Warthington
Feb 2 at 20:46
$begingroup$
@JamesWharthington You may want to learn about complex exponentials and logarithms.
$endgroup$
– J.G.
Feb 2 at 20:48
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It might to better to first study the well known convergent series
$$ln cos left(frac{x}{2}right)=-ln 2 + sum_{k=1}^infty frac{ (-1)^{k-1} cos(kx)}{k}tag{1}$$
You can then naively differentiate this to give the divergent series
$$frac{1}{2}tanleft(frac{x}{2}right)=sum_{k=1}^infty (-1)^{k-1}sin(kx)$$
You can examine the nature of the divergences through looking at the partial sums. They appear to be correlated plus and minus going divergences. If you want to claim that this function can be integrated with rigour (which I've read somewhere is possible [possibly in a pamphlet by Hardy on Fourier Series] ) I think you must prove that these "correlated" plus and minus going divergences somehow cancel each other out in the integration process.
At $x=pi/2$ you have the series.
$$frac{1}{2}=1-1+1-1+...$$
A proof for series (1) is here and you should be able to prove your series in a similar fashion by changing to the exponential form.
answered Feb 2 at 21:35
James ArathoonJames Arathoon
1,608423
$endgroup$
$begingroup$
I have found these series in chapter 3 of Fourier's Analytical Theory of Heat:books.google.com/…
$endgroup$
– James Warthington
Feb 2 at 21:46
add a comment |
$begingroup$
It might to better to first study the well known convergent series
$$ln cos left(frac{x}{2}right)=-ln 2 + sum_{k=1}^infty frac{ (-1)^{k-1} cos(kx)}{k}tag{1}$$
You can then naively differentiate this to give the divergent series
$$frac{1}{2}tanleft(frac{x}{2}right)=sum_{k=1}^infty (-1)^{k-1}sin(kx)$$
You can examine the nature of the divergences through looking at the partial sums. They appear to be correlated plus and minus going divergences. If you want to claim that this function can be integrated with rigour (which I've read somewhere is possible [possibly in a pamphlet by Hardy on Fourier Series] ) I think you must prove that these "correlated" plus and minus going divergences somehow cancel each other out in the integration process.
At $x=pi/2$ you have the series.
$$frac{1}{2}=1-1+1-1+...$$
A proof for series (1) is here and you should be able to prove your series in a similar fashion by changing to the exponential form.
answered Feb 2 at 21:35
James ArathoonJames Arathoon
1,608423
$endgroup$
$begingroup$
I have found these series in chapter 3 of Fourier's Analytical Theory of Heat:books.google.com/…
$endgroup$
– James Warthington
Feb 2 at 21:46
add a comment |
$begingroup$
It might to better to first study the well known convergent series
$$ln cos left(frac{x}{2}right)=-ln 2 + sum_{k=1}^infty frac{ (-1)^{k-1} cos(kx)}{k}tag{1}$$
You can then naively differentiate this to give the divergent series
$$frac{1}{2}tanleft(frac{x}{2}right)=sum_{k=1}^infty (-1)^{k-1}sin(kx)$$
You can examine the nature of the divergences through looking at the partial sums. They appear to be correlated plus and minus going divergences. If you want to claim that this function can be integrated with rigour (which I've read somewhere is possible [possibly in a pamphlet by Hardy on Fourier Series] ) I think you must prove that these "correlated" plus and minus going divergences somehow cancel each other out in the integration process.
At $x=pi/2$ you have the series.
$$frac{1}{2}=1-1+1-1+...$$
A proof for series (1) is here and you should be able to prove your series in a similar fashion by changing to the exponential form.
answered Feb 2 at 21:35
James ArathoonJames Arathoon
1,608423
$endgroup$
It might to better to first study the well known convergent series
$$ln cos left(frac{x}{2}right)=-ln 2 + sum_{k=1}^infty frac{ (-1)^{k-1} cos(kx)}{k}tag{1}$$
You can then naively differentiate this to give the divergent series
$$frac{1}{2}tanleft(frac{x}{2}right)=sum_{k=1}^infty (-1)^{k-1}sin(kx)$$
You can examine the nature of the divergences through looking at the partial sums. They appear to be correlated plus and minus going divergences. If you want to claim that this function can be integrated with rigour (which I've read somewhere is possible [possibly in a pamphlet by Hardy on Fourier Series] ) I think you must prove that these "correlated" plus and minus going divergences somehow cancel each other out in the integration process.
At $x=pi/2$ you have the series.
$$frac{1}{2}=1-1+1-1+...$$
A proof for series (1) is here and you should be able to prove your series in a similar fashion by changing to the exponential form.
answered Feb 2 at 21:35
James ArathoonJames Arathoon
1,608423
edited Feb 2 at 21:41
answered Feb 2 at 21:35
James ArathoonJames Arathoon
1,608423
answered Feb 2 at 21:35
James ArathoonJames Arathoon
1,608423
answered Feb 2 at 21:35
James ArathoonJames Arathoon
1,608423
1,608423
$begingroup$
I have found these series in chapter 3 of Fourier's Analytical Theory of Heat:books.google.com/…
$endgroup$
– James Warthington
Feb 2 at 21:46
add a comment |
$begingroup$
I have found these series in chapter 3 of Fourier's Analytical Theory of Heat:books.google.com/…
$endgroup$
– James Warthington
Feb 2 at 21:46
$begingroup$
I have found these series in chapter 3 of Fourier's Analytical Theory of Heat:books.google.com/…
$endgroup$
– James Warthington
Feb 2 at 21:46
$begingroup$
I have found these series in chapter 3 of Fourier's Analytical Theory of Heat:books.google.com/…
$endgroup$
– James Warthington
Feb 2 at 21:46
add a comment |
$begingroup$
The first series is nothing more than the Fourier series of the $2pi$-periodic odd function
$$
y=frac{x}{2},quadpi<x<pi.
$$
Indeed the coefficients of the expansion $frac{x}{2}=sumlimits_{n=1}^infty b_nsin nx$ are:
$$
b_n=frac{1}{pi}int_{-pi}^pifrac{x}{2}sin nx;dx=frac{1}{pi}left[-frac{x}{2n}cos nxright]_{-pi}^pi=-frac{1}{npi}frac{pi(-1)^n+pi(-1)^n}{2}=frac{(-1)^{n+1}}{n}.
$$
The answer to the second question is most easily obtained through exponential representation of cosine and well-known expression for the partial sums of geometrical series:
$$begin {array}{}
S_N(x)&=sum_{n=1}^N (-1)^{n+1}cos nx\
&=sum_{n=1}^N(-1)^{n+1}frac {e^{inx}+e^{-inx}}{2}\
&=frac12left [frac {e^{ix}-(-1)^{N+2}e^{(N+1)ix}}{1+e^{ix}}+
frac {e^{-ix}-(-1)^{N+2}e^{-(N+1)ix}}{1+e^{-ix}}right]\
&=frac12left [frac {e^{ix/2}-(-1)^{N+2}e^{(N+1/2)ix}}{e^{-ix/2}+e^{ix/2}}+
frac {e^{-ix/2}-(-1)^{N+2}e^{-(N+1/2)ix}}{e^{ix/2}+e^{-ix/2}}right]\
&=frac12left [1-(-1)^{N}frac{cos(N+1/2)x}{cos (x/2)}right].
end {array}
$$
From this one can conclude that $S_N (x) $ does not converge for any value of $x $.
answered Feb 2 at 22:43
useruser
6,56011031
$endgroup$
add a comment |
$begingroup$
The first series is nothing more than the Fourier series of the $2pi$-periodic odd function
$$
y=frac{x}{2},quadpi<x<pi.
$$
Indeed the coefficients of the expansion $frac{x}{2}=sumlimits_{n=1}^infty b_nsin nx$ are:
$$
b_n=frac{1}{pi}int_{-pi}^pifrac{x}{2}sin nx;dx=frac{1}{pi}left[-frac{x}{2n}cos nxright]_{-pi}^pi=-frac{1}{npi}frac{pi(-1)^n+pi(-1)^n}{2}=frac{(-1)^{n+1}}{n}.
$$
The answer to the second question is most easily obtained through exponential representation of cosine and well-known expression for the partial sums of geometrical series:
$$begin {array}{}
S_N(x)&=sum_{n=1}^N (-1)^{n+1}cos nx\
&=sum_{n=1}^N(-1)^{n+1}frac {e^{inx}+e^{-inx}}{2}\
&=frac12left [frac {e^{ix}-(-1)^{N+2}e^{(N+1)ix}}{1+e^{ix}}+
frac {e^{-ix}-(-1)^{N+2}e^{-(N+1)ix}}{1+e^{-ix}}right]\
&=frac12left [frac {e^{ix/2}-(-1)^{N+2}e^{(N+1/2)ix}}{e^{-ix/2}+e^{ix/2}}+
frac {e^{-ix/2}-(-1)^{N+2}e^{-(N+1/2)ix}}{e^{ix/2}+e^{-ix/2}}right]\
&=frac12left [1-(-1)^{N}frac{cos(N+1/2)x}{cos (x/2)}right].
end {array}
$$
From this one can conclude that $S_N (x) $ does not converge for any value of $x $.
answered Feb 2 at 22:43
useruser
6,56011031
$endgroup$
add a comment |
$begingroup$
The first series is nothing more than the Fourier series of the $2pi$-periodic odd function
$$
y=frac{x}{2},quadpi<x<pi.
$$
Indeed the coefficients of the expansion $frac{x}{2}=sumlimits_{n=1}^infty b_nsin nx$ are:
$$
b_n=frac{1}{pi}int_{-pi}^pifrac{x}{2}sin nx;dx=frac{1}{pi}left[-frac{x}{2n}cos nxright]_{-pi}^pi=-frac{1}{npi}frac{pi(-1)^n+pi(-1)^n}{2}=frac{(-1)^{n+1}}{n}.
$$
The answer to the second question is most easily obtained through exponential representation of cosine and well-known expression for the partial sums of geometrical series:
$$begin {array}{}
S_N(x)&=sum_{n=1}^N (-1)^{n+1}cos nx\
&=sum_{n=1}^N(-1)^{n+1}frac {e^{inx}+e^{-inx}}{2}\
&=frac12left [frac {e^{ix}-(-1)^{N+2}e^{(N+1)ix}}{1+e^{ix}}+
frac {e^{-ix}-(-1)^{N+2}e^{-(N+1)ix}}{1+e^{-ix}}right]\
&=frac12left [frac {e^{ix/2}-(-1)^{N+2}e^{(N+1/2)ix}}{e^{-ix/2}+e^{ix/2}}+
frac {e^{-ix/2}-(-1)^{N+2}e^{-(N+1/2)ix}}{e^{ix/2}+e^{-ix/2}}right]\
&=frac12left [1-(-1)^{N}frac{cos(N+1/2)x}{cos (x/2)}right].
end {array}
$$
From this one can conclude that $S_N (x) $ does not converge for any value of $x $.
answered Feb 2 at 22:43
useruser
6,56011031
$endgroup$
The first series is nothing more than the Fourier series of the $2pi$-periodic odd function
$$
y=frac{x}{2},quadpi<x<pi.
$$
Indeed the coefficients of the expansion $frac{x}{2}=sumlimits_{n=1}^infty b_nsin nx$ are:
$$
b_n=frac{1}{pi}int_{-pi}^pifrac{x}{2}sin nx;dx=frac{1}{pi}left[-frac{x}{2n}cos nxright]_{-pi}^pi=-frac{1}{npi}frac{pi(-1)^n+pi(-1)^n}{2}=frac{(-1)^{n+1}}{n}.
$$
The answer to the second question is most easily obtained through exponential representation of cosine and well-known expression for the partial sums of geometrical series:
$$begin {array}{}
S_N(x)&=sum_{n=1}^N (-1)^{n+1}cos nx\
&=sum_{n=1}^N(-1)^{n+1}frac {e^{inx}+e^{-inx}}{2}\
&=frac12left [frac {e^{ix}-(-1)^{N+2}e^{(N+1)ix}}{1+e^{ix}}+
frac {e^{-ix}-(-1)^{N+2}e^{-(N+1)ix}}{1+e^{-ix}}right]\
&=frac12left [frac {e^{ix/2}-(-1)^{N+2}e^{(N+1/2)ix}}{e^{-ix/2}+e^{ix/2}}+
frac {e^{-ix/2}-(-1)^{N+2}e^{-(N+1/2)ix}}{e^{ix/2}+e^{-ix/2}}right]\
&=frac12left [1-(-1)^{N}frac{cos(N+1/2)x}{cos (x/2)}right].
end {array}
$$
From this one can conclude that $S_N (x) $ does not converge for any value of $x $.
answered Feb 2 at 22:43
useruser
6,56011031
edited Feb 3 at 21:55
answered Feb 2 at 22:43
useruser
6,56011031
answered Feb 2 at 22:43
useruser
6,56011031
answered Feb 2 at 22:43
useruser
6,56011031
6,56011031
add a comment |
add a comment |
$begingroup$
For the first series, consider that you look for the imaginary part of
$$sum_{n=1}^infty (-1)^{n-1} frac {e^{i n x}} n=sum_{n=1}^infty (-1)^{n-1} frac {left( e^{i x}right)^n} n=log left(1+e^{i x}right)$$So
$$sum_{n=1}^infty (-1)^{n-1} frac {sin(n x)} n=frac{1}{2} i left(log left(1+e^{-i x}right)-log left(1+e^{i x}right)right)=frac{1}{2} left(arg left(1+e^{i x}right)- arg left(1+e^{-i x}right)right)=frac x 2$$ provided $-pi leq x leq pi$ .
answered Feb 3 at 5:26
Claude LeiboviciClaude Leibovici
126k1158135
$endgroup$
$begingroup$
Do you know anyone to obtain this result without recoursing to complex analysis? I haven't studied complex analysis yet so I don't understand your answer.
$endgroup$
– James Warthington
Feb 27 at 1:30
add a comment |
$begingroup$
For the first series, consider that you look for the imaginary part of
$$sum_{n=1}^infty (-1)^{n-1} frac {e^{i n x}} n=sum_{n=1}^infty (-1)^{n-1} frac {left( e^{i x}right)^n} n=log left(1+e^{i x}right)$$So
$$sum_{n=1}^infty (-1)^{n-1} frac {sin(n x)} n=frac{1}{2} i left(log left(1+e^{-i x}right)-log left(1+e^{i x}right)right)=frac{1}{2} left(arg left(1+e^{i x}right)- arg left(1+e^{-i x}right)right)=frac x 2$$ provided $-pi leq x leq pi$ .
answered Feb 3 at 5:26
Claude LeiboviciClaude Leibovici
126k1158135
$endgroup$
$begingroup$
Do you know anyone to obtain this result without recoursing to complex analysis? I haven't studied complex analysis yet so I don't understand your answer.
$endgroup$
– James Warthington
Feb 27 at 1:30
add a comment |
$begingroup$
For the first series, consider that you look for the imaginary part of
$$sum_{n=1}^infty (-1)^{n-1} frac {e^{i n x}} n=sum_{n=1}^infty (-1)^{n-1} frac {left( e^{i x}right)^n} n=log left(1+e^{i x}right)$$So
$$sum_{n=1}^infty (-1)^{n-1} frac {sin(n x)} n=frac{1}{2} i left(log left(1+e^{-i x}right)-log left(1+e^{i x}right)right)=frac{1}{2} left(arg left(1+e^{i x}right)- arg left(1+e^{-i x}right)right)=frac x 2$$ provided $-pi leq x leq pi$ .
answered Feb 3 at 5:26
Claude LeiboviciClaude Leibovici
126k1158135
$endgroup$
For the first series, consider that you look for the imaginary part of
$$sum_{n=1}^infty (-1)^{n-1} frac {e^{i n x}} n=sum_{n=1}^infty (-1)^{n-1} frac {left( e^{i x}right)^n} n=log left(1+e^{i x}right)$$So
$$sum_{n=1}^infty (-1)^{n-1} frac {sin(n x)} n=frac{1}{2} i left(log left(1+e^{-i x}right)-log left(1+e^{i x}right)right)=frac{1}{2} left(arg left(1+e^{i x}right)- arg left(1+e^{-i x}right)right)=frac x 2$$ provided $-pi leq x leq pi$ .
answered Feb 3 at 5:26
Claude LeiboviciClaude Leibovici
126k1158135
answered Feb 3 at 5:26
Claude LeiboviciClaude Leibovici
126k1158135
answered Feb 3 at 5:26
Claude LeiboviciClaude Leibovici
126k1158135
answered Feb 3 at 5:26
Claude LeiboviciClaude Leibovici
126k1158135
126k1158135
$begingroup$
Do you know anyone to obtain this result without recoursing to complex analysis? I haven't studied complex analysis yet so I don't understand your answer.
$endgroup$
– James Warthington
Feb 27 at 1:30
add a comment |
$begingroup$
Do you know anyone to obtain this result without recoursing to complex analysis? I haven't studied complex analysis yet so I don't understand your answer.
$endgroup$
– James Warthington
Feb 27 at 1:30
$begingroup$
Do you know anyone to obtain this result without recoursing to complex analysis? I haven't studied complex analysis yet so I don't understand your answer.
$endgroup$
– James Warthington
Feb 27 at 1:30
$begingroup$
Do you know anyone to obtain this result without recoursing to complex analysis? I haven't studied complex analysis yet so I don't understand your answer.
$endgroup$
– James Warthington
Feb 27 at 1:30
add a comment |
$begingroup$
The first equation's right-hand side is $Imln(1+exp ix)$. The second equation's right-hand side can only have convergent partial sums if $lim_{ntoinfty}cos nx=0$, but this clearly fails for $x/pi$ rational. A famous but less obvious result is that other real $x$ also do not obtain such a limit.
answered Feb 2 at 20:44
J.G.J.G.
33.5k23252
$endgroup$
$begingroup$
I have never seen $ln(1+expix)$ before, what is it?
$endgroup$
– James Warthington
Feb 2 at 20:46
$begingroup$
@JamesWharthington You may want to learn about complex exponentials and logarithms.
$endgroup$
– J.G.
Feb 2 at 20:48
add a comment |
$begingroup$
The first equation's right-hand side is $Imln(1+exp ix)$. The second equation's right-hand side can only have convergent partial sums if $lim_{ntoinfty}cos nx=0$, but this clearly fails for $x/pi$ rational. A famous but less obvious result is that other real $x$ also do not obtain such a limit.
answered Feb 2 at 20:44
J.G.J.G.
33.5k23252
$endgroup$
$begingroup$
I have never seen $ln(1+expix)$ before, what is it?
$endgroup$
– James Warthington
Feb 2 at 20:46
$begingroup$
@JamesWharthington You may want to learn about complex exponentials and logarithms.
$endgroup$
– J.G.
Feb 2 at 20:48
add a comment |
$begingroup$
The first equation's right-hand side is $Imln(1+exp ix)$. The second equation's right-hand side can only have convergent partial sums if $lim_{ntoinfty}cos nx=0$, but this clearly fails for $x/pi$ rational. A famous but less obvious result is that other real $x$ also do not obtain such a limit.
answered Feb 2 at 20:44
J.G.J.G.
33.5k23252
$endgroup$
The first equation's right-hand side is $Imln(1+exp ix)$. The second equation's right-hand side can only have convergent partial sums if $lim_{ntoinfty}cos nx=0$, but this clearly fails for $x/pi$ rational. A famous but less obvious result is that other real $x$ also do not obtain such a limit.
answered Feb 2 at 20:44
J.G.J.G.
33.5k23252
answered Feb 2 at 20:44
J.G.J.G.
33.5k23252
answered Feb 2 at 20:44
J.G.J.G.
33.5k23252
answered Feb 2 at 20:44
J.G.J.G.
33.5k23252
33.5k23252
$begingroup$
I have never seen $ln(1+expix)$ before, what is it?
$endgroup$
– James Warthington
Feb 2 at 20:46
$begingroup$
@JamesWharthington You may want to learn about complex exponentials and logarithms.
$endgroup$
– J.G.
Feb 2 at 20:48
add a comment |
$begingroup$
I have never seen $ln(1+expix)$ before, what is it?
$endgroup$
– James Warthington
Feb 2 at 20:46
$begingroup$
@JamesWharthington You may want to learn about complex exponentials and logarithms.
$endgroup$
– J.G.
Feb 2 at 20:48
$begingroup$
I have never seen $ln(1+expix)$ before, what is it?
$endgroup$
– James Warthington
Feb 2 at 20:46
$begingroup$
I have never seen $ln(1+expix)$ before, what is it?
$endgroup$
– James Warthington
Feb 2 at 20:46
$begingroup$
@JamesWharthington You may want to learn about complex exponentials and logarithms.
$endgroup$
– J.G.
Feb 2 at 20:48
$begingroup$
@JamesWharthington You may want to learn about complex exponentials and logarithms.
$endgroup$
– J.G.
Feb 2 at 20:48
add a comment |
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1
$begingroup$
Do you know hay Fourier series are.?
$endgroup$
– Julián Aguirre
Feb 2 at 20:22
$begingroup$
Your second question is easy enough: $cos(nx)$ does not tend to $0$ as $n to infty$, which means the series fails the divergence test.
$endgroup$
– Theo Bendit
Feb 2 at 20:24
$begingroup$
I haven't learnt Fourier series yet. But I am eager to learn, sir.
$endgroup$
– James Warthington
Feb 2 at 20:24
$begingroup$
@Theo Bendit Which test should you perform?
$endgroup$
– James Warthington
Feb 2 at 20:25
$begingroup$
@James Warthington, it is just a basic theorem about series. If $sum a_n$ converges then $a_nto 0$.
$endgroup$
– Mark
Feb 2 at 20:26