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“Bad†Fourier Series derivation
Clash Royale CLAN TAG#URR8PPP
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Let $f(theta)$ $2pi$-periodic such that $f(theta)=e^theta$ for $-pi<0<pi$, and $$e^theta=sum_n=-infty^inftyc_ne^intheta,,, mathrmfor,, |theta|<pi $$
it's Fourier series. If we formally differentiate this equation, we obtain
$$e^theta=sum_n=-infty^inftyinc_ne^intheta. $$
But this implies $c_n=inc_n$ or $(1-in)c_n=0$, so, $c_n=0,forall ninmathbbZ$. This is obviously wrong.
Where is the mistake?
The only thing that I could contest is the derivative of $f$. I know a theorem saying that a sufficient condition is $f$ continuous and piecewise smooth, and $f$ is not continous. But it don't implies that I can't derivate term by term.
sequences-and-series fourier-analysis fourier-series
asked 4 hours ago
Mateus Rocha
557115
add a comment |Â
up vote
4
down vote
favorite
Let $f(theta)$ $2pi$-periodic such that $f(theta)=e^theta$ for $-pi<0<pi$, and $$e^theta=sum_n=-infty^inftyc_ne^intheta,,, mathrmfor,, |theta|<pi $$
it's Fourier series. If we formally differentiate this equation, we obtain
$$e^theta=sum_n=-infty^inftyinc_ne^intheta. $$
But this implies $c_n=inc_n$ or $(1-in)c_n=0$, so, $c_n=0,forall ninmathbbZ$. This is obviously wrong.
Where is the mistake?
The only thing that I could contest is the derivative of $f$. I know a theorem saying that a sufficient condition is $f$ continuous and piecewise smooth, and $f$ is not continous. But it don't implies that I can't derivate term by term.
sequences-and-series fourier-analysis fourier-series
asked 4 hours ago
Mateus Rocha
557115
If you calculate the $c_n$ you can check that the derivative series does not converge uniformly---indeed, it does not converge at all.
– Driver 8
3 hours ago
@Driver8 Uniform convergence is a sufficient condition, not a necessary one. The series obtained by differentiating under the summation does not converge. Hence, the interchange is not legitimate.
– Mark Viola
3 hours ago
Yes, indeed. I will delete my (dumb) comment
– Driver 8
3 hours ago
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Let $f(theta)$ $2pi$-periodic such that $f(theta)=e^theta$ for $-pi<0<pi$, and $$e^theta=sum_n=-infty^inftyc_ne^intheta,,, mathrmfor,, |theta|<pi $$
it's Fourier series. If we formally differentiate this equation, we obtain
$$e^theta=sum_n=-infty^inftyinc_ne^intheta. $$
But this implies $c_n=inc_n$ or $(1-in)c_n=0$, so, $c_n=0,forall ninmathbbZ$. This is obviously wrong.
Where is the mistake?
The only thing that I could contest is the derivative of $f$. I know a theorem saying that a sufficient condition is $f$ continuous and piecewise smooth, and $f$ is not continous. But it don't implies that I can't derivate term by term.
sequences-and-series fourier-analysis fourier-series
asked 4 hours ago
Mateus Rocha
557115
Let $f(theta)$ $2pi$-periodic such that $f(theta)=e^theta$ for $-pi<0<pi$, and $$e^theta=sum_n=-infty^inftyc_ne^intheta,,, mathrmfor,, |theta|<pi $$
it's Fourier series. If we formally differentiate this equation, we obtain
$$e^theta=sum_n=-infty^inftyinc_ne^intheta. $$
But this implies $c_n=inc_n$ or $(1-in)c_n=0$, so, $c_n=0,forall ninmathbbZ$. This is obviously wrong.
Where is the mistake?
The only thing that I could contest is the derivative of $f$. I know a theorem saying that a sufficient condition is $f$ continuous and piecewise smooth, and $f$ is not continous. But it don't implies that I can't derivate term by term.
sequences-and-series fourier-analysis fourier-series
sequences-and-series fourier-analysis fourier-series
asked 4 hours ago
Mateus Rocha
557115
asked 4 hours ago
Mateus Rocha
557115
edited 4 hours ago
asked 4 hours ago
Mateus Rocha
557115
asked 4 hours ago
Mateus Rocha
557115
asked 4 hours ago
Mateus Rocha
557115
557115
If you calculate the $c_n$ you can check that the derivative series does not converge uniformly---indeed, it does not converge at all.
– Driver 8
3 hours ago
@Driver8 Uniform convergence is a sufficient condition, not a necessary one. The series obtained by differentiating under the summation does not converge. Hence, the interchange is not legitimate.
– Mark Viola
3 hours ago
Yes, indeed. I will delete my (dumb) comment
– Driver 8
3 hours ago
add a comment |Â
If you calculate the $c_n$ you can check that the derivative series does not converge uniformly---indeed, it does not converge at all.
– Driver 8
3 hours ago
@Driver8 Uniform convergence is a sufficient condition, not a necessary one. The series obtained by differentiating under the summation does not converge. Hence, the interchange is not legitimate.
– Mark Viola
3 hours ago
Yes, indeed. I will delete my (dumb) comment
– Driver 8
3 hours ago
If you calculate the $c_n$ you can check that the derivative series does not converge uniformly---indeed, it does not converge at all.
– Driver 8
3 hours ago
If you calculate the $c_n$ you can check that the derivative series does not converge uniformly---indeed, it does not converge at all.
– Driver 8
3 hours ago
@Driver8 Uniform convergence is a sufficient condition, not a necessary one. The series obtained by differentiating under the summation does not converge. Hence, the interchange is not legitimate.
– Mark Viola
3 hours ago
@Driver8 Uniform convergence is a sufficient condition, not a necessary one. The series obtained by differentiating under the summation does not converge. Hence, the interchange is not legitimate.
– Mark Viola
3 hours ago
Yes, indeed. I will delete my (dumb) comment
– Driver 8
3 hours ago
Yes, indeed. I will delete my (dumb) comment
– Driver 8
3 hours ago
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
4
down vote
accepted
You have to think in terms of distributional derivatives, because in this sense you can switch series and derivative. Think $f$ as defined on the 1-torus. Then the distributional derivative of $f$ it is not $f$ itself, but $f-(e^pi-e^-pi)delta_[pi]$, where $delta_[pi]$ is the delta Dirac in the point $[pi]$ of the 1-torus. So, in distributional sense: $$sum_n=-infty^infty inc_ne_n = f'=f-(e^pi-e^-pi)delta_[pi]=sum_n=-infty^infty c_ne_n - (e^pi-e^-pi)delta_[pi]$$
i.e.:
$$sum_n=-infty^infty (in-1)c_ne_n=-(e^pi-e^-pi)delta_[pi]=-(e^pi-e^-pi)sum_n=-infty^infty e^inpie_n \ = -(e^pi-e^-pi)sum_n=-infty^infty frac(-1)^n2pie_n.$$
Then:
$$forall nin mathbbZ, (in-1)c_n=-frac(-1)^n2pi(e^pi-e^-pi)$$
i.e.:
$$forall ninmathbbZ, c_n=frac(-1)^n(e^pi-e^-pi)2pi(1-in).$$
answered 3 hours ago
Bob
1,3181622
The series $sum_nin mathbbZinc_ne^intheta$ diverges.
– Mark Viola
3 hours ago
@MarkViola not in distributional sense: check Gel'fand - Shilov, generalized functions, vol. 1
– Bob
3 hours ago
@MarkViola I already gave you a reference
– Bob
3 hours ago
1
Chapter 1, pag.30, example 2.: "any series of the form $sum_n=-infty^infty a_n e^inx$ whose coefficients increas no faster than some power of $n$ as $|n|rightarrowinfty$ converges in the sense of generalized functions"
– Bob
3 hours ago
I understand this, but I was not under the impression that the question in the OP was asked broadly in the sense of generalized functions, but rather classical analysis only.
– Mark Viola
3 hours ago
 |Â
show 4 more comments
up vote
0
down vote
Note that $c_n=fracsinh(pi)pifrac(-1)^n1-in$ and the Fourier series for $e^theta$ on $-pi<theta<pi$ is
$$e^theta =fracsinh(pi)pisum_-infty^infty frac(-1)^n1-in,e^intheta$$
Clearly, the series $sum_n=-infty^infty inc_ne^intheta=fracsinh(pi)pisum_-infty^infty frac(-1)^n,in1-in,e^intheta$, which is obtained by formal differentiation under the summation, is divergent. Hence, differentiation under the summation is illegitimate and does not lead to a Fourier series representation of $e^theta$ for $-pi<theta<pi$.
answered 3 hours ago
Mark Viola
127k1172167
Mateus, please let me know how I can improve my answer. I really want to give you the best answer I can.
– Mark Viola
3 hours ago
Why was this answer down voted??
– Mark Viola
3 hours ago
Mark, your answer is good actually. I gave you 1+
– Mateus Rocha
30 mins ago
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
You have to think in terms of distributional derivatives, because in this sense you can switch series and derivative. Think $f$ as defined on the 1-torus. Then the distributional derivative of $f$ it is not $f$ itself, but $f-(e^pi-e^-pi)delta_[pi]$, where $delta_[pi]$ is the delta Dirac in the point $[pi]$ of the 1-torus. So, in distributional sense: $$sum_n=-infty^infty inc_ne_n = f'=f-(e^pi-e^-pi)delta_[pi]=sum_n=-infty^infty c_ne_n - (e^pi-e^-pi)delta_[pi]$$
i.e.:
$$sum_n=-infty^infty (in-1)c_ne_n=-(e^pi-e^-pi)delta_[pi]=-(e^pi-e^-pi)sum_n=-infty^infty e^inpie_n \ = -(e^pi-e^-pi)sum_n=-infty^infty frac(-1)^n2pie_n.$$
Then:
$$forall nin mathbbZ, (in-1)c_n=-frac(-1)^n2pi(e^pi-e^-pi)$$
i.e.:
$$forall ninmathbbZ, c_n=frac(-1)^n(e^pi-e^-pi)2pi(1-in).$$
answered 3 hours ago
Bob
1,3181622
The series $sum_nin mathbbZinc_ne^intheta$ diverges.
– Mark Viola
3 hours ago
@MarkViola not in distributional sense: check Gel'fand - Shilov, generalized functions, vol. 1
– Bob
3 hours ago
@MarkViola I already gave you a reference
– Bob
3 hours ago
1
Chapter 1, pag.30, example 2.: "any series of the form $sum_n=-infty^infty a_n e^inx$ whose coefficients increas no faster than some power of $n$ as $|n|rightarrowinfty$ converges in the sense of generalized functions"
– Bob
3 hours ago
I understand this, but I was not under the impression that the question in the OP was asked broadly in the sense of generalized functions, but rather classical analysis only.
– Mark Viola
3 hours ago
 |Â
show 4 more comments
up vote
4
down vote
accepted
You have to think in terms of distributional derivatives, because in this sense you can switch series and derivative. Think $f$ as defined on the 1-torus. Then the distributional derivative of $f$ it is not $f$ itself, but $f-(e^pi-e^-pi)delta_[pi]$, where $delta_[pi]$ is the delta Dirac in the point $[pi]$ of the 1-torus. So, in distributional sense: $$sum_n=-infty^infty inc_ne_n = f'=f-(e^pi-e^-pi)delta_[pi]=sum_n=-infty^infty c_ne_n - (e^pi-e^-pi)delta_[pi]$$
i.e.:
$$sum_n=-infty^infty (in-1)c_ne_n=-(e^pi-e^-pi)delta_[pi]=-(e^pi-e^-pi)sum_n=-infty^infty e^inpie_n \ = -(e^pi-e^-pi)sum_n=-infty^infty frac(-1)^n2pie_n.$$
Then:
$$forall nin mathbbZ, (in-1)c_n=-frac(-1)^n2pi(e^pi-e^-pi)$$
i.e.:
$$forall ninmathbbZ, c_n=frac(-1)^n(e^pi-e^-pi)2pi(1-in).$$
answered 3 hours ago
Bob
1,3181622
The series $sum_nin mathbbZinc_ne^intheta$ diverges.
– Mark Viola
3 hours ago
@MarkViola not in distributional sense: check Gel'fand - Shilov, generalized functions, vol. 1
– Bob
3 hours ago
@MarkViola I already gave you a reference
– Bob
3 hours ago
1
Chapter 1, pag.30, example 2.: "any series of the form $sum_n=-infty^infty a_n e^inx$ whose coefficients increas no faster than some power of $n$ as $|n|rightarrowinfty$ converges in the sense of generalized functions"
– Bob
3 hours ago
I understand this, but I was not under the impression that the question in the OP was asked broadly in the sense of generalized functions, but rather classical analysis only.
– Mark Viola
3 hours ago
 |Â
show 4 more comments
up vote
4
down vote
accepted
up vote
4
down vote
accepted
You have to think in terms of distributional derivatives, because in this sense you can switch series and derivative. Think $f$ as defined on the 1-torus. Then the distributional derivative of $f$ it is not $f$ itself, but $f-(e^pi-e^-pi)delta_[pi]$, where $delta_[pi]$ is the delta Dirac in the point $[pi]$ of the 1-torus. So, in distributional sense: $$sum_n=-infty^infty inc_ne_n = f'=f-(e^pi-e^-pi)delta_[pi]=sum_n=-infty^infty c_ne_n - (e^pi-e^-pi)delta_[pi]$$
i.e.:
$$sum_n=-infty^infty (in-1)c_ne_n=-(e^pi-e^-pi)delta_[pi]=-(e^pi-e^-pi)sum_n=-infty^infty e^inpie_n \ = -(e^pi-e^-pi)sum_n=-infty^infty frac(-1)^n2pie_n.$$
Then:
$$forall nin mathbbZ, (in-1)c_n=-frac(-1)^n2pi(e^pi-e^-pi)$$
i.e.:
$$forall ninmathbbZ, c_n=frac(-1)^n(e^pi-e^-pi)2pi(1-in).$$
answered 3 hours ago
Bob
1,3181622
You have to think in terms of distributional derivatives, because in this sense you can switch series and derivative. Think $f$ as defined on the 1-torus. Then the distributional derivative of $f$ it is not $f$ itself, but $f-(e^pi-e^-pi)delta_[pi]$, where $delta_[pi]$ is the delta Dirac in the point $[pi]$ of the 1-torus. So, in distributional sense: $$sum_n=-infty^infty inc_ne_n = f'=f-(e^pi-e^-pi)delta_[pi]=sum_n=-infty^infty c_ne_n - (e^pi-e^-pi)delta_[pi]$$
i.e.:
$$sum_n=-infty^infty (in-1)c_ne_n=-(e^pi-e^-pi)delta_[pi]=-(e^pi-e^-pi)sum_n=-infty^infty e^inpie_n \ = -(e^pi-e^-pi)sum_n=-infty^infty frac(-1)^n2pie_n.$$
Then:
$$forall nin mathbbZ, (in-1)c_n=-frac(-1)^n2pi(e^pi-e^-pi)$$
i.e.:
$$forall ninmathbbZ, c_n=frac(-1)^n(e^pi-e^-pi)2pi(1-in).$$
answered 3 hours ago
Bob
1,3181622
edited 1 hour ago
answered 3 hours ago
Bob
1,3181622
answered 3 hours ago
Bob
1,3181622
answered 3 hours ago
Bob
1,3181622
1,3181622
The series $sum_nin mathbbZinc_ne^intheta$ diverges.
– Mark Viola
3 hours ago
@MarkViola not in distributional sense: check Gel'fand - Shilov, generalized functions, vol. 1
– Bob
3 hours ago
@MarkViola I already gave you a reference
– Bob
3 hours ago
1
Chapter 1, pag.30, example 2.: "any series of the form $sum_n=-infty^infty a_n e^inx$ whose coefficients increas no faster than some power of $n$ as $|n|rightarrowinfty$ converges in the sense of generalized functions"
– Bob
3 hours ago
I understand this, but I was not under the impression that the question in the OP was asked broadly in the sense of generalized functions, but rather classical analysis only.
– Mark Viola
3 hours ago
 |Â
show 4 more comments
The series $sum_nin mathbbZinc_ne^intheta$ diverges.
– Mark Viola
3 hours ago
@MarkViola not in distributional sense: check Gel'fand - Shilov, generalized functions, vol. 1
– Bob
3 hours ago
@MarkViola I already gave you a reference
– Bob
3 hours ago
1
Chapter 1, pag.30, example 2.: "any series of the form $sum_n=-infty^infty a_n e^inx$ whose coefficients increas no faster than some power of $n$ as $|n|rightarrowinfty$ converges in the sense of generalized functions"
– Bob
3 hours ago
I understand this, but I was not under the impression that the question in the OP was asked broadly in the sense of generalized functions, but rather classical analysis only.
– Mark Viola
3 hours ago
The series $sum_nin mathbbZinc_ne^intheta$ diverges.
– Mark Viola
3 hours ago
The series $sum_nin mathbbZinc_ne^intheta$ diverges.
– Mark Viola
3 hours ago
@MarkViola not in distributional sense: check Gel'fand - Shilov, generalized functions, vol. 1
– Bob
3 hours ago
@MarkViola not in distributional sense: check Gel'fand - Shilov, generalized functions, vol. 1
– Bob
3 hours ago
@MarkViola I already gave you a reference
– Bob
3 hours ago
@MarkViola I already gave you a reference
– Bob
3 hours ago
1
1
Chapter 1, pag.30, example 2.: "any series of the form $sum_n=-infty^infty a_n e^inx$ whose coefficients increas no faster than some power of $n$ as $|n|rightarrowinfty$ converges in the sense of generalized functions"
– Bob
3 hours ago
Chapter 1, pag.30, example 2.: "any series of the form $sum_n=-infty^infty a_n e^inx$ whose coefficients increas no faster than some power of $n$ as $|n|rightarrowinfty$ converges in the sense of generalized functions"
– Bob
3 hours ago
I understand this, but I was not under the impression that the question in the OP was asked broadly in the sense of generalized functions, but rather classical analysis only.
– Mark Viola
3 hours ago
I understand this, but I was not under the impression that the question in the OP was asked broadly in the sense of generalized functions, but rather classical analysis only.
– Mark Viola
3 hours ago
 |Â
show 4 more comments
up vote
0
down vote
Note that $c_n=fracsinh(pi)pifrac(-1)^n1-in$ and the Fourier series for $e^theta$ on $-pi<theta<pi$ is
$$e^theta =fracsinh(pi)pisum_-infty^infty frac(-1)^n1-in,e^intheta$$
Clearly, the series $sum_n=-infty^infty inc_ne^intheta=fracsinh(pi)pisum_-infty^infty frac(-1)^n,in1-in,e^intheta$, which is obtained by formal differentiation under the summation, is divergent. Hence, differentiation under the summation is illegitimate and does not lead to a Fourier series representation of $e^theta$ for $-pi<theta<pi$.
answered 3 hours ago
Mark Viola
127k1172167
Mateus, please let me know how I can improve my answer. I really want to give you the best answer I can.
– Mark Viola
3 hours ago
Why was this answer down voted??
– Mark Viola
3 hours ago
Mark, your answer is good actually. I gave you 1+
– Mateus Rocha
30 mins ago
add a comment |Â
up vote
0
down vote
Note that $c_n=fracsinh(pi)pifrac(-1)^n1-in$ and the Fourier series for $e^theta$ on $-pi<theta<pi$ is
$$e^theta =fracsinh(pi)pisum_-infty^infty frac(-1)^n1-in,e^intheta$$
Clearly, the series $sum_n=-infty^infty inc_ne^intheta=fracsinh(pi)pisum_-infty^infty frac(-1)^n,in1-in,e^intheta$, which is obtained by formal differentiation under the summation, is divergent. Hence, differentiation under the summation is illegitimate and does not lead to a Fourier series representation of $e^theta$ for $-pi<theta<pi$.
answered 3 hours ago
Mark Viola
127k1172167
Mateus, please let me know how I can improve my answer. I really want to give you the best answer I can.
– Mark Viola
3 hours ago
Why was this answer down voted??
– Mark Viola
3 hours ago
Mark, your answer is good actually. I gave you 1+
– Mateus Rocha
30 mins ago
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Note that $c_n=fracsinh(pi)pifrac(-1)^n1-in$ and the Fourier series for $e^theta$ on $-pi<theta<pi$ is
$$e^theta =fracsinh(pi)pisum_-infty^infty frac(-1)^n1-in,e^intheta$$
Clearly, the series $sum_n=-infty^infty inc_ne^intheta=fracsinh(pi)pisum_-infty^infty frac(-1)^n,in1-in,e^intheta$, which is obtained by formal differentiation under the summation, is divergent. Hence, differentiation under the summation is illegitimate and does not lead to a Fourier series representation of $e^theta$ for $-pi<theta<pi$.
answered 3 hours ago
Mark Viola
127k1172167
Note that $c_n=fracsinh(pi)pifrac(-1)^n1-in$ and the Fourier series for $e^theta$ on $-pi<theta<pi$ is
$$e^theta =fracsinh(pi)pisum_-infty^infty frac(-1)^n1-in,e^intheta$$
Clearly, the series $sum_n=-infty^infty inc_ne^intheta=fracsinh(pi)pisum_-infty^infty frac(-1)^n,in1-in,e^intheta$, which is obtained by formal differentiation under the summation, is divergent. Hence, differentiation under the summation is illegitimate and does not lead to a Fourier series representation of $e^theta$ for $-pi<theta<pi$.
answered 3 hours ago
Mark Viola
127k1172167
edited 3 hours ago
answered 3 hours ago
Mark Viola
127k1172167
answered 3 hours ago
Mark Viola
127k1172167
answered 3 hours ago
Mark Viola
127k1172167
127k1172167
Mateus, please let me know how I can improve my answer. I really want to give you the best answer I can.
– Mark Viola
3 hours ago
Why was this answer down voted??
– Mark Viola
3 hours ago
Mark, your answer is good actually. I gave you 1+
– Mateus Rocha
30 mins ago
add a comment |Â
Mateus, please let me know how I can improve my answer. I really want to give you the best answer I can.
– Mark Viola
3 hours ago
Why was this answer down voted??
– Mark Viola
3 hours ago
Mark, your answer is good actually. I gave you 1+
– Mateus Rocha
30 mins ago
Mateus, please let me know how I can improve my answer. I really want to give you the best answer I can.
– Mark Viola
3 hours ago
Mateus, please let me know how I can improve my answer. I really want to give you the best answer I can.
– Mark Viola
3 hours ago
Why was this answer down voted??
– Mark Viola
3 hours ago
Why was this answer down voted??
– Mark Viola
3 hours ago
Mark, your answer is good actually. I gave you 1+
– Mateus Rocha
30 mins ago
Mark, your answer is good actually. I gave you 1+
– Mateus Rocha
30 mins ago
add a comment |Â
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If you calculate the $c_n$ you can check that the derivative series does not converge uniformly---indeed, it does not converge at all.
– Driver 8
3 hours ago
@Driver8 Uniform convergence is a sufficient condition, not a necessary one. The series obtained by differentiating under the summation does not converge. Hence, the interchange is not legitimate.
– Mark Viola
3 hours ago
Yes, indeed. I will delete my (dumb) comment
– Driver 8
3 hours ago