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Differentiability of Fourier series
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Consider the function defined by the Fourier series
$$ f(x;alpha) = sum_n=1^infty frac1n^alpha exp(i n^2 x ) , $$
where $alpha >1 $.
For what values of $alpha $ is $f$ differentiable? Based on numerics, it is conjectured that $alpha = 2$ is a critical value. For $alpha <2 $, the function is nowhere differentiable; while for $alpha >2 $, the function is differentiable almost everywhere.
fourier-analysis harmonic-analysis
edited 23 mins ago
Josiah Park
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up vote
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Consider the function defined by the Fourier series
$$ f(x;alpha) = sum_n=1^infty frac1n^alpha exp(i n^2 x ) , $$
where $alpha >1 $.
For what values of $alpha $ is $f$ differentiable? Based on numerics, it is conjectured that $alpha = 2$ is a critical value. For $alpha <2 $, the function is nowhere differentiable; while for $alpha >2 $, the function is differentiable almost everywhere.
fourier-analysis harmonic-analysis
edited 23 mins ago
Josiah Park
3789
asked 2 hours ago
pie
111
New contributor
pie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
The $alpha$th power of the popcorn function has similar differentiability properties, essentially by Thue–Siegel–Roth. I can imagine your result having something to do with rational approximation (although that's very far from a proof).
– LSpice
2 hours ago
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up vote
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up vote
2
down vote
favorite
Consider the function defined by the Fourier series
$$ f(x;alpha) = sum_n=1^infty frac1n^alpha exp(i n^2 x ) , $$
where $alpha >1 $.
For what values of $alpha $ is $f$ differentiable? Based on numerics, it is conjectured that $alpha = 2$ is a critical value. For $alpha <2 $, the function is nowhere differentiable; while for $alpha >2 $, the function is differentiable almost everywhere.
fourier-analysis harmonic-analysis
edited 23 mins ago
Josiah Park
3789
asked 2 hours ago
pie
111
New contributor
pie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Consider the function defined by the Fourier series
$$ f(x;alpha) = sum_n=1^infty frac1n^alpha exp(i n^2 x ) , $$
where $alpha >1 $.
For what values of $alpha $ is $f$ differentiable? Based on numerics, it is conjectured that $alpha = 2$ is a critical value. For $alpha <2 $, the function is nowhere differentiable; while for $alpha >2 $, the function is differentiable almost everywhere.
fourier-analysis harmonic-analysis
fourier-analysis harmonic-analysis
edited 23 mins ago
Josiah Park
3789
asked 2 hours ago
pie
111
New contributor
pie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 23 mins ago
Josiah Park
3789
asked 2 hours ago
pie
111
New contributor
pie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 23 mins ago
Josiah Park
3789
edited 23 mins ago
Josiah Park
3789
edited 23 mins ago
Josiah Park
3789
3789
asked 2 hours ago
pie
111
New contributor
pie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
pie
111
asked 2 hours ago
pie
111
111
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pie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
pie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
pie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
The $alpha$th power of the popcorn function has similar differentiability properties, essentially by Thue–Siegel–Roth. I can imagine your result having something to do with rational approximation (although that's very far from a proof).
– LSpice
2 hours ago
add a comment |Â
The $alpha$th power of the popcorn function has similar differentiability properties, essentially by Thue–Siegel–Roth. I can imagine your result having something to do with rational approximation (although that's very far from a proof).
– LSpice
2 hours ago
The $alpha$th power of the popcorn function has similar differentiability properties, essentially by Thue–Siegel–Roth. I can imagine your result having something to do with rational approximation (although that's very far from a proof).
– LSpice
2 hours ago
The $alpha$th power of the popcorn function has similar differentiability properties, essentially by Thue–Siegel–Roth. I can imagine your result having something to do with rational approximation (although that's very far from a proof).
– LSpice
2 hours ago
add a comment |Â
1 Answer
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The fact that $f(x,alpha)$ is everywhere non-differentiable for $alpha<2$ follows by Theorem 2.1 here but earlier versions known to Freud and Hardy suffice
Theorem [Hardy]: An integrable periodic function $f$ with Fourier series
$sum a_k sin(n_k x)$, satisfying $inflimits_kfracn_k+1n_k > 1$ is differentiable at a point only if $limlimits_krightarrowinfty a_kn_k = 0$.
For the question at the endpoint one has a mix of non-differentiability and differentiability results known, one interesting answer being the following of Gerver
Theorem [Gerver, 1970]: $f(x)=sumlimits_n=1^infty fracsinn^2 xn^2$ is differentiable at points $kpi$ if $k=frac2p+12q+1$, $p,qinmathbbZ$.
answered 45 mins ago
Josiah Park
3789
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
The fact that $f(x,alpha)$ is everywhere non-differentiable for $alpha<2$ follows by Theorem 2.1 here but earlier versions known to Freud and Hardy suffice
Theorem [Hardy]: An integrable periodic function $f$ with Fourier series
$sum a_k sin(n_k x)$, satisfying $inflimits_kfracn_k+1n_k > 1$ is differentiable at a point only if $limlimits_krightarrowinfty a_kn_k = 0$.
For the question at the endpoint one has a mix of non-differentiability and differentiability results known, one interesting answer being the following of Gerver
Theorem [Gerver, 1970]: $f(x)=sumlimits_n=1^infty fracsinn^2 xn^2$ is differentiable at points $kpi$ if $k=frac2p+12q+1$, $p,qinmathbbZ$.
answered 45 mins ago
Josiah Park
3789
add a comment |Â
up vote
3
down vote
The fact that $f(x,alpha)$ is everywhere non-differentiable for $alpha<2$ follows by Theorem 2.1 here but earlier versions known to Freud and Hardy suffice
Theorem [Hardy]: An integrable periodic function $f$ with Fourier series
$sum a_k sin(n_k x)$, satisfying $inflimits_kfracn_k+1n_k > 1$ is differentiable at a point only if $limlimits_krightarrowinfty a_kn_k = 0$.
For the question at the endpoint one has a mix of non-differentiability and differentiability results known, one interesting answer being the following of Gerver
Theorem [Gerver, 1970]: $f(x)=sumlimits_n=1^infty fracsinn^2 xn^2$ is differentiable at points $kpi$ if $k=frac2p+12q+1$, $p,qinmathbbZ$.
answered 45 mins ago
Josiah Park
3789
add a comment |Â
up vote
3
down vote
up vote
3
down vote
The fact that $f(x,alpha)$ is everywhere non-differentiable for $alpha<2$ follows by Theorem 2.1 here but earlier versions known to Freud and Hardy suffice
Theorem [Hardy]: An integrable periodic function $f$ with Fourier series
$sum a_k sin(n_k x)$, satisfying $inflimits_kfracn_k+1n_k > 1$ is differentiable at a point only if $limlimits_krightarrowinfty a_kn_k = 0$.
For the question at the endpoint one has a mix of non-differentiability and differentiability results known, one interesting answer being the following of Gerver
Theorem [Gerver, 1970]: $f(x)=sumlimits_n=1^infty fracsinn^2 xn^2$ is differentiable at points $kpi$ if $k=frac2p+12q+1$, $p,qinmathbbZ$.
answered 45 mins ago
Josiah Park
3789
The fact that $f(x,alpha)$ is everywhere non-differentiable for $alpha<2$ follows by Theorem 2.1 here but earlier versions known to Freud and Hardy suffice
Theorem [Hardy]: An integrable periodic function $f$ with Fourier series
$sum a_k sin(n_k x)$, satisfying $inflimits_kfracn_k+1n_k > 1$ is differentiable at a point only if $limlimits_krightarrowinfty a_kn_k = 0$.
For the question at the endpoint one has a mix of non-differentiability and differentiability results known, one interesting answer being the following of Gerver
Theorem [Gerver, 1970]: $f(x)=sumlimits_n=1^infty fracsinn^2 xn^2$ is differentiable at points $kpi$ if $k=frac2p+12q+1$, $p,qinmathbbZ$.
answered 45 mins ago
Josiah Park
3789
answered 45 mins ago
Josiah Park
3789
answered 45 mins ago
Josiah Park
3789
answered 45 mins ago
Josiah Park
3789
3789
add a comment |Â
add a comment |Â
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The $alpha$th power of the popcorn function has similar differentiability properties, essentially by Thue–Siegel–Roth. I can imagine your result having something to do with rational approximation (although that's very far from a proof).
– LSpice
2 hours ago