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$f(x) = sum_n=1^infty sin (n x) / n^2 $ not differentiable at $x=0 $.
Is $sum_n=1^inftyfracsin(nx)n$ continuous?Uniform convergence of the series $sum_n=1^infty fraccos(2nt)4 n^2 - 1 $Convergence of Fourier series for a sum which is not uniform convergentIs f necessarily continuously differentiable under given conditions?Absolute convergence of fourier seriesConvergence of the Fourier series of a continuously differentiable functionIs $sum_n=1^infty fracsin(2^nx)2^n$ continuous? or differentiable anywhere?Disproving uniform convergence of $sum_n=1^infty frac2pi frac1n Big (1-cosBig(nfracpi2Big) Big)sin(nx)$Differentiability of the sum of $sum_n=1^inftyfracsin(nx)n^2$A proof of a result in Fourier analysis
$begingroup$
Apparently, by Weierstrass' test, this function is continuous. How to prove that it is not differentiable at $x=0$? I plotted its graph. It seems that
$$lim_xrightarrow 0 fracfx =infty . $$
How to prove it?
Term-by-term differentiation is not necessarily valid here.
fourier-analysis fourier-series
asked Mar 21 at 12:48
poissonpoisson
32018
$endgroup$
add a comment |
$begingroup$
Apparently, by Weierstrass' test, this function is continuous. How to prove that it is not differentiable at $x=0$? I plotted its graph. It seems that
$$lim_xrightarrow 0 fracfx =infty . $$
How to prove it?
Term-by-term differentiation is not necessarily valid here.
fourier-analysis fourier-series
asked Mar 21 at 12:48
poissonpoisson
32018
$endgroup$
add a comment |
$begingroup$
Apparently, by Weierstrass' test, this function is continuous. How to prove that it is not differentiable at $x=0$? I plotted its graph. It seems that
$$lim_xrightarrow 0 fracfx =infty . $$
How to prove it?
Term-by-term differentiation is not necessarily valid here.
fourier-analysis fourier-series
asked Mar 21 at 12:48
poissonpoisson
32018
$endgroup$
Apparently, by Weierstrass' test, this function is continuous. How to prove that it is not differentiable at $x=0$? I plotted its graph. It seems that
$$lim_xrightarrow 0 fracfx =infty . $$
How to prove it?
Term-by-term differentiation is not necessarily valid here.
fourier-analysis fourier-series
fourier-analysis fourier-series
asked Mar 21 at 12:48
poissonpoisson
32018
asked Mar 21 at 12:48
poissonpoisson
32018
asked Mar 21 at 12:48
poissonpoisson
32018
asked Mar 21 at 12:48
poissonpoisson
32018
asked Mar 21 at 12:48
poissonpoisson
32018
32018
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
One can show (by using Abel's summation and the Taylor series of the main branch of the logarithm) that
$$sum_n=1^infty fraccos(nx)n = -log(2sin(x/2))$$
for $ 0 < x < pi$ and the convergence is uniform on any interval $[a,b]$ with $0<a<b<pi$. Thus we find for $x>0$ that (by interchanging integration and summation)
$$fracf(x+h)-f(x)h = frac1h int_x^x+h -ln(2 sin(t/2)) dx.$$
One can check that the integrand on the right-hand side is absolute integrable, say on $[0,pi]$. Thus the the integral as a function of $x$ is contionous in $x=0$. We get
$$fracf(h)-f(0)h = frac1h int_0^h - ln(2 sin(t/2)) , dt.$$
Since $t/2 le 2 sin(t/2) le t$ for $h in [0,pi/2]$ and $x mapsto log(x)$ is monotone on $(0,1]$, we can compare the growth of the last integral with
$$frac1h int_0^h ln(t) , dt = ln(h)-1.$$
Thus
$$fracf(h)-f(0)h sim -ln(h)$$
for $h downarrow 0$.
answered Mar 21 at 13:14
p4schp4sch
5,425318
$endgroup$
$begingroup$
Thanks a lot! Is it possible to get the asymptotic behavior of f as $x rightarrow 0$?
$endgroup$
– poisson
Mar 21 at 23:23
1
$begingroup$
$-xlog x$ by integrating the series above and noting that near $0$, $2sinfracx2$~$x$
$endgroup$
– Conrad
Mar 22 at 1:24
add a comment |
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1 Answer
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1 Answer
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votes
$begingroup$
One can show (by using Abel's summation and the Taylor series of the main branch of the logarithm) that
$$sum_n=1^infty fraccos(nx)n = -log(2sin(x/2))$$
for $ 0 < x < pi$ and the convergence is uniform on any interval $[a,b]$ with $0<a<b<pi$. Thus we find for $x>0$ that (by interchanging integration and summation)
$$fracf(x+h)-f(x)h = frac1h int_x^x+h -ln(2 sin(t/2)) dx.$$
One can check that the integrand on the right-hand side is absolute integrable, say on $[0,pi]$. Thus the the integral as a function of $x$ is contionous in $x=0$. We get
$$fracf(h)-f(0)h = frac1h int_0^h - ln(2 sin(t/2)) , dt.$$
Since $t/2 le 2 sin(t/2) le t$ for $h in [0,pi/2]$ and $x mapsto log(x)$ is monotone on $(0,1]$, we can compare the growth of the last integral with
$$frac1h int_0^h ln(t) , dt = ln(h)-1.$$
Thus
$$fracf(h)-f(0)h sim -ln(h)$$
for $h downarrow 0$.
answered Mar 21 at 13:14
p4schp4sch
5,425318
$endgroup$
$begingroup$
Thanks a lot! Is it possible to get the asymptotic behavior of f as $x rightarrow 0$?
$endgroup$
– poisson
Mar 21 at 23:23
1
$begingroup$
$-xlog x$ by integrating the series above and noting that near $0$, $2sinfracx2$~$x$
$endgroup$
– Conrad
Mar 22 at 1:24
add a comment |
$begingroup$
One can show (by using Abel's summation and the Taylor series of the main branch of the logarithm) that
$$sum_n=1^infty fraccos(nx)n = -log(2sin(x/2))$$
for $ 0 < x < pi$ and the convergence is uniform on any interval $[a,b]$ with $0<a<b<pi$. Thus we find for $x>0$ that (by interchanging integration and summation)
$$fracf(x+h)-f(x)h = frac1h int_x^x+h -ln(2 sin(t/2)) dx.$$
One can check that the integrand on the right-hand side is absolute integrable, say on $[0,pi]$. Thus the the integral as a function of $x$ is contionous in $x=0$. We get
$$fracf(h)-f(0)h = frac1h int_0^h - ln(2 sin(t/2)) , dt.$$
Since $t/2 le 2 sin(t/2) le t$ for $h in [0,pi/2]$ and $x mapsto log(x)$ is monotone on $(0,1]$, we can compare the growth of the last integral with
$$frac1h int_0^h ln(t) , dt = ln(h)-1.$$
Thus
$$fracf(h)-f(0)h sim -ln(h)$$
for $h downarrow 0$.
answered Mar 21 at 13:14
p4schp4sch
5,425318
$endgroup$
$begingroup$
Thanks a lot! Is it possible to get the asymptotic behavior of f as $x rightarrow 0$?
$endgroup$
– poisson
Mar 21 at 23:23
1
$begingroup$
$-xlog x$ by integrating the series above and noting that near $0$, $2sinfracx2$~$x$
$endgroup$
– Conrad
Mar 22 at 1:24
add a comment |
$begingroup$
One can show (by using Abel's summation and the Taylor series of the main branch of the logarithm) that
$$sum_n=1^infty fraccos(nx)n = -log(2sin(x/2))$$
for $ 0 < x < pi$ and the convergence is uniform on any interval $[a,b]$ with $0<a<b<pi$. Thus we find for $x>0$ that (by interchanging integration and summation)
$$fracf(x+h)-f(x)h = frac1h int_x^x+h -ln(2 sin(t/2)) dx.$$
One can check that the integrand on the right-hand side is absolute integrable, say on $[0,pi]$. Thus the the integral as a function of $x$ is contionous in $x=0$. We get
$$fracf(h)-f(0)h = frac1h int_0^h - ln(2 sin(t/2)) , dt.$$
Since $t/2 le 2 sin(t/2) le t$ for $h in [0,pi/2]$ and $x mapsto log(x)$ is monotone on $(0,1]$, we can compare the growth of the last integral with
$$frac1h int_0^h ln(t) , dt = ln(h)-1.$$
Thus
$$fracf(h)-f(0)h sim -ln(h)$$
for $h downarrow 0$.
answered Mar 21 at 13:14
p4schp4sch
5,425318
$endgroup$
One can show (by using Abel's summation and the Taylor series of the main branch of the logarithm) that
$$sum_n=1^infty fraccos(nx)n = -log(2sin(x/2))$$
for $ 0 < x < pi$ and the convergence is uniform on any interval $[a,b]$ with $0<a<b<pi$. Thus we find for $x>0$ that (by interchanging integration and summation)
$$fracf(x+h)-f(x)h = frac1h int_x^x+h -ln(2 sin(t/2)) dx.$$
One can check that the integrand on the right-hand side is absolute integrable, say on $[0,pi]$. Thus the the integral as a function of $x$ is contionous in $x=0$. We get
$$fracf(h)-f(0)h = frac1h int_0^h - ln(2 sin(t/2)) , dt.$$
Since $t/2 le 2 sin(t/2) le t$ for $h in [0,pi/2]$ and $x mapsto log(x)$ is monotone on $(0,1]$, we can compare the growth of the last integral with
$$frac1h int_0^h ln(t) , dt = ln(h)-1.$$
Thus
$$fracf(h)-f(0)h sim -ln(h)$$
for $h downarrow 0$.
answered Mar 21 at 13:14
p4schp4sch
5,425318
answered Mar 21 at 13:14
p4schp4sch
5,425318
answered Mar 21 at 13:14
p4schp4sch
5,425318
answered Mar 21 at 13:14
p4schp4sch
5,425318
5,425318
$begingroup$
Thanks a lot! Is it possible to get the asymptotic behavior of f as $x rightarrow 0$?
$endgroup$
– poisson
Mar 21 at 23:23
1
$begingroup$
$-xlog x$ by integrating the series above and noting that near $0$, $2sinfracx2$~$x$
$endgroup$
– Conrad
Mar 22 at 1:24
add a comment |
$begingroup$
Thanks a lot! Is it possible to get the asymptotic behavior of f as $x rightarrow 0$?
$endgroup$
– poisson
Mar 21 at 23:23
1
$begingroup$
$-xlog x$ by integrating the series above and noting that near $0$, $2sinfracx2$~$x$
$endgroup$
– Conrad
Mar 22 at 1:24
$begingroup$
Thanks a lot! Is it possible to get the asymptotic behavior of f as $x rightarrow 0$?
$endgroup$
– poisson
Mar 21 at 23:23
$begingroup$
Thanks a lot! Is it possible to get the asymptotic behavior of f as $x rightarrow 0$?
$endgroup$
– poisson
Mar 21 at 23:23
1
1
$begingroup$
$-xlog x$ by integrating the series above and noting that near $0$, $2sinfracx2$~$x$
$endgroup$
– Conrad
Mar 22 at 1:24
$begingroup$
$-xlog x$ by integrating the series above and noting that near $0$, $2sinfracx2$~$x$
$endgroup$
– Conrad
Mar 22 at 1:24
add a comment |
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