Estimating Maximum of Truncated Fourier Series
$begingroup$
Let $L$ be an integer. Is there any good way to estimate the $L^{infty}$ norm of
$$f(x)=a_0+sum_{n=1}^{2L}Big(a_ncosBig(frac{npi x}{L}Big)+b_nsinBig(frac{npi x}{L}Big)Big)?$$
The trivial way would be to use the triangle inequality and the fact that $|cos(x)|, |sin(x)|le 1$ to get
$$|f|_{infty}le |a_0|+sum_{n=1}^{2L}(|a_n|+|b_n|).$$
One can do slightly better with Holder to get
$$begin{aligned}
|f|_{infty} &le |a_0|+sup_{xinmathbb{R}}left(sum_{n=1}^{2L}a_n^2sum_{n=1}^{2L}cos^2Big(frac{npi x}{L}Big)right)^{1/2}+sup_{xinmathbb{R}}left(sum_{n=1}^{2L}b_n^2sum_{n=1}^{2L}sin^2Big(frac{npi x}{L}Big)right)^{1/2} \
&le |a_0|+sqrt{2L}left(sum_{n=1}^{2L}a_n^2right)^{1/2}+sqrt{2L}left(sum_{n=1}^{2L}b_n^2right)^{1/2}.
end{aligned}$$
Is there a way to do better?
By differentiating we also know that at whichever $x_0$ the maximum occurs, we'll have
$$sum_{n=1}^{2L}a_nsinBig(frac{npi x_0}{L}Big)=sum_{n=1}^{2L}b_ncosBig(frac{npi x_0}{L}Big).$$
calculus real-analysis fourier-analysis
asked Dec 1 '18 at 19:09
Will FisherWill Fisher
4,0231032
$endgroup$
add a comment |
$begingroup$
Let $L$ be an integer. Is there any good way to estimate the $L^{infty}$ norm of
$$f(x)=a_0+sum_{n=1}^{2L}Big(a_ncosBig(frac{npi x}{L}Big)+b_nsinBig(frac{npi x}{L}Big)Big)?$$
The trivial way would be to use the triangle inequality and the fact that $|cos(x)|, |sin(x)|le 1$ to get
$$|f|_{infty}le |a_0|+sum_{n=1}^{2L}(|a_n|+|b_n|).$$
One can do slightly better with Holder to get
$$begin{aligned}
|f|_{infty} &le |a_0|+sup_{xinmathbb{R}}left(sum_{n=1}^{2L}a_n^2sum_{n=1}^{2L}cos^2Big(frac{npi x}{L}Big)right)^{1/2}+sup_{xinmathbb{R}}left(sum_{n=1}^{2L}b_n^2sum_{n=1}^{2L}sin^2Big(frac{npi x}{L}Big)right)^{1/2} \
&le |a_0|+sqrt{2L}left(sum_{n=1}^{2L}a_n^2right)^{1/2}+sqrt{2L}left(sum_{n=1}^{2L}b_n^2right)^{1/2}.
end{aligned}$$
Is there a way to do better?
By differentiating we also know that at whichever $x_0$ the maximum occurs, we'll have
$$sum_{n=1}^{2L}a_nsinBig(frac{npi x_0}{L}Big)=sum_{n=1}^{2L}b_ncosBig(frac{npi x_0}{L}Big).$$
calculus real-analysis fourier-analysis
asked Dec 1 '18 at 19:09
Will FisherWill Fisher
4,0231032
$endgroup$
add a comment |
$begingroup$
Let $L$ be an integer. Is there any good way to estimate the $L^{infty}$ norm of
$$f(x)=a_0+sum_{n=1}^{2L}Big(a_ncosBig(frac{npi x}{L}Big)+b_nsinBig(frac{npi x}{L}Big)Big)?$$
The trivial way would be to use the triangle inequality and the fact that $|cos(x)|, |sin(x)|le 1$ to get
$$|f|_{infty}le |a_0|+sum_{n=1}^{2L}(|a_n|+|b_n|).$$
One can do slightly better with Holder to get
$$begin{aligned}
|f|_{infty} &le |a_0|+sup_{xinmathbb{R}}left(sum_{n=1}^{2L}a_n^2sum_{n=1}^{2L}cos^2Big(frac{npi x}{L}Big)right)^{1/2}+sup_{xinmathbb{R}}left(sum_{n=1}^{2L}b_n^2sum_{n=1}^{2L}sin^2Big(frac{npi x}{L}Big)right)^{1/2} \
&le |a_0|+sqrt{2L}left(sum_{n=1}^{2L}a_n^2right)^{1/2}+sqrt{2L}left(sum_{n=1}^{2L}b_n^2right)^{1/2}.
end{aligned}$$
Is there a way to do better?
By differentiating we also know that at whichever $x_0$ the maximum occurs, we'll have
$$sum_{n=1}^{2L}a_nsinBig(frac{npi x_0}{L}Big)=sum_{n=1}^{2L}b_ncosBig(frac{npi x_0}{L}Big).$$
calculus real-analysis fourier-analysis
asked Dec 1 '18 at 19:09
Will FisherWill Fisher
4,0231032
$endgroup$
Let $L$ be an integer. Is there any good way to estimate the $L^{infty}$ norm of
$$f(x)=a_0+sum_{n=1}^{2L}Big(a_ncosBig(frac{npi x}{L}Big)+b_nsinBig(frac{npi x}{L}Big)Big)?$$
The trivial way would be to use the triangle inequality and the fact that $|cos(x)|, |sin(x)|le 1$ to get
$$|f|_{infty}le |a_0|+sum_{n=1}^{2L}(|a_n|+|b_n|).$$
One can do slightly better with Holder to get
$$begin{aligned}
|f|_{infty} &le |a_0|+sup_{xinmathbb{R}}left(sum_{n=1}^{2L}a_n^2sum_{n=1}^{2L}cos^2Big(frac{npi x}{L}Big)right)^{1/2}+sup_{xinmathbb{R}}left(sum_{n=1}^{2L}b_n^2sum_{n=1}^{2L}sin^2Big(frac{npi x}{L}Big)right)^{1/2} \
&le |a_0|+sqrt{2L}left(sum_{n=1}^{2L}a_n^2right)^{1/2}+sqrt{2L}left(sum_{n=1}^{2L}b_n^2right)^{1/2}.
end{aligned}$$
Is there a way to do better?
By differentiating we also know that at whichever $x_0$ the maximum occurs, we'll have
$$sum_{n=1}^{2L}a_nsinBig(frac{npi x_0}{L}Big)=sum_{n=1}^{2L}b_ncosBig(frac{npi x_0}{L}Big).$$
calculus real-analysis fourier-analysis
calculus real-analysis fourier-analysis
asked Dec 1 '18 at 19:09
Will FisherWill Fisher
4,0231032
asked Dec 1 '18 at 19:09
Will FisherWill Fisher
4,0231032
edited Dec 1 '18 at 19:18
Will Fisher
asked Dec 1 '18 at 19:09
Will FisherWill Fisher
4,0231032
asked Dec 1 '18 at 19:09
Will FisherWill Fisher
4,0231032
asked Dec 1 '18 at 19:09
Will FisherWill Fisher
4,0231032
4,0231032
add a comment |
add a comment |
1 Answer
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No, in general there's no "reasonable" way to get a "good" estimate on $||f||_infty$ from the Fourier coefficients. If there were a lot of things would be much simpler.
That trivial estimate is very simple, but it's typically a very bad estimate. You should note that your "better" estimate is actually never any better and is usually much worse, since $$sum_{j=1}^{2L}|c_j|lesqrt{2L}left(sum_{j=1}^{2L}|c_j|^2right)^{1/2}.$$
answered Dec 2 '18 at 19:15
David C. UllrichDavid C. Ullrich
59.5k43893
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
No, in general there's no "reasonable" way to get a "good" estimate on $||f||_infty$ from the Fourier coefficients. If there were a lot of things would be much simpler.
That trivial estimate is very simple, but it's typically a very bad estimate. You should note that your "better" estimate is actually never any better and is usually much worse, since $$sum_{j=1}^{2L}|c_j|lesqrt{2L}left(sum_{j=1}^{2L}|c_j|^2right)^{1/2}.$$
answered Dec 2 '18 at 19:15
David C. UllrichDavid C. Ullrich
59.5k43893
$endgroup$
add a comment |
$begingroup$
No, in general there's no "reasonable" way to get a "good" estimate on $||f||_infty$ from the Fourier coefficients. If there were a lot of things would be much simpler.
That trivial estimate is very simple, but it's typically a very bad estimate. You should note that your "better" estimate is actually never any better and is usually much worse, since $$sum_{j=1}^{2L}|c_j|lesqrt{2L}left(sum_{j=1}^{2L}|c_j|^2right)^{1/2}.$$
answered Dec 2 '18 at 19:15
David C. UllrichDavid C. Ullrich
59.5k43893
$endgroup$
add a comment |
$begingroup$
No, in general there's no "reasonable" way to get a "good" estimate on $||f||_infty$ from the Fourier coefficients. If there were a lot of things would be much simpler.
That trivial estimate is very simple, but it's typically a very bad estimate. You should note that your "better" estimate is actually never any better and is usually much worse, since $$sum_{j=1}^{2L}|c_j|lesqrt{2L}left(sum_{j=1}^{2L}|c_j|^2right)^{1/2}.$$
answered Dec 2 '18 at 19:15
David C. UllrichDavid C. Ullrich
59.5k43893
$endgroup$
No, in general there's no "reasonable" way to get a "good" estimate on $||f||_infty$ from the Fourier coefficients. If there were a lot of things would be much simpler.
That trivial estimate is very simple, but it's typically a very bad estimate. You should note that your "better" estimate is actually never any better and is usually much worse, since $$sum_{j=1}^{2L}|c_j|lesqrt{2L}left(sum_{j=1}^{2L}|c_j|^2right)^{1/2}.$$
answered Dec 2 '18 at 19:15
David C. UllrichDavid C. Ullrich
59.5k43893
answered Dec 2 '18 at 19:15
David C. UllrichDavid C. Ullrich
59.5k43893
answered Dec 2 '18 at 19:15
David C. UllrichDavid C. Ullrich
59.5k43893
answered Dec 2 '18 at 19:15
David C. UllrichDavid C. Ullrich
59.5k43893
59.5k43893
add a comment |
add a comment |
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