Is $f(x) leq x$ for $0 leq x leq pi$ when sine series $f(x)$ are used to approximate $x$ based on derivatives...
$begingroup$
This is a simpler "cousin" question to Would sine trigonometric series $f(x)$ for approximating $g(x) = x$ always be $f(x) leq x$ for $0 leq x leq pi$? . I am asking this as a separate question, since this may be relatively easier to answer than the linked question.
I am trying to use set of $c_k$ with $k in mathbb{N}$ such that $f(x) = sum_{k=1}^{M} c_k sin kx approx x$.
$c_k$ is determined by setting
$$f^{(2k+1)}(0) =0,,, k=1,..,M-1$$
$$f'(0) =1$$
These conditions provide $M$ linear equations to solve for $c_k$.
Question is, is it guaranteed that $f(x) leq x$ for $0 leq x leq pi$ for any $Mgeq 2$?
real-analysis linear-algebra interpolation trigonometric-series
asked Jan 9 at 15:11
Jacob MacherovJacob Macherov
212
$endgroup$
add a comment |
$begingroup$
This is a simpler "cousin" question to Would sine trigonometric series $f(x)$ for approximating $g(x) = x$ always be $f(x) leq x$ for $0 leq x leq pi$? . I am asking this as a separate question, since this may be relatively easier to answer than the linked question.
I am trying to use set of $c_k$ with $k in mathbb{N}$ such that $f(x) = sum_{k=1}^{M} c_k sin kx approx x$.
$c_k$ is determined by setting
$$f^{(2k+1)}(0) =0,,, k=1,..,M-1$$
$$f'(0) =1$$
These conditions provide $M$ linear equations to solve for $c_k$.
Question is, is it guaranteed that $f(x) leq x$ for $0 leq x leq pi$ for any $Mgeq 2$?
real-analysis linear-algebra interpolation trigonometric-series
asked Jan 9 at 15:11
Jacob MacherovJacob Macherov
212
$endgroup$
add a comment |
$begingroup$
This is a simpler "cousin" question to Would sine trigonometric series $f(x)$ for approximating $g(x) = x$ always be $f(x) leq x$ for $0 leq x leq pi$? . I am asking this as a separate question, since this may be relatively easier to answer than the linked question.
I am trying to use set of $c_k$ with $k in mathbb{N}$ such that $f(x) = sum_{k=1}^{M} c_k sin kx approx x$.
$c_k$ is determined by setting
$$f^{(2k+1)}(0) =0,,, k=1,..,M-1$$
$$f'(0) =1$$
These conditions provide $M$ linear equations to solve for $c_k$.
Question is, is it guaranteed that $f(x) leq x$ for $0 leq x leq pi$ for any $Mgeq 2$?
real-analysis linear-algebra interpolation trigonometric-series
asked Jan 9 at 15:11
Jacob MacherovJacob Macherov
212
$endgroup$
This is a simpler "cousin" question to Would sine trigonometric series $f(x)$ for approximating $g(x) = x$ always be $f(x) leq x$ for $0 leq x leq pi$? . I am asking this as a separate question, since this may be relatively easier to answer than the linked question.
I am trying to use set of $c_k$ with $k in mathbb{N}$ such that $f(x) = sum_{k=1}^{M} c_k sin kx approx x$.
$c_k$ is determined by setting
$$f^{(2k+1)}(0) =0,,, k=1,..,M-1$$
$$f'(0) =1$$
These conditions provide $M$ linear equations to solve for $c_k$.
Question is, is it guaranteed that $f(x) leq x$ for $0 leq x leq pi$ for any $Mgeq 2$?
real-analysis linear-algebra interpolation trigonometric-series
real-analysis linear-algebra interpolation trigonometric-series
asked Jan 9 at 15:11
Jacob MacherovJacob Macherov
212
asked Jan 9 at 15:11
Jacob MacherovJacob Macherov
212
asked Jan 9 at 15:11
Jacob MacherovJacob Macherov
212
asked Jan 9 at 15:11
Jacob MacherovJacob Macherov
212
asked Jan 9 at 15:11
Jacob MacherovJacob Macherov
212
212
add a comment |
add a comment |
1 Answer
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$begingroup$
The matrix of your system is
$$
begin{bmatrix}
1&1&1&cdots&1\
1&2^3&2^5&cdots&2^{2M-1}\
1&3^3&3^5&cdots&2^{2M-1}\
vdots&vdots&vdots&ddots&vdots\
1&M^3&M^5&cdots&M^{2M-1}.
end{bmatrix}
$$
It is not hard to check that this determinant is never zero, so your system has unique solution. That means that $c_1=1$, $c_2=cdots=c_M=0$ is the only solution, and thus
$$
f(x)=sin x
$$
for all $M$.
answered Jan 31 at 18:27
Martin ArgeramiMartin Argerami
129k1184185
$endgroup$
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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oldest
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votes
$begingroup$
The matrix of your system is
$$
begin{bmatrix}
1&1&1&cdots&1\
1&2^3&2^5&cdots&2^{2M-1}\
1&3^3&3^5&cdots&2^{2M-1}\
vdots&vdots&vdots&ddots&vdots\
1&M^3&M^5&cdots&M^{2M-1}.
end{bmatrix}
$$
It is not hard to check that this determinant is never zero, so your system has unique solution. That means that $c_1=1$, $c_2=cdots=c_M=0$ is the only solution, and thus
$$
f(x)=sin x
$$
for all $M$.
answered Jan 31 at 18:27
Martin ArgeramiMartin Argerami
129k1184185
$endgroup$
add a comment |
$begingroup$
The matrix of your system is
$$
begin{bmatrix}
1&1&1&cdots&1\
1&2^3&2^5&cdots&2^{2M-1}\
1&3^3&3^5&cdots&2^{2M-1}\
vdots&vdots&vdots&ddots&vdots\
1&M^3&M^5&cdots&M^{2M-1}.
end{bmatrix}
$$
It is not hard to check that this determinant is never zero, so your system has unique solution. That means that $c_1=1$, $c_2=cdots=c_M=0$ is the only solution, and thus
$$
f(x)=sin x
$$
for all $M$.
answered Jan 31 at 18:27
Martin ArgeramiMartin Argerami
129k1184185
$endgroup$
add a comment |
$begingroup$
The matrix of your system is
$$
begin{bmatrix}
1&1&1&cdots&1\
1&2^3&2^5&cdots&2^{2M-1}\
1&3^3&3^5&cdots&2^{2M-1}\
vdots&vdots&vdots&ddots&vdots\
1&M^3&M^5&cdots&M^{2M-1}.
end{bmatrix}
$$
It is not hard to check that this determinant is never zero, so your system has unique solution. That means that $c_1=1$, $c_2=cdots=c_M=0$ is the only solution, and thus
$$
f(x)=sin x
$$
for all $M$.
answered Jan 31 at 18:27
Martin ArgeramiMartin Argerami
129k1184185
$endgroup$
The matrix of your system is
$$
begin{bmatrix}
1&1&1&cdots&1\
1&2^3&2^5&cdots&2^{2M-1}\
1&3^3&3^5&cdots&2^{2M-1}\
vdots&vdots&vdots&ddots&vdots\
1&M^3&M^5&cdots&M^{2M-1}.
end{bmatrix}
$$
It is not hard to check that this determinant is never zero, so your system has unique solution. That means that $c_1=1$, $c_2=cdots=c_M=0$ is the only solution, and thus
$$
f(x)=sin x
$$
for all $M$.
answered Jan 31 at 18:27
Martin ArgeramiMartin Argerami
129k1184185
answered Jan 31 at 18:27
Martin ArgeramiMartin Argerami
129k1184185
answered Jan 31 at 18:27
Martin ArgeramiMartin Argerami
129k1184185
answered Jan 31 at 18:27
Martin ArgeramiMartin Argerami
129k1184185
129k1184185
add a comment |
add a comment |
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