Would sine trigonometric series $f(x)$ for approximating $g(x) = x$ always be $f(x) leq x$ for $0 leq x leq...
$begingroup$
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
$$f^{(2k+1)}(0) =0,,, k=1,..,n$$
$$f'(0) =1$$
$$f^{(2k-1)}(pi) =0,,, k=1,..,M-n-1$$
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 $n,M$?
(the last $f^{(2k-1)}(pi)$ condition does not exist when $M = n+1$ - and this is allowed as well. Also, $Mgeq 2$, $ngeq 1$.)
real-analysis linear-algebra interpolation trigonometric-series
asked Jan 8 at 13:06
Jacob MacherovJacob Macherov
212
$endgroup$
add a comment |
$begingroup$
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
$$f^{(2k+1)}(0) =0,,, k=1,..,n$$
$$f'(0) =1$$
$$f^{(2k-1)}(pi) =0,,, k=1,..,M-n-1$$
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 $n,M$?
(the last $f^{(2k-1)}(pi)$ condition does not exist when $M = n+1$ - and this is allowed as well. Also, $Mgeq 2$, $ngeq 1$.)
real-analysis linear-algebra interpolation trigonometric-series
asked Jan 8 at 13:06
Jacob MacherovJacob Macherov
212
$endgroup$
$begingroup$
I’m inclined to say yes because $sin’x=cos x$ is always $le 1$ on that interval
$endgroup$
– Chase Ryan Taylor
Jan 8 at 13:24
add a comment |
$begingroup$
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
$$f^{(2k+1)}(0) =0,,, k=1,..,n$$
$$f'(0) =1$$
$$f^{(2k-1)}(pi) =0,,, k=1,..,M-n-1$$
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 $n,M$?
(the last $f^{(2k-1)}(pi)$ condition does not exist when $M = n+1$ - and this is allowed as well. Also, $Mgeq 2$, $ngeq 1$.)
real-analysis linear-algebra interpolation trigonometric-series
asked Jan 8 at 13:06
Jacob MacherovJacob Macherov
212
$endgroup$
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
$$f^{(2k+1)}(0) =0,,, k=1,..,n$$
$$f'(0) =1$$
$$f^{(2k-1)}(pi) =0,,, k=1,..,M-n-1$$
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 $n,M$?
(the last $f^{(2k-1)}(pi)$ condition does not exist when $M = n+1$ - and this is allowed as well. Also, $Mgeq 2$, $ngeq 1$.)
real-analysis linear-algebra interpolation trigonometric-series
real-analysis linear-algebra interpolation trigonometric-series
asked Jan 8 at 13:06
Jacob MacherovJacob Macherov
212
asked Jan 8 at 13:06
Jacob MacherovJacob Macherov
212
edited Jan 8 at 13:23
Jacob Macherov
asked Jan 8 at 13:06
Jacob MacherovJacob Macherov
212
asked Jan 8 at 13:06
Jacob MacherovJacob Macherov
212
asked Jan 8 at 13:06
Jacob MacherovJacob Macherov
212
212
$begingroup$
I’m inclined to say yes because $sin’x=cos x$ is always $le 1$ on that interval
$endgroup$
– Chase Ryan Taylor
Jan 8 at 13:24
add a comment |
$begingroup$
I’m inclined to say yes because $sin’x=cos x$ is always $le 1$ on that interval
$endgroup$
– Chase Ryan Taylor
Jan 8 at 13:24
$begingroup$
I’m inclined to say yes because $sin’x=cos x$ is always $le 1$ on that interval
$endgroup$
– Chase Ryan Taylor
Jan 8 at 13:24
$begingroup$
I’m inclined to say yes because $sin’x=cos x$ is always $le 1$ on that interval
$endgroup$
– Chase Ryan Taylor
Jan 8 at 13:24
add a comment |
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$begingroup$
I’m inclined to say yes because $sin’x=cos x$ is always $le 1$ on that interval
$endgroup$
– Chase Ryan Taylor
Jan 8 at 13:24