Maximum of the leading coefficient when $text{deg}P=6$, $0leq P(x)leq1$ for $-1leq xleq 1$.
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Let $P(x)$ be a real polynomial of degree 6 with the following property:
For all $-1leq xleq 1$, we have $0leq P(x)leq 1$.
what is the maximum possible value of the leading coefficient of $P(x)$?
I obtained obvious answers when the degree is 1 or 2, but couldn't solve when the degree is 6.
I have a hunch that this has something to do with the Legendre polynomials, but I'm not quite sure. Any advice is welcome.
inequality polynomials real-numbers
asked Dec 31 '18 at 8:23
DilemianDilemian
454514
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add a comment |
$begingroup$
Let $P(x)$ be a real polynomial of degree 6 with the following property:
For all $-1leq xleq 1$, we have $0leq P(x)leq 1$.
what is the maximum possible value of the leading coefficient of $P(x)$?
I obtained obvious answers when the degree is 1 or 2, but couldn't solve when the degree is 6.
I have a hunch that this has something to do with the Legendre polynomials, but I'm not quite sure. Any advice is welcome.
inequality polynomials real-numbers
asked Dec 31 '18 at 8:23
DilemianDilemian
454514
$endgroup$
add a comment |
$begingroup$
Let $P(x)$ be a real polynomial of degree 6 with the following property:
For all $-1leq xleq 1$, we have $0leq P(x)leq 1$.
what is the maximum possible value of the leading coefficient of $P(x)$?
I obtained obvious answers when the degree is 1 or 2, but couldn't solve when the degree is 6.
I have a hunch that this has something to do with the Legendre polynomials, but I'm not quite sure. Any advice is welcome.
inequality polynomials real-numbers
asked Dec 31 '18 at 8:23
DilemianDilemian
454514
$endgroup$
Let $P(x)$ be a real polynomial of degree 6 with the following property:
For all $-1leq xleq 1$, we have $0leq P(x)leq 1$.
what is the maximum possible value of the leading coefficient of $P(x)$?
I obtained obvious answers when the degree is 1 or 2, but couldn't solve when the degree is 6.
I have a hunch that this has something to do with the Legendre polynomials, but I'm not quite sure. Any advice is welcome.
inequality polynomials real-numbers
inequality polynomials real-numbers
asked Dec 31 '18 at 8:23
DilemianDilemian
454514
asked Dec 31 '18 at 8:23
DilemianDilemian
454514
asked Dec 31 '18 at 8:23
DilemianDilemian
454514
asked Dec 31 '18 at 8:23
DilemianDilemian
454514
asked Dec 31 '18 at 8:23
DilemianDilemian
454514
454514
add a comment |
add a comment |
2 Answers
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Let $T(x)=cos 6theta$ $(xin[-1,1])$ where $x=costheta$. The leading coefficient of $T$ is $2^{6-1}$ The maximal difference to $0$ occurs at $theta_k=kpi/6,k=0,1,ldots,6$. Let $x_k=costheta_k$, when $k$ is even, $T(x_k)=1$, when $k$ is odd, $T(x_k)=-1$. Suppose $deg R=6$ and the leading coefficient of $R$ is also $2^5$, if $max|R|<1$, at $x_k$, we have $T(x_0)-R(x_0)>0$, $T(x_1)-R(x_1)<0$ ..., it changes sign at least $6$ times, contradicts with $deg T-R<6$. Hence for all polynomial $Q$ s.t. $deg Q=6$ and the leading coefficient is $2^{6-1}$,
the minimal of $max|Q|$, which is $1$, occurs when $Q(x)=cos6arccos x$.
Now, denote $L(x)=frac12(Q(x)+1)/max|Q|$, we can see the maximal leading coefficient of $L$ is $2^4=16$.
answered Dec 31 '18 at 8:45
Kemono ChenKemono Chen
3,3401844
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Hint
The value you’re looking for is the same as the one of the problem ... For all $-1 le x le 1$, we have $-1/2 le Q(x) le 1/2$... as we’re just substracting $1/2$ to the polynomial which doesn’t change the leading coefficient.
Now the map $f(x) = frac{1}{2^{6-1}}T_6(x)$ where $T_n$ is the n-th Chebyshev polynomials is the one having $1$ for leading coefficient and of minimal $infty$-norm (see Wikipedia article for more details). And this norm is equal to $Vert f Vert_infty =frac{1}{2^{6-1}}$. In your problem you allow the norm to be equal to $1/2$.
Hence the value you’re looking for is equal to $frac{1/2}{1/2^{6-1}}=2^4=16$.
answered Dec 31 '18 at 8:47
mathcounterexamples.netmathcounterexamples.net
26.9k22158
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2 Answers
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2 Answers
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$begingroup$
Let $T(x)=cos 6theta$ $(xin[-1,1])$ where $x=costheta$. The leading coefficient of $T$ is $2^{6-1}$ The maximal difference to $0$ occurs at $theta_k=kpi/6,k=0,1,ldots,6$. Let $x_k=costheta_k$, when $k$ is even, $T(x_k)=1$, when $k$ is odd, $T(x_k)=-1$. Suppose $deg R=6$ and the leading coefficient of $R$ is also $2^5$, if $max|R|<1$, at $x_k$, we have $T(x_0)-R(x_0)>0$, $T(x_1)-R(x_1)<0$ ..., it changes sign at least $6$ times, contradicts with $deg T-R<6$. Hence for all polynomial $Q$ s.t. $deg Q=6$ and the leading coefficient is $2^{6-1}$,
the minimal of $max|Q|$, which is $1$, occurs when $Q(x)=cos6arccos x$.
Now, denote $L(x)=frac12(Q(x)+1)/max|Q|$, we can see the maximal leading coefficient of $L$ is $2^4=16$.
answered Dec 31 '18 at 8:45
Kemono ChenKemono Chen
3,3401844
$endgroup$
add a comment |
$begingroup$
Let $T(x)=cos 6theta$ $(xin[-1,1])$ where $x=costheta$. The leading coefficient of $T$ is $2^{6-1}$ The maximal difference to $0$ occurs at $theta_k=kpi/6,k=0,1,ldots,6$. Let $x_k=costheta_k$, when $k$ is even, $T(x_k)=1$, when $k$ is odd, $T(x_k)=-1$. Suppose $deg R=6$ and the leading coefficient of $R$ is also $2^5$, if $max|R|<1$, at $x_k$, we have $T(x_0)-R(x_0)>0$, $T(x_1)-R(x_1)<0$ ..., it changes sign at least $6$ times, contradicts with $deg T-R<6$. Hence for all polynomial $Q$ s.t. $deg Q=6$ and the leading coefficient is $2^{6-1}$,
the minimal of $max|Q|$, which is $1$, occurs when $Q(x)=cos6arccos x$.
Now, denote $L(x)=frac12(Q(x)+1)/max|Q|$, we can see the maximal leading coefficient of $L$ is $2^4=16$.
answered Dec 31 '18 at 8:45
Kemono ChenKemono Chen
3,3401844
$endgroup$
add a comment |
$begingroup$
Let $T(x)=cos 6theta$ $(xin[-1,1])$ where $x=costheta$. The leading coefficient of $T$ is $2^{6-1}$ The maximal difference to $0$ occurs at $theta_k=kpi/6,k=0,1,ldots,6$. Let $x_k=costheta_k$, when $k$ is even, $T(x_k)=1$, when $k$ is odd, $T(x_k)=-1$. Suppose $deg R=6$ and the leading coefficient of $R$ is also $2^5$, if $max|R|<1$, at $x_k$, we have $T(x_0)-R(x_0)>0$, $T(x_1)-R(x_1)<0$ ..., it changes sign at least $6$ times, contradicts with $deg T-R<6$. Hence for all polynomial $Q$ s.t. $deg Q=6$ and the leading coefficient is $2^{6-1}$,
the minimal of $max|Q|$, which is $1$, occurs when $Q(x)=cos6arccos x$.
Now, denote $L(x)=frac12(Q(x)+1)/max|Q|$, we can see the maximal leading coefficient of $L$ is $2^4=16$.
answered Dec 31 '18 at 8:45
Kemono ChenKemono Chen
3,3401844
$endgroup$
Let $T(x)=cos 6theta$ $(xin[-1,1])$ where $x=costheta$. The leading coefficient of $T$ is $2^{6-1}$ The maximal difference to $0$ occurs at $theta_k=kpi/6,k=0,1,ldots,6$. Let $x_k=costheta_k$, when $k$ is even, $T(x_k)=1$, when $k$ is odd, $T(x_k)=-1$. Suppose $deg R=6$ and the leading coefficient of $R$ is also $2^5$, if $max|R|<1$, at $x_k$, we have $T(x_0)-R(x_0)>0$, $T(x_1)-R(x_1)<0$ ..., it changes sign at least $6$ times, contradicts with $deg T-R<6$. Hence for all polynomial $Q$ s.t. $deg Q=6$ and the leading coefficient is $2^{6-1}$,
the minimal of $max|Q|$, which is $1$, occurs when $Q(x)=cos6arccos x$.
Now, denote $L(x)=frac12(Q(x)+1)/max|Q|$, we can see the maximal leading coefficient of $L$ is $2^4=16$.
answered Dec 31 '18 at 8:45
Kemono ChenKemono Chen
3,3401844
edited Dec 31 '18 at 9:03
answered Dec 31 '18 at 8:45
Kemono ChenKemono Chen
3,3401844
answered Dec 31 '18 at 8:45
Kemono ChenKemono Chen
3,3401844
answered Dec 31 '18 at 8:45
Kemono ChenKemono Chen
3,3401844
3,3401844
add a comment |
add a comment |
$begingroup$
Hint
The value you’re looking for is the same as the one of the problem ... For all $-1 le x le 1$, we have $-1/2 le Q(x) le 1/2$... as we’re just substracting $1/2$ to the polynomial which doesn’t change the leading coefficient.
Now the map $f(x) = frac{1}{2^{6-1}}T_6(x)$ where $T_n$ is the n-th Chebyshev polynomials is the one having $1$ for leading coefficient and of minimal $infty$-norm (see Wikipedia article for more details). And this norm is equal to $Vert f Vert_infty =frac{1}{2^{6-1}}$. In your problem you allow the norm to be equal to $1/2$.
Hence the value you’re looking for is equal to $frac{1/2}{1/2^{6-1}}=2^4=16$.
answered Dec 31 '18 at 8:47
mathcounterexamples.netmathcounterexamples.net
26.9k22158
$endgroup$
add a comment |
$begingroup$
Hint
The value you’re looking for is the same as the one of the problem ... For all $-1 le x le 1$, we have $-1/2 le Q(x) le 1/2$... as we’re just substracting $1/2$ to the polynomial which doesn’t change the leading coefficient.
Now the map $f(x) = frac{1}{2^{6-1}}T_6(x)$ where $T_n$ is the n-th Chebyshev polynomials is the one having $1$ for leading coefficient and of minimal $infty$-norm (see Wikipedia article for more details). And this norm is equal to $Vert f Vert_infty =frac{1}{2^{6-1}}$. In your problem you allow the norm to be equal to $1/2$.
Hence the value you’re looking for is equal to $frac{1/2}{1/2^{6-1}}=2^4=16$.
answered Dec 31 '18 at 8:47
mathcounterexamples.netmathcounterexamples.net
26.9k22158
$endgroup$
add a comment |
$begingroup$
Hint
The value you’re looking for is the same as the one of the problem ... For all $-1 le x le 1$, we have $-1/2 le Q(x) le 1/2$... as we’re just substracting $1/2$ to the polynomial which doesn’t change the leading coefficient.
Now the map $f(x) = frac{1}{2^{6-1}}T_6(x)$ where $T_n$ is the n-th Chebyshev polynomials is the one having $1$ for leading coefficient and of minimal $infty$-norm (see Wikipedia article for more details). And this norm is equal to $Vert f Vert_infty =frac{1}{2^{6-1}}$. In your problem you allow the norm to be equal to $1/2$.
Hence the value you’re looking for is equal to $frac{1/2}{1/2^{6-1}}=2^4=16$.
answered Dec 31 '18 at 8:47
mathcounterexamples.netmathcounterexamples.net
26.9k22158
$endgroup$
Hint
The value you’re looking for is the same as the one of the problem ... For all $-1 le x le 1$, we have $-1/2 le Q(x) le 1/2$... as we’re just substracting $1/2$ to the polynomial which doesn’t change the leading coefficient.
Now the map $f(x) = frac{1}{2^{6-1}}T_6(x)$ where $T_n$ is the n-th Chebyshev polynomials is the one having $1$ for leading coefficient and of minimal $infty$-norm (see Wikipedia article for more details). And this norm is equal to $Vert f Vert_infty =frac{1}{2^{6-1}}$. In your problem you allow the norm to be equal to $1/2$.
Hence the value you’re looking for is equal to $frac{1/2}{1/2^{6-1}}=2^4=16$.
answered Dec 31 '18 at 8:47
mathcounterexamples.netmathcounterexamples.net
26.9k22158
edited Dec 31 '18 at 8:54
answered Dec 31 '18 at 8:47
mathcounterexamples.netmathcounterexamples.net
26.9k22158
answered Dec 31 '18 at 8:47
mathcounterexamples.netmathcounterexamples.net
26.9k22158
answered Dec 31 '18 at 8:47
mathcounterexamples.netmathcounterexamples.net
26.9k22158
26.9k22158
add a comment |
add a comment |
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