How to construct a polynomial with certain restraints
$begingroup$
Construct a fifth-degree polynomial $p(x)$ in the interval ${cal I} = [-3,3]$ that meets these conditions:
$p(x)$ has 5 different roots, each in the interval ${cal I}$.- The coefficient of the leading term, $x^5$, must be equal to 1.
- $max [ |p(x)| ] leq 1/5, x in {cal I}$
I am not sure if this polynomial exists.
real-analysis complex-analysis polynomials
asked Jan 22 at 23:37
Kwnstantinos NikoloutsosKwnstantinos Nikoloutsos
267310
$endgroup$
|
show 2 more comments
$begingroup$
Construct a fifth-degree polynomial $p(x)$ in the interval ${cal I} = [-3,3]$ that meets these conditions:
$p(x)$ has 5 different roots, each in the interval ${cal I}$.- The coefficient of the leading term, $x^5$, must be equal to 1.
- $max [ |p(x)| ] leq 1/5, x in {cal I}$
I am not sure if this polynomial exists.
real-analysis complex-analysis polynomials
asked Jan 22 at 23:37
Kwnstantinos NikoloutsosKwnstantinos Nikoloutsos
267310
$endgroup$
$begingroup$
Is that $E$ meant to be $in$?
$endgroup$
– Rhys Hughes
Jan 22 at 23:39
$begingroup$
@RhysHughes Yes, I am sorry I am not familiar with Latex.You can edit it if you want
$endgroup$
– Kwnstantinos Nikoloutsos
Jan 22 at 23:41
$begingroup$
@EricWofsey Yes the roots must be in [-3,3] and should be real numbers
$endgroup$
– Kwnstantinos Nikoloutsos
Jan 22 at 23:42
1
$begingroup$
Suggestion: $f(x) = (x-a)(x-b)(x-c)(x-d)(x-e)$ where ${ a, b, c, d, e } in [-3,3]$. Also, $f(0) = -a b c d e$.
$endgroup$
– David G. Stork
Jan 22 at 23:56
$begingroup$
@DavidG.Stork Suggestion $(x-3)(x-a)x(x+a)(x+3)$, find the maximum as a function of $a in (0,3)$ and hope there's an $a$ that works.
$endgroup$
– Ethan Bolker
Jan 23 at 0:09
|
show 2 more comments
$begingroup$
Construct a fifth-degree polynomial $p(x)$ in the interval ${cal I} = [-3,3]$ that meets these conditions:
$p(x)$ has 5 different roots, each in the interval ${cal I}$.- The coefficient of the leading term, $x^5$, must be equal to 1.
- $max [ |p(x)| ] leq 1/5, x in {cal I}$
I am not sure if this polynomial exists.
real-analysis complex-analysis polynomials
asked Jan 22 at 23:37
Kwnstantinos NikoloutsosKwnstantinos Nikoloutsos
267310
$endgroup$
Construct a fifth-degree polynomial $p(x)$ in the interval ${cal I} = [-3,3]$ that meets these conditions:
$p(x)$ has 5 different roots, each in the interval ${cal I}$.- The coefficient of the leading term, $x^5$, must be equal to 1.
- $max [ |p(x)| ] leq 1/5, x in {cal I}$
I am not sure if this polynomial exists.
real-analysis complex-analysis polynomials
real-analysis complex-analysis polynomials
asked Jan 22 at 23:37
Kwnstantinos NikoloutsosKwnstantinos Nikoloutsos
267310
asked Jan 22 at 23:37
Kwnstantinos NikoloutsosKwnstantinos Nikoloutsos
267310
edited Jan 23 at 0:31
Kwnstantinos Nikoloutsos
asked Jan 22 at 23:37
Kwnstantinos NikoloutsosKwnstantinos Nikoloutsos
267310
asked Jan 22 at 23:37
Kwnstantinos NikoloutsosKwnstantinos Nikoloutsos
267310
asked Jan 22 at 23:37
Kwnstantinos NikoloutsosKwnstantinos Nikoloutsos
267310
267310
$begingroup$
Is that $E$ meant to be $in$?
$endgroup$
– Rhys Hughes
Jan 22 at 23:39
$begingroup$
@RhysHughes Yes, I am sorry I am not familiar with Latex.You can edit it if you want
$endgroup$
– Kwnstantinos Nikoloutsos
Jan 22 at 23:41
$begingroup$
@EricWofsey Yes the roots must be in [-3,3] and should be real numbers
$endgroup$
– Kwnstantinos Nikoloutsos
Jan 22 at 23:42
1
$begingroup$
Suggestion: $f(x) = (x-a)(x-b)(x-c)(x-d)(x-e)$ where ${ a, b, c, d, e } in [-3,3]$. Also, $f(0) = -a b c d e$.
$endgroup$
– David G. Stork
Jan 22 at 23:56
$begingroup$
@DavidG.Stork Suggestion $(x-3)(x-a)x(x+a)(x+3)$, find the maximum as a function of $a in (0,3)$ and hope there's an $a$ that works.
$endgroup$
– Ethan Bolker
Jan 23 at 0:09
|
show 2 more comments
$begingroup$
Is that $E$ meant to be $in$?
$endgroup$
– Rhys Hughes
Jan 22 at 23:39
$begingroup$
@RhysHughes Yes, I am sorry I am not familiar with Latex.You can edit it if you want
$endgroup$
– Kwnstantinos Nikoloutsos
Jan 22 at 23:41
$begingroup$
@EricWofsey Yes the roots must be in [-3,3] and should be real numbers
$endgroup$
– Kwnstantinos Nikoloutsos
Jan 22 at 23:42
1
$begingroup$
Suggestion: $f(x) = (x-a)(x-b)(x-c)(x-d)(x-e)$ where ${ a, b, c, d, e } in [-3,3]$. Also, $f(0) = -a b c d e$.
$endgroup$
– David G. Stork
Jan 22 at 23:56
$begingroup$
@DavidG.Stork Suggestion $(x-3)(x-a)x(x+a)(x+3)$, find the maximum as a function of $a in (0,3)$ and hope there's an $a$ that works.
$endgroup$
– Ethan Bolker
Jan 23 at 0:09
$begingroup$
Is that $E$ meant to be $in$?
$endgroup$
– Rhys Hughes
Jan 22 at 23:39
$begingroup$
Is that $E$ meant to be $in$?
$endgroup$
– Rhys Hughes
Jan 22 at 23:39
$begingroup$
@RhysHughes Yes, I am sorry I am not familiar with Latex.You can edit it if you want
$endgroup$
– Kwnstantinos Nikoloutsos
Jan 22 at 23:41
$begingroup$
@RhysHughes Yes, I am sorry I am not familiar with Latex.You can edit it if you want
$endgroup$
– Kwnstantinos Nikoloutsos
Jan 22 at 23:41
$begingroup$
@EricWofsey Yes the roots must be in [-3,3] and should be real numbers
$endgroup$
– Kwnstantinos Nikoloutsos
Jan 22 at 23:42
$begingroup$
@EricWofsey Yes the roots must be in [-3,3] and should be real numbers
$endgroup$
– Kwnstantinos Nikoloutsos
Jan 22 at 23:42
1
1
$begingroup$
Suggestion: $f(x) = (x-a)(x-b)(x-c)(x-d)(x-e)$ where ${ a, b, c, d, e } in [-3,3]$. Also, $f(0) = -a b c d e$.
$endgroup$
– David G. Stork
Jan 22 at 23:56
$begingroup$
Suggestion: $f(x) = (x-a)(x-b)(x-c)(x-d)(x-e)$ where ${ a, b, c, d, e } in [-3,3]$. Also, $f(0) = -a b c d e$.
$endgroup$
– David G. Stork
Jan 22 at 23:56
$begingroup$
@DavidG.Stork Suggestion $(x-3)(x-a)x(x+a)(x+3)$, find the maximum as a function of $a in (0,3)$ and hope there's an $a$ that works.
$endgroup$
– Ethan Bolker
Jan 23 at 0:09
$begingroup$
@DavidG.Stork Suggestion $(x-3)(x-a)x(x+a)(x+3)$, find the maximum as a function of $a in (0,3)$ and hope there's an $a$ that works.
$endgroup$
– Ethan Bolker
Jan 23 at 0:09
|
show 2 more comments
2 Answers
2
active
oldest
votes
$begingroup$
No such polynomial exists.
Let $p(x)$ be a monic polynomial of degree $5$. Let $q(t)=dfrac{1}{3^5}p(3t)$.
Then $q(t)$ is a monic polynomial of degree $5$ and
$$
max_{x in [-3,3]} |p(x)| = 3^5 max_{t in [-1,1]} |q(t)| geq dfrac{3^5}{2^{4}} > frac15
$$
because the monic polynomial of degree $5$ with minimal norm in $[-1,1]$ is $dfrac{1}{2^{4}}T_{5}(t)$, where $T_5$ is the $5$-th Chebyshev polynomial. The minimal norm is $dfrac{1}{2^{4}}$.
answered Jan 23 at 1:15
lhflhf
166k10171396
$endgroup$
add a comment |
$begingroup$
If the roots of $p$ are $a<b<c<d<e$, then the subintervals of $[-3,3]$ where $p(x)=(x-a)(x-b)(x-c)(x-d)(x-e)$ is positive are $(a,b),(c,d),$ and $(e,3]$. To make the maximum value of $p(x)$ as small as possible, the idea is to make the intervals where it is positive very short. That means we want $a$ and $b$ to be very close together, $c$ and $d$ to be very close together, and $e$ to be very close to $3$.
So, fix $epsilon>0$ and whatever values of $a$ and $c$ you want (with $a<c-epsilon$ and $c+epsilon<3-epsilon$) and consider $p(x)=(x-a)(x-b)(x-c)(x-d)(x-e)$ where $b=a+epsilon$, $d=c+epsilon$, and $e=3-epsilon$. Note that then if $xin (a,b)$, $|x-a|$ and $|x-b|$ are at most $epsilon$, so $|p(x)|<6^3epsilon^2$ (the other factors in $p(x)$ have absolute value less than $6$, the diameter of $[-3,3]$). Similarly we have $|p(x)|<6^3epsilon^2$ for all $xin (c,d)$ and $|p(x)|<6^4epsilon$ for all $xin (e,3]$. So, as long as we pick $epsilon$ small enough, $|p(x)|$ will always be at most $1/5$ on any interval where $p(x)$ is positive, and so $p(x)leq 1/5$ on all of $[-3,3]$.
answered Jan 23 at 0:21
Eric WofseyEric Wofsey
187k14216344
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
No such polynomial exists.
Let $p(x)$ be a monic polynomial of degree $5$. Let $q(t)=dfrac{1}{3^5}p(3t)$.
Then $q(t)$ is a monic polynomial of degree $5$ and
$$
max_{x in [-3,3]} |p(x)| = 3^5 max_{t in [-1,1]} |q(t)| geq dfrac{3^5}{2^{4}} > frac15
$$
because the monic polynomial of degree $5$ with minimal norm in $[-1,1]$ is $dfrac{1}{2^{4}}T_{5}(t)$, where $T_5$ is the $5$-th Chebyshev polynomial. The minimal norm is $dfrac{1}{2^{4}}$.
answered Jan 23 at 1:15
lhflhf
166k10171396
$endgroup$
add a comment |
$begingroup$
No such polynomial exists.
Let $p(x)$ be a monic polynomial of degree $5$. Let $q(t)=dfrac{1}{3^5}p(3t)$.
Then $q(t)$ is a monic polynomial of degree $5$ and
$$
max_{x in [-3,3]} |p(x)| = 3^5 max_{t in [-1,1]} |q(t)| geq dfrac{3^5}{2^{4}} > frac15
$$
because the monic polynomial of degree $5$ with minimal norm in $[-1,1]$ is $dfrac{1}{2^{4}}T_{5}(t)$, where $T_5$ is the $5$-th Chebyshev polynomial. The minimal norm is $dfrac{1}{2^{4}}$.
answered Jan 23 at 1:15
lhflhf
166k10171396
$endgroup$
add a comment |
$begingroup$
No such polynomial exists.
Let $p(x)$ be a monic polynomial of degree $5$. Let $q(t)=dfrac{1}{3^5}p(3t)$.
Then $q(t)$ is a monic polynomial of degree $5$ and
$$
max_{x in [-3,3]} |p(x)| = 3^5 max_{t in [-1,1]} |q(t)| geq dfrac{3^5}{2^{4}} > frac15
$$
because the monic polynomial of degree $5$ with minimal norm in $[-1,1]$ is $dfrac{1}{2^{4}}T_{5}(t)$, where $T_5$ is the $5$-th Chebyshev polynomial. The minimal norm is $dfrac{1}{2^{4}}$.
answered Jan 23 at 1:15
lhflhf
166k10171396
$endgroup$
No such polynomial exists.
Let $p(x)$ be a monic polynomial of degree $5$. Let $q(t)=dfrac{1}{3^5}p(3t)$.
Then $q(t)$ is a monic polynomial of degree $5$ and
$$
max_{x in [-3,3]} |p(x)| = 3^5 max_{t in [-1,1]} |q(t)| geq dfrac{3^5}{2^{4}} > frac15
$$
because the monic polynomial of degree $5$ with minimal norm in $[-1,1]$ is $dfrac{1}{2^{4}}T_{5}(t)$, where $T_5$ is the $5$-th Chebyshev polynomial. The minimal norm is $dfrac{1}{2^{4}}$.
answered Jan 23 at 1:15
lhflhf
166k10171396
answered Jan 23 at 1:15
lhflhf
166k10171396
answered Jan 23 at 1:15
lhflhf
166k10171396
answered Jan 23 at 1:15
lhflhf
166k10171396
166k10171396
add a comment |
add a comment |
$begingroup$
If the roots of $p$ are $a<b<c<d<e$, then the subintervals of $[-3,3]$ where $p(x)=(x-a)(x-b)(x-c)(x-d)(x-e)$ is positive are $(a,b),(c,d),$ and $(e,3]$. To make the maximum value of $p(x)$ as small as possible, the idea is to make the intervals where it is positive very short. That means we want $a$ and $b$ to be very close together, $c$ and $d$ to be very close together, and $e$ to be very close to $3$.
So, fix $epsilon>0$ and whatever values of $a$ and $c$ you want (with $a<c-epsilon$ and $c+epsilon<3-epsilon$) and consider $p(x)=(x-a)(x-b)(x-c)(x-d)(x-e)$ where $b=a+epsilon$, $d=c+epsilon$, and $e=3-epsilon$. Note that then if $xin (a,b)$, $|x-a|$ and $|x-b|$ are at most $epsilon$, so $|p(x)|<6^3epsilon^2$ (the other factors in $p(x)$ have absolute value less than $6$, the diameter of $[-3,3]$). Similarly we have $|p(x)|<6^3epsilon^2$ for all $xin (c,d)$ and $|p(x)|<6^4epsilon$ for all $xin (e,3]$. So, as long as we pick $epsilon$ small enough, $|p(x)|$ will always be at most $1/5$ on any interval where $p(x)$ is positive, and so $p(x)leq 1/5$ on all of $[-3,3]$.
answered Jan 23 at 0:21
Eric WofseyEric Wofsey
187k14216344
$endgroup$
add a comment |
$begingroup$
If the roots of $p$ are $a<b<c<d<e$, then the subintervals of $[-3,3]$ where $p(x)=(x-a)(x-b)(x-c)(x-d)(x-e)$ is positive are $(a,b),(c,d),$ and $(e,3]$. To make the maximum value of $p(x)$ as small as possible, the idea is to make the intervals where it is positive very short. That means we want $a$ and $b$ to be very close together, $c$ and $d$ to be very close together, and $e$ to be very close to $3$.
So, fix $epsilon>0$ and whatever values of $a$ and $c$ you want (with $a<c-epsilon$ and $c+epsilon<3-epsilon$) and consider $p(x)=(x-a)(x-b)(x-c)(x-d)(x-e)$ where $b=a+epsilon$, $d=c+epsilon$, and $e=3-epsilon$. Note that then if $xin (a,b)$, $|x-a|$ and $|x-b|$ are at most $epsilon$, so $|p(x)|<6^3epsilon^2$ (the other factors in $p(x)$ have absolute value less than $6$, the diameter of $[-3,3]$). Similarly we have $|p(x)|<6^3epsilon^2$ for all $xin (c,d)$ and $|p(x)|<6^4epsilon$ for all $xin (e,3]$. So, as long as we pick $epsilon$ small enough, $|p(x)|$ will always be at most $1/5$ on any interval where $p(x)$ is positive, and so $p(x)leq 1/5$ on all of $[-3,3]$.
answered Jan 23 at 0:21
Eric WofseyEric Wofsey
187k14216344
$endgroup$
add a comment |
$begingroup$
If the roots of $p$ are $a<b<c<d<e$, then the subintervals of $[-3,3]$ where $p(x)=(x-a)(x-b)(x-c)(x-d)(x-e)$ is positive are $(a,b),(c,d),$ and $(e,3]$. To make the maximum value of $p(x)$ as small as possible, the idea is to make the intervals where it is positive very short. That means we want $a$ and $b$ to be very close together, $c$ and $d$ to be very close together, and $e$ to be very close to $3$.
So, fix $epsilon>0$ and whatever values of $a$ and $c$ you want (with $a<c-epsilon$ and $c+epsilon<3-epsilon$) and consider $p(x)=(x-a)(x-b)(x-c)(x-d)(x-e)$ where $b=a+epsilon$, $d=c+epsilon$, and $e=3-epsilon$. Note that then if $xin (a,b)$, $|x-a|$ and $|x-b|$ are at most $epsilon$, so $|p(x)|<6^3epsilon^2$ (the other factors in $p(x)$ have absolute value less than $6$, the diameter of $[-3,3]$). Similarly we have $|p(x)|<6^3epsilon^2$ for all $xin (c,d)$ and $|p(x)|<6^4epsilon$ for all $xin (e,3]$. So, as long as we pick $epsilon$ small enough, $|p(x)|$ will always be at most $1/5$ on any interval where $p(x)$ is positive, and so $p(x)leq 1/5$ on all of $[-3,3]$.
answered Jan 23 at 0:21
Eric WofseyEric Wofsey
187k14216344
$endgroup$
If the roots of $p$ are $a<b<c<d<e$, then the subintervals of $[-3,3]$ where $p(x)=(x-a)(x-b)(x-c)(x-d)(x-e)$ is positive are $(a,b),(c,d),$ and $(e,3]$. To make the maximum value of $p(x)$ as small as possible, the idea is to make the intervals where it is positive very short. That means we want $a$ and $b$ to be very close together, $c$ and $d$ to be very close together, and $e$ to be very close to $3$.
So, fix $epsilon>0$ and whatever values of $a$ and $c$ you want (with $a<c-epsilon$ and $c+epsilon<3-epsilon$) and consider $p(x)=(x-a)(x-b)(x-c)(x-d)(x-e)$ where $b=a+epsilon$, $d=c+epsilon$, and $e=3-epsilon$. Note that then if $xin (a,b)$, $|x-a|$ and $|x-b|$ are at most $epsilon$, so $|p(x)|<6^3epsilon^2$ (the other factors in $p(x)$ have absolute value less than $6$, the diameter of $[-3,3]$). Similarly we have $|p(x)|<6^3epsilon^2$ for all $xin (c,d)$ and $|p(x)|<6^4epsilon$ for all $xin (e,3]$. So, as long as we pick $epsilon$ small enough, $|p(x)|$ will always be at most $1/5$ on any interval where $p(x)$ is positive, and so $p(x)leq 1/5$ on all of $[-3,3]$.
answered Jan 23 at 0:21
Eric WofseyEric Wofsey
187k14216344
answered Jan 23 at 0:21
Eric WofseyEric Wofsey
187k14216344
answered Jan 23 at 0:21
Eric WofseyEric Wofsey
187k14216344
answered Jan 23 at 0:21
Eric WofseyEric Wofsey
187k14216344
187k14216344
add a comment |
add a comment |
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$begingroup$
Is that $E$ meant to be $in$?
$endgroup$
– Rhys Hughes
Jan 22 at 23:39
$begingroup$
@RhysHughes Yes, I am sorry I am not familiar with Latex.You can edit it if you want
$endgroup$
– Kwnstantinos Nikoloutsos
Jan 22 at 23:41
$begingroup$
@EricWofsey Yes the roots must be in [-3,3] and should be real numbers
$endgroup$
– Kwnstantinos Nikoloutsos
Jan 22 at 23:42
1
$begingroup$
Suggestion: $f(x) = (x-a)(x-b)(x-c)(x-d)(x-e)$ where ${ a, b, c, d, e } in [-3,3]$. Also, $f(0) = -a b c d e$.
$endgroup$
– David G. Stork
Jan 22 at 23:56
$begingroup$
@DavidG.Stork Suggestion $(x-3)(x-a)x(x+a)(x+3)$, find the maximum as a function of $a in (0,3)$ and hope there's an $a$ that works.
$endgroup$
– Ethan Bolker
Jan 23 at 0:09