Pointwise convergence of a sequence of polynomials
$begingroup$
Consider the identity function $f(x) = x$ and let ${h_n;n in mathbb{N}}$ be a sequence of polynomials, which are defined on $[0,a]$ with some fixed $a<1$, and are of the form $h_n(x) = sum_{i=1}^n c_{i,n} x^i$ where $c_{i,n}$ is the $i$th coefficient in the $n$th polynomial $h_n$. Further assume that:
(1) $0 leq h_n(x) leq x$ and $h_n(0) = 0$ for every $n$;
(2) each $h_n$ is monotonically increasing;
(3) $lim_{n to infty} h_n(x) = x$ uniformly for all $x in [0,a]$.
From (1) we know that $c_{1,n} = h_n'(0) leq x'|_{x=0} = 1$ for every $n$. My question is: from conditions (1), (2) and (3), is it necessarily true that $lim_{n to infty} c_{1,n} = 1$? If I impose the extra condition $sup_{n in mathbb{N}} {sum_{i=2}^n |c_{i,n}|} < M$ for some absolute constant $M > 0$, then the claim holds. But this is a too stringent condition. But I am not sure if the claim is unconditionally true. If not, what could be a minimal set of conditions that I need to impose? Thanks very much.
calculus real-analysis measure-theory polynomials convergence
asked Jul 28 '17 at 8:50
RichieRichie
394210
$endgroup$
|
show 3 more comments
$begingroup$
Consider the identity function $f(x) = x$ and let ${h_n;n in mathbb{N}}$ be a sequence of polynomials, which are defined on $[0,a]$ with some fixed $a<1$, and are of the form $h_n(x) = sum_{i=1}^n c_{i,n} x^i$ where $c_{i,n}$ is the $i$th coefficient in the $n$th polynomial $h_n$. Further assume that:
(1) $0 leq h_n(x) leq x$ and $h_n(0) = 0$ for every $n$;
(2) each $h_n$ is monotonically increasing;
(3) $lim_{n to infty} h_n(x) = x$ uniformly for all $x in [0,a]$.
From (1) we know that $c_{1,n} = h_n'(0) leq x'|_{x=0} = 1$ for every $n$. My question is: from conditions (1), (2) and (3), is it necessarily true that $lim_{n to infty} c_{1,n} = 1$? If I impose the extra condition $sup_{n in mathbb{N}} {sum_{i=2}^n |c_{i,n}|} < M$ for some absolute constant $M > 0$, then the claim holds. But this is a too stringent condition. But I am not sure if the claim is unconditionally true. If not, what could be a minimal set of conditions that I need to impose? Thanks very much.
calculus real-analysis measure-theory polynomials convergence
asked Jul 28 '17 at 8:50
RichieRichie
394210
$endgroup$
$begingroup$
You forgot to mention what's the purpose of this, since $x$ is a polynomial. So why don't you just set $h_n(x)=x$?
$endgroup$
– Professor Vector
Jul 28 '17 at 9:36
$begingroup$
@Professor Vector The purpose is that given a sequence of functions ${f_n}$ on some $[a,b]$ that converge uniformly to a function $f$, in general we cannot say anything about how the sequence of derivatives ${f_n '}$ behave. So I am considering an example where we can get some result on the sequence of derivatives.
$endgroup$
– Richie
Jul 28 '17 at 10:15
$begingroup$
Well, you've just added more confusion, because in your question, there was no "uniform convergence". Of course, there are examples where the derivatives converge (I gave a trivial one), but you evaded the question of the purpose. It's pointless, imho.
$endgroup$
– Professor Vector
Jul 28 '17 at 10:21
$begingroup$
Ok, fine. The title is misleading. It is not a question on approximation, but a question on convergence. As you said, there are examples where the derivatives converge, and I am constructing this particular one, because the polynomials are "well-behaved", so that I think it should be true that the derivatives converge. But maybe we need certain condition on how the coefficients behave, for example, an upper bound on $c_{i,n}$ in terms of $n$. But I am not sure about this.
$endgroup$
– Richie
Jul 28 '17 at 10:35
$begingroup$
Is there even a non-trivial sequence $h_n$ with the desired properties? It seems to me that (1) will never be satisfied by any polynomial other than $x$. Am I missing something?
$endgroup$
– Jose27
Jul 28 '17 at 13:14
|
show 3 more comments
$begingroup$
Consider the identity function $f(x) = x$ and let ${h_n;n in mathbb{N}}$ be a sequence of polynomials, which are defined on $[0,a]$ with some fixed $a<1$, and are of the form $h_n(x) = sum_{i=1}^n c_{i,n} x^i$ where $c_{i,n}$ is the $i$th coefficient in the $n$th polynomial $h_n$. Further assume that:
(1) $0 leq h_n(x) leq x$ and $h_n(0) = 0$ for every $n$;
(2) each $h_n$ is monotonically increasing;
(3) $lim_{n to infty} h_n(x) = x$ uniformly for all $x in [0,a]$.
From (1) we know that $c_{1,n} = h_n'(0) leq x'|_{x=0} = 1$ for every $n$. My question is: from conditions (1), (2) and (3), is it necessarily true that $lim_{n to infty} c_{1,n} = 1$? If I impose the extra condition $sup_{n in mathbb{N}} {sum_{i=2}^n |c_{i,n}|} < M$ for some absolute constant $M > 0$, then the claim holds. But this is a too stringent condition. But I am not sure if the claim is unconditionally true. If not, what could be a minimal set of conditions that I need to impose? Thanks very much.
calculus real-analysis measure-theory polynomials convergence
asked Jul 28 '17 at 8:50
RichieRichie
394210
$endgroup$
Consider the identity function $f(x) = x$ and let ${h_n;n in mathbb{N}}$ be a sequence of polynomials, which are defined on $[0,a]$ with some fixed $a<1$, and are of the form $h_n(x) = sum_{i=1}^n c_{i,n} x^i$ where $c_{i,n}$ is the $i$th coefficient in the $n$th polynomial $h_n$. Further assume that:
(1) $0 leq h_n(x) leq x$ and $h_n(0) = 0$ for every $n$;
(2) each $h_n$ is monotonically increasing;
(3) $lim_{n to infty} h_n(x) = x$ uniformly for all $x in [0,a]$.
From (1) we know that $c_{1,n} = h_n'(0) leq x'|_{x=0} = 1$ for every $n$. My question is: from conditions (1), (2) and (3), is it necessarily true that $lim_{n to infty} c_{1,n} = 1$? If I impose the extra condition $sup_{n in mathbb{N}} {sum_{i=2}^n |c_{i,n}|} < M$ for some absolute constant $M > 0$, then the claim holds. But this is a too stringent condition. But I am not sure if the claim is unconditionally true. If not, what could be a minimal set of conditions that I need to impose? Thanks very much.
calculus real-analysis measure-theory polynomials convergence
calculus real-analysis measure-theory polynomials convergence
asked Jul 28 '17 at 8:50
RichieRichie
394210
asked Jul 28 '17 at 8:50
RichieRichie
394210
edited Jul 29 '17 at 7:04
Richie
asked Jul 28 '17 at 8:50
RichieRichie
394210
asked Jul 28 '17 at 8:50
RichieRichie
394210
asked Jul 28 '17 at 8:50
RichieRichie
394210
394210
$begingroup$
You forgot to mention what's the purpose of this, since $x$ is a polynomial. So why don't you just set $h_n(x)=x$?
$endgroup$
– Professor Vector
Jul 28 '17 at 9:36
$begingroup$
@Professor Vector The purpose is that given a sequence of functions ${f_n}$ on some $[a,b]$ that converge uniformly to a function $f$, in general we cannot say anything about how the sequence of derivatives ${f_n '}$ behave. So I am considering an example where we can get some result on the sequence of derivatives.
$endgroup$
– Richie
Jul 28 '17 at 10:15
$begingroup$
Well, you've just added more confusion, because in your question, there was no "uniform convergence". Of course, there are examples where the derivatives converge (I gave a trivial one), but you evaded the question of the purpose. It's pointless, imho.
$endgroup$
– Professor Vector
Jul 28 '17 at 10:21
$begingroup$
Ok, fine. The title is misleading. It is not a question on approximation, but a question on convergence. As you said, there are examples where the derivatives converge, and I am constructing this particular one, because the polynomials are "well-behaved", so that I think it should be true that the derivatives converge. But maybe we need certain condition on how the coefficients behave, for example, an upper bound on $c_{i,n}$ in terms of $n$. But I am not sure about this.
$endgroup$
– Richie
Jul 28 '17 at 10:35
$begingroup$
Is there even a non-trivial sequence $h_n$ with the desired properties? It seems to me that (1) will never be satisfied by any polynomial other than $x$. Am I missing something?
$endgroup$
– Jose27
Jul 28 '17 at 13:14
|
show 3 more comments
$begingroup$
You forgot to mention what's the purpose of this, since $x$ is a polynomial. So why don't you just set $h_n(x)=x$?
$endgroup$
– Professor Vector
Jul 28 '17 at 9:36
$begingroup$
@Professor Vector The purpose is that given a sequence of functions ${f_n}$ on some $[a,b]$ that converge uniformly to a function $f$, in general we cannot say anything about how the sequence of derivatives ${f_n '}$ behave. So I am considering an example where we can get some result on the sequence of derivatives.
$endgroup$
– Richie
Jul 28 '17 at 10:15
$begingroup$
Well, you've just added more confusion, because in your question, there was no "uniform convergence". Of course, there are examples where the derivatives converge (I gave a trivial one), but you evaded the question of the purpose. It's pointless, imho.
$endgroup$
– Professor Vector
Jul 28 '17 at 10:21
$begingroup$
Ok, fine. The title is misleading. It is not a question on approximation, but a question on convergence. As you said, there are examples where the derivatives converge, and I am constructing this particular one, because the polynomials are "well-behaved", so that I think it should be true that the derivatives converge. But maybe we need certain condition on how the coefficients behave, for example, an upper bound on $c_{i,n}$ in terms of $n$. But I am not sure about this.
$endgroup$
– Richie
Jul 28 '17 at 10:35
$begingroup$
Is there even a non-trivial sequence $h_n$ with the desired properties? It seems to me that (1) will never be satisfied by any polynomial other than $x$. Am I missing something?
$endgroup$
– Jose27
Jul 28 '17 at 13:14
$begingroup$
You forgot to mention what's the purpose of this, since $x$ is a polynomial. So why don't you just set $h_n(x)=x$?
$endgroup$
– Professor Vector
Jul 28 '17 at 9:36
$begingroup$
You forgot to mention what's the purpose of this, since $x$ is a polynomial. So why don't you just set $h_n(x)=x$?
$endgroup$
– Professor Vector
Jul 28 '17 at 9:36
$begingroup$
@Professor Vector The purpose is that given a sequence of functions ${f_n}$ on some $[a,b]$ that converge uniformly to a function $f$, in general we cannot say anything about how the sequence of derivatives ${f_n '}$ behave. So I am considering an example where we can get some result on the sequence of derivatives.
$endgroup$
– Richie
Jul 28 '17 at 10:15
$begingroup$
@Professor Vector The purpose is that given a sequence of functions ${f_n}$ on some $[a,b]$ that converge uniformly to a function $f$, in general we cannot say anything about how the sequence of derivatives ${f_n '}$ behave. So I am considering an example where we can get some result on the sequence of derivatives.
$endgroup$
– Richie
Jul 28 '17 at 10:15
$begingroup$
Well, you've just added more confusion, because in your question, there was no "uniform convergence". Of course, there are examples where the derivatives converge (I gave a trivial one), but you evaded the question of the purpose. It's pointless, imho.
$endgroup$
– Professor Vector
Jul 28 '17 at 10:21
$begingroup$
Well, you've just added more confusion, because in your question, there was no "uniform convergence". Of course, there are examples where the derivatives converge (I gave a trivial one), but you evaded the question of the purpose. It's pointless, imho.
$endgroup$
– Professor Vector
Jul 28 '17 at 10:21
$begingroup$
Ok, fine. The title is misleading. It is not a question on approximation, but a question on convergence. As you said, there are examples where the derivatives converge, and I am constructing this particular one, because the polynomials are "well-behaved", so that I think it should be true that the derivatives converge. But maybe we need certain condition on how the coefficients behave, for example, an upper bound on $c_{i,n}$ in terms of $n$. But I am not sure about this.
$endgroup$
– Richie
Jul 28 '17 at 10:35
$begingroup$
Ok, fine. The title is misleading. It is not a question on approximation, but a question on convergence. As you said, there are examples where the derivatives converge, and I am constructing this particular one, because the polynomials are "well-behaved", so that I think it should be true that the derivatives converge. But maybe we need certain condition on how the coefficients behave, for example, an upper bound on $c_{i,n}$ in terms of $n$. But I am not sure about this.
$endgroup$
– Richie
Jul 28 '17 at 10:35
$begingroup$
Is there even a non-trivial sequence $h_n$ with the desired properties? It seems to me that (1) will never be satisfied by any polynomial other than $x$. Am I missing something?
$endgroup$
– Jose27
Jul 28 '17 at 13:14
$begingroup$
Is there even a non-trivial sequence $h_n$ with the desired properties? It seems to me that (1) will never be satisfied by any polynomial other than $x$. Am I missing something?
$endgroup$
– Jose27
Jul 28 '17 at 13:14
|
show 3 more comments
1 Answer
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No. For $epsilon>0$, let $d_epsilon(x)=epsilon$ for $xleqepsilon$, $d_epsilon(x)=1-epsilon$ for $xgeq 2epsilon$, and interpolate linearly on $[epsilon,2epsilon]$. By the Weierstrass approximation theorem, let $p_epsilon$ be a polynomial that uniformly approximates $d_epsilon$ within $epsilon/2$ on $[0,a]$, and let $q_epsilon$ be the antiderivative of $p_epsilon$ with $q_epsilon(0)=0$. Note that $q_epsilon$ satisfies your conditions (1) and (2) (since $p_epsilon$ is always between $epsilon/2$ and $1-epsilon/2$), and $q_epsilon(x)to x$ uniformly on $[0,a]$ as $epsilonto 0$. However, $q_epsilon'(0)=p_epsilon(0)to 0$ as $epsilonto 0$.
In particular, then, we can pick a sequence $epsilon_kto 0$ for which the polynomials $q_{epsilon_k}$ have increasing degree, and take them to be a subsequence of your sequence $(h_n)$. Then $c_{1,n}$ will not converge to $1$ since it converges to $0$ on a subsequence.
answered Jan 9 at 5:22
Eric WofseyEric Wofsey
193k14221352
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add a comment |
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$begingroup$
No. For $epsilon>0$, let $d_epsilon(x)=epsilon$ for $xleqepsilon$, $d_epsilon(x)=1-epsilon$ for $xgeq 2epsilon$, and interpolate linearly on $[epsilon,2epsilon]$. By the Weierstrass approximation theorem, let $p_epsilon$ be a polynomial that uniformly approximates $d_epsilon$ within $epsilon/2$ on $[0,a]$, and let $q_epsilon$ be the antiderivative of $p_epsilon$ with $q_epsilon(0)=0$. Note that $q_epsilon$ satisfies your conditions (1) and (2) (since $p_epsilon$ is always between $epsilon/2$ and $1-epsilon/2$), and $q_epsilon(x)to x$ uniformly on $[0,a]$ as $epsilonto 0$. However, $q_epsilon'(0)=p_epsilon(0)to 0$ as $epsilonto 0$.
In particular, then, we can pick a sequence $epsilon_kto 0$ for which the polynomials $q_{epsilon_k}$ have increasing degree, and take them to be a subsequence of your sequence $(h_n)$. Then $c_{1,n}$ will not converge to $1$ since it converges to $0$ on a subsequence.
answered Jan 9 at 5:22
Eric WofseyEric Wofsey
193k14221352
$endgroup$
add a comment |
$begingroup$
No. For $epsilon>0$, let $d_epsilon(x)=epsilon$ for $xleqepsilon$, $d_epsilon(x)=1-epsilon$ for $xgeq 2epsilon$, and interpolate linearly on $[epsilon,2epsilon]$. By the Weierstrass approximation theorem, let $p_epsilon$ be a polynomial that uniformly approximates $d_epsilon$ within $epsilon/2$ on $[0,a]$, and let $q_epsilon$ be the antiderivative of $p_epsilon$ with $q_epsilon(0)=0$. Note that $q_epsilon$ satisfies your conditions (1) and (2) (since $p_epsilon$ is always between $epsilon/2$ and $1-epsilon/2$), and $q_epsilon(x)to x$ uniformly on $[0,a]$ as $epsilonto 0$. However, $q_epsilon'(0)=p_epsilon(0)to 0$ as $epsilonto 0$.
In particular, then, we can pick a sequence $epsilon_kto 0$ for which the polynomials $q_{epsilon_k}$ have increasing degree, and take them to be a subsequence of your sequence $(h_n)$. Then $c_{1,n}$ will not converge to $1$ since it converges to $0$ on a subsequence.
answered Jan 9 at 5:22
Eric WofseyEric Wofsey
193k14221352
$endgroup$
add a comment |
$begingroup$
No. For $epsilon>0$, let $d_epsilon(x)=epsilon$ for $xleqepsilon$, $d_epsilon(x)=1-epsilon$ for $xgeq 2epsilon$, and interpolate linearly on $[epsilon,2epsilon]$. By the Weierstrass approximation theorem, let $p_epsilon$ be a polynomial that uniformly approximates $d_epsilon$ within $epsilon/2$ on $[0,a]$, and let $q_epsilon$ be the antiderivative of $p_epsilon$ with $q_epsilon(0)=0$. Note that $q_epsilon$ satisfies your conditions (1) and (2) (since $p_epsilon$ is always between $epsilon/2$ and $1-epsilon/2$), and $q_epsilon(x)to x$ uniformly on $[0,a]$ as $epsilonto 0$. However, $q_epsilon'(0)=p_epsilon(0)to 0$ as $epsilonto 0$.
In particular, then, we can pick a sequence $epsilon_kto 0$ for which the polynomials $q_{epsilon_k}$ have increasing degree, and take them to be a subsequence of your sequence $(h_n)$. Then $c_{1,n}$ will not converge to $1$ since it converges to $0$ on a subsequence.
answered Jan 9 at 5:22
Eric WofseyEric Wofsey
193k14221352
$endgroup$
No. For $epsilon>0$, let $d_epsilon(x)=epsilon$ for $xleqepsilon$, $d_epsilon(x)=1-epsilon$ for $xgeq 2epsilon$, and interpolate linearly on $[epsilon,2epsilon]$. By the Weierstrass approximation theorem, let $p_epsilon$ be a polynomial that uniformly approximates $d_epsilon$ within $epsilon/2$ on $[0,a]$, and let $q_epsilon$ be the antiderivative of $p_epsilon$ with $q_epsilon(0)=0$. Note that $q_epsilon$ satisfies your conditions (1) and (2) (since $p_epsilon$ is always between $epsilon/2$ and $1-epsilon/2$), and $q_epsilon(x)to x$ uniformly on $[0,a]$ as $epsilonto 0$. However, $q_epsilon'(0)=p_epsilon(0)to 0$ as $epsilonto 0$.
In particular, then, we can pick a sequence $epsilon_kto 0$ for which the polynomials $q_{epsilon_k}$ have increasing degree, and take them to be a subsequence of your sequence $(h_n)$. Then $c_{1,n}$ will not converge to $1$ since it converges to $0$ on a subsequence.
answered Jan 9 at 5:22
Eric WofseyEric Wofsey
193k14221352
answered Jan 9 at 5:22
Eric WofseyEric Wofsey
193k14221352
answered Jan 9 at 5:22
Eric WofseyEric Wofsey
193k14221352
answered Jan 9 at 5:22
Eric WofseyEric Wofsey
193k14221352
193k14221352
add a comment |
add a comment |
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$begingroup$
You forgot to mention what's the purpose of this, since $x$ is a polynomial. So why don't you just set $h_n(x)=x$?
$endgroup$
– Professor Vector
Jul 28 '17 at 9:36
$begingroup$
@Professor Vector The purpose is that given a sequence of functions ${f_n}$ on some $[a,b]$ that converge uniformly to a function $f$, in general we cannot say anything about how the sequence of derivatives ${f_n '}$ behave. So I am considering an example where we can get some result on the sequence of derivatives.
$endgroup$
– Richie
Jul 28 '17 at 10:15
$begingroup$
Well, you've just added more confusion, because in your question, there was no "uniform convergence". Of course, there are examples where the derivatives converge (I gave a trivial one), but you evaded the question of the purpose. It's pointless, imho.
$endgroup$
– Professor Vector
Jul 28 '17 at 10:21
$begingroup$
Ok, fine. The title is misleading. It is not a question on approximation, but a question on convergence. As you said, there are examples where the derivatives converge, and I am constructing this particular one, because the polynomials are "well-behaved", so that I think it should be true that the derivatives converge. But maybe we need certain condition on how the coefficients behave, for example, an upper bound on $c_{i,n}$ in terms of $n$. But I am not sure about this.
$endgroup$
– Richie
Jul 28 '17 at 10:35
$begingroup$
Is there even a non-trivial sequence $h_n$ with the desired properties? It seems to me that (1) will never be satisfied by any polynomial other than $x$. Am I missing something?
$endgroup$
– Jose27
Jul 28 '17 at 13:14