Why is a polynomial with infinite zeropoints the zeropolymomial? [duplicate]
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This question already has an answer here:
How to prove that a polynomial of degree $n$ has at most $n$ roots?
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This was given us as a fact, but why is this true? The zeropolynomial is the polynomial where all the coefficients are equal to $0$ if $R(x)$ is a polynomial over $mathbb{C}$ and every $xin mathbb{N}_0$ is a zeropoint then one can rewrite the polynomial $R(x)=x(x-1)(x-2)…(x-n)…$ but if we put an $xin mathbb{C}-mathbb{N}_0$ in $R(x)$, how does one know that $R(x)=0$?
algebra-precalculus polynomials
asked Dec 16 '18 at 12:32
RM777RM777
4029
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marked as duplicate by Dietrich Burde, Community♦ Dec 16 '18 at 12:54
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
How to prove that a polynomial of degree $n$ has at most $n$ roots?
8 answers
This was given us as a fact, but why is this true? The zeropolynomial is the polynomial where all the coefficients are equal to $0$ if $R(x)$ is a polynomial over $mathbb{C}$ and every $xin mathbb{N}_0$ is a zeropoint then one can rewrite the polynomial $R(x)=x(x-1)(x-2)…(x-n)…$ but if we put an $xin mathbb{C}-mathbb{N}_0$ in $R(x)$, how does one know that $R(x)=0$?
algebra-precalculus polynomials
asked Dec 16 '18 at 12:32
RM777RM777
4029
$endgroup$
marked as duplicate by Dietrich Burde, Community♦ Dec 16 '18 at 12:54
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
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A polynomial of n-th degree has only n zero points.
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– Aretino
Dec 16 '18 at 12:40
add a comment |
$begingroup$
This question already has an answer here:
How to prove that a polynomial of degree $n$ has at most $n$ roots?
8 answers
This was given us as a fact, but why is this true? The zeropolynomial is the polynomial where all the coefficients are equal to $0$ if $R(x)$ is a polynomial over $mathbb{C}$ and every $xin mathbb{N}_0$ is a zeropoint then one can rewrite the polynomial $R(x)=x(x-1)(x-2)…(x-n)…$ but if we put an $xin mathbb{C}-mathbb{N}_0$ in $R(x)$, how does one know that $R(x)=0$?
algebra-precalculus polynomials
asked Dec 16 '18 at 12:32
RM777RM777
4029
$endgroup$
This question already has an answer here:
How to prove that a polynomial of degree $n$ has at most $n$ roots?
8 answers
This was given us as a fact, but why is this true? The zeropolynomial is the polynomial where all the coefficients are equal to $0$ if $R(x)$ is a polynomial over $mathbb{C}$ and every $xin mathbb{N}_0$ is a zeropoint then one can rewrite the polynomial $R(x)=x(x-1)(x-2)…(x-n)…$ but if we put an $xin mathbb{C}-mathbb{N}_0$ in $R(x)$, how does one know that $R(x)=0$?
This question already has an answer here:
How to prove that a polynomial of degree $n$ has at most $n$ roots?
8 answers
algebra-precalculus polynomials
algebra-precalculus polynomials
asked Dec 16 '18 at 12:32
RM777RM777
4029
asked Dec 16 '18 at 12:32
RM777RM777
4029
asked Dec 16 '18 at 12:32
RM777RM777
4029
asked Dec 16 '18 at 12:32
RM777RM777
4029
asked Dec 16 '18 at 12:32
RM777RM777
4029
4029
marked as duplicate by Dietrich Burde, Community♦ Dec 16 '18 at 12:54
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Dietrich Burde, Community♦ Dec 16 '18 at 12:54
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
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A polynomial of n-th degree has only n zero points.
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– Aretino
Dec 16 '18 at 12:40
add a comment |
1
$begingroup$
A polynomial of n-th degree has only n zero points.
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– Aretino
Dec 16 '18 at 12:40
1
1
$begingroup$
A polynomial of n-th degree has only n zero points.
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– Aretino
Dec 16 '18 at 12:40
$begingroup$
A polynomial of n-th degree has only n zero points.
$endgroup$
– Aretino
Dec 16 '18 at 12:40
add a comment |
2 Answers
2
active
oldest
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A polynomial $f(x)$ has by definition a finite degree $n$ which is given by the highest degree $n$ of the variable $x$ involved in the polynomial. If you multiply two polynomials, the degrees add (if the underlying coefficient ring has no zero divisors as in the case of a field). Only the zero polynomial has all elements (in the coefficient ring or an extension ring) as zeros. A polynomial as described cannot exist.
answered Dec 16 '18 at 12:39
WuestenfuxWuestenfux
4,2161413
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By the Fundamental theorem of algebra, a polynomial $p$ of degree $n>0$ has exactly $n$ complex roots (counting multiplicity). Let $ r_1,dots,r_n$ be its root, then we must have $p(z)ne 0$ for all $z$ such that
$$
|z| > max{|r_1|,dots,|r_n|} =: R
$$
since all the roots lie in the set ${zinBbb C: |z|le R}$.
This means that the only polynomial with infinitely many zeroes is the consstant polynomial $pequiv0$.
answered Dec 16 '18 at 12:47
BigbearZzzBigbearZzz
8,49221652
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
A polynomial $f(x)$ has by definition a finite degree $n$ which is given by the highest degree $n$ of the variable $x$ involved in the polynomial. If you multiply two polynomials, the degrees add (if the underlying coefficient ring has no zero divisors as in the case of a field). Only the zero polynomial has all elements (in the coefficient ring or an extension ring) as zeros. A polynomial as described cannot exist.
answered Dec 16 '18 at 12:39
WuestenfuxWuestenfux
4,2161413
$endgroup$
add a comment |
$begingroup$
A polynomial $f(x)$ has by definition a finite degree $n$ which is given by the highest degree $n$ of the variable $x$ involved in the polynomial. If you multiply two polynomials, the degrees add (if the underlying coefficient ring has no zero divisors as in the case of a field). Only the zero polynomial has all elements (in the coefficient ring or an extension ring) as zeros. A polynomial as described cannot exist.
answered Dec 16 '18 at 12:39
WuestenfuxWuestenfux
4,2161413
$endgroup$
add a comment |
$begingroup$
A polynomial $f(x)$ has by definition a finite degree $n$ which is given by the highest degree $n$ of the variable $x$ involved in the polynomial. If you multiply two polynomials, the degrees add (if the underlying coefficient ring has no zero divisors as in the case of a field). Only the zero polynomial has all elements (in the coefficient ring or an extension ring) as zeros. A polynomial as described cannot exist.
answered Dec 16 '18 at 12:39
WuestenfuxWuestenfux
4,2161413
$endgroup$
A polynomial $f(x)$ has by definition a finite degree $n$ which is given by the highest degree $n$ of the variable $x$ involved in the polynomial. If you multiply two polynomials, the degrees add (if the underlying coefficient ring has no zero divisors as in the case of a field). Only the zero polynomial has all elements (in the coefficient ring or an extension ring) as zeros. A polynomial as described cannot exist.
answered Dec 16 '18 at 12:39
WuestenfuxWuestenfux
4,2161413
answered Dec 16 '18 at 12:39
WuestenfuxWuestenfux
4,2161413
answered Dec 16 '18 at 12:39
WuestenfuxWuestenfux
4,2161413
answered Dec 16 '18 at 12:39
WuestenfuxWuestenfux
4,2161413
4,2161413
add a comment |
add a comment |
$begingroup$
By the Fundamental theorem of algebra, a polynomial $p$ of degree $n>0$ has exactly $n$ complex roots (counting multiplicity). Let $ r_1,dots,r_n$ be its root, then we must have $p(z)ne 0$ for all $z$ such that
$$
|z| > max{|r_1|,dots,|r_n|} =: R
$$
since all the roots lie in the set ${zinBbb C: |z|le R}$.
This means that the only polynomial with infinitely many zeroes is the consstant polynomial $pequiv0$.
answered Dec 16 '18 at 12:47
BigbearZzzBigbearZzz
8,49221652
$endgroup$
add a comment |
$begingroup$
By the Fundamental theorem of algebra, a polynomial $p$ of degree $n>0$ has exactly $n$ complex roots (counting multiplicity). Let $ r_1,dots,r_n$ be its root, then we must have $p(z)ne 0$ for all $z$ such that
$$
|z| > max{|r_1|,dots,|r_n|} =: R
$$
since all the roots lie in the set ${zinBbb C: |z|le R}$.
This means that the only polynomial with infinitely many zeroes is the consstant polynomial $pequiv0$.
answered Dec 16 '18 at 12:47
BigbearZzzBigbearZzz
8,49221652
$endgroup$
add a comment |
$begingroup$
By the Fundamental theorem of algebra, a polynomial $p$ of degree $n>0$ has exactly $n$ complex roots (counting multiplicity). Let $ r_1,dots,r_n$ be its root, then we must have $p(z)ne 0$ for all $z$ such that
$$
|z| > max{|r_1|,dots,|r_n|} =: R
$$
since all the roots lie in the set ${zinBbb C: |z|le R}$.
This means that the only polynomial with infinitely many zeroes is the consstant polynomial $pequiv0$.
answered Dec 16 '18 at 12:47
BigbearZzzBigbearZzz
8,49221652
$endgroup$
By the Fundamental theorem of algebra, a polynomial $p$ of degree $n>0$ has exactly $n$ complex roots (counting multiplicity). Let $ r_1,dots,r_n$ be its root, then we must have $p(z)ne 0$ for all $z$ such that
$$
|z| > max{|r_1|,dots,|r_n|} =: R
$$
since all the roots lie in the set ${zinBbb C: |z|le R}$.
This means that the only polynomial with infinitely many zeroes is the consstant polynomial $pequiv0$.
answered Dec 16 '18 at 12:47
BigbearZzzBigbearZzz
8,49221652
answered Dec 16 '18 at 12:47
BigbearZzzBigbearZzz
8,49221652
answered Dec 16 '18 at 12:47
BigbearZzzBigbearZzz
8,49221652
answered Dec 16 '18 at 12:47
BigbearZzzBigbearZzz
8,49221652
8,49221652
add a comment |
add a comment |
1
$begingroup$
A polynomial of n-th degree has only n zero points.
$endgroup$
– Aretino
Dec 16 '18 at 12:40