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$begingroup$
I've read the definition of a polynomial on Wikipedia, and got quite a different understanding from what was explained to me in a Khan Academy video.
In the Khan Academy video it says that:
$6$
is a polynomial, specifically a monomial because it's the same as:
$6x^0$
I was wondering if this was true, and also whether:
$6 + 1$
is a binomial or
6 + 15 - 2
is a trinomial
The Wikipedia articles defines a polynomial as an:
...expression consisting of variables (also called indeterminates) and coefficients...
And gives examples of:
$x^2 − 4x + 7$
and
$x3 + 2xyz2 − yz + 1$
Something like I've shown higher up in my question, such as adding or subtracting 3 integers doesn't look like your typical polynomial. If 6 is a monomial and therefore a polynomial, then also 6 + 15 - 2 is a trinomial and polynomial, even if there are no explicitly written variables or indeterminates, as Wikipedia refers to them?
polynomials definition
asked Mar 13 at 23:07
ZebrafishZebrafish
21018
$endgroup$
|
show 4 more comments
$begingroup$
I've read the definition of a polynomial on Wikipedia, and got quite a different understanding from what was explained to me in a Khan Academy video.
In the Khan Academy video it says that:
$6$
is a polynomial, specifically a monomial because it's the same as:
$6x^0$
I was wondering if this was true, and also whether:
$6 + 1$
is a binomial or
6 + 15 - 2
is a trinomial
The Wikipedia articles defines a polynomial as an:
...expression consisting of variables (also called indeterminates) and coefficients...
And gives examples of:
$x^2 − 4x + 7$
and
$x3 + 2xyz2 − yz + 1$
Something like I've shown higher up in my question, such as adding or subtracting 3 integers doesn't look like your typical polynomial. If 6 is a monomial and therefore a polynomial, then also 6 + 15 - 2 is a trinomial and polynomial, even if there are no explicitly written variables or indeterminates, as Wikipedia refers to them?
polynomials definition
asked Mar 13 at 23:07
ZebrafishZebrafish
21018
$endgroup$
1
$begingroup$
No, $6+1=7$ is $7x_0$ as before, and $6+15-2=19$ is $19x_0$.
$endgroup$
– Dietrich Burde
Mar 13 at 23:08
2
$begingroup$
A polynomial can also be constant, so $6$ is a polynomial.
$endgroup$
– Peter
Mar 13 at 23:10
$begingroup$
@Dietrich Oh I see, 6 is a polynomial and 1 is a polynomial, but 6 + 1 isn't because it can be simplified and combined?
$endgroup$
– Zebrafish
Mar 13 at 23:11
1
$begingroup$
It is still a polynomial, but not a binomial
$endgroup$
– Peter
Mar 13 at 23:12
1
$begingroup$
I would say $6+1$ is a monomial. The same as $7x^0$. A polynomial has one coefficient for a power. If you have two for the same power, say $5x + x^2 + 3x$ then they are added and combined to one coefficient. So $5x + x^2 + 3x$ is the same as $x^2 + (5+3)x$. I, personally would say "$5x + x^2 + 3x$ is a polynomial" but I'd also say "it is of degree $2$ and it has two non-zero coefficients: $1$ is the coefficient of $x^2$ and $5+3$ is the coefficient of $x$". That's waht i'd say. Your instructor may say otherwise.
$endgroup$
– fleablood
Mar 13 at 23:29
|
show 4 more comments
$begingroup$
I've read the definition of a polynomial on Wikipedia, and got quite a different understanding from what was explained to me in a Khan Academy video.
In the Khan Academy video it says that:
$6$
is a polynomial, specifically a monomial because it's the same as:
$6x^0$
I was wondering if this was true, and also whether:
$6 + 1$
is a binomial or
6 + 15 - 2
is a trinomial
The Wikipedia articles defines a polynomial as an:
...expression consisting of variables (also called indeterminates) and coefficients...
And gives examples of:
$x^2 − 4x + 7$
and
$x3 + 2xyz2 − yz + 1$
Something like I've shown higher up in my question, such as adding or subtracting 3 integers doesn't look like your typical polynomial. If 6 is a monomial and therefore a polynomial, then also 6 + 15 - 2 is a trinomial and polynomial, even if there are no explicitly written variables or indeterminates, as Wikipedia refers to them?
polynomials definition
asked Mar 13 at 23:07
ZebrafishZebrafish
21018
$endgroup$
I've read the definition of a polynomial on Wikipedia, and got quite a different understanding from what was explained to me in a Khan Academy video.
In the Khan Academy video it says that:
$6$
is a polynomial, specifically a monomial because it's the same as:
$6x^0$
I was wondering if this was true, and also whether:
$6 + 1$
is a binomial or
6 + 15 - 2
is a trinomial
The Wikipedia articles defines a polynomial as an:
...expression consisting of variables (also called indeterminates) and coefficients...
And gives examples of:
$x^2 − 4x + 7$
and
$x3 + 2xyz2 − yz + 1$
Something like I've shown higher up in my question, such as adding or subtracting 3 integers doesn't look like your typical polynomial. If 6 is a monomial and therefore a polynomial, then also 6 + 15 - 2 is a trinomial and polynomial, even if there are no explicitly written variables or indeterminates, as Wikipedia refers to them?
polynomials definition
polynomials definition
asked Mar 13 at 23:07
ZebrafishZebrafish
21018
asked Mar 13 at 23:07
ZebrafishZebrafish
21018
asked Mar 13 at 23:07
ZebrafishZebrafish
21018
asked Mar 13 at 23:07
ZebrafishZebrafish
21018
asked Mar 13 at 23:07
ZebrafishZebrafish
21018
21018
1
$begingroup$
No, $6+1=7$ is $7x_0$ as before, and $6+15-2=19$ is $19x_0$.
$endgroup$
– Dietrich Burde
Mar 13 at 23:08
2
$begingroup$
A polynomial can also be constant, so $6$ is a polynomial.
$endgroup$
– Peter
Mar 13 at 23:10
$begingroup$
@Dietrich Oh I see, 6 is a polynomial and 1 is a polynomial, but 6 + 1 isn't because it can be simplified and combined?
$endgroup$
– Zebrafish
Mar 13 at 23:11
1
$begingroup$
It is still a polynomial, but not a binomial
$endgroup$
– Peter
Mar 13 at 23:12
1
$begingroup$
I would say $6+1$ is a monomial. The same as $7x^0$. A polynomial has one coefficient for a power. If you have two for the same power, say $5x + x^2 + 3x$ then they are added and combined to one coefficient. So $5x + x^2 + 3x$ is the same as $x^2 + (5+3)x$. I, personally would say "$5x + x^2 + 3x$ is a polynomial" but I'd also say "it is of degree $2$ and it has two non-zero coefficients: $1$ is the coefficient of $x^2$ and $5+3$ is the coefficient of $x$". That's waht i'd say. Your instructor may say otherwise.
$endgroup$
– fleablood
Mar 13 at 23:29
|
show 4 more comments
1
$begingroup$
No, $6+1=7$ is $7x_0$ as before, and $6+15-2=19$ is $19x_0$.
$endgroup$
– Dietrich Burde
Mar 13 at 23:08
2
$begingroup$
A polynomial can also be constant, so $6$ is a polynomial.
$endgroup$
– Peter
Mar 13 at 23:10
$begingroup$
@Dietrich Oh I see, 6 is a polynomial and 1 is a polynomial, but 6 + 1 isn't because it can be simplified and combined?
$endgroup$
– Zebrafish
Mar 13 at 23:11
1
$begingroup$
It is still a polynomial, but not a binomial
$endgroup$
– Peter
Mar 13 at 23:12
1
$begingroup$
I would say $6+1$ is a monomial. The same as $7x^0$. A polynomial has one coefficient for a power. If you have two for the same power, say $5x + x^2 + 3x$ then they are added and combined to one coefficient. So $5x + x^2 + 3x$ is the same as $x^2 + (5+3)x$. I, personally would say "$5x + x^2 + 3x$ is a polynomial" but I'd also say "it is of degree $2$ and it has two non-zero coefficients: $1$ is the coefficient of $x^2$ and $5+3$ is the coefficient of $x$". That's waht i'd say. Your instructor may say otherwise.
$endgroup$
– fleablood
Mar 13 at 23:29
1
1
$begingroup$
No, $6+1=7$ is $7x_0$ as before, and $6+15-2=19$ is $19x_0$.
$endgroup$
– Dietrich Burde
Mar 13 at 23:08
$begingroup$
No, $6+1=7$ is $7x_0$ as before, and $6+15-2=19$ is $19x_0$.
$endgroup$
– Dietrich Burde
Mar 13 at 23:08
2
2
$begingroup$
A polynomial can also be constant, so $6$ is a polynomial.
$endgroup$
– Peter
Mar 13 at 23:10
$begingroup$
A polynomial can also be constant, so $6$ is a polynomial.
$endgroup$
– Peter
Mar 13 at 23:10
$begingroup$
@Dietrich Oh I see, 6 is a polynomial and 1 is a polynomial, but 6 + 1 isn't because it can be simplified and combined?
$endgroup$
– Zebrafish
Mar 13 at 23:11
$begingroup$
@Dietrich Oh I see, 6 is a polynomial and 1 is a polynomial, but 6 + 1 isn't because it can be simplified and combined?
$endgroup$
– Zebrafish
Mar 13 at 23:11
1
1
$begingroup$
It is still a polynomial, but not a binomial
$endgroup$
– Peter
Mar 13 at 23:12
$begingroup$
It is still a polynomial, but not a binomial
$endgroup$
– Peter
Mar 13 at 23:12
1
1
$begingroup$
I would say $6+1$ is a monomial. The same as $7x^0$. A polynomial has one coefficient for a power. If you have two for the same power, say $5x + x^2 + 3x$ then they are added and combined to one coefficient. So $5x + x^2 + 3x$ is the same as $x^2 + (5+3)x$. I, personally would say "$5x + x^2 + 3x$ is a polynomial" but I'd also say "it is of degree $2$ and it has two non-zero coefficients: $1$ is the coefficient of $x^2$ and $5+3$ is the coefficient of $x$". That's waht i'd say. Your instructor may say otherwise.
$endgroup$
– fleablood
Mar 13 at 23:29
$begingroup$
I would say $6+1$ is a monomial. The same as $7x^0$. A polynomial has one coefficient for a power. If you have two for the same power, say $5x + x^2 + 3x$ then they are added and combined to one coefficient. So $5x + x^2 + 3x$ is the same as $x^2 + (5+3)x$. I, personally would say "$5x + x^2 + 3x$ is a polynomial" but I'd also say "it is of degree $2$ and it has two non-zero coefficients: $1$ is the coefficient of $x^2$ and $5+3$ is the coefficient of $x$". That's waht i'd say. Your instructor may say otherwise.
$endgroup$
– fleablood
Mar 13 at 23:29
|
show 4 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Regarding the one variable case, if you want to be extra rigorous, one can define polynomials as lists of numbers $(a_0,a_1, dots)$ such that the $a_i$ eventually start to be all zero. This is just a fancy way of saying that $3$ can be defined as $(3,0,0,dots)$ and $3X+X^5$ as $(0,3,0,0,0,1,0,0,dots)$, i.e. grouping all the coefficients that belong to a same power together.
What this means is that even if we informally have different terms with the same variable, at the time of assessing certain properties of a polynomial, one should write it into this "canonical" form. For example, we could write $7$ as $6+1$ or $X^2 + 4X$ as $X^2 + X + 3X$ or even $X(X+4)$. But $6+1$ is a monomial, because after regrouping, it corresponds to $(7,0,0, dots)$. In the same way, $X^2 + X + 3X$ consists of two monomials, because it corresponds to $(0,4,1,0,0,dots)$.
answered Mar 13 at 23:23
Guido A.Guido A.
8,0701730
$endgroup$
1
$begingroup$
We could also say that $6+1=6X^0 + 1Y^0$.
$endgroup$
– enedil
Mar 14 at 0:03
$begingroup$
Yeah, I avoided the multivariate case explicitly but still, one could index $mathbb{N}_0^k$ for $R[X_1,dots,X_k]$ as a countable list and regroup coefficients that way, so that in your example these belong to the same group (namely the place of the zero $k$-uple).
$endgroup$
– Guido A.
Mar 14 at 2:07
add a comment |
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$begingroup$
Regarding the one variable case, if you want to be extra rigorous, one can define polynomials as lists of numbers $(a_0,a_1, dots)$ such that the $a_i$ eventually start to be all zero. This is just a fancy way of saying that $3$ can be defined as $(3,0,0,dots)$ and $3X+X^5$ as $(0,3,0,0,0,1,0,0,dots)$, i.e. grouping all the coefficients that belong to a same power together.
What this means is that even if we informally have different terms with the same variable, at the time of assessing certain properties of a polynomial, one should write it into this "canonical" form. For example, we could write $7$ as $6+1$ or $X^2 + 4X$ as $X^2 + X + 3X$ or even $X(X+4)$. But $6+1$ is a monomial, because after regrouping, it corresponds to $(7,0,0, dots)$. In the same way, $X^2 + X + 3X$ consists of two monomials, because it corresponds to $(0,4,1,0,0,dots)$.
answered Mar 13 at 23:23
Guido A.Guido A.
8,0701730
$endgroup$
1
$begingroup$
We could also say that $6+1=6X^0 + 1Y^0$.
$endgroup$
– enedil
Mar 14 at 0:03
$begingroup$
Yeah, I avoided the multivariate case explicitly but still, one could index $mathbb{N}_0^k$ for $R[X_1,dots,X_k]$ as a countable list and regroup coefficients that way, so that in your example these belong to the same group (namely the place of the zero $k$-uple).
$endgroup$
– Guido A.
Mar 14 at 2:07
add a comment |
$begingroup$
Regarding the one variable case, if you want to be extra rigorous, one can define polynomials as lists of numbers $(a_0,a_1, dots)$ such that the $a_i$ eventually start to be all zero. This is just a fancy way of saying that $3$ can be defined as $(3,0,0,dots)$ and $3X+X^5$ as $(0,3,0,0,0,1,0,0,dots)$, i.e. grouping all the coefficients that belong to a same power together.
What this means is that even if we informally have different terms with the same variable, at the time of assessing certain properties of a polynomial, one should write it into this "canonical" form. For example, we could write $7$ as $6+1$ or $X^2 + 4X$ as $X^2 + X + 3X$ or even $X(X+4)$. But $6+1$ is a monomial, because after regrouping, it corresponds to $(7,0,0, dots)$. In the same way, $X^2 + X + 3X$ consists of two monomials, because it corresponds to $(0,4,1,0,0,dots)$.
answered Mar 13 at 23:23
Guido A.Guido A.
8,0701730
$endgroup$
1
$begingroup$
We could also say that $6+1=6X^0 + 1Y^0$.
$endgroup$
– enedil
Mar 14 at 0:03
$begingroup$
Yeah, I avoided the multivariate case explicitly but still, one could index $mathbb{N}_0^k$ for $R[X_1,dots,X_k]$ as a countable list and regroup coefficients that way, so that in your example these belong to the same group (namely the place of the zero $k$-uple).
$endgroup$
– Guido A.
Mar 14 at 2:07
add a comment |
$begingroup$
Regarding the one variable case, if you want to be extra rigorous, one can define polynomials as lists of numbers $(a_0,a_1, dots)$ such that the $a_i$ eventually start to be all zero. This is just a fancy way of saying that $3$ can be defined as $(3,0,0,dots)$ and $3X+X^5$ as $(0,3,0,0,0,1,0,0,dots)$, i.e. grouping all the coefficients that belong to a same power together.
What this means is that even if we informally have different terms with the same variable, at the time of assessing certain properties of a polynomial, one should write it into this "canonical" form. For example, we could write $7$ as $6+1$ or $X^2 + 4X$ as $X^2 + X + 3X$ or even $X(X+4)$. But $6+1$ is a monomial, because after regrouping, it corresponds to $(7,0,0, dots)$. In the same way, $X^2 + X + 3X$ consists of two monomials, because it corresponds to $(0,4,1,0,0,dots)$.
answered Mar 13 at 23:23
Guido A.Guido A.
8,0701730
$endgroup$
Regarding the one variable case, if you want to be extra rigorous, one can define polynomials as lists of numbers $(a_0,a_1, dots)$ such that the $a_i$ eventually start to be all zero. This is just a fancy way of saying that $3$ can be defined as $(3,0,0,dots)$ and $3X+X^5$ as $(0,3,0,0,0,1,0,0,dots)$, i.e. grouping all the coefficients that belong to a same power together.
What this means is that even if we informally have different terms with the same variable, at the time of assessing certain properties of a polynomial, one should write it into this "canonical" form. For example, we could write $7$ as $6+1$ or $X^2 + 4X$ as $X^2 + X + 3X$ or even $X(X+4)$. But $6+1$ is a monomial, because after regrouping, it corresponds to $(7,0,0, dots)$. In the same way, $X^2 + X + 3X$ consists of two monomials, because it corresponds to $(0,4,1,0,0,dots)$.
answered Mar 13 at 23:23
Guido A.Guido A.
8,0701730
answered Mar 13 at 23:23
Guido A.Guido A.
8,0701730
answered Mar 13 at 23:23
Guido A.Guido A.
8,0701730
answered Mar 13 at 23:23
Guido A.Guido A.
8,0701730
8,0701730
1
$begingroup$
We could also say that $6+1=6X^0 + 1Y^0$.
$endgroup$
– enedil
Mar 14 at 0:03
$begingroup$
Yeah, I avoided the multivariate case explicitly but still, one could index $mathbb{N}_0^k$ for $R[X_1,dots,X_k]$ as a countable list and regroup coefficients that way, so that in your example these belong to the same group (namely the place of the zero $k$-uple).
$endgroup$
– Guido A.
Mar 14 at 2:07
add a comment |
1
$begingroup$
We could also say that $6+1=6X^0 + 1Y^0$.
$endgroup$
– enedil
Mar 14 at 0:03
$begingroup$
Yeah, I avoided the multivariate case explicitly but still, one could index $mathbb{N}_0^k$ for $R[X_1,dots,X_k]$ as a countable list and regroup coefficients that way, so that in your example these belong to the same group (namely the place of the zero $k$-uple).
$endgroup$
– Guido A.
Mar 14 at 2:07
1
1
$begingroup$
We could also say that $6+1=6X^0 + 1Y^0$.
$endgroup$
– enedil
Mar 14 at 0:03
$begingroup$
We could also say that $6+1=6X^0 + 1Y^0$.
$endgroup$
– enedil
Mar 14 at 0:03
$begingroup$
Yeah, I avoided the multivariate case explicitly but still, one could index $mathbb{N}_0^k$ for $R[X_1,dots,X_k]$ as a countable list and regroup coefficients that way, so that in your example these belong to the same group (namely the place of the zero $k$-uple).
$endgroup$
– Guido A.
Mar 14 at 2:07
$begingroup$
Yeah, I avoided the multivariate case explicitly but still, one could index $mathbb{N}_0^k$ for $R[X_1,dots,X_k]$ as a countable list and regroup coefficients that way, so that in your example these belong to the same group (namely the place of the zero $k$-uple).
$endgroup$
– Guido A.
Mar 14 at 2:07
add a comment |
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1
$begingroup$
No, $6+1=7$ is $7x_0$ as before, and $6+15-2=19$ is $19x_0$.
$endgroup$
– Dietrich Burde
Mar 13 at 23:08
2
$begingroup$
A polynomial can also be constant, so $6$ is a polynomial.
$endgroup$
– Peter
Mar 13 at 23:10
$begingroup$
@Dietrich Oh I see, 6 is a polynomial and 1 is a polynomial, but 6 + 1 isn't because it can be simplified and combined?
$endgroup$
– Zebrafish
Mar 13 at 23:11
1
$begingroup$
It is still a polynomial, but not a binomial
$endgroup$
– Peter
Mar 13 at 23:12
1
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I would say $6+1$ is a monomial. The same as $7x^0$. A polynomial has one coefficient for a power. If you have two for the same power, say $5x + x^2 + 3x$ then they are added and combined to one coefficient. So $5x + x^2 + 3x$ is the same as $x^2 + (5+3)x$. I, personally would say "$5x + x^2 + 3x$ is a polynomial" but I'd also say "it is of degree $2$ and it has two non-zero coefficients: $1$ is the coefficient of $x^2$ and $5+3$ is the coefficient of $x$". That's waht i'd say. Your instructor may say otherwise.
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– fleablood
Mar 13 at 23:29