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What exactly is in the polynomial vector space?
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I am having some trouble understanding the polynomial vector space notation: $P_m(F)$ which means the set of all polynomials that take inputs from the space $F$ and output to the space $F$ with coefficients in $F$ if there exits $a_0, ...,a_m$ $epsilon F$ such that $p(z)=a_0+a_1z+a_2z^2+...+a_mz^m$
What does it really mean "if there exist $a_0, ...,a_m$ $epsilon F $? Can't we always include some arbitrary coefficient values to a polynomial?
And, since $p(z)=a_0+a_1z+a_2z^2+...+a_mz^m$ is just one huge sum, is it just one huge polynomial? I'm also confused as to whether or not it includes just the polynomials (i.e. x^2) or the polynomials evaluated at all possible values as well? (i.e. $(0)^2, (1)^2, (2)^2, etc.$)
And, following polynomials of this form, it doesn't seem to inlclude something like "$x+5$", as it seems to refer to functions in the form of $x^0, x^1,...x^n$".
I'm new to linear algebra so please bear with me. Can someone please explain this notation and the contents of this set?
linear-algebra polynomials vector-spaces
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Jaigus
1576
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I am having some trouble understanding the polynomial vector space notation: $P_m(F)$ which means the set of all polynomials that take inputs from the space $F$ and output to the space $F$ with coefficients in $F$ if there exits $a_0, ...,a_m$ $epsilon F$ such that $p(z)=a_0+a_1z+a_2z^2+...+a_mz^m$
What does it really mean "if there exist $a_0, ...,a_m$ $epsilon F $? Can't we always include some arbitrary coefficient values to a polynomial?
And, since $p(z)=a_0+a_1z+a_2z^2+...+a_mz^m$ is just one huge sum, is it just one huge polynomial? I'm also confused as to whether or not it includes just the polynomials (i.e. x^2) or the polynomials evaluated at all possible values as well? (i.e. $(0)^2, (1)^2, (2)^2, etc.$)
And, following polynomials of this form, it doesn't seem to inlclude something like "$x+5$", as it seems to refer to functions in the form of $x^0, x^1,...x^n$".
I'm new to linear algebra so please bear with me. Can someone please explain this notation and the contents of this set?
linear-algebra polynomials vector-spaces
asked 1 hour ago
Jaigus
1576
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I am having some trouble understanding the polynomial vector space notation: $P_m(F)$ which means the set of all polynomials that take inputs from the space $F$ and output to the space $F$ with coefficients in $F$ if there exits $a_0, ...,a_m$ $epsilon F$ such that $p(z)=a_0+a_1z+a_2z^2+...+a_mz^m$
What does it really mean "if there exist $a_0, ...,a_m$ $epsilon F $? Can't we always include some arbitrary coefficient values to a polynomial?
And, since $p(z)=a_0+a_1z+a_2z^2+...+a_mz^m$ is just one huge sum, is it just one huge polynomial? I'm also confused as to whether or not it includes just the polynomials (i.e. x^2) or the polynomials evaluated at all possible values as well? (i.e. $(0)^2, (1)^2, (2)^2, etc.$)
And, following polynomials of this form, it doesn't seem to inlclude something like "$x+5$", as it seems to refer to functions in the form of $x^0, x^1,...x^n$".
I'm new to linear algebra so please bear with me. Can someone please explain this notation and the contents of this set?
linear-algebra polynomials vector-spaces
asked 1 hour ago
Jaigus
1576
I am having some trouble understanding the polynomial vector space notation: $P_m(F)$ which means the set of all polynomials that take inputs from the space $F$ and output to the space $F$ with coefficients in $F$ if there exits $a_0, ...,a_m$ $epsilon F$ such that $p(z)=a_0+a_1z+a_2z^2+...+a_mz^m$
What does it really mean "if there exist $a_0, ...,a_m$ $epsilon F $? Can't we always include some arbitrary coefficient values to a polynomial?
And, since $p(z)=a_0+a_1z+a_2z^2+...+a_mz^m$ is just one huge sum, is it just one huge polynomial? I'm also confused as to whether or not it includes just the polynomials (i.e. x^2) or the polynomials evaluated at all possible values as well? (i.e. $(0)^2, (1)^2, (2)^2, etc.$)
And, following polynomials of this form, it doesn't seem to inlclude something like "$x+5$", as it seems to refer to functions in the form of $x^0, x^1,...x^n$".
I'm new to linear algebra so please bear with me. Can someone please explain this notation and the contents of this set?
linear-algebra polynomials vector-spaces
linear-algebra polynomials vector-spaces
asked 1 hour ago
Jaigus
1576
asked 1 hour ago
Jaigus
1576
asked 1 hour ago
Jaigus
1576
asked 1 hour ago
Jaigus
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Jaigus
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1576
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3 Answers
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A polynomial with coefficients in the field $F$ in the variable $X$ is an expression of the sort
$$
P(X)=a_o+a_1X+a_2X^2+cdots+a_dX^d
$$
where the coefficients $a_iin F$. Polynomials can be added and multiplied in the usual way to give other polynomials as result. A polynomial can also be multiplied by a scalar (i.e. an element of $F$) and a result is a polynomial.
(In fact there's a more formal definition of polynomials, but let's ignore it for now)
There's a notion of degree of a polynomial. The polynomial $P(X)$ written above has degree $deg(P(X))=d$ provided $a_dneq0$. The degree of a polynomial is well-behaved under the operations described above, namely
$$
deg(P+Q)leqmax(deg(P),deg(Q)), deg(PQ)=deg(P)+deg(Q), deg(aP)=deg(P)
$$
as long as $aneq0$. The first and last such relations for the degree imply that if we let
$$
P_m(X)=textpolynomials $P(X)$ such that $deg(P)leq m$
$$
then $P_m(X)$ is closed under addition and multiplication by scalars. A straightforward exercise shows then that $P_m(X)$ is a linear space.
answered 1 hour ago
Andrea Mori
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You should not interprete polynomials over a field $F$ as a map of a function from $F$ to $F$.
A polynomial is a formal linear combination over $F$ of the monomials $x^i$. So, first of all, all monomials are also polynomials. Examples are $x^2, x^3, x$ or $0$. Linear combinations of those monomials are finite sums of such monomials, where you are allowed to multiply each monomial with a coefficient $a_i in F$. For $F=mathbb R$, examples are $x^2, 3x^2-5, x^3-7x$ or $x^6+x^2-4$. One important point is that you do not allow infinite sums of monomials.
You can add polynomials (as it is neccesary for a vector space). If you have polynomials $p(x)=a_0 + a_1 x + a_2 x^2 + dots a_n x^n$ and $q(x) = b_0 + b_1x + dots +b_mx^m$ (so all $a_i$ and $b_i$ are elements of $F$), you can add those two to obtain
$$(p+q)(x) = sum_i=0^max(m,n) (a_i+b_i)x^i$$
where we set all $a_i$ and $b_i$ not already defined to $0$.
You can also multiply $p$ with a scalar $alpha in F$:
$$(alpha p)(x) = sum_i=1^n (alpha a_i)x^i$$
So why is a polynomial not a map?
Given a polynomial $p(x)in F[x]$, we can define a map $sigma_pcolon F to F, a mapsto p(a)$. If you think of the polynomial $x^2$ as a parabola, you consider this map. But this map is not the same as $p(x)$. For example, take $F=mathbb Z / 2mathbb Z$, the field with two elements. Then the polynomials $p(x) = x^2$ and $q(x) = x$ are different (since they have different coefficients), but the evaluation maps $sigma_pcolon mathbb Z / 2mathbb Z to mathbb Z / 2mathbb Z, xmapsto x^2$ and $sigma_q colon mathbb Z / 2mathbb Z to mathbb Z / 2mathbb Z, xmapsto x$ are the same (since $x^2=x$ for all $x in mathbb Z / 2mathbb Z$).
answered 1 hour ago
Babelfish
1,026115
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There are a lot of questions and potential confusions here, I will try to clarify some of them with the use of examples.
$P_1(mathbbR)$ is the set of all linear functions with real coefficients, i.e. $ax^1+b x^0$ where $a$ and $b$ are real. The polynomial $x+5$ is an element of this space because $x+5= 1x^1+ 5 x^0$.
On the other hand, $P_2(mathbbC)$ is the set of all quadratic polynomials with complex coefficients. An example element in this space is the polynomial $3x^2+(2+i)x+3$. Another example is $3x^2+0x+1$, which is usually written as $3x^2+1$. This example is actually also an element of $P_2(mathbbR)$ too, because all the coefficients are real.
answered 1 hour ago
Manano
1217
Thank you! The examples you've shown made it click. So we should simply look at this huge sum as an descriptive tool we use to express other polynomials just like conventional vectors. So, $P_n(mathbbF)$ can express anything in $P_n-1(mathbbF)$ by simply placing a 0 (or 0's if necessary) as a coefficient(s) in the term having nth degree (or all terms having less than a degree of n).
– Jaigus
26 mins ago
How about the statement: "$p_0, p_1, ..., p_m$ in $P_m(mathbb F)$ such that $p_j(5) = 0$ for each j"? This refers to all the polynomials in the set that evaluate to 0 at 5, yet does the subscript "j" in this statement mean each polynomial of degree j that evaluates to 0 at 5 or just each of the polynomials that evaluate to 0 at 5?
– Jaigus
9 mins ago
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
A polynomial with coefficients in the field $F$ in the variable $X$ is an expression of the sort
$$
P(X)=a_o+a_1X+a_2X^2+cdots+a_dX^d
$$
where the coefficients $a_iin F$. Polynomials can be added and multiplied in the usual way to give other polynomials as result. A polynomial can also be multiplied by a scalar (i.e. an element of $F$) and a result is a polynomial.
(In fact there's a more formal definition of polynomials, but let's ignore it for now)
There's a notion of degree of a polynomial. The polynomial $P(X)$ written above has degree $deg(P(X))=d$ provided $a_dneq0$. The degree of a polynomial is well-behaved under the operations described above, namely
$$
deg(P+Q)leqmax(deg(P),deg(Q)), deg(PQ)=deg(P)+deg(Q), deg(aP)=deg(P)
$$
as long as $aneq0$. The first and last such relations for the degree imply that if we let
$$
P_m(X)=textpolynomials $P(X)$ such that $deg(P)leq m$
$$
then $P_m(X)$ is closed under addition and multiplication by scalars. A straightforward exercise shows then that $P_m(X)$ is a linear space.
answered 1 hour ago
Andrea Mori
19k13464
add a comment |Â
up vote
3
down vote
A polynomial with coefficients in the field $F$ in the variable $X$ is an expression of the sort
$$
P(X)=a_o+a_1X+a_2X^2+cdots+a_dX^d
$$
where the coefficients $a_iin F$. Polynomials can be added and multiplied in the usual way to give other polynomials as result. A polynomial can also be multiplied by a scalar (i.e. an element of $F$) and a result is a polynomial.
(In fact there's a more formal definition of polynomials, but let's ignore it for now)
There's a notion of degree of a polynomial. The polynomial $P(X)$ written above has degree $deg(P(X))=d$ provided $a_dneq0$. The degree of a polynomial is well-behaved under the operations described above, namely
$$
deg(P+Q)leqmax(deg(P),deg(Q)), deg(PQ)=deg(P)+deg(Q), deg(aP)=deg(P)
$$
as long as $aneq0$. The first and last such relations for the degree imply that if we let
$$
P_m(X)=textpolynomials $P(X)$ such that $deg(P)leq m$
$$
then $P_m(X)$ is closed under addition and multiplication by scalars. A straightforward exercise shows then that $P_m(X)$ is a linear space.
answered 1 hour ago
Andrea Mori
19k13464
add a comment |Â
up vote
3
down vote
up vote
3
down vote
A polynomial with coefficients in the field $F$ in the variable $X$ is an expression of the sort
$$
P(X)=a_o+a_1X+a_2X^2+cdots+a_dX^d
$$
where the coefficients $a_iin F$. Polynomials can be added and multiplied in the usual way to give other polynomials as result. A polynomial can also be multiplied by a scalar (i.e. an element of $F$) and a result is a polynomial.
(In fact there's a more formal definition of polynomials, but let's ignore it for now)
There's a notion of degree of a polynomial. The polynomial $P(X)$ written above has degree $deg(P(X))=d$ provided $a_dneq0$. The degree of a polynomial is well-behaved under the operations described above, namely
$$
deg(P+Q)leqmax(deg(P),deg(Q)), deg(PQ)=deg(P)+deg(Q), deg(aP)=deg(P)
$$
as long as $aneq0$. The first and last such relations for the degree imply that if we let
$$
P_m(X)=textpolynomials $P(X)$ such that $deg(P)leq m$
$$
then $P_m(X)$ is closed under addition and multiplication by scalars. A straightforward exercise shows then that $P_m(X)$ is a linear space.
answered 1 hour ago
Andrea Mori
19k13464
A polynomial with coefficients in the field $F$ in the variable $X$ is an expression of the sort
$$
P(X)=a_o+a_1X+a_2X^2+cdots+a_dX^d
$$
where the coefficients $a_iin F$. Polynomials can be added and multiplied in the usual way to give other polynomials as result. A polynomial can also be multiplied by a scalar (i.e. an element of $F$) and a result is a polynomial.
(In fact there's a more formal definition of polynomials, but let's ignore it for now)
There's a notion of degree of a polynomial. The polynomial $P(X)$ written above has degree $deg(P(X))=d$ provided $a_dneq0$. The degree of a polynomial is well-behaved under the operations described above, namely
$$
deg(P+Q)leqmax(deg(P),deg(Q)), deg(PQ)=deg(P)+deg(Q), deg(aP)=deg(P)
$$
as long as $aneq0$. The first and last such relations for the degree imply that if we let
$$
P_m(X)=textpolynomials $P(X)$ such that $deg(P)leq m$
$$
then $P_m(X)$ is closed under addition and multiplication by scalars. A straightforward exercise shows then that $P_m(X)$ is a linear space.
answered 1 hour ago
Andrea Mori
19k13464
answered 1 hour ago
Andrea Mori
19k13464
answered 1 hour ago
Andrea Mori
19k13464
answered 1 hour ago
Andrea Mori
19k13464
19k13464
add a comment |Â
add a comment |Â
up vote
2
down vote
You should not interprete polynomials over a field $F$ as a map of a function from $F$ to $F$.
A polynomial is a formal linear combination over $F$ of the monomials $x^i$. So, first of all, all monomials are also polynomials. Examples are $x^2, x^3, x$ or $0$. Linear combinations of those monomials are finite sums of such monomials, where you are allowed to multiply each monomial with a coefficient $a_i in F$. For $F=mathbb R$, examples are $x^2, 3x^2-5, x^3-7x$ or $x^6+x^2-4$. One important point is that you do not allow infinite sums of monomials.
You can add polynomials (as it is neccesary for a vector space). If you have polynomials $p(x)=a_0 + a_1 x + a_2 x^2 + dots a_n x^n$ and $q(x) = b_0 + b_1x + dots +b_mx^m$ (so all $a_i$ and $b_i$ are elements of $F$), you can add those two to obtain
$$(p+q)(x) = sum_i=0^max(m,n) (a_i+b_i)x^i$$
where we set all $a_i$ and $b_i$ not already defined to $0$.
You can also multiply $p$ with a scalar $alpha in F$:
$$(alpha p)(x) = sum_i=1^n (alpha a_i)x^i$$
So why is a polynomial not a map?
Given a polynomial $p(x)in F[x]$, we can define a map $sigma_pcolon F to F, a mapsto p(a)$. If you think of the polynomial $x^2$ as a parabola, you consider this map. But this map is not the same as $p(x)$. For example, take $F=mathbb Z / 2mathbb Z$, the field with two elements. Then the polynomials $p(x) = x^2$ and $q(x) = x$ are different (since they have different coefficients), but the evaluation maps $sigma_pcolon mathbb Z / 2mathbb Z to mathbb Z / 2mathbb Z, xmapsto x^2$ and $sigma_q colon mathbb Z / 2mathbb Z to mathbb Z / 2mathbb Z, xmapsto x$ are the same (since $x^2=x$ for all $x in mathbb Z / 2mathbb Z$).
answered 1 hour ago
Babelfish
1,026115
add a comment |Â
up vote
2
down vote
You should not interprete polynomials over a field $F$ as a map of a function from $F$ to $F$.
A polynomial is a formal linear combination over $F$ of the monomials $x^i$. So, first of all, all monomials are also polynomials. Examples are $x^2, x^3, x$ or $0$. Linear combinations of those monomials are finite sums of such monomials, where you are allowed to multiply each monomial with a coefficient $a_i in F$. For $F=mathbb R$, examples are $x^2, 3x^2-5, x^3-7x$ or $x^6+x^2-4$. One important point is that you do not allow infinite sums of monomials.
You can add polynomials (as it is neccesary for a vector space). If you have polynomials $p(x)=a_0 + a_1 x + a_2 x^2 + dots a_n x^n$ and $q(x) = b_0 + b_1x + dots +b_mx^m$ (so all $a_i$ and $b_i$ are elements of $F$), you can add those two to obtain
$$(p+q)(x) = sum_i=0^max(m,n) (a_i+b_i)x^i$$
where we set all $a_i$ and $b_i$ not already defined to $0$.
You can also multiply $p$ with a scalar $alpha in F$:
$$(alpha p)(x) = sum_i=1^n (alpha a_i)x^i$$
So why is a polynomial not a map?
Given a polynomial $p(x)in F[x]$, we can define a map $sigma_pcolon F to F, a mapsto p(a)$. If you think of the polynomial $x^2$ as a parabola, you consider this map. But this map is not the same as $p(x)$. For example, take $F=mathbb Z / 2mathbb Z$, the field with two elements. Then the polynomials $p(x) = x^2$ and $q(x) = x$ are different (since they have different coefficients), but the evaluation maps $sigma_pcolon mathbb Z / 2mathbb Z to mathbb Z / 2mathbb Z, xmapsto x^2$ and $sigma_q colon mathbb Z / 2mathbb Z to mathbb Z / 2mathbb Z, xmapsto x$ are the same (since $x^2=x$ for all $x in mathbb Z / 2mathbb Z$).
answered 1 hour ago
Babelfish
1,026115
add a comment |Â
up vote
2
down vote
up vote
2
down vote
You should not interprete polynomials over a field $F$ as a map of a function from $F$ to $F$.
A polynomial is a formal linear combination over $F$ of the monomials $x^i$. So, first of all, all monomials are also polynomials. Examples are $x^2, x^3, x$ or $0$. Linear combinations of those monomials are finite sums of such monomials, where you are allowed to multiply each monomial with a coefficient $a_i in F$. For $F=mathbb R$, examples are $x^2, 3x^2-5, x^3-7x$ or $x^6+x^2-4$. One important point is that you do not allow infinite sums of monomials.
You can add polynomials (as it is neccesary for a vector space). If you have polynomials $p(x)=a_0 + a_1 x + a_2 x^2 + dots a_n x^n$ and $q(x) = b_0 + b_1x + dots +b_mx^m$ (so all $a_i$ and $b_i$ are elements of $F$), you can add those two to obtain
$$(p+q)(x) = sum_i=0^max(m,n) (a_i+b_i)x^i$$
where we set all $a_i$ and $b_i$ not already defined to $0$.
You can also multiply $p$ with a scalar $alpha in F$:
$$(alpha p)(x) = sum_i=1^n (alpha a_i)x^i$$
So why is a polynomial not a map?
Given a polynomial $p(x)in F[x]$, we can define a map $sigma_pcolon F to F, a mapsto p(a)$. If you think of the polynomial $x^2$ as a parabola, you consider this map. But this map is not the same as $p(x)$. For example, take $F=mathbb Z / 2mathbb Z$, the field with two elements. Then the polynomials $p(x) = x^2$ and $q(x) = x$ are different (since they have different coefficients), but the evaluation maps $sigma_pcolon mathbb Z / 2mathbb Z to mathbb Z / 2mathbb Z, xmapsto x^2$ and $sigma_q colon mathbb Z / 2mathbb Z to mathbb Z / 2mathbb Z, xmapsto x$ are the same (since $x^2=x$ for all $x in mathbb Z / 2mathbb Z$).
answered 1 hour ago
Babelfish
1,026115
You should not interprete polynomials over a field $F$ as a map of a function from $F$ to $F$.
A polynomial is a formal linear combination over $F$ of the monomials $x^i$. So, first of all, all monomials are also polynomials. Examples are $x^2, x^3, x$ or $0$. Linear combinations of those monomials are finite sums of such monomials, where you are allowed to multiply each monomial with a coefficient $a_i in F$. For $F=mathbb R$, examples are $x^2, 3x^2-5, x^3-7x$ or $x^6+x^2-4$. One important point is that you do not allow infinite sums of monomials.
You can add polynomials (as it is neccesary for a vector space). If you have polynomials $p(x)=a_0 + a_1 x + a_2 x^2 + dots a_n x^n$ and $q(x) = b_0 + b_1x + dots +b_mx^m$ (so all $a_i$ and $b_i$ are elements of $F$), you can add those two to obtain
$$(p+q)(x) = sum_i=0^max(m,n) (a_i+b_i)x^i$$
where we set all $a_i$ and $b_i$ not already defined to $0$.
You can also multiply $p$ with a scalar $alpha in F$:
$$(alpha p)(x) = sum_i=1^n (alpha a_i)x^i$$
So why is a polynomial not a map?
Given a polynomial $p(x)in F[x]$, we can define a map $sigma_pcolon F to F, a mapsto p(a)$. If you think of the polynomial $x^2$ as a parabola, you consider this map. But this map is not the same as $p(x)$. For example, take $F=mathbb Z / 2mathbb Z$, the field with two elements. Then the polynomials $p(x) = x^2$ and $q(x) = x$ are different (since they have different coefficients), but the evaluation maps $sigma_pcolon mathbb Z / 2mathbb Z to mathbb Z / 2mathbb Z, xmapsto x^2$ and $sigma_q colon mathbb Z / 2mathbb Z to mathbb Z / 2mathbb Z, xmapsto x$ are the same (since $x^2=x$ for all $x in mathbb Z / 2mathbb Z$).
answered 1 hour ago
Babelfish
1,026115
edited 1 hour ago
answered 1 hour ago
Babelfish
1,026115
answered 1 hour ago
Babelfish
1,026115
answered 1 hour ago
Babelfish
1,026115
1,026115
add a comment |Â
add a comment |Â
up vote
1
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There are a lot of questions and potential confusions here, I will try to clarify some of them with the use of examples.
$P_1(mathbbR)$ is the set of all linear functions with real coefficients, i.e. $ax^1+b x^0$ where $a$ and $b$ are real. The polynomial $x+5$ is an element of this space because $x+5= 1x^1+ 5 x^0$.
On the other hand, $P_2(mathbbC)$ is the set of all quadratic polynomials with complex coefficients. An example element in this space is the polynomial $3x^2+(2+i)x+3$. Another example is $3x^2+0x+1$, which is usually written as $3x^2+1$. This example is actually also an element of $P_2(mathbbR)$ too, because all the coefficients are real.
answered 1 hour ago
Manano
1217
Thank you! The examples you've shown made it click. So we should simply look at this huge sum as an descriptive tool we use to express other polynomials just like conventional vectors. So, $P_n(mathbbF)$ can express anything in $P_n-1(mathbbF)$ by simply placing a 0 (or 0's if necessary) as a coefficient(s) in the term having nth degree (or all terms having less than a degree of n).
– Jaigus
26 mins ago
How about the statement: "$p_0, p_1, ..., p_m$ in $P_m(mathbb F)$ such that $p_j(5) = 0$ for each j"? This refers to all the polynomials in the set that evaluate to 0 at 5, yet does the subscript "j" in this statement mean each polynomial of degree j that evaluates to 0 at 5 or just each of the polynomials that evaluate to 0 at 5?
– Jaigus
9 mins ago
add a comment |Â
up vote
1
down vote
There are a lot of questions and potential confusions here, I will try to clarify some of them with the use of examples.
$P_1(mathbbR)$ is the set of all linear functions with real coefficients, i.e. $ax^1+b x^0$ where $a$ and $b$ are real. The polynomial $x+5$ is an element of this space because $x+5= 1x^1+ 5 x^0$.
On the other hand, $P_2(mathbbC)$ is the set of all quadratic polynomials with complex coefficients. An example element in this space is the polynomial $3x^2+(2+i)x+3$. Another example is $3x^2+0x+1$, which is usually written as $3x^2+1$. This example is actually also an element of $P_2(mathbbR)$ too, because all the coefficients are real.
answered 1 hour ago
Manano
1217
Thank you! The examples you've shown made it click. So we should simply look at this huge sum as an descriptive tool we use to express other polynomials just like conventional vectors. So, $P_n(mathbbF)$ can express anything in $P_n-1(mathbbF)$ by simply placing a 0 (or 0's if necessary) as a coefficient(s) in the term having nth degree (or all terms having less than a degree of n).
– Jaigus
26 mins ago
How about the statement: "$p_0, p_1, ..., p_m$ in $P_m(mathbb F)$ such that $p_j(5) = 0$ for each j"? This refers to all the polynomials in the set that evaluate to 0 at 5, yet does the subscript "j" in this statement mean each polynomial of degree j that evaluates to 0 at 5 or just each of the polynomials that evaluate to 0 at 5?
– Jaigus
9 mins ago
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There are a lot of questions and potential confusions here, I will try to clarify some of them with the use of examples.
$P_1(mathbbR)$ is the set of all linear functions with real coefficients, i.e. $ax^1+b x^0$ where $a$ and $b$ are real. The polynomial $x+5$ is an element of this space because $x+5= 1x^1+ 5 x^0$.
On the other hand, $P_2(mathbbC)$ is the set of all quadratic polynomials with complex coefficients. An example element in this space is the polynomial $3x^2+(2+i)x+3$. Another example is $3x^2+0x+1$, which is usually written as $3x^2+1$. This example is actually also an element of $P_2(mathbbR)$ too, because all the coefficients are real.
answered 1 hour ago
Manano
1217
There are a lot of questions and potential confusions here, I will try to clarify some of them with the use of examples.
$P_1(mathbbR)$ is the set of all linear functions with real coefficients, i.e. $ax^1+b x^0$ where $a$ and $b$ are real. The polynomial $x+5$ is an element of this space because $x+5= 1x^1+ 5 x^0$.
On the other hand, $P_2(mathbbC)$ is the set of all quadratic polynomials with complex coefficients. An example element in this space is the polynomial $3x^2+(2+i)x+3$. Another example is $3x^2+0x+1$, which is usually written as $3x^2+1$. This example is actually also an element of $P_2(mathbbR)$ too, because all the coefficients are real.
answered 1 hour ago
Manano
1217
answered 1 hour ago
Manano
1217
answered 1 hour ago
Manano
1217
answered 1 hour ago
Manano
1217
1217
Thank you! The examples you've shown made it click. So we should simply look at this huge sum as an descriptive tool we use to express other polynomials just like conventional vectors. So, $P_n(mathbbF)$ can express anything in $P_n-1(mathbbF)$ by simply placing a 0 (or 0's if necessary) as a coefficient(s) in the term having nth degree (or all terms having less than a degree of n).
– Jaigus
26 mins ago
How about the statement: "$p_0, p_1, ..., p_m$ in $P_m(mathbb F)$ such that $p_j(5) = 0$ for each j"? This refers to all the polynomials in the set that evaluate to 0 at 5, yet does the subscript "j" in this statement mean each polynomial of degree j that evaluates to 0 at 5 or just each of the polynomials that evaluate to 0 at 5?
– Jaigus
9 mins ago
add a comment |Â
Thank you! The examples you've shown made it click. So we should simply look at this huge sum as an descriptive tool we use to express other polynomials just like conventional vectors. So, $P_n(mathbbF)$ can express anything in $P_n-1(mathbbF)$ by simply placing a 0 (or 0's if necessary) as a coefficient(s) in the term having nth degree (or all terms having less than a degree of n).
– Jaigus
26 mins ago
How about the statement: "$p_0, p_1, ..., p_m$ in $P_m(mathbb F)$ such that $p_j(5) = 0$ for each j"? This refers to all the polynomials in the set that evaluate to 0 at 5, yet does the subscript "j" in this statement mean each polynomial of degree j that evaluates to 0 at 5 or just each of the polynomials that evaluate to 0 at 5?
– Jaigus
9 mins ago
Thank you! The examples you've shown made it click. So we should simply look at this huge sum as an descriptive tool we use to express other polynomials just like conventional vectors. So, $P_n(mathbbF)$ can express anything in $P_n-1(mathbbF)$ by simply placing a 0 (or 0's if necessary) as a coefficient(s) in the term having nth degree (or all terms having less than a degree of n).
– Jaigus
26 mins ago
Thank you! The examples you've shown made it click. So we should simply look at this huge sum as an descriptive tool we use to express other polynomials just like conventional vectors. So, $P_n(mathbbF)$ can express anything in $P_n-1(mathbbF)$ by simply placing a 0 (or 0's if necessary) as a coefficient(s) in the term having nth degree (or all terms having less than a degree of n).
– Jaigus
26 mins ago
How about the statement: "$p_0, p_1, ..., p_m$ in $P_m(mathbb F)$ such that $p_j(5) = 0$ for each j"? This refers to all the polynomials in the set that evaluate to 0 at 5, yet does the subscript "j" in this statement mean each polynomial of degree j that evaluates to 0 at 5 or just each of the polynomials that evaluate to 0 at 5?
– Jaigus
9 mins ago
How about the statement: "$p_0, p_1, ..., p_m$ in $P_m(mathbb F)$ such that $p_j(5) = 0$ for each j"? This refers to all the polynomials in the set that evaluate to 0 at 5, yet does the subscript "j" in this statement mean each polynomial of degree j that evaluates to 0 at 5 or just each of the polynomials that evaluate to 0 at 5?
– Jaigus
9 mins ago
add a comment |Â
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