What does “the vector space of all polynomials on the interval of $[0,1]$” refer to? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)What is happening in a linear algebra computation?Textbook that brings together linear algebra and PDEs?How do you learn from MITOpenCourseWare 18.06 Linear Algebra Course?What are differences between affine space and vector space?Is an endomorphism always an isomorphism?Is Hoffman-Kunze a good book to read next?What exactly is antieigenvalue analysis?Determining if the transformation $(x_1,x_2,x_3)mapsto(2x_2,x_2+2,2x_2)$ is linear or notWhat are the differences between equivalence, row equivalence and similarity in matrices?What is the meaning of the difference between the equations of two non-intersecting circles represent?
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What does “the vector space of all polynomials on the interval of $[0,1]$” refer to?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)What is happening in a linear algebra computation?Textbook that brings together linear algebra and PDEs?How do you learn from MITOpenCourseWare 18.06 Linear Algebra Course?What are differences between affine space and vector space?Is an endomorphism always an isomorphism?Is Hoffman-Kunze a good book to read next?What exactly is antieigenvalue analysis?Determining if the transformation $(x_1,x_2,x_3)mapsto(2x_2,x_2+2,2x_2)$ is linear or notWhat are the differences between equivalence, row equivalence and similarity in matrices?What is the meaning of the difference between the equations of two non-intersecting circles represent?
$begingroup$
What might interval refer to in this context? I can't find a common quality between $x^2$ and $x+1$ that would make sense to qualify them as "being on the interval $[0,1]$". I've only ever heard of intervals in the context of intervals on a number line in $R^2$ for example. I read this in a linear algebra book so it's possible that it is related to linear algebra in some way.
linear-algebra polynomials
asked Mar 27 at 21:02
Nest Doberman Nest Doberman
311
$endgroup$
add a comment |
$begingroup$
What might interval refer to in this context? I can't find a common quality between $x^2$ and $x+1$ that would make sense to qualify them as "being on the interval $[0,1]$". I've only ever heard of intervals in the context of intervals on a number line in $R^2$ for example. I read this in a linear algebra book so it's possible that it is related to linear algebra in some way.
linear-algebra polynomials
asked Mar 27 at 21:02
Nest Doberman Nest Doberman
311
$endgroup$
7
$begingroup$
On the interval of $[0,1]$ refers to the domain of such polynomials...
$endgroup$
– Don Thousand
Mar 27 at 21:04
3
$begingroup$
I'd take it to mean "we care about every real polynomial, but only care about them on the domain $[0,1]$.
$endgroup$
– Jason DeVito
Mar 27 at 21:04
$begingroup$
every polynomial whose domain is defined in [0,1]. Any polynomial is continuous, though.
$endgroup$
– user29418
Mar 27 at 21:07
1
$begingroup$
Intervals and linear algebra, or algebra are no contradiction.
$endgroup$
– Dietrich Burde
Mar 27 at 21:13
add a comment |
$begingroup$
What might interval refer to in this context? I can't find a common quality between $x^2$ and $x+1$ that would make sense to qualify them as "being on the interval $[0,1]$". I've only ever heard of intervals in the context of intervals on a number line in $R^2$ for example. I read this in a linear algebra book so it's possible that it is related to linear algebra in some way.
linear-algebra polynomials
asked Mar 27 at 21:02
Nest Doberman Nest Doberman
311
$endgroup$
What might interval refer to in this context? I can't find a common quality between $x^2$ and $x+1$ that would make sense to qualify them as "being on the interval $[0,1]$". I've only ever heard of intervals in the context of intervals on a number line in $R^2$ for example. I read this in a linear algebra book so it's possible that it is related to linear algebra in some way.
linear-algebra polynomials
linear-algebra polynomials
asked Mar 27 at 21:02
Nest Doberman Nest Doberman
311
asked Mar 27 at 21:02
Nest Doberman Nest Doberman
311
asked Mar 27 at 21:02
Nest Doberman Nest Doberman
311
asked Mar 27 at 21:02
Nest Doberman Nest Doberman
311
asked Mar 27 at 21:02
Nest Doberman Nest Doberman
311
311
7
$begingroup$
On the interval of $[0,1]$ refers to the domain of such polynomials...
$endgroup$
– Don Thousand
Mar 27 at 21:04
3
$begingroup$
I'd take it to mean "we care about every real polynomial, but only care about them on the domain $[0,1]$.
$endgroup$
– Jason DeVito
Mar 27 at 21:04
$begingroup$
every polynomial whose domain is defined in [0,1]. Any polynomial is continuous, though.
$endgroup$
– user29418
Mar 27 at 21:07
1
$begingroup$
Intervals and linear algebra, or algebra are no contradiction.
$endgroup$
– Dietrich Burde
Mar 27 at 21:13
add a comment |
7
$begingroup$
On the interval of $[0,1]$ refers to the domain of such polynomials...
$endgroup$
– Don Thousand
Mar 27 at 21:04
3
$begingroup$
I'd take it to mean "we care about every real polynomial, but only care about them on the domain $[0,1]$.
$endgroup$
– Jason DeVito
Mar 27 at 21:04
$begingroup$
every polynomial whose domain is defined in [0,1]. Any polynomial is continuous, though.
$endgroup$
– user29418
Mar 27 at 21:07
1
$begingroup$
Intervals and linear algebra, or algebra are no contradiction.
$endgroup$
– Dietrich Burde
Mar 27 at 21:13
7
7
$begingroup$
On the interval of $[0,1]$ refers to the domain of such polynomials...
$endgroup$
– Don Thousand
Mar 27 at 21:04
$begingroup$
On the interval of $[0,1]$ refers to the domain of such polynomials...
$endgroup$
– Don Thousand
Mar 27 at 21:04
3
3
$begingroup$
I'd take it to mean "we care about every real polynomial, but only care about them on the domain $[0,1]$.
$endgroup$
– Jason DeVito
Mar 27 at 21:04
$begingroup$
I'd take it to mean "we care about every real polynomial, but only care about them on the domain $[0,1]$.
$endgroup$
– Jason DeVito
Mar 27 at 21:04
$begingroup$
every polynomial whose domain is defined in [0,1]. Any polynomial is continuous, though.
$endgroup$
– user29418
Mar 27 at 21:07
$begingroup$
every polynomial whose domain is defined in [0,1]. Any polynomial is continuous, though.
$endgroup$
– user29418
Mar 27 at 21:07
1
1
$begingroup$
Intervals and linear algebra, or algebra are no contradiction.
$endgroup$
– Dietrich Burde
Mar 27 at 21:13
$begingroup$
Intervals and linear algebra, or algebra are no contradiction.
$endgroup$
– Dietrich Burde
Mar 27 at 21:13
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
There is a distinction between polynomials, defined as linear combinations of the abstract objects $1,x,x^2,x^3,cdots$, and polynomial functions, which are functions from some domain $D$ that happen to have a polynomial formula. There is a natural map from the former to the latter; we take a polynomial $f$ to the function $xto f(x)$.
By specifying "on the interval $[0,1]$", we imply that we're talking about the latter space of polynomial functions.
Now, if that (one-variable, in a field) domain is infinite, the two spaces are isomorphic; the map from polynomials to polynomial functions is injective, since a polynomial with infinitely many zeros is the zero polynomial. This is the case for our example domain $[0,1]$.
Why would we specify that interval rather than the whole field $mathbbR$? Most likely, we're going to go on to introduce some additional structure on the space related to that interval, such as the inner product $langle f,grangle = int_0^1 f(x)g(x),dx$ and the $2$-norm $|f|_2 = sqrtint_0^1 f(x)^2,dx$. These wouldn't work on $mathbbR$, since the integrals diverge for nonzero polynomials.
answered Mar 27 at 22:26
jmerryjmerry
17k11633
$endgroup$
add a comment |
$begingroup$
Consider, for example, the set $V$ of differentiable functions, which map from $[0,1]$ to the real line. It's not hard to see that $V$ is a vector space with the zero function as its identity (intuition behind that being that linear combinations of differentiable functions are also differentiable).
Now, if you look at the $P subset V$, which only considers the elements of $V$ which are polynomials, you will end up with the set your text is describing.
answered Mar 27 at 21:08
gt6989bgt6989b
36k22557
$endgroup$
$begingroup$
Did you mean to say 'it is not hard to see' or 'it is hard to see'? Both seem defendable, but I was curious
$endgroup$
– Vincent
Mar 27 at 22:40
$begingroup$
@Vincent typo thanks corrected
$endgroup$
– gt6989b
Mar 28 at 14:38
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
There is a distinction between polynomials, defined as linear combinations of the abstract objects $1,x,x^2,x^3,cdots$, and polynomial functions, which are functions from some domain $D$ that happen to have a polynomial formula. There is a natural map from the former to the latter; we take a polynomial $f$ to the function $xto f(x)$.
By specifying "on the interval $[0,1]$", we imply that we're talking about the latter space of polynomial functions.
Now, if that (one-variable, in a field) domain is infinite, the two spaces are isomorphic; the map from polynomials to polynomial functions is injective, since a polynomial with infinitely many zeros is the zero polynomial. This is the case for our example domain $[0,1]$.
Why would we specify that interval rather than the whole field $mathbbR$? Most likely, we're going to go on to introduce some additional structure on the space related to that interval, such as the inner product $langle f,grangle = int_0^1 f(x)g(x),dx$ and the $2$-norm $|f|_2 = sqrtint_0^1 f(x)^2,dx$. These wouldn't work on $mathbbR$, since the integrals diverge for nonzero polynomials.
answered Mar 27 at 22:26
jmerryjmerry
17k11633
$endgroup$
add a comment |
$begingroup$
There is a distinction between polynomials, defined as linear combinations of the abstract objects $1,x,x^2,x^3,cdots$, and polynomial functions, which are functions from some domain $D$ that happen to have a polynomial formula. There is a natural map from the former to the latter; we take a polynomial $f$ to the function $xto f(x)$.
By specifying "on the interval $[0,1]$", we imply that we're talking about the latter space of polynomial functions.
Now, if that (one-variable, in a field) domain is infinite, the two spaces are isomorphic; the map from polynomials to polynomial functions is injective, since a polynomial with infinitely many zeros is the zero polynomial. This is the case for our example domain $[0,1]$.
Why would we specify that interval rather than the whole field $mathbbR$? Most likely, we're going to go on to introduce some additional structure on the space related to that interval, such as the inner product $langle f,grangle = int_0^1 f(x)g(x),dx$ and the $2$-norm $|f|_2 = sqrtint_0^1 f(x)^2,dx$. These wouldn't work on $mathbbR$, since the integrals diverge for nonzero polynomials.
answered Mar 27 at 22:26
jmerryjmerry
17k11633
$endgroup$
add a comment |
$begingroup$
There is a distinction between polynomials, defined as linear combinations of the abstract objects $1,x,x^2,x^3,cdots$, and polynomial functions, which are functions from some domain $D$ that happen to have a polynomial formula. There is a natural map from the former to the latter; we take a polynomial $f$ to the function $xto f(x)$.
By specifying "on the interval $[0,1]$", we imply that we're talking about the latter space of polynomial functions.
Now, if that (one-variable, in a field) domain is infinite, the two spaces are isomorphic; the map from polynomials to polynomial functions is injective, since a polynomial with infinitely many zeros is the zero polynomial. This is the case for our example domain $[0,1]$.
Why would we specify that interval rather than the whole field $mathbbR$? Most likely, we're going to go on to introduce some additional structure on the space related to that interval, such as the inner product $langle f,grangle = int_0^1 f(x)g(x),dx$ and the $2$-norm $|f|_2 = sqrtint_0^1 f(x)^2,dx$. These wouldn't work on $mathbbR$, since the integrals diverge for nonzero polynomials.
answered Mar 27 at 22:26
jmerryjmerry
17k11633
$endgroup$
There is a distinction between polynomials, defined as linear combinations of the abstract objects $1,x,x^2,x^3,cdots$, and polynomial functions, which are functions from some domain $D$ that happen to have a polynomial formula. There is a natural map from the former to the latter; we take a polynomial $f$ to the function $xto f(x)$.
By specifying "on the interval $[0,1]$", we imply that we're talking about the latter space of polynomial functions.
Now, if that (one-variable, in a field) domain is infinite, the two spaces are isomorphic; the map from polynomials to polynomial functions is injective, since a polynomial with infinitely many zeros is the zero polynomial. This is the case for our example domain $[0,1]$.
Why would we specify that interval rather than the whole field $mathbbR$? Most likely, we're going to go on to introduce some additional structure on the space related to that interval, such as the inner product $langle f,grangle = int_0^1 f(x)g(x),dx$ and the $2$-norm $|f|_2 = sqrtint_0^1 f(x)^2,dx$. These wouldn't work on $mathbbR$, since the integrals diverge for nonzero polynomials.
answered Mar 27 at 22:26
jmerryjmerry
17k11633
answered Mar 27 at 22:26
jmerryjmerry
17k11633
answered Mar 27 at 22:26
jmerryjmerry
17k11633
answered Mar 27 at 22:26
jmerryjmerry
17k11633
17k11633
add a comment |
add a comment |
$begingroup$
Consider, for example, the set $V$ of differentiable functions, which map from $[0,1]$ to the real line. It's not hard to see that $V$ is a vector space with the zero function as its identity (intuition behind that being that linear combinations of differentiable functions are also differentiable).
Now, if you look at the $P subset V$, which only considers the elements of $V$ which are polynomials, you will end up with the set your text is describing.
answered Mar 27 at 21:08
gt6989bgt6989b
36k22557
$endgroup$
$begingroup$
Did you mean to say 'it is not hard to see' or 'it is hard to see'? Both seem defendable, but I was curious
$endgroup$
– Vincent
Mar 27 at 22:40
$begingroup$
@Vincent typo thanks corrected
$endgroup$
– gt6989b
Mar 28 at 14:38
add a comment |
$begingroup$
Consider, for example, the set $V$ of differentiable functions, which map from $[0,1]$ to the real line. It's not hard to see that $V$ is a vector space with the zero function as its identity (intuition behind that being that linear combinations of differentiable functions are also differentiable).
Now, if you look at the $P subset V$, which only considers the elements of $V$ which are polynomials, you will end up with the set your text is describing.
answered Mar 27 at 21:08
gt6989bgt6989b
36k22557
$endgroup$
$begingroup$
Did you mean to say 'it is not hard to see' or 'it is hard to see'? Both seem defendable, but I was curious
$endgroup$
– Vincent
Mar 27 at 22:40
$begingroup$
@Vincent typo thanks corrected
$endgroup$
– gt6989b
Mar 28 at 14:38
add a comment |
$begingroup$
Consider, for example, the set $V$ of differentiable functions, which map from $[0,1]$ to the real line. It's not hard to see that $V$ is a vector space with the zero function as its identity (intuition behind that being that linear combinations of differentiable functions are also differentiable).
Now, if you look at the $P subset V$, which only considers the elements of $V$ which are polynomials, you will end up with the set your text is describing.
answered Mar 27 at 21:08
gt6989bgt6989b
36k22557
$endgroup$
Consider, for example, the set $V$ of differentiable functions, which map from $[0,1]$ to the real line. It's not hard to see that $V$ is a vector space with the zero function as its identity (intuition behind that being that linear combinations of differentiable functions are also differentiable).
Now, if you look at the $P subset V$, which only considers the elements of $V$ which are polynomials, you will end up with the set your text is describing.
answered Mar 27 at 21:08
gt6989bgt6989b
36k22557
edited Mar 28 at 14:37
answered Mar 27 at 21:08
gt6989bgt6989b
36k22557
answered Mar 27 at 21:08
gt6989bgt6989b
36k22557
answered Mar 27 at 21:08
gt6989bgt6989b
36k22557
36k22557
$begingroup$
Did you mean to say 'it is not hard to see' or 'it is hard to see'? Both seem defendable, but I was curious
$endgroup$
– Vincent
Mar 27 at 22:40
$begingroup$
@Vincent typo thanks corrected
$endgroup$
– gt6989b
Mar 28 at 14:38
add a comment |
$begingroup$
Did you mean to say 'it is not hard to see' or 'it is hard to see'? Both seem defendable, but I was curious
$endgroup$
– Vincent
Mar 27 at 22:40
$begingroup$
@Vincent typo thanks corrected
$endgroup$
– gt6989b
Mar 28 at 14:38
$begingroup$
Did you mean to say 'it is not hard to see' or 'it is hard to see'? Both seem defendable, but I was curious
$endgroup$
– Vincent
Mar 27 at 22:40
$begingroup$
Did you mean to say 'it is not hard to see' or 'it is hard to see'? Both seem defendable, but I was curious
$endgroup$
– Vincent
Mar 27 at 22:40
$begingroup$
@Vincent typo thanks corrected
$endgroup$
– gt6989b
Mar 28 at 14:38
$begingroup$
@Vincent typo thanks corrected
$endgroup$
– gt6989b
Mar 28 at 14:38
add a comment |
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$begingroup$
On the interval of $[0,1]$ refers to the domain of such polynomials...
$endgroup$
– Don Thousand
Mar 27 at 21:04
3
$begingroup$
I'd take it to mean "we care about every real polynomial, but only care about them on the domain $[0,1]$.
$endgroup$
– Jason DeVito
Mar 27 at 21:04
$begingroup$
every polynomial whose domain is defined in [0,1]. Any polynomial is continuous, though.
$endgroup$
– user29418
Mar 27 at 21:07
1
$begingroup$
Intervals and linear algebra, or algebra are no contradiction.
$endgroup$
– Dietrich Burde
Mar 27 at 21:13