Intuition of a Function as an Infinite Dimensional vector
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Functions as infinite dimensional vectors make sense intuitively after studying much linear algebra, but I only recently realized I've been taught two ways of thinking about them.
For a one dimensional function $f(x)$ you can expand in the basis of polynomials (a taylor series) and you get an infinite dimensional vector where the $i$th component is attached to the basis $x^{i}$, or you can think of the function as a list of all its values over its domain e.g. $ f = ...f(-0.01),f(0),f(0.01)... $ where 0.01 goes to 0.
In the former case, differentiation and integration become matrix multiplication, which is a nice property, and the latter case makes things like the inner product of functions feel a lot more natural.
But in the former case the length of the vector is the cardinality of the natural numbers, as opposed to the latter where it has the size of the reals, so these clearly aren't equivalent representations.
Is there any relationship between these two ways of thinking of a function as an infinite dimensional vector?
functions vectors
edited Nov 28 at 15:54
Aweygan
13.3k21441
asked Nov 28 at 15:46
Craig
526
add a comment |
up vote
2
down vote
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Functions as infinite dimensional vectors make sense intuitively after studying much linear algebra, but I only recently realized I've been taught two ways of thinking about them.
For a one dimensional function $f(x)$ you can expand in the basis of polynomials (a taylor series) and you get an infinite dimensional vector where the $i$th component is attached to the basis $x^{i}$, or you can think of the function as a list of all its values over its domain e.g. $ f = ...f(-0.01),f(0),f(0.01)... $ where 0.01 goes to 0.
In the former case, differentiation and integration become matrix multiplication, which is a nice property, and the latter case makes things like the inner product of functions feel a lot more natural.
But in the former case the length of the vector is the cardinality of the natural numbers, as opposed to the latter where it has the size of the reals, so these clearly aren't equivalent representations.
Is there any relationship between these two ways of thinking of a function as an infinite dimensional vector?
functions vectors
edited Nov 28 at 15:54
Aweygan
13.3k21441
asked Nov 28 at 15:46
Craig
526
1
The first approach (taylor series) only works for analytic functions. So it is clearly much more restricted than the second approach.
– Ramiro
Nov 28 at 16:03
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Functions as infinite dimensional vectors make sense intuitively after studying much linear algebra, but I only recently realized I've been taught two ways of thinking about them.
For a one dimensional function $f(x)$ you can expand in the basis of polynomials (a taylor series) and you get an infinite dimensional vector where the $i$th component is attached to the basis $x^{i}$, or you can think of the function as a list of all its values over its domain e.g. $ f = ...f(-0.01),f(0),f(0.01)... $ where 0.01 goes to 0.
In the former case, differentiation and integration become matrix multiplication, which is a nice property, and the latter case makes things like the inner product of functions feel a lot more natural.
But in the former case the length of the vector is the cardinality of the natural numbers, as opposed to the latter where it has the size of the reals, so these clearly aren't equivalent representations.
Is there any relationship between these two ways of thinking of a function as an infinite dimensional vector?
functions vectors
edited Nov 28 at 15:54
Aweygan
13.3k21441
asked Nov 28 at 15:46
Craig
526
Functions as infinite dimensional vectors make sense intuitively after studying much linear algebra, but I only recently realized I've been taught two ways of thinking about them.
For a one dimensional function $f(x)$ you can expand in the basis of polynomials (a taylor series) and you get an infinite dimensional vector where the $i$th component is attached to the basis $x^{i}$, or you can think of the function as a list of all its values over its domain e.g. $ f = ...f(-0.01),f(0),f(0.01)... $ where 0.01 goes to 0.
In the former case, differentiation and integration become matrix multiplication, which is a nice property, and the latter case makes things like the inner product of functions feel a lot more natural.
But in the former case the length of the vector is the cardinality of the natural numbers, as opposed to the latter where it has the size of the reals, so these clearly aren't equivalent representations.
Is there any relationship between these two ways of thinking of a function as an infinite dimensional vector?
functions vectors
functions vectors
edited Nov 28 at 15:54
Aweygan
13.3k21441
asked Nov 28 at 15:46
Craig
526
edited Nov 28 at 15:54
Aweygan
13.3k21441
asked Nov 28 at 15:46
Craig
526
edited Nov 28 at 15:54
Aweygan
13.3k21441
edited Nov 28 at 15:54
Aweygan
13.3k21441
edited Nov 28 at 15:54
Aweygan
13.3k21441
13.3k21441
asked Nov 28 at 15:46
Craig
526
asked Nov 28 at 15:46
Craig
526
asked Nov 28 at 15:46
Craig
526
526
1
The first approach (taylor series) only works for analytic functions. So it is clearly much more restricted than the second approach.
– Ramiro
Nov 28 at 16:03
add a comment |
1
The first approach (taylor series) only works for analytic functions. So it is clearly much more restricted than the second approach.
– Ramiro
Nov 28 at 16:03
1
1
The first approach (taylor series) only works for analytic functions. So it is clearly much more restricted than the second approach.
– Ramiro
Nov 28 at 16:03
The first approach (taylor series) only works for analytic functions. So it is clearly much more restricted than the second approach.
– Ramiro
Nov 28 at 16:03
add a comment |
1 Answer
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I think the short answer is "no", the two ways are not related.
The first, where you think of the functions as vectors in a space of functions you care about the formal properties of vectors (adding them and multiplying by scalars). Then you enrich that formality by introducing ideas of limits and convergence. To use the Taylor series to represent a function you have to make sense of infinite sums. That goes well beyond vector space axioms.
In the second view you are thinking of a vector as an abstraction of a list of values, and then thinking of a function as a way to describe a list of values. In that model ordinary real $n$ space is the set of real values functions from ${1, 2, ldots, n}$, the space of real sequences is the set of functions $mathbb{N} to mathbb{R}$ and the space of real functions is the set of functions $mathbb{R} to mathbb{R}$.
You can extend this second view even further. There's no reason the codomain must be $mathbb{R}$, or even numerical. In computer science you often refer to essentially arbitrary arrays or lists as "vectors".
answered Nov 28 at 15:56
Ethan Bolker
40.7k546108
add a comment |
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I think the short answer is "no", the two ways are not related.
The first, where you think of the functions as vectors in a space of functions you care about the formal properties of vectors (adding them and multiplying by scalars). Then you enrich that formality by introducing ideas of limits and convergence. To use the Taylor series to represent a function you have to make sense of infinite sums. That goes well beyond vector space axioms.
In the second view you are thinking of a vector as an abstraction of a list of values, and then thinking of a function as a way to describe a list of values. In that model ordinary real $n$ space is the set of real values functions from ${1, 2, ldots, n}$, the space of real sequences is the set of functions $mathbb{N} to mathbb{R}$ and the space of real functions is the set of functions $mathbb{R} to mathbb{R}$.
You can extend this second view even further. There's no reason the codomain must be $mathbb{R}$, or even numerical. In computer science you often refer to essentially arbitrary arrays or lists as "vectors".
answered Nov 28 at 15:56
Ethan Bolker
40.7k546108
add a comment |
up vote
1
down vote
accepted
I think the short answer is "no", the two ways are not related.
The first, where you think of the functions as vectors in a space of functions you care about the formal properties of vectors (adding them and multiplying by scalars). Then you enrich that formality by introducing ideas of limits and convergence. To use the Taylor series to represent a function you have to make sense of infinite sums. That goes well beyond vector space axioms.
In the second view you are thinking of a vector as an abstraction of a list of values, and then thinking of a function as a way to describe a list of values. In that model ordinary real $n$ space is the set of real values functions from ${1, 2, ldots, n}$, the space of real sequences is the set of functions $mathbb{N} to mathbb{R}$ and the space of real functions is the set of functions $mathbb{R} to mathbb{R}$.
You can extend this second view even further. There's no reason the codomain must be $mathbb{R}$, or even numerical. In computer science you often refer to essentially arbitrary arrays or lists as "vectors".
answered Nov 28 at 15:56
Ethan Bolker
40.7k546108
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
I think the short answer is "no", the two ways are not related.
The first, where you think of the functions as vectors in a space of functions you care about the formal properties of vectors (adding them and multiplying by scalars). Then you enrich that formality by introducing ideas of limits and convergence. To use the Taylor series to represent a function you have to make sense of infinite sums. That goes well beyond vector space axioms.
In the second view you are thinking of a vector as an abstraction of a list of values, and then thinking of a function as a way to describe a list of values. In that model ordinary real $n$ space is the set of real values functions from ${1, 2, ldots, n}$, the space of real sequences is the set of functions $mathbb{N} to mathbb{R}$ and the space of real functions is the set of functions $mathbb{R} to mathbb{R}$.
You can extend this second view even further. There's no reason the codomain must be $mathbb{R}$, or even numerical. In computer science you often refer to essentially arbitrary arrays or lists as "vectors".
answered Nov 28 at 15:56
Ethan Bolker
40.7k546108
I think the short answer is "no", the two ways are not related.
The first, where you think of the functions as vectors in a space of functions you care about the formal properties of vectors (adding them and multiplying by scalars). Then you enrich that formality by introducing ideas of limits and convergence. To use the Taylor series to represent a function you have to make sense of infinite sums. That goes well beyond vector space axioms.
In the second view you are thinking of a vector as an abstraction of a list of values, and then thinking of a function as a way to describe a list of values. In that model ordinary real $n$ space is the set of real values functions from ${1, 2, ldots, n}$, the space of real sequences is the set of functions $mathbb{N} to mathbb{R}$ and the space of real functions is the set of functions $mathbb{R} to mathbb{R}$.
You can extend this second view even further. There's no reason the codomain must be $mathbb{R}$, or even numerical. In computer science you often refer to essentially arbitrary arrays or lists as "vectors".
answered Nov 28 at 15:56
Ethan Bolker
40.7k546108
answered Nov 28 at 15:56
Ethan Bolker
40.7k546108
answered Nov 28 at 15:56
Ethan Bolker
40.7k546108
answered Nov 28 at 15:56
Ethan Bolker
40.7k546108
40.7k546108
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The first approach (taylor series) only works for analytic functions. So it is clearly much more restricted than the second approach.
– Ramiro
Nov 28 at 16:03