Distance between polynomials in euclidean space
In Euclidean space $R[x]_{leq n}$consisting of polynomials with scalar product i need to find distance between polynomial $f=2$ and subspace of polynomials with zero constant term.
How can i approach to this problem?
linear-algebra polynomials
asked Nov 29 at 7:12
chaseperfection
172
|
show 1 more comment
In Euclidean space $R[x]_{leq n}$consisting of polynomials with scalar product i need to find distance between polynomial $f=2$ and subspace of polynomials with zero constant term.
How can i approach to this problem?
linear-algebra polynomials
asked Nov 29 at 7:12
chaseperfection
172
Did you mean a polynomial of degree 2? Or the constant 2?
– twnly
Nov 29 at 7:18
If $f = 2$ denotes the constant polynomial $2$, then it belongs to the subspace mentioned. So, I think the distance from the subspace is $0$.
– Aniruddha Deshmukh
Nov 29 at 7:20
Remark: the 'subspace of polynomials with non zero constant term' is not a subspace
– Charles Madeline
Nov 29 at 7:20
@twnly, constant
– chaseperfection
Nov 29 at 7:26
@CharlesMadeline, yes, sorry, i edited. With zero constant term
– chaseperfection
Nov 29 at 7:27
|
show 1 more comment
In Euclidean space $R[x]_{leq n}$consisting of polynomials with scalar product i need to find distance between polynomial $f=2$ and subspace of polynomials with zero constant term.
How can i approach to this problem?
linear-algebra polynomials
asked Nov 29 at 7:12
chaseperfection
172
In Euclidean space $R[x]_{leq n}$consisting of polynomials with scalar product i need to find distance between polynomial $f=2$ and subspace of polynomials with zero constant term.
How can i approach to this problem?
linear-algebra polynomials
linear-algebra polynomials
asked Nov 29 at 7:12
chaseperfection
172
asked Nov 29 at 7:12
chaseperfection
172
edited Nov 29 at 7:26
asked Nov 29 at 7:12
chaseperfection
172
asked Nov 29 at 7:12
chaseperfection
172
asked Nov 29 at 7:12
chaseperfection
172
172
Did you mean a polynomial of degree 2? Or the constant 2?
– twnly
Nov 29 at 7:18
If $f = 2$ denotes the constant polynomial $2$, then it belongs to the subspace mentioned. So, I think the distance from the subspace is $0$.
– Aniruddha Deshmukh
Nov 29 at 7:20
Remark: the 'subspace of polynomials with non zero constant term' is not a subspace
– Charles Madeline
Nov 29 at 7:20
@twnly, constant
– chaseperfection
Nov 29 at 7:26
@CharlesMadeline, yes, sorry, i edited. With zero constant term
– chaseperfection
Nov 29 at 7:27
|
show 1 more comment
Did you mean a polynomial of degree 2? Or the constant 2?
– twnly
Nov 29 at 7:18
If $f = 2$ denotes the constant polynomial $2$, then it belongs to the subspace mentioned. So, I think the distance from the subspace is $0$.
– Aniruddha Deshmukh
Nov 29 at 7:20
Remark: the 'subspace of polynomials with non zero constant term' is not a subspace
– Charles Madeline
Nov 29 at 7:20
@twnly, constant
– chaseperfection
Nov 29 at 7:26
@CharlesMadeline, yes, sorry, i edited. With zero constant term
– chaseperfection
Nov 29 at 7:27
Did you mean a polynomial of degree 2? Or the constant 2?
– twnly
Nov 29 at 7:18
Did you mean a polynomial of degree 2? Or the constant 2?
– twnly
Nov 29 at 7:18
If $f = 2$ denotes the constant polynomial $2$, then it belongs to the subspace mentioned. So, I think the distance from the subspace is $0$.
– Aniruddha Deshmukh
Nov 29 at 7:20
If $f = 2$ denotes the constant polynomial $2$, then it belongs to the subspace mentioned. So, I think the distance from the subspace is $0$.
– Aniruddha Deshmukh
Nov 29 at 7:20
Remark: the 'subspace of polynomials with non zero constant term' is not a subspace
– Charles Madeline
Nov 29 at 7:20
Remark: the 'subspace of polynomials with non zero constant term' is not a subspace
– Charles Madeline
Nov 29 at 7:20
@twnly, constant
– chaseperfection
Nov 29 at 7:26
@twnly, constant
– chaseperfection
Nov 29 at 7:26
@CharlesMadeline, yes, sorry, i edited. With zero constant term
– chaseperfection
Nov 29 at 7:27
@CharlesMadeline, yes, sorry, i edited. With zero constant term
– chaseperfection
Nov 29 at 7:27
|
show 1 more comment
1 Answer
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If your inner product is defined by $langle p,q rangle=int_0^{1} p(x)q(x)dx$ then the answer is $2$. To see this let $f_n(x)=2n(frac 1 n -x)$ for $x <frac 1 n$ and $0$ otherwise. it is easy to see that the distance between $f_n$ and $2$ tends to $2$. Now Weirstrass approximation shows that required distance is also $2$. [Some details: the distance is obviously $leq 2$ so we only have to produce some sequence in the subspace whose distance to $2$ tends to $2$. By Weirstrass approximation any continuous function vanishing at $0$ can be approximated uniformly by polynomials vanishing at $0$ and I am using this for $f_n$].
answered Nov 29 at 7:35
Kavi Rama Murthy
48.9k31854
add a comment |
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If your inner product is defined by $langle p,q rangle=int_0^{1} p(x)q(x)dx$ then the answer is $2$. To see this let $f_n(x)=2n(frac 1 n -x)$ for $x <frac 1 n$ and $0$ otherwise. it is easy to see that the distance between $f_n$ and $2$ tends to $2$. Now Weirstrass approximation shows that required distance is also $2$. [Some details: the distance is obviously $leq 2$ so we only have to produce some sequence in the subspace whose distance to $2$ tends to $2$. By Weirstrass approximation any continuous function vanishing at $0$ can be approximated uniformly by polynomials vanishing at $0$ and I am using this for $f_n$].
answered Nov 29 at 7:35
Kavi Rama Murthy
48.9k31854
add a comment |
If your inner product is defined by $langle p,q rangle=int_0^{1} p(x)q(x)dx$ then the answer is $2$. To see this let $f_n(x)=2n(frac 1 n -x)$ for $x <frac 1 n$ and $0$ otherwise. it is easy to see that the distance between $f_n$ and $2$ tends to $2$. Now Weirstrass approximation shows that required distance is also $2$. [Some details: the distance is obviously $leq 2$ so we only have to produce some sequence in the subspace whose distance to $2$ tends to $2$. By Weirstrass approximation any continuous function vanishing at $0$ can be approximated uniformly by polynomials vanishing at $0$ and I am using this for $f_n$].
answered Nov 29 at 7:35
Kavi Rama Murthy
48.9k31854
add a comment |
If your inner product is defined by $langle p,q rangle=int_0^{1} p(x)q(x)dx$ then the answer is $2$. To see this let $f_n(x)=2n(frac 1 n -x)$ for $x <frac 1 n$ and $0$ otherwise. it is easy to see that the distance between $f_n$ and $2$ tends to $2$. Now Weirstrass approximation shows that required distance is also $2$. [Some details: the distance is obviously $leq 2$ so we only have to produce some sequence in the subspace whose distance to $2$ tends to $2$. By Weirstrass approximation any continuous function vanishing at $0$ can be approximated uniformly by polynomials vanishing at $0$ and I am using this for $f_n$].
answered Nov 29 at 7:35
Kavi Rama Murthy
48.9k31854
If your inner product is defined by $langle p,q rangle=int_0^{1} p(x)q(x)dx$ then the answer is $2$. To see this let $f_n(x)=2n(frac 1 n -x)$ for $x <frac 1 n$ and $0$ otherwise. it is easy to see that the distance between $f_n$ and $2$ tends to $2$. Now Weirstrass approximation shows that required distance is also $2$. [Some details: the distance is obviously $leq 2$ so we only have to produce some sequence in the subspace whose distance to $2$ tends to $2$. By Weirstrass approximation any continuous function vanishing at $0$ can be approximated uniformly by polynomials vanishing at $0$ and I am using this for $f_n$].
answered Nov 29 at 7:35
Kavi Rama Murthy
48.9k31854
answered Nov 29 at 7:35
Kavi Rama Murthy
48.9k31854
answered Nov 29 at 7:35
Kavi Rama Murthy
48.9k31854
answered Nov 29 at 7:35
Kavi Rama Murthy
48.9k31854
48.9k31854
add a comment |
add a comment |
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Did you mean a polynomial of degree 2? Or the constant 2?
– twnly
Nov 29 at 7:18
If $f = 2$ denotes the constant polynomial $2$, then it belongs to the subspace mentioned. So, I think the distance from the subspace is $0$.
– Aniruddha Deshmukh
Nov 29 at 7:20
Remark: the 'subspace of polynomials with non zero constant term' is not a subspace
– Charles Madeline
Nov 29 at 7:20
@twnly, constant
– chaseperfection
Nov 29 at 7:26
@CharlesMadeline, yes, sorry, i edited. With zero constant term
– chaseperfection
Nov 29 at 7:27