Inner Product, Definite Integral
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Does the map $<f, g>$ $=$ $int _0^1:left(left(fleft(xright)-frac{d}{dx}fleft(xright)right)left(gleft(xright)-frac{d}{dx}gleft(xright)right)right)dx$ define an inner product on the set of all polynomial functions of order less than or equal to $n$?
Although I felt that this map was symmetric and bilinear by the properties of definite integrals, I did not think it was positive definite. This was because, say, $<f, f>$ would become $int _0^1:left(left(fleft(xright)-frac{d}{dx}fleft(xright)right)^2right)dx$, which has value greater than zero for values other than the zero function. For example, if $f(x)$ was $2x$, then $f'(x)$ would be $2$ and the value of the above definite integral would be $4/3$. This is why I don't think this space defines an inner product.
So what I am worried about is that I am misunderstanding how the concept of zero vectors with regards to inner products work as this seems to obvious.
If anyone can verify what I have done or tell me how I went wrong, I would greatly appreciate it!
linear-algebra derivatives definite-integrals inner-product-space
asked Nov 15 at 5:03
sktsasus
968414
add a comment |
up vote
0
down vote
favorite
Does the map $<f, g>$ $=$ $int _0^1:left(left(fleft(xright)-frac{d}{dx}fleft(xright)right)left(gleft(xright)-frac{d}{dx}gleft(xright)right)right)dx$ define an inner product on the set of all polynomial functions of order less than or equal to $n$?
Although I felt that this map was symmetric and bilinear by the properties of definite integrals, I did not think it was positive definite. This was because, say, $<f, f>$ would become $int _0^1:left(left(fleft(xright)-frac{d}{dx}fleft(xright)right)^2right)dx$, which has value greater than zero for values other than the zero function. For example, if $f(x)$ was $2x$, then $f'(x)$ would be $2$ and the value of the above definite integral would be $4/3$. This is why I don't think this space defines an inner product.
So what I am worried about is that I am misunderstanding how the concept of zero vectors with regards to inner products work as this seems to obvious.
If anyone can verify what I have done or tell me how I went wrong, I would greatly appreciate it!
linear-algebra derivatives definite-integrals inner-product-space
asked Nov 15 at 5:03
sktsasus
968414
1
Positive definite means exactly that $langle f,franglegeq 0$ with equality if and only if $f=0$ for all $f$ in your vector space. Your bilinear form has this property, so it should be an inner product as long as you're considering polynomials over $mathbb{R}$.
– Melody
Nov 15 at 5:11
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Does the map $<f, g>$ $=$ $int _0^1:left(left(fleft(xright)-frac{d}{dx}fleft(xright)right)left(gleft(xright)-frac{d}{dx}gleft(xright)right)right)dx$ define an inner product on the set of all polynomial functions of order less than or equal to $n$?
Although I felt that this map was symmetric and bilinear by the properties of definite integrals, I did not think it was positive definite. This was because, say, $<f, f>$ would become $int _0^1:left(left(fleft(xright)-frac{d}{dx}fleft(xright)right)^2right)dx$, which has value greater than zero for values other than the zero function. For example, if $f(x)$ was $2x$, then $f'(x)$ would be $2$ and the value of the above definite integral would be $4/3$. This is why I don't think this space defines an inner product.
So what I am worried about is that I am misunderstanding how the concept of zero vectors with regards to inner products work as this seems to obvious.
If anyone can verify what I have done or tell me how I went wrong, I would greatly appreciate it!
linear-algebra derivatives definite-integrals inner-product-space
asked Nov 15 at 5:03
sktsasus
968414
Does the map $<f, g>$ $=$ $int _0^1:left(left(fleft(xright)-frac{d}{dx}fleft(xright)right)left(gleft(xright)-frac{d}{dx}gleft(xright)right)right)dx$ define an inner product on the set of all polynomial functions of order less than or equal to $n$?
Although I felt that this map was symmetric and bilinear by the properties of definite integrals, I did not think it was positive definite. This was because, say, $<f, f>$ would become $int _0^1:left(left(fleft(xright)-frac{d}{dx}fleft(xright)right)^2right)dx$, which has value greater than zero for values other than the zero function. For example, if $f(x)$ was $2x$, then $f'(x)$ would be $2$ and the value of the above definite integral would be $4/3$. This is why I don't think this space defines an inner product.
So what I am worried about is that I am misunderstanding how the concept of zero vectors with regards to inner products work as this seems to obvious.
If anyone can verify what I have done or tell me how I went wrong, I would greatly appreciate it!
linear-algebra derivatives definite-integrals inner-product-space
linear-algebra derivatives definite-integrals inner-product-space
asked Nov 15 at 5:03
sktsasus
968414
asked Nov 15 at 5:03
sktsasus
968414
edited Nov 15 at 5:08
asked Nov 15 at 5:03
sktsasus
968414
asked Nov 15 at 5:03
sktsasus
968414
asked Nov 15 at 5:03
sktsasus
968414
968414
1
Positive definite means exactly that $langle f,franglegeq 0$ with equality if and only if $f=0$ for all $f$ in your vector space. Your bilinear form has this property, so it should be an inner product as long as you're considering polynomials over $mathbb{R}$.
– Melody
Nov 15 at 5:11
add a comment |
1
Positive definite means exactly that $langle f,franglegeq 0$ with equality if and only if $f=0$ for all $f$ in your vector space. Your bilinear form has this property, so it should be an inner product as long as you're considering polynomials over $mathbb{R}$.
– Melody
Nov 15 at 5:11
1
1
Positive definite means exactly that $langle f,franglegeq 0$ with equality if and only if $f=0$ for all $f$ in your vector space. Your bilinear form has this property, so it should be an inner product as long as you're considering polynomials over $mathbb{R}$.
– Melody
Nov 15 at 5:11
Positive definite means exactly that $langle f,franglegeq 0$ with equality if and only if $f=0$ for all $f$ in your vector space. Your bilinear form has this property, so it should be an inner product as long as you're considering polynomials over $mathbb{R}$.
– Melody
Nov 15 at 5:11
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
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accepted
We have $<f, f>=int _0^1:left(fleft(xright)-f'left(xright)right)^2dx ge 0$ and
$<f, f>=0 iff f=f'$ on $[0,1] $.
Since $f$ is a polynomial, we get
$<f, f>=0 iff f=0$ on $[0,1] $.
answered Nov 15 at 5:57
Fred
42.2k1642
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
We have $<f, f>=int _0^1:left(fleft(xright)-f'left(xright)right)^2dx ge 0$ and
$<f, f>=0 iff f=f'$ on $[0,1] $.
Since $f$ is a polynomial, we get
$<f, f>=0 iff f=0$ on $[0,1] $.
answered Nov 15 at 5:57
Fred
42.2k1642
add a comment |
up vote
1
down vote
accepted
We have $<f, f>=int _0^1:left(fleft(xright)-f'left(xright)right)^2dx ge 0$ and
$<f, f>=0 iff f=f'$ on $[0,1] $.
Since $f$ is a polynomial, we get
$<f, f>=0 iff f=0$ on $[0,1] $.
answered Nov 15 at 5:57
Fred
42.2k1642
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
We have $<f, f>=int _0^1:left(fleft(xright)-f'left(xright)right)^2dx ge 0$ and
$<f, f>=0 iff f=f'$ on $[0,1] $.
Since $f$ is a polynomial, we get
$<f, f>=0 iff f=0$ on $[0,1] $.
answered Nov 15 at 5:57
Fred
42.2k1642
We have $<f, f>=int _0^1:left(fleft(xright)-f'left(xright)right)^2dx ge 0$ and
$<f, f>=0 iff f=f'$ on $[0,1] $.
Since $f$ is a polynomial, we get
$<f, f>=0 iff f=0$ on $[0,1] $.
answered Nov 15 at 5:57
Fred
42.2k1642
answered Nov 15 at 5:57
Fred
42.2k1642
answered Nov 15 at 5:57
Fred
42.2k1642
answered Nov 15 at 5:57
Fred
42.2k1642
42.2k1642
add a comment |
add a comment |
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1
Positive definite means exactly that $langle f,franglegeq 0$ with equality if and only if $f=0$ for all $f$ in your vector space. Your bilinear form has this property, so it should be an inner product as long as you're considering polynomials over $mathbb{R}$.
– Melody
Nov 15 at 5:11