Let $int_0^1 p(x)q(x)dx$ be an inner product. If $f(p(x)) = p'(-15) +8p(-1)$ is a functional, find $r(x)$...
$begingroup$
Let $langle p,q rangle = int_0^1 p(x)q(x)dx$ be an inner product on $mathbb{P}^2$. If $f(p(x)) = p'(-15) +8p(-1)$ is a linear functional, find the unique $r(x)$ such that $f(p(x)) = langle p,rrangle $ for all $p(x)in mathbb{P}^2$.
Note that $mathbb{P}^2$ is the space of polynomial of degree less than two. I know that this problem is an application of the Riesz Representation Theorem, but I can't seem to solve it. So far, I found an orthonormal basis for the space and tried to write $r(x) = sum_{n=1}^2{f(b_i)b_i}$ where $B_perp = {1,12x-6}$, but the answer I got was incorrect. Suggestions?
linear-algebra functional-analysis
asked Dec 5 '18 at 4:07
johnny133253johnny133253
18510
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add a comment |
$begingroup$
Let $langle p,q rangle = int_0^1 p(x)q(x)dx$ be an inner product on $mathbb{P}^2$. If $f(p(x)) = p'(-15) +8p(-1)$ is a linear functional, find the unique $r(x)$ such that $f(p(x)) = langle p,rrangle $ for all $p(x)in mathbb{P}^2$.
Note that $mathbb{P}^2$ is the space of polynomial of degree less than two. I know that this problem is an application of the Riesz Representation Theorem, but I can't seem to solve it. So far, I found an orthonormal basis for the space and tried to write $r(x) = sum_{n=1}^2{f(b_i)b_i}$ where $B_perp = {1,12x-6}$, but the answer I got was incorrect. Suggestions?
linear-algebra functional-analysis
asked Dec 5 '18 at 4:07
johnny133253johnny133253
18510
$endgroup$
1
$begingroup$
Haven't checked in detail but here is a possibility for you to investigate. Your basis is not orthonormal since $langle 12x-6,12x-6rangle=12ne1$. You should be able to use this basis anyway as it is orthogonal, but I think it would be easy to make a careless error.
$endgroup$
– David
Dec 5 '18 at 4:13
add a comment |
$begingroup$
Let $langle p,q rangle = int_0^1 p(x)q(x)dx$ be an inner product on $mathbb{P}^2$. If $f(p(x)) = p'(-15) +8p(-1)$ is a linear functional, find the unique $r(x)$ such that $f(p(x)) = langle p,rrangle $ for all $p(x)in mathbb{P}^2$.
Note that $mathbb{P}^2$ is the space of polynomial of degree less than two. I know that this problem is an application of the Riesz Representation Theorem, but I can't seem to solve it. So far, I found an orthonormal basis for the space and tried to write $r(x) = sum_{n=1}^2{f(b_i)b_i}$ where $B_perp = {1,12x-6}$, but the answer I got was incorrect. Suggestions?
linear-algebra functional-analysis
asked Dec 5 '18 at 4:07
johnny133253johnny133253
18510
$endgroup$
Let $langle p,q rangle = int_0^1 p(x)q(x)dx$ be an inner product on $mathbb{P}^2$. If $f(p(x)) = p'(-15) +8p(-1)$ is a linear functional, find the unique $r(x)$ such that $f(p(x)) = langle p,rrangle $ for all $p(x)in mathbb{P}^2$.
Note that $mathbb{P}^2$ is the space of polynomial of degree less than two. I know that this problem is an application of the Riesz Representation Theorem, but I can't seem to solve it. So far, I found an orthonormal basis for the space and tried to write $r(x) = sum_{n=1}^2{f(b_i)b_i}$ where $B_perp = {1,12x-6}$, but the answer I got was incorrect. Suggestions?
linear-algebra functional-analysis
linear-algebra functional-analysis
asked Dec 5 '18 at 4:07
johnny133253johnny133253
18510
asked Dec 5 '18 at 4:07
johnny133253johnny133253
18510
asked Dec 5 '18 at 4:07
johnny133253johnny133253
18510
asked Dec 5 '18 at 4:07
johnny133253johnny133253
18510
asked Dec 5 '18 at 4:07
johnny133253johnny133253
18510
18510
1
$begingroup$
Haven't checked in detail but here is a possibility for you to investigate. Your basis is not orthonormal since $langle 12x-6,12x-6rangle=12ne1$. You should be able to use this basis anyway as it is orthogonal, but I think it would be easy to make a careless error.
$endgroup$
– David
Dec 5 '18 at 4:13
add a comment |
1
$begingroup$
Haven't checked in detail but here is a possibility for you to investigate. Your basis is not orthonormal since $langle 12x-6,12x-6rangle=12ne1$. You should be able to use this basis anyway as it is orthogonal, but I think it would be easy to make a careless error.
$endgroup$
– David
Dec 5 '18 at 4:13
1
1
$begingroup$
Haven't checked in detail but here is a possibility for you to investigate. Your basis is not orthonormal since $langle 12x-6,12x-6rangle=12ne1$. You should be able to use this basis anyway as it is orthogonal, but I think it would be easy to make a careless error.
$endgroup$
– David
Dec 5 '18 at 4:13
$begingroup$
Haven't checked in detail but here is a possibility for you to investigate. Your basis is not orthonormal since $langle 12x-6,12x-6rangle=12ne1$. You should be able to use this basis anyway as it is orthogonal, but I think it would be easy to make a careless error.
$endgroup$
– David
Dec 5 '18 at 4:13
add a comment |
1 Answer
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$begingroup$
Your basis is not orthonormal. Since $1$ and $12x-6$ are orthogonal, what you need to do is to normalize $12x-6$. If your normalize $12x-6$, you should get the right result.
In this case, though, it is much easier to just find $r$ directly. That is, $r(x)=cx+d$, so
$$
langle ax+b,cx+drangle=int_0^1(ax+b)(cx+d)=aleft(frac c3+frac d2right)+bleft(frac c2+dright).
$$
As $f(ax+b)=a+8(-a+b)=-7a+8b$, we need $c/3+d/2=-7$, $c/2+d=8$. This gives $c=-132$, $d=74$. So $r(x)=-132x+74$.
answered Dec 5 '18 at 5:16
Martin ArgeramiMartin Argerami
125k1181180
$endgroup$
$begingroup$
Thank you. I did try this process and got lost at the very last step. Now I understand!
$endgroup$
– johnny133253
Dec 5 '18 at 5:34
add a comment |
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$begingroup$
Your basis is not orthonormal. Since $1$ and $12x-6$ are orthogonal, what you need to do is to normalize $12x-6$. If your normalize $12x-6$, you should get the right result.
In this case, though, it is much easier to just find $r$ directly. That is, $r(x)=cx+d$, so
$$
langle ax+b,cx+drangle=int_0^1(ax+b)(cx+d)=aleft(frac c3+frac d2right)+bleft(frac c2+dright).
$$
As $f(ax+b)=a+8(-a+b)=-7a+8b$, we need $c/3+d/2=-7$, $c/2+d=8$. This gives $c=-132$, $d=74$. So $r(x)=-132x+74$.
answered Dec 5 '18 at 5:16
Martin ArgeramiMartin Argerami
125k1181180
$endgroup$
$begingroup$
Thank you. I did try this process and got lost at the very last step. Now I understand!
$endgroup$
– johnny133253
Dec 5 '18 at 5:34
add a comment |
$begingroup$
Your basis is not orthonormal. Since $1$ and $12x-6$ are orthogonal, what you need to do is to normalize $12x-6$. If your normalize $12x-6$, you should get the right result.
In this case, though, it is much easier to just find $r$ directly. That is, $r(x)=cx+d$, so
$$
langle ax+b,cx+drangle=int_0^1(ax+b)(cx+d)=aleft(frac c3+frac d2right)+bleft(frac c2+dright).
$$
As $f(ax+b)=a+8(-a+b)=-7a+8b$, we need $c/3+d/2=-7$, $c/2+d=8$. This gives $c=-132$, $d=74$. So $r(x)=-132x+74$.
answered Dec 5 '18 at 5:16
Martin ArgeramiMartin Argerami
125k1181180
$endgroup$
$begingroup$
Thank you. I did try this process and got lost at the very last step. Now I understand!
$endgroup$
– johnny133253
Dec 5 '18 at 5:34
add a comment |
$begingroup$
Your basis is not orthonormal. Since $1$ and $12x-6$ are orthogonal, what you need to do is to normalize $12x-6$. If your normalize $12x-6$, you should get the right result.
In this case, though, it is much easier to just find $r$ directly. That is, $r(x)=cx+d$, so
$$
langle ax+b,cx+drangle=int_0^1(ax+b)(cx+d)=aleft(frac c3+frac d2right)+bleft(frac c2+dright).
$$
As $f(ax+b)=a+8(-a+b)=-7a+8b$, we need $c/3+d/2=-7$, $c/2+d=8$. This gives $c=-132$, $d=74$. So $r(x)=-132x+74$.
answered Dec 5 '18 at 5:16
Martin ArgeramiMartin Argerami
125k1181180
$endgroup$
Your basis is not orthonormal. Since $1$ and $12x-6$ are orthogonal, what you need to do is to normalize $12x-6$. If your normalize $12x-6$, you should get the right result.
In this case, though, it is much easier to just find $r$ directly. That is, $r(x)=cx+d$, so
$$
langle ax+b,cx+drangle=int_0^1(ax+b)(cx+d)=aleft(frac c3+frac d2right)+bleft(frac c2+dright).
$$
As $f(ax+b)=a+8(-a+b)=-7a+8b$, we need $c/3+d/2=-7$, $c/2+d=8$. This gives $c=-132$, $d=74$. So $r(x)=-132x+74$.
answered Dec 5 '18 at 5:16
Martin ArgeramiMartin Argerami
125k1181180
answered Dec 5 '18 at 5:16
Martin ArgeramiMartin Argerami
125k1181180
answered Dec 5 '18 at 5:16
Martin ArgeramiMartin Argerami
125k1181180
answered Dec 5 '18 at 5:16
Martin ArgeramiMartin Argerami
125k1181180
125k1181180
$begingroup$
Thank you. I did try this process and got lost at the very last step. Now I understand!
$endgroup$
– johnny133253
Dec 5 '18 at 5:34
add a comment |
$begingroup$
Thank you. I did try this process and got lost at the very last step. Now I understand!
$endgroup$
– johnny133253
Dec 5 '18 at 5:34
$begingroup$
Thank you. I did try this process and got lost at the very last step. Now I understand!
$endgroup$
– johnny133253
Dec 5 '18 at 5:34
$begingroup$
Thank you. I did try this process and got lost at the very last step. Now I understand!
$endgroup$
– johnny133253
Dec 5 '18 at 5:34
add a comment |
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$begingroup$
Haven't checked in detail but here is a possibility for you to investigate. Your basis is not orthonormal since $langle 12x-6,12x-6rangle=12ne1$. You should be able to use this basis anyway as it is orthogonal, but I think it would be easy to make a careless error.
$endgroup$
– David
Dec 5 '18 at 4:13