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Show that the constant polynomial $g $ closest to $f$
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In the vector space $C(1,3)$, endowed with the internal product $$langle f, g rangle = int_ 1^3 f (x) g (x) dx $$ for all $f, g in C (1,3) $. For $f (x) = dfrac1x $ with $x in (1,3) $, show that the constant polynomial $g $ closest to $f $ is $g (x) = dfracln (3)2 $. Calculate $|| f-g || ^ 2 $
What do you mean with a closer function?
calculus real-analysis linear-algebra multivariable-calculus
asked Jul 14 at 23:12
Santiago Seeker
597
add a comment |Â
up vote
1
down vote
favorite
In the vector space $C(1,3)$, endowed with the internal product $$langle f, g rangle = int_ 1^3 f (x) g (x) dx $$ for all $f, g in C (1,3) $. For $f (x) = dfrac1x $ with $x in (1,3) $, show that the constant polynomial $g $ closest to $f $ is $g (x) = dfracln (3)2 $. Calculate $|| f-g || ^ 2 $
What do you mean with a closer function?
calculus real-analysis linear-algebra multivariable-calculus
asked Jul 14 at 23:12
Santiago Seeker
597
This is an approximation problem. We show that the polynomial that minimizes the error in approximation is the given $g$.
– Sean Roberson
Jul 14 at 23:15
And how can I solve it? because I found some exercises right here on the platform, but I do not understand them very well.
– Santiago Seeker
Jul 14 at 23:19
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
In the vector space $C(1,3)$, endowed with the internal product $$langle f, g rangle = int_ 1^3 f (x) g (x) dx $$ for all $f, g in C (1,3) $. For $f (x) = dfrac1x $ with $x in (1,3) $, show that the constant polynomial $g $ closest to $f $ is $g (x) = dfracln (3)2 $. Calculate $|| f-g || ^ 2 $
What do you mean with a closer function?
calculus real-analysis linear-algebra multivariable-calculus
asked Jul 14 at 23:12
Santiago Seeker
597
In the vector space $C(1,3)$, endowed with the internal product $$langle f, g rangle = int_ 1^3 f (x) g (x) dx $$ for all $f, g in C (1,3) $. For $f (x) = dfrac1x $ with $x in (1,3) $, show that the constant polynomial $g $ closest to $f $ is $g (x) = dfracln (3)2 $. Calculate $|| f-g || ^ 2 $
What do you mean with a closer function?
calculus real-analysis linear-algebra multivariable-calculus
asked Jul 14 at 23:12
Santiago Seeker
597
asked Jul 14 at 23:12
Santiago Seeker
597
asked Jul 14 at 23:12
Santiago Seeker
597
asked Jul 14 at 23:12
Santiago Seeker
597
597
This is an approximation problem. We show that the polynomial that minimizes the error in approximation is the given $g$.
– Sean Roberson
Jul 14 at 23:15
And how can I solve it? because I found some exercises right here on the platform, but I do not understand them very well.
– Santiago Seeker
Jul 14 at 23:19
add a comment |Â
This is an approximation problem. We show that the polynomial that minimizes the error in approximation is the given $g$.
– Sean Roberson
Jul 14 at 23:15
And how can I solve it? because I found some exercises right here on the platform, but I do not understand them very well.
– Santiago Seeker
Jul 14 at 23:19
This is an approximation problem. We show that the polynomial that minimizes the error in approximation is the given $g$.
– Sean Roberson
Jul 14 at 23:15
This is an approximation problem. We show that the polynomial that minimizes the error in approximation is the given $g$.
– Sean Roberson
Jul 14 at 23:15
And how can I solve it? because I found some exercises right here on the platform, but I do not understand them very well.
– Santiago Seeker
Jul 14 at 23:19
And how can I solve it? because I found some exercises right here on the platform, but I do not understand them very well.
– Santiago Seeker
Jul 14 at 23:19
add a comment |Â
1 Answer
1
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oldest
votes
up vote
3
down vote
accepted
We have
$$
||f-g||^2 = int_1^3 (f-g)^2(x)dx = int_1^3 (1/x-g)^2dx = 2 g^2 - g log(9) + 2/3 =p(g)
$$
The parabola $p(g)$ has a minimum at $g^*=log(3)/2$ and $||f-g||^2$ is precisely $$p(g^*)=frac4 - 3 log^2 36$$
answered Jul 14 at 23:22
Rafael Gonzalez Lopez
652112
Excellent! thank you very much ... apologize, you have a material where you can do the justifications of the procedure, that is, a little theory about the subject.
– Santiago Seeker
Jul 14 at 23:28
Not sorry. You ask for the constant $g$ "closest" to $f$. This is by definition the constant $g$ that minimize (as a function of $g$) $||f-g||$ and, since $||f-g||geq 0$ and $h(x)=x^2$ is an increasing function in $mathbbR_geq 0$, this is equivalente to minimize $||f-g||^2 = langle f-g,f-grangle$.
– Rafael Gonzalez Lopez
Jul 14 at 23:32
For instance, if you ask by the linear function $g$ closest to $f$, now you substitute $g=ax+b$ before the integration and minimize using 2-variables optimization $a$ and $b$.
– Rafael Gonzalez Lopez
Jul 14 at 23:34
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
We have
$$
||f-g||^2 = int_1^3 (f-g)^2(x)dx = int_1^3 (1/x-g)^2dx = 2 g^2 - g log(9) + 2/3 =p(g)
$$
The parabola $p(g)$ has a minimum at $g^*=log(3)/2$ and $||f-g||^2$ is precisely $$p(g^*)=frac4 - 3 log^2 36$$
answered Jul 14 at 23:22
Rafael Gonzalez Lopez
652112
Excellent! thank you very much ... apologize, you have a material where you can do the justifications of the procedure, that is, a little theory about the subject.
– Santiago Seeker
Jul 14 at 23:28
Not sorry. You ask for the constant $g$ "closest" to $f$. This is by definition the constant $g$ that minimize (as a function of $g$) $||f-g||$ and, since $||f-g||geq 0$ and $h(x)=x^2$ is an increasing function in $mathbbR_geq 0$, this is equivalente to minimize $||f-g||^2 = langle f-g,f-grangle$.
– Rafael Gonzalez Lopez
Jul 14 at 23:32
For instance, if you ask by the linear function $g$ closest to $f$, now you substitute $g=ax+b$ before the integration and minimize using 2-variables optimization $a$ and $b$.
– Rafael Gonzalez Lopez
Jul 14 at 23:34
add a comment |Â
up vote
3
down vote
accepted
We have
$$
||f-g||^2 = int_1^3 (f-g)^2(x)dx = int_1^3 (1/x-g)^2dx = 2 g^2 - g log(9) + 2/3 =p(g)
$$
The parabola $p(g)$ has a minimum at $g^*=log(3)/2$ and $||f-g||^2$ is precisely $$p(g^*)=frac4 - 3 log^2 36$$
answered Jul 14 at 23:22
Rafael Gonzalez Lopez
652112
Excellent! thank you very much ... apologize, you have a material where you can do the justifications of the procedure, that is, a little theory about the subject.
– Santiago Seeker
Jul 14 at 23:28
Not sorry. You ask for the constant $g$ "closest" to $f$. This is by definition the constant $g$ that minimize (as a function of $g$) $||f-g||$ and, since $||f-g||geq 0$ and $h(x)=x^2$ is an increasing function in $mathbbR_geq 0$, this is equivalente to minimize $||f-g||^2 = langle f-g,f-grangle$.
– Rafael Gonzalez Lopez
Jul 14 at 23:32
For instance, if you ask by the linear function $g$ closest to $f$, now you substitute $g=ax+b$ before the integration and minimize using 2-variables optimization $a$ and $b$.
– Rafael Gonzalez Lopez
Jul 14 at 23:34
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
We have
$$
||f-g||^2 = int_1^3 (f-g)^2(x)dx = int_1^3 (1/x-g)^2dx = 2 g^2 - g log(9) + 2/3 =p(g)
$$
The parabola $p(g)$ has a minimum at $g^*=log(3)/2$ and $||f-g||^2$ is precisely $$p(g^*)=frac4 - 3 log^2 36$$
answered Jul 14 at 23:22
Rafael Gonzalez Lopez
652112
We have
$$
||f-g||^2 = int_1^3 (f-g)^2(x)dx = int_1^3 (1/x-g)^2dx = 2 g^2 - g log(9) + 2/3 =p(g)
$$
The parabola $p(g)$ has a minimum at $g^*=log(3)/2$ and $||f-g||^2$ is precisely $$p(g^*)=frac4 - 3 log^2 36$$
answered Jul 14 at 23:22
Rafael Gonzalez Lopez
652112
answered Jul 14 at 23:22
Rafael Gonzalez Lopez
652112
answered Jul 14 at 23:22
Rafael Gonzalez Lopez
652112
answered Jul 14 at 23:22
Rafael Gonzalez Lopez
652112
652112
Excellent! thank you very much ... apologize, you have a material where you can do the justifications of the procedure, that is, a little theory about the subject.
– Santiago Seeker
Jul 14 at 23:28
Not sorry. You ask for the constant $g$ "closest" to $f$. This is by definition the constant $g$ that minimize (as a function of $g$) $||f-g||$ and, since $||f-g||geq 0$ and $h(x)=x^2$ is an increasing function in $mathbbR_geq 0$, this is equivalente to minimize $||f-g||^2 = langle f-g,f-grangle$.
– Rafael Gonzalez Lopez
Jul 14 at 23:32
For instance, if you ask by the linear function $g$ closest to $f$, now you substitute $g=ax+b$ before the integration and minimize using 2-variables optimization $a$ and $b$.
– Rafael Gonzalez Lopez
Jul 14 at 23:34
add a comment |Â
Excellent! thank you very much ... apologize, you have a material where you can do the justifications of the procedure, that is, a little theory about the subject.
– Santiago Seeker
Jul 14 at 23:28
Not sorry. You ask for the constant $g$ "closest" to $f$. This is by definition the constant $g$ that minimize (as a function of $g$) $||f-g||$ and, since $||f-g||geq 0$ and $h(x)=x^2$ is an increasing function in $mathbbR_geq 0$, this is equivalente to minimize $||f-g||^2 = langle f-g,f-grangle$.
– Rafael Gonzalez Lopez
Jul 14 at 23:32
For instance, if you ask by the linear function $g$ closest to $f$, now you substitute $g=ax+b$ before the integration and minimize using 2-variables optimization $a$ and $b$.
– Rafael Gonzalez Lopez
Jul 14 at 23:34
Excellent! thank you very much ... apologize, you have a material where you can do the justifications of the procedure, that is, a little theory about the subject.
– Santiago Seeker
Jul 14 at 23:28
Excellent! thank you very much ... apologize, you have a material where you can do the justifications of the procedure, that is, a little theory about the subject.
– Santiago Seeker
Jul 14 at 23:28
Not sorry. You ask for the constant $g$ "closest" to $f$. This is by definition the constant $g$ that minimize (as a function of $g$) $||f-g||$ and, since $||f-g||geq 0$ and $h(x)=x^2$ is an increasing function in $mathbbR_geq 0$, this is equivalente to minimize $||f-g||^2 = langle f-g,f-grangle$.
– Rafael Gonzalez Lopez
Jul 14 at 23:32
Not sorry. You ask for the constant $g$ "closest" to $f$. This is by definition the constant $g$ that minimize (as a function of $g$) $||f-g||$ and, since $||f-g||geq 0$ and $h(x)=x^2$ is an increasing function in $mathbbR_geq 0$, this is equivalente to minimize $||f-g||^2 = langle f-g,f-grangle$.
– Rafael Gonzalez Lopez
Jul 14 at 23:32
For instance, if you ask by the linear function $g$ closest to $f$, now you substitute $g=ax+b$ before the integration and minimize using 2-variables optimization $a$ and $b$.
– Rafael Gonzalez Lopez
Jul 14 at 23:34
For instance, if you ask by the linear function $g$ closest to $f$, now you substitute $g=ax+b$ before the integration and minimize using 2-variables optimization $a$ and $b$.
– Rafael Gonzalez Lopez
Jul 14 at 23:34
add a comment |Â
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This is an approximation problem. We show that the polynomial that minimizes the error in approximation is the given $g$.
– Sean Roberson
Jul 14 at 23:15
And how can I solve it? because I found some exercises right here on the platform, but I do not understand them very well.
– Santiago Seeker
Jul 14 at 23:19