Approximation and sum of integrals.
$begingroup$
Consider the sum of integrals:
$$
int_0^2 left|f(x)-x^4right|^2 dx+ int_{-1}^1 left|f(x)-x^4right|^2 dx.$$
Find a polynomial of degree at most two, such that the sum above is the smallest.
Ok. I know that $f(x)=ax^2+bx+c$, but count it
$$
int_0^2 left|ax^2+bx+c-x^4right|^2 dx
+ int_{-1}^1 left|ax^2+bx+c-x^4right|^2 dx
$$
isn't effective.
integration approximation
edited Jan 9 at 23:18
gt6989b
34.5k22456
asked Jan 9 at 22:46
pawelKpawelK
527
$endgroup$
add a comment |
$begingroup$
Consider the sum of integrals:
$$
int_0^2 left|f(x)-x^4right|^2 dx+ int_{-1}^1 left|f(x)-x^4right|^2 dx.$$
Find a polynomial of degree at most two, such that the sum above is the smallest.
Ok. I know that $f(x)=ax^2+bx+c$, but count it
$$
int_0^2 left|ax^2+bx+c-x^4right|^2 dx
+ int_{-1}^1 left|ax^2+bx+c-x^4right|^2 dx
$$
isn't effective.
integration approximation
edited Jan 9 at 23:18
gt6989b
34.5k22456
asked Jan 9 at 22:46
pawelKpawelK
527
$endgroup$
5
$begingroup$
Why isn't it effective? The squaring removes any worry about the absolute value signs. Expand the square and you get an integral with an eighth degree polynomial. Do the integral and evaluate it. Now take the derivative with respect to $a,b,c$ and set to zero....
$endgroup$
– Ross Millikan
Jan 9 at 22:53
$begingroup$
So, I have :$int_{0}^{2}|ax^2+bx+c-x^4|^2 dx+ int_{-1}^{1}|ax^2+bx+c-x^4|^2 dx=frac{34a^2}{5}+8ab+frac{20ac}{3}-frac{260a}{7}+frac{10b^2}{3}+4bc-frac{64b}{3}+4c^2-frac{68c}{5}+frac{514}{9}$ And it does not look nice.
$endgroup$
– pawelK
Jan 9 at 23:06
$begingroup$
@pawelK why not nice? it is a quadratic in 3 variables, can you minimize this?
$endgroup$
– gt6989b
Jan 9 at 23:19
1
$begingroup$
I didn't check the computations, but that is exactly what I was suggesting. Yes, it is not a cute solution but it will get there.
$endgroup$
– Ross Millikan
Jan 9 at 23:26
add a comment |
$begingroup$
Consider the sum of integrals:
$$
int_0^2 left|f(x)-x^4right|^2 dx+ int_{-1}^1 left|f(x)-x^4right|^2 dx.$$
Find a polynomial of degree at most two, such that the sum above is the smallest.
Ok. I know that $f(x)=ax^2+bx+c$, but count it
$$
int_0^2 left|ax^2+bx+c-x^4right|^2 dx
+ int_{-1}^1 left|ax^2+bx+c-x^4right|^2 dx
$$
isn't effective.
integration approximation
edited Jan 9 at 23:18
gt6989b
34.5k22456
asked Jan 9 at 22:46
pawelKpawelK
527
$endgroup$
Consider the sum of integrals:
$$
int_0^2 left|f(x)-x^4right|^2 dx+ int_{-1}^1 left|f(x)-x^4right|^2 dx.$$
Find a polynomial of degree at most two, such that the sum above is the smallest.
Ok. I know that $f(x)=ax^2+bx+c$, but count it
$$
int_0^2 left|ax^2+bx+c-x^4right|^2 dx
+ int_{-1}^1 left|ax^2+bx+c-x^4right|^2 dx
$$
isn't effective.
integration approximation
integration approximation
edited Jan 9 at 23:18
gt6989b
34.5k22456
asked Jan 9 at 22:46
pawelKpawelK
527
edited Jan 9 at 23:18
gt6989b
34.5k22456
asked Jan 9 at 22:46
pawelKpawelK
527
edited Jan 9 at 23:18
gt6989b
34.5k22456
edited Jan 9 at 23:18
gt6989b
34.5k22456
edited Jan 9 at 23:18
gt6989b
34.5k22456
34.5k22456
asked Jan 9 at 22:46
pawelKpawelK
527
asked Jan 9 at 22:46
pawelKpawelK
527
asked Jan 9 at 22:46
pawelKpawelK
527
527
5
$begingroup$
Why isn't it effective? The squaring removes any worry about the absolute value signs. Expand the square and you get an integral with an eighth degree polynomial. Do the integral and evaluate it. Now take the derivative with respect to $a,b,c$ and set to zero....
$endgroup$
– Ross Millikan
Jan 9 at 22:53
$begingroup$
So, I have :$int_{0}^{2}|ax^2+bx+c-x^4|^2 dx+ int_{-1}^{1}|ax^2+bx+c-x^4|^2 dx=frac{34a^2}{5}+8ab+frac{20ac}{3}-frac{260a}{7}+frac{10b^2}{3}+4bc-frac{64b}{3}+4c^2-frac{68c}{5}+frac{514}{9}$ And it does not look nice.
$endgroup$
– pawelK
Jan 9 at 23:06
$begingroup$
@pawelK why not nice? it is a quadratic in 3 variables, can you minimize this?
$endgroup$
– gt6989b
Jan 9 at 23:19
1
$begingroup$
I didn't check the computations, but that is exactly what I was suggesting. Yes, it is not a cute solution but it will get there.
$endgroup$
– Ross Millikan
Jan 9 at 23:26
add a comment |
5
$begingroup$
Why isn't it effective? The squaring removes any worry about the absolute value signs. Expand the square and you get an integral with an eighth degree polynomial. Do the integral and evaluate it. Now take the derivative with respect to $a,b,c$ and set to zero....
$endgroup$
– Ross Millikan
Jan 9 at 22:53
$begingroup$
So, I have :$int_{0}^{2}|ax^2+bx+c-x^4|^2 dx+ int_{-1}^{1}|ax^2+bx+c-x^4|^2 dx=frac{34a^2}{5}+8ab+frac{20ac}{3}-frac{260a}{7}+frac{10b^2}{3}+4bc-frac{64b}{3}+4c^2-frac{68c}{5}+frac{514}{9}$ And it does not look nice.
$endgroup$
– pawelK
Jan 9 at 23:06
$begingroup$
@pawelK why not nice? it is a quadratic in 3 variables, can you minimize this?
$endgroup$
– gt6989b
Jan 9 at 23:19
1
$begingroup$
I didn't check the computations, but that is exactly what I was suggesting. Yes, it is not a cute solution but it will get there.
$endgroup$
– Ross Millikan
Jan 9 at 23:26
5
5
$begingroup$
Why isn't it effective? The squaring removes any worry about the absolute value signs. Expand the square and you get an integral with an eighth degree polynomial. Do the integral and evaluate it. Now take the derivative with respect to $a,b,c$ and set to zero....
$endgroup$
– Ross Millikan
Jan 9 at 22:53
$begingroup$
Why isn't it effective? The squaring removes any worry about the absolute value signs. Expand the square and you get an integral with an eighth degree polynomial. Do the integral and evaluate it. Now take the derivative with respect to $a,b,c$ and set to zero....
$endgroup$
– Ross Millikan
Jan 9 at 22:53
$begingroup$
So, I have :$int_{0}^{2}|ax^2+bx+c-x^4|^2 dx+ int_{-1}^{1}|ax^2+bx+c-x^4|^2 dx=frac{34a^2}{5}+8ab+frac{20ac}{3}-frac{260a}{7}+frac{10b^2}{3}+4bc-frac{64b}{3}+4c^2-frac{68c}{5}+frac{514}{9}$ And it does not look nice.
$endgroup$
– pawelK
Jan 9 at 23:06
$begingroup$
So, I have :$int_{0}^{2}|ax^2+bx+c-x^4|^2 dx+ int_{-1}^{1}|ax^2+bx+c-x^4|^2 dx=frac{34a^2}{5}+8ab+frac{20ac}{3}-frac{260a}{7}+frac{10b^2}{3}+4bc-frac{64b}{3}+4c^2-frac{68c}{5}+frac{514}{9}$ And it does not look nice.
$endgroup$
– pawelK
Jan 9 at 23:06
$begingroup$
@pawelK why not nice? it is a quadratic in 3 variables, can you minimize this?
$endgroup$
– gt6989b
Jan 9 at 23:19
$begingroup$
@pawelK why not nice? it is a quadratic in 3 variables, can you minimize this?
$endgroup$
– gt6989b
Jan 9 at 23:19
1
1
$begingroup$
I didn't check the computations, but that is exactly what I was suggesting. Yes, it is not a cute solution but it will get there.
$endgroup$
– Ross Millikan
Jan 9 at 23:26
$begingroup$
I didn't check the computations, but that is exactly what I was suggesting. Yes, it is not a cute solution but it will get there.
$endgroup$
– Ross Millikan
Jan 9 at 23:26
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
A « maybe » easier solution is to notice that you are asked to minimize $|f-X^4|^2$, over the $f in V$, where $V$ is a subspace of the ambient space $W=mathbb{R}_4[X]$ and $|cdot|$ is a Euclidean norm over $W$.
So there is an automatic procedure:
- Compute an orthonormal basis of $V$.
- Compute the inner products of $X^4$ with the elements of said basis.
- Take the corresponding element in $V$.
answered Jan 9 at 23:41
MindlackMindlack
4,780210
$endgroup$
$begingroup$
Ok, I have: $int_{0}^{2}(ax^2+bx+c-x^4)1dx=0$ etc. But I we have two integrals and I will get different values $a,b,c$ for each integral.
$endgroup$
– pawelK
Jan 13 at 22:04
$begingroup$
Will it be $int_{0}^{2}(ax^2+bx+c-x^4)1dx+int_{-1}^{1}(ax^2+bx+c-x^4)1dx=0$?
$endgroup$
– pawelK
Jan 13 at 22:11
$begingroup$
Yes, and as well when you replace $1$ with $x$ and $x^2$.
$endgroup$
– Mindlack
Jan 13 at 22:52
add a comment |
$begingroup$
$F(a,b,c) = int_0^2 (ax^2+bx + c - x^4)^2 dx + int_{-1}^{1} (ax^2 + bx + c - x^4)^2 dx$
Differentiating under the integral sign.
$frac {F(a,b,c)}{partial a} = int_0^2 2(ax^2+bx + c - x^4)x^2 dx + int_{-1}^{1} 2(ax^2 + bx + c - x^4)x^2 dx = 0$
$frac {a}{5}x^5 + frac {b}{4} x^4 +frac {c}{3} x^2 + frac {1}{7} x^7|_0^2|_{-1}^1 = 0\
frac {34}{5}a + frac {16}{4} b +frac {10}{3} c = frac {130}{7}$
Do the same for the other variables, and you will get a system of linear equations.
answered Jan 9 at 23:58
Doug MDoug M
45.3k31954
$endgroup$
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
A « maybe » easier solution is to notice that you are asked to minimize $|f-X^4|^2$, over the $f in V$, where $V$ is a subspace of the ambient space $W=mathbb{R}_4[X]$ and $|cdot|$ is a Euclidean norm over $W$.
So there is an automatic procedure:
- Compute an orthonormal basis of $V$.
- Compute the inner products of $X^4$ with the elements of said basis.
- Take the corresponding element in $V$.
answered Jan 9 at 23:41
MindlackMindlack
4,780210
$endgroup$
$begingroup$
Ok, I have: $int_{0}^{2}(ax^2+bx+c-x^4)1dx=0$ etc. But I we have two integrals and I will get different values $a,b,c$ for each integral.
$endgroup$
– pawelK
Jan 13 at 22:04
$begingroup$
Will it be $int_{0}^{2}(ax^2+bx+c-x^4)1dx+int_{-1}^{1}(ax^2+bx+c-x^4)1dx=0$?
$endgroup$
– pawelK
Jan 13 at 22:11
$begingroup$
Yes, and as well when you replace $1$ with $x$ and $x^2$.
$endgroup$
– Mindlack
Jan 13 at 22:52
add a comment |
$begingroup$
A « maybe » easier solution is to notice that you are asked to minimize $|f-X^4|^2$, over the $f in V$, where $V$ is a subspace of the ambient space $W=mathbb{R}_4[X]$ and $|cdot|$ is a Euclidean norm over $W$.
So there is an automatic procedure:
- Compute an orthonormal basis of $V$.
- Compute the inner products of $X^4$ with the elements of said basis.
- Take the corresponding element in $V$.
answered Jan 9 at 23:41
MindlackMindlack
4,780210
$endgroup$
$begingroup$
Ok, I have: $int_{0}^{2}(ax^2+bx+c-x^4)1dx=0$ etc. But I we have two integrals and I will get different values $a,b,c$ for each integral.
$endgroup$
– pawelK
Jan 13 at 22:04
$begingroup$
Will it be $int_{0}^{2}(ax^2+bx+c-x^4)1dx+int_{-1}^{1}(ax^2+bx+c-x^4)1dx=0$?
$endgroup$
– pawelK
Jan 13 at 22:11
$begingroup$
Yes, and as well when you replace $1$ with $x$ and $x^2$.
$endgroup$
– Mindlack
Jan 13 at 22:52
add a comment |
$begingroup$
A « maybe » easier solution is to notice that you are asked to minimize $|f-X^4|^2$, over the $f in V$, where $V$ is a subspace of the ambient space $W=mathbb{R}_4[X]$ and $|cdot|$ is a Euclidean norm over $W$.
So there is an automatic procedure:
- Compute an orthonormal basis of $V$.
- Compute the inner products of $X^4$ with the elements of said basis.
- Take the corresponding element in $V$.
answered Jan 9 at 23:41
MindlackMindlack
4,780210
$endgroup$
A « maybe » easier solution is to notice that you are asked to minimize $|f-X^4|^2$, over the $f in V$, where $V$ is a subspace of the ambient space $W=mathbb{R}_4[X]$ and $|cdot|$ is a Euclidean norm over $W$.
So there is an automatic procedure:
- Compute an orthonormal basis of $V$.
- Compute the inner products of $X^4$ with the elements of said basis.
- Take the corresponding element in $V$.
answered Jan 9 at 23:41
MindlackMindlack
4,780210
answered Jan 9 at 23:41
MindlackMindlack
4,780210
answered Jan 9 at 23:41
MindlackMindlack
4,780210
answered Jan 9 at 23:41
MindlackMindlack
4,780210
4,780210
$begingroup$
Ok, I have: $int_{0}^{2}(ax^2+bx+c-x^4)1dx=0$ etc. But I we have two integrals and I will get different values $a,b,c$ for each integral.
$endgroup$
– pawelK
Jan 13 at 22:04
$begingroup$
Will it be $int_{0}^{2}(ax^2+bx+c-x^4)1dx+int_{-1}^{1}(ax^2+bx+c-x^4)1dx=0$?
$endgroup$
– pawelK
Jan 13 at 22:11
$begingroup$
Yes, and as well when you replace $1$ with $x$ and $x^2$.
$endgroup$
– Mindlack
Jan 13 at 22:52
add a comment |
$begingroup$
Ok, I have: $int_{0}^{2}(ax^2+bx+c-x^4)1dx=0$ etc. But I we have two integrals and I will get different values $a,b,c$ for each integral.
$endgroup$
– pawelK
Jan 13 at 22:04
$begingroup$
Will it be $int_{0}^{2}(ax^2+bx+c-x^4)1dx+int_{-1}^{1}(ax^2+bx+c-x^4)1dx=0$?
$endgroup$
– pawelK
Jan 13 at 22:11
$begingroup$
Yes, and as well when you replace $1$ with $x$ and $x^2$.
$endgroup$
– Mindlack
Jan 13 at 22:52
$begingroup$
Ok, I have: $int_{0}^{2}(ax^2+bx+c-x^4)1dx=0$ etc. But I we have two integrals and I will get different values $a,b,c$ for each integral.
$endgroup$
– pawelK
Jan 13 at 22:04
$begingroup$
Ok, I have: $int_{0}^{2}(ax^2+bx+c-x^4)1dx=0$ etc. But I we have two integrals and I will get different values $a,b,c$ for each integral.
$endgroup$
– pawelK
Jan 13 at 22:04
$begingroup$
Will it be $int_{0}^{2}(ax^2+bx+c-x^4)1dx+int_{-1}^{1}(ax^2+bx+c-x^4)1dx=0$?
$endgroup$
– pawelK
Jan 13 at 22:11
$begingroup$
Will it be $int_{0}^{2}(ax^2+bx+c-x^4)1dx+int_{-1}^{1}(ax^2+bx+c-x^4)1dx=0$?
$endgroup$
– pawelK
Jan 13 at 22:11
$begingroup$
Yes, and as well when you replace $1$ with $x$ and $x^2$.
$endgroup$
– Mindlack
Jan 13 at 22:52
$begingroup$
Yes, and as well when you replace $1$ with $x$ and $x^2$.
$endgroup$
– Mindlack
Jan 13 at 22:52
add a comment |
$begingroup$
$F(a,b,c) = int_0^2 (ax^2+bx + c - x^4)^2 dx + int_{-1}^{1} (ax^2 + bx + c - x^4)^2 dx$
Differentiating under the integral sign.
$frac {F(a,b,c)}{partial a} = int_0^2 2(ax^2+bx + c - x^4)x^2 dx + int_{-1}^{1} 2(ax^2 + bx + c - x^4)x^2 dx = 0$
$frac {a}{5}x^5 + frac {b}{4} x^4 +frac {c}{3} x^2 + frac {1}{7} x^7|_0^2|_{-1}^1 = 0\
frac {34}{5}a + frac {16}{4} b +frac {10}{3} c = frac {130}{7}$
Do the same for the other variables, and you will get a system of linear equations.
answered Jan 9 at 23:58
Doug MDoug M
45.3k31954
$endgroup$
add a comment |
$begingroup$
$F(a,b,c) = int_0^2 (ax^2+bx + c - x^4)^2 dx + int_{-1}^{1} (ax^2 + bx + c - x^4)^2 dx$
Differentiating under the integral sign.
$frac {F(a,b,c)}{partial a} = int_0^2 2(ax^2+bx + c - x^4)x^2 dx + int_{-1}^{1} 2(ax^2 + bx + c - x^4)x^2 dx = 0$
$frac {a}{5}x^5 + frac {b}{4} x^4 +frac {c}{3} x^2 + frac {1}{7} x^7|_0^2|_{-1}^1 = 0\
frac {34}{5}a + frac {16}{4} b +frac {10}{3} c = frac {130}{7}$
Do the same for the other variables, and you will get a system of linear equations.
answered Jan 9 at 23:58
Doug MDoug M
45.3k31954
$endgroup$
add a comment |
$begingroup$
$F(a,b,c) = int_0^2 (ax^2+bx + c - x^4)^2 dx + int_{-1}^{1} (ax^2 + bx + c - x^4)^2 dx$
Differentiating under the integral sign.
$frac {F(a,b,c)}{partial a} = int_0^2 2(ax^2+bx + c - x^4)x^2 dx + int_{-1}^{1} 2(ax^2 + bx + c - x^4)x^2 dx = 0$
$frac {a}{5}x^5 + frac {b}{4} x^4 +frac {c}{3} x^2 + frac {1}{7} x^7|_0^2|_{-1}^1 = 0\
frac {34}{5}a + frac {16}{4} b +frac {10}{3} c = frac {130}{7}$
Do the same for the other variables, and you will get a system of linear equations.
answered Jan 9 at 23:58
Doug MDoug M
45.3k31954
$endgroup$
$F(a,b,c) = int_0^2 (ax^2+bx + c - x^4)^2 dx + int_{-1}^{1} (ax^2 + bx + c - x^4)^2 dx$
Differentiating under the integral sign.
$frac {F(a,b,c)}{partial a} = int_0^2 2(ax^2+bx + c - x^4)x^2 dx + int_{-1}^{1} 2(ax^2 + bx + c - x^4)x^2 dx = 0$
$frac {a}{5}x^5 + frac {b}{4} x^4 +frac {c}{3} x^2 + frac {1}{7} x^7|_0^2|_{-1}^1 = 0\
frac {34}{5}a + frac {16}{4} b +frac {10}{3} c = frac {130}{7}$
Do the same for the other variables, and you will get a system of linear equations.
answered Jan 9 at 23:58
Doug MDoug M
45.3k31954
answered Jan 9 at 23:58
Doug MDoug M
45.3k31954
answered Jan 9 at 23:58
Doug MDoug M
45.3k31954
answered Jan 9 at 23:58
Doug MDoug M
45.3k31954
45.3k31954
add a comment |
add a comment |
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5
$begingroup$
Why isn't it effective? The squaring removes any worry about the absolute value signs. Expand the square and you get an integral with an eighth degree polynomial. Do the integral and evaluate it. Now take the derivative with respect to $a,b,c$ and set to zero....
$endgroup$
– Ross Millikan
Jan 9 at 22:53
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So, I have :$int_{0}^{2}|ax^2+bx+c-x^4|^2 dx+ int_{-1}^{1}|ax^2+bx+c-x^4|^2 dx=frac{34a^2}{5}+8ab+frac{20ac}{3}-frac{260a}{7}+frac{10b^2}{3}+4bc-frac{64b}{3}+4c^2-frac{68c}{5}+frac{514}{9}$ And it does not look nice.
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– pawelK
Jan 9 at 23:06
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@pawelK why not nice? it is a quadratic in 3 variables, can you minimize this?
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– gt6989b
Jan 9 at 23:19
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I didn't check the computations, but that is exactly what I was suggesting. Yes, it is not a cute solution but it will get there.
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– Ross Millikan
Jan 9 at 23:26