Approximation and sum of integrals.












0












$begingroup$



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







share|cite|improve this question











$endgroup$








  • 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
















0












$begingroup$



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







share|cite|improve this question











$endgroup$








  • 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














0












0








0





$begingroup$



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







share|cite|improve this question











$endgroup$





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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 9 at 23:18









gt6989b

34.5k22456




34.5k22456










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














  • 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










2 Answers
2






active

oldest

votes


















2












$begingroup$

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:




  1. Compute an orthonormal basis of $V$.

  2. Compute the inner products of $X^4$ with the elements of said basis.

  3. Take the corresponding element in $V$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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



















0












$begingroup$

$F(a,b,c) = int_0^2 (ax^2+bx + c - x^4)^2 dx + int_{-1}^{1} (ax^2 + bx + c - x^4)^2 dx$



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.






share|cite|improve this answer









$endgroup$













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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    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:




    1. Compute an orthonormal basis of $V$.

    2. Compute the inner products of $X^4$ with the elements of said basis.

    3. Take the corresponding element in $V$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      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
















    2












    $begingroup$

    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:




    1. Compute an orthonormal basis of $V$.

    2. Compute the inner products of $X^4$ with the elements of said basis.

    3. Take the corresponding element in $V$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      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














    2












    2








    2





    $begingroup$

    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:




    1. Compute an orthonormal basis of $V$.

    2. Compute the inner products of $X^4$ with the elements of said basis.

    3. Take the corresponding element in $V$.






    share|cite|improve this answer









    $endgroup$



    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:




    1. Compute an orthonormal basis of $V$.

    2. Compute the inner products of $X^4$ with the elements of said basis.

    3. Take the corresponding element in $V$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    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


















    • $begingroup$
      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











    0












    $begingroup$

    $F(a,b,c) = int_0^2 (ax^2+bx + c - x^4)^2 dx + int_{-1}^{1} (ax^2 + bx + c - x^4)^2 dx$



    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.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      $F(a,b,c) = int_0^2 (ax^2+bx + c - x^4)^2 dx + int_{-1}^{1} (ax^2 + bx + c - x^4)^2 dx$



      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.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        $F(a,b,c) = int_0^2 (ax^2+bx + c - x^4)^2 dx + int_{-1}^{1} (ax^2 + bx + c - x^4)^2 dx$



        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.






        share|cite|improve this answer









        $endgroup$



        $F(a,b,c) = int_0^2 (ax^2+bx + c - x^4)^2 dx + int_{-1}^{1} (ax^2 + bx + c - x^4)^2 dx$



        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.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 9 at 23:58









        Doug MDoug M

        45.3k31954




        45.3k31954






























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