First derivative approximation using function values in equidistant points












1












$begingroup$


Given a function $f: [x_0,x_4] → Bbb R$ and equidistant points $x_0, x_1, x_2, x_3, x_4$ so that $h=x_{i+1} - x_i > 0$.



Normally I would do, $f'(x) approx dfrac{f(x+h)-f(x)}{h}$, but here I want to find approximation for $f'(x_2)$ as linear combination of function values in these points as $f'(x_2) approx sum_{j=0}^4 alpha_j f(x_j)$. There are no boundary conditions so I can't make a system of linear equations, at least I see no straight forward way.



Are there any standard methods of how to determine alphas related to the finite elements method, so that the formula will be as precise as possible for polynomials of as high order as possible?



I see that this is closely related to the five-point midpoint formula
https://www3.nd.edu/~zxu2/acms40390F15/Lec-4.1.pdf if one sets x_0= x_2 -2h and so on, but how to derive this formula?



How does that generalise to higher order derivatives?






finite-element-method numerical-calculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    What about $f'(x_2) approx frac{f(x_3)-f(x_1)}{2h}$
    $endgroup$
    – Jakobian
    Jan 13 at 12:54










  • $begingroup$
    This only uses two points.
    $endgroup$
    – random guy
    Jan 13 at 13:00










  • $begingroup$
    And it makes sense, considering how derivative is a local property
    $endgroup$
    – Jakobian
    Jan 13 at 13:49










  • $begingroup$
    But still, how can this be derived?
    $endgroup$
    – random guy
    Jan 13 at 14:23
















1












$begingroup$


Given a function $f: [x_0,x_4] → Bbb R$ and equidistant points $x_0, x_1, x_2, x_3, x_4$ so that $h=x_{i+1} - x_i > 0$.



Normally I would do, $f'(x) approx dfrac{f(x+h)-f(x)}{h}$, but here I want to find approximation for $f'(x_2)$ as linear combination of function values in these points as $f'(x_2) approx sum_{j=0}^4 alpha_j f(x_j)$. There are no boundary conditions so I can't make a system of linear equations, at least I see no straight forward way.



Are there any standard methods of how to determine alphas related to the finite elements method, so that the formula will be as precise as possible for polynomials of as high order as possible?



I see that this is closely related to the five-point midpoint formula
https://www3.nd.edu/~zxu2/acms40390F15/Lec-4.1.pdf if one sets x_0= x_2 -2h and so on, but how to derive this formula?



How does that generalise to higher order derivatives?






finite-element-method numerical-calculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    What about $f'(x_2) approx frac{f(x_3)-f(x_1)}{2h}$
    $endgroup$
    – Jakobian
    Jan 13 at 12:54










  • $begingroup$
    This only uses two points.
    $endgroup$
    – random guy
    Jan 13 at 13:00










  • $begingroup$
    And it makes sense, considering how derivative is a local property
    $endgroup$
    – Jakobian
    Jan 13 at 13:49










  • $begingroup$
    But still, how can this be derived?
    $endgroup$
    – random guy
    Jan 13 at 14:23














1












1








1





$begingroup$


Given a function $f: [x_0,x_4] → Bbb R$ and equidistant points $x_0, x_1, x_2, x_3, x_4$ so that $h=x_{i+1} - x_i > 0$.



Normally I would do, $f'(x) approx dfrac{f(x+h)-f(x)}{h}$, but here I want to find approximation for $f'(x_2)$ as linear combination of function values in these points as $f'(x_2) approx sum_{j=0}^4 alpha_j f(x_j)$. There are no boundary conditions so I can't make a system of linear equations, at least I see no straight forward way.



Are there any standard methods of how to determine alphas related to the finite elements method, so that the formula will be as precise as possible for polynomials of as high order as possible?



I see that this is closely related to the five-point midpoint formula
https://www3.nd.edu/~zxu2/acms40390F15/Lec-4.1.pdf if one sets x_0= x_2 -2h and so on, but how to derive this formula?



How does that generalise to higher order derivatives?






finite-element-method numerical-calculus







share|cite|improve this question











$endgroup$




Given a function $f: [x_0,x_4] → Bbb R$ and equidistant points $x_0, x_1, x_2, x_3, x_4$ so that $h=x_{i+1} - x_i > 0$.



Normally I would do, $f'(x) approx dfrac{f(x+h)-f(x)}{h}$, but here I want to find approximation for $f'(x_2)$ as linear combination of function values in these points as $f'(x_2) approx sum_{j=0}^4 alpha_j f(x_j)$. There are no boundary conditions so I can't make a system of linear equations, at least I see no straight forward way.



Are there any standard methods of how to determine alphas related to the finite elements method, so that the formula will be as precise as possible for polynomials of as high order as possible?



I see that this is closely related to the five-point midpoint formula
https://www3.nd.edu/~zxu2/acms40390F15/Lec-4.1.pdf if one sets x_0= x_2 -2h and so on, but how to derive this formula?



How does that generalise to higher order derivatives?






finite-element-method numerical-calculus




finite-element-method numerical-calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 13 at 15:33







random guy

















asked Jan 13 at 12:35









random guyrandom guy

302110




302110












  • $begingroup$
    What about $f'(x_2) approx frac{f(x_3)-f(x_1)}{2h}$
    $endgroup$
    – Jakobian
    Jan 13 at 12:54










  • $begingroup$
    This only uses two points.
    $endgroup$
    – random guy
    Jan 13 at 13:00










  • $begingroup$
    And it makes sense, considering how derivative is a local property
    $endgroup$
    – Jakobian
    Jan 13 at 13:49










  • $begingroup$
    But still, how can this be derived?
    $endgroup$
    – random guy
    Jan 13 at 14:23


















  • $begingroup$
    What about $f'(x_2) approx frac{f(x_3)-f(x_1)}{2h}$
    $endgroup$
    – Jakobian
    Jan 13 at 12:54










  • $begingroup$
    This only uses two points.
    $endgroup$
    – random guy
    Jan 13 at 13:00










  • $begingroup$
    And it makes sense, considering how derivative is a local property
    $endgroup$
    – Jakobian
    Jan 13 at 13:49










  • $begingroup$
    But still, how can this be derived?
    $endgroup$
    – random guy
    Jan 13 at 14:23
















$begingroup$
What about $f'(x_2) approx frac{f(x_3)-f(x_1)}{2h}$
$endgroup$
– Jakobian
Jan 13 at 12:54




$begingroup$
What about $f'(x_2) approx frac{f(x_3)-f(x_1)}{2h}$
$endgroup$
– Jakobian
Jan 13 at 12:54












$begingroup$
This only uses two points.
$endgroup$
– random guy
Jan 13 at 13:00




$begingroup$
This only uses two points.
$endgroup$
– random guy
Jan 13 at 13:00












$begingroup$
And it makes sense, considering how derivative is a local property
$endgroup$
– Jakobian
Jan 13 at 13:49




$begingroup$
And it makes sense, considering how derivative is a local property
$endgroup$
– Jakobian
Jan 13 at 13:49












$begingroup$
But still, how can this be derived?
$endgroup$
– random guy
Jan 13 at 14:23




$begingroup$
But still, how can this be derived?
$endgroup$
– random guy
Jan 13 at 14:23










1 Answer
1






active

oldest

votes


















2












$begingroup$

Use Taylor series for small values of $h$
$$f(x)=f(x+(k-2) h)=f(x)+h (k-2) f'(x)+frac{1}{2} h^2 (k-2)^2 f''(x)+frac{1}{6} h^3 (k-2)^3
f^{(3)}(x)+frac{1}{24} h^4 (k-2)^4 f^{(4)}(x)+frac{1}{120} h^5 (k-2)^5
f^{(5)}(x)+Oleft(h^6right)$$



Now, write $$h f'(x)=a f(x-2h)+b f(x-h)+c f(x)+d f(x+h)+e f(x+2h)$$ and replace to get
$$h f'(x)=f(x) (a+b+c+d+e)+h f'(x) (-2 a-b+d+2 e)+frac{1}{2} h^2 f''(x) (4 a+b+d+4
e)+frac{1}{6} h^3 f^{(3)}(x) (-8 a-b+d+8 e)+frac{1}{24} h^4 f^{(4)}(x) (16
a+b+d+16 e)+frac{1}{120} h^5 f^{(5)}(x) (-32 a-b+d+32 e)+Oleft(h^6right)$$
Then, you have the equations
$$a+b+c+d+e=0 tag 1$$
$$-2 a-b+d+2 e=1 tag 2$$
$$4 a+b+d+4 e=0 tag 3$$
$$-8 a-b+d+8 e=0 tag 4$$
$$16 a+b+d+16 e=0 tag 5$$



from which $a=frac 1 {12}$, $b=-frac 2 {3}$, $c=0$, $d=frac 2 {3}$, $e=frac 1 {12}$.



So, using your notations
$$f'(x_2)=frac{f(x_0)-8f(x_1)+8f(x_3)-f(x_4) }{12 h}$$ and the "error" is $frac 1 {30} h^4 f^{(5)}(x_2)$



Let us try for $$f(x)=e^x tan (x) log (sin (x)+1)$$ at $x=2$ using $h=0.1$. This would give for the derivative $f'(2)=20.671695$ while the exact value would be $20.671717$






share|cite|improve this answer











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    1 Answer
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    1 Answer
    1






    active

    oldest

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    active

    oldest

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    active

    oldest

    votes









    2












    $begingroup$

    Use Taylor series for small values of $h$
    $$f(x)=f(x+(k-2) h)=f(x)+h (k-2) f'(x)+frac{1}{2} h^2 (k-2)^2 f''(x)+frac{1}{6} h^3 (k-2)^3
    f^{(3)}(x)+frac{1}{24} h^4 (k-2)^4 f^{(4)}(x)+frac{1}{120} h^5 (k-2)^5
    f^{(5)}(x)+Oleft(h^6right)$$



    Now, write $$h f'(x)=a f(x-2h)+b f(x-h)+c f(x)+d f(x+h)+e f(x+2h)$$ and replace to get
    $$h f'(x)=f(x) (a+b+c+d+e)+h f'(x) (-2 a-b+d+2 e)+frac{1}{2} h^2 f''(x) (4 a+b+d+4
    e)+frac{1}{6} h^3 f^{(3)}(x) (-8 a-b+d+8 e)+frac{1}{24} h^4 f^{(4)}(x) (16
    a+b+d+16 e)+frac{1}{120} h^5 f^{(5)}(x) (-32 a-b+d+32 e)+Oleft(h^6right)$$
    Then, you have the equations
    $$a+b+c+d+e=0 tag 1$$
    $$-2 a-b+d+2 e=1 tag 2$$
    $$4 a+b+d+4 e=0 tag 3$$
    $$-8 a-b+d+8 e=0 tag 4$$
    $$16 a+b+d+16 e=0 tag 5$$



    from which $a=frac 1 {12}$, $b=-frac 2 {3}$, $c=0$, $d=frac 2 {3}$, $e=frac 1 {12}$.



    So, using your notations
    $$f'(x_2)=frac{f(x_0)-8f(x_1)+8f(x_3)-f(x_4) }{12 h}$$ and the "error" is $frac 1 {30} h^4 f^{(5)}(x_2)$



    Let us try for $$f(x)=e^x tan (x) log (sin (x)+1)$$ at $x=2$ using $h=0.1$. This would give for the derivative $f'(2)=20.671695$ while the exact value would be $20.671717$






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$

      Use Taylor series for small values of $h$
      $$f(x)=f(x+(k-2) h)=f(x)+h (k-2) f'(x)+frac{1}{2} h^2 (k-2)^2 f''(x)+frac{1}{6} h^3 (k-2)^3
      f^{(3)}(x)+frac{1}{24} h^4 (k-2)^4 f^{(4)}(x)+frac{1}{120} h^5 (k-2)^5
      f^{(5)}(x)+Oleft(h^6right)$$



      Now, write $$h f'(x)=a f(x-2h)+b f(x-h)+c f(x)+d f(x+h)+e f(x+2h)$$ and replace to get
      $$h f'(x)=f(x) (a+b+c+d+e)+h f'(x) (-2 a-b+d+2 e)+frac{1}{2} h^2 f''(x) (4 a+b+d+4
      e)+frac{1}{6} h^3 f^{(3)}(x) (-8 a-b+d+8 e)+frac{1}{24} h^4 f^{(4)}(x) (16
      a+b+d+16 e)+frac{1}{120} h^5 f^{(5)}(x) (-32 a-b+d+32 e)+Oleft(h^6right)$$
      Then, you have the equations
      $$a+b+c+d+e=0 tag 1$$
      $$-2 a-b+d+2 e=1 tag 2$$
      $$4 a+b+d+4 e=0 tag 3$$
      $$-8 a-b+d+8 e=0 tag 4$$
      $$16 a+b+d+16 e=0 tag 5$$



      from which $a=frac 1 {12}$, $b=-frac 2 {3}$, $c=0$, $d=frac 2 {3}$, $e=frac 1 {12}$.



      So, using your notations
      $$f'(x_2)=frac{f(x_0)-8f(x_1)+8f(x_3)-f(x_4) }{12 h}$$ and the "error" is $frac 1 {30} h^4 f^{(5)}(x_2)$



      Let us try for $$f(x)=e^x tan (x) log (sin (x)+1)$$ at $x=2$ using $h=0.1$. This would give for the derivative $f'(2)=20.671695$ while the exact value would be $20.671717$






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        Use Taylor series for small values of $h$
        $$f(x)=f(x+(k-2) h)=f(x)+h (k-2) f'(x)+frac{1}{2} h^2 (k-2)^2 f''(x)+frac{1}{6} h^3 (k-2)^3
        f^{(3)}(x)+frac{1}{24} h^4 (k-2)^4 f^{(4)}(x)+frac{1}{120} h^5 (k-2)^5
        f^{(5)}(x)+Oleft(h^6right)$$



        Now, write $$h f'(x)=a f(x-2h)+b f(x-h)+c f(x)+d f(x+h)+e f(x+2h)$$ and replace to get
        $$h f'(x)=f(x) (a+b+c+d+e)+h f'(x) (-2 a-b+d+2 e)+frac{1}{2} h^2 f''(x) (4 a+b+d+4
        e)+frac{1}{6} h^3 f^{(3)}(x) (-8 a-b+d+8 e)+frac{1}{24} h^4 f^{(4)}(x) (16
        a+b+d+16 e)+frac{1}{120} h^5 f^{(5)}(x) (-32 a-b+d+32 e)+Oleft(h^6right)$$
        Then, you have the equations
        $$a+b+c+d+e=0 tag 1$$
        $$-2 a-b+d+2 e=1 tag 2$$
        $$4 a+b+d+4 e=0 tag 3$$
        $$-8 a-b+d+8 e=0 tag 4$$
        $$16 a+b+d+16 e=0 tag 5$$



        from which $a=frac 1 {12}$, $b=-frac 2 {3}$, $c=0$, $d=frac 2 {3}$, $e=frac 1 {12}$.



        So, using your notations
        $$f'(x_2)=frac{f(x_0)-8f(x_1)+8f(x_3)-f(x_4) }{12 h}$$ and the "error" is $frac 1 {30} h^4 f^{(5)}(x_2)$



        Let us try for $$f(x)=e^x tan (x) log (sin (x)+1)$$ at $x=2$ using $h=0.1$. This would give for the derivative $f'(2)=20.671695$ while the exact value would be $20.671717$






        share|cite|improve this answer











        $endgroup$



        Use Taylor series for small values of $h$
        $$f(x)=f(x+(k-2) h)=f(x)+h (k-2) f'(x)+frac{1}{2} h^2 (k-2)^2 f''(x)+frac{1}{6} h^3 (k-2)^3
        f^{(3)}(x)+frac{1}{24} h^4 (k-2)^4 f^{(4)}(x)+frac{1}{120} h^5 (k-2)^5
        f^{(5)}(x)+Oleft(h^6right)$$



        Now, write $$h f'(x)=a f(x-2h)+b f(x-h)+c f(x)+d f(x+h)+e f(x+2h)$$ and replace to get
        $$h f'(x)=f(x) (a+b+c+d+e)+h f'(x) (-2 a-b+d+2 e)+frac{1}{2} h^2 f''(x) (4 a+b+d+4
        e)+frac{1}{6} h^3 f^{(3)}(x) (-8 a-b+d+8 e)+frac{1}{24} h^4 f^{(4)}(x) (16
        a+b+d+16 e)+frac{1}{120} h^5 f^{(5)}(x) (-32 a-b+d+32 e)+Oleft(h^6right)$$
        Then, you have the equations
        $$a+b+c+d+e=0 tag 1$$
        $$-2 a-b+d+2 e=1 tag 2$$
        $$4 a+b+d+4 e=0 tag 3$$
        $$-8 a-b+d+8 e=0 tag 4$$
        $$16 a+b+d+16 e=0 tag 5$$



        from which $a=frac 1 {12}$, $b=-frac 2 {3}$, $c=0$, $d=frac 2 {3}$, $e=frac 1 {12}$.



        So, using your notations
        $$f'(x_2)=frac{f(x_0)-8f(x_1)+8f(x_3)-f(x_4) }{12 h}$$ and the "error" is $frac 1 {30} h^4 f^{(5)}(x_2)$



        Let us try for $$f(x)=e^x tan (x) log (sin (x)+1)$$ at $x=2$ using $h=0.1$. This would give for the derivative $f'(2)=20.671695$ while the exact value would be $20.671717$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 13 at 14:58

























        answered Jan 13 at 14:44









        Claude LeiboviciClaude Leibovici

        126k1158135




        126k1158135






























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