Why is the error of $O(h^2)$ when using Taylor expansion and centered approximation for the first derivative












0












$begingroup$


We know that the approximation of the first derivative by centered approximation is given by
$$ f'(x) =
frac{f(t+h) - f(x-h)}{2h} + O(h^2)$$
The quality of the above approximation is determined by the error. For smaller values of $h>0$ the error should be as small as possible. To determine this error we use the Taylor expansion



begin{align*}
y(t+h) &= y(t) + y'(t)h + frac{y''(t)}{2}h^2+frac{y'''(t)}{3!}h^3+ O(h^4)\
y(t-h) &= y(t) - y'(t)h + frac{y''(t)}{2}h^2-frac{y'''(t)}{3!}h^3+ O(h^4)\
y(t+h) - y(t-h) &= 2y'(t)h + 2frac{y'''(t)}{3!}h^3 + O(h^4)\
frac{y(t+h)-y(t-h)}{2h}-y'(t) &= frac{y'''(t)}{3!}h^2 + O(h^3) = O(h^2)
end{align*}



an thus we conclude
$$
y'(t) = frac{y(t+h)-y(t-h)}{2h}+O(h^2)
$$



Here is my question



We could have arbitrarily decided to take the Taylor expansion with the polynomial of fourth and fifth order namely writing the first line as follows:
$$
y(t+h) = y(t) + y'(t)h + frac{y''(t)}{2}h^2+frac{y'''(t)}{3!}h^3+frac{y''''(t)}{4!}h^4+frac{y'''''(t)}{5!}h^5+ O(h^6)
$$
Thus the last line would have been:
$$
frac{y(t+h)-y(t-h)}{2h}-y'(t) = frac{y'''(t)}{3!}h^2 + frac{y'''''(t)}{5!}h^4 + O(h^5) = O(h^4)
$$
and conclude that the error is of order $= O(h^4)$



Why is this not true?






derivatives taylor-expansion finite-differences rounding-error







share|cite|improve this question











$endgroup$












  • $begingroup$
    You are dropping several $h^n$ factors. This explains your mistake. And the developments involves $o$, not $O$.
    $endgroup$
    – Yves Daoust
    Jan 8 at 13:34


















0












$begingroup$


We know that the approximation of the first derivative by centered approximation is given by
$$ f'(x) =
frac{f(t+h) - f(x-h)}{2h} + O(h^2)$$
The quality of the above approximation is determined by the error. For smaller values of $h>0$ the error should be as small as possible. To determine this error we use the Taylor expansion



begin{align*}
y(t+h) &= y(t) + y'(t)h + frac{y''(t)}{2}h^2+frac{y'''(t)}{3!}h^3+ O(h^4)\
y(t-h) &= y(t) - y'(t)h + frac{y''(t)}{2}h^2-frac{y'''(t)}{3!}h^3+ O(h^4)\
y(t+h) - y(t-h) &= 2y'(t)h + 2frac{y'''(t)}{3!}h^3 + O(h^4)\
frac{y(t+h)-y(t-h)}{2h}-y'(t) &= frac{y'''(t)}{3!}h^2 + O(h^3) = O(h^2)
end{align*}



an thus we conclude
$$
y'(t) = frac{y(t+h)-y(t-h)}{2h}+O(h^2)
$$



Here is my question



We could have arbitrarily decided to take the Taylor expansion with the polynomial of fourth and fifth order namely writing the first line as follows:
$$
y(t+h) = y(t) + y'(t)h + frac{y''(t)}{2}h^2+frac{y'''(t)}{3!}h^3+frac{y''''(t)}{4!}h^4+frac{y'''''(t)}{5!}h^5+ O(h^6)
$$
Thus the last line would have been:
$$
frac{y(t+h)-y(t-h)}{2h}-y'(t) = frac{y'''(t)}{3!}h^2 + frac{y'''''(t)}{5!}h^4 + O(h^5) = O(h^4)
$$
and conclude that the error is of order $= O(h^4)$



Why is this not true?






derivatives taylor-expansion finite-differences rounding-error







share|cite|improve this question











$endgroup$












  • $begingroup$
    You are dropping several $h^n$ factors. This explains your mistake. And the developments involves $o$, not $O$.
    $endgroup$
    – Yves Daoust
    Jan 8 at 13:34
















0












0








0





$begingroup$


We know that the approximation of the first derivative by centered approximation is given by
$$ f'(x) =
frac{f(t+h) - f(x-h)}{2h} + O(h^2)$$
The quality of the above approximation is determined by the error. For smaller values of $h>0$ the error should be as small as possible. To determine this error we use the Taylor expansion



begin{align*}
y(t+h) &= y(t) + y'(t)h + frac{y''(t)}{2}h^2+frac{y'''(t)}{3!}h^3+ O(h^4)\
y(t-h) &= y(t) - y'(t)h + frac{y''(t)}{2}h^2-frac{y'''(t)}{3!}h^3+ O(h^4)\
y(t+h) - y(t-h) &= 2y'(t)h + 2frac{y'''(t)}{3!}h^3 + O(h^4)\
frac{y(t+h)-y(t-h)}{2h}-y'(t) &= frac{y'''(t)}{3!}h^2 + O(h^3) = O(h^2)
end{align*}



an thus we conclude
$$
y'(t) = frac{y(t+h)-y(t-h)}{2h}+O(h^2)
$$



Here is my question



We could have arbitrarily decided to take the Taylor expansion with the polynomial of fourth and fifth order namely writing the first line as follows:
$$
y(t+h) = y(t) + y'(t)h + frac{y''(t)}{2}h^2+frac{y'''(t)}{3!}h^3+frac{y''''(t)}{4!}h^4+frac{y'''''(t)}{5!}h^5+ O(h^6)
$$
Thus the last line would have been:
$$
frac{y(t+h)-y(t-h)}{2h}-y'(t) = frac{y'''(t)}{3!}h^2 + frac{y'''''(t)}{5!}h^4 + O(h^5) = O(h^4)
$$
and conclude that the error is of order $= O(h^4)$



Why is this not true?






derivatives taylor-expansion finite-differences rounding-error







share|cite|improve this question











$endgroup$




We know that the approximation of the first derivative by centered approximation is given by
$$ f'(x) =
frac{f(t+h) - f(x-h)}{2h} + O(h^2)$$
The quality of the above approximation is determined by the error. For smaller values of $h>0$ the error should be as small as possible. To determine this error we use the Taylor expansion



begin{align*}
y(t+h) &= y(t) + y'(t)h + frac{y''(t)}{2}h^2+frac{y'''(t)}{3!}h^3+ O(h^4)\
y(t-h) &= y(t) - y'(t)h + frac{y''(t)}{2}h^2-frac{y'''(t)}{3!}h^3+ O(h^4)\
y(t+h) - y(t-h) &= 2y'(t)h + 2frac{y'''(t)}{3!}h^3 + O(h^4)\
frac{y(t+h)-y(t-h)}{2h}-y'(t) &= frac{y'''(t)}{3!}h^2 + O(h^3) = O(h^2)
end{align*}



an thus we conclude
$$
y'(t) = frac{y(t+h)-y(t-h)}{2h}+O(h^2)
$$



Here is my question



We could have arbitrarily decided to take the Taylor expansion with the polynomial of fourth and fifth order namely writing the first line as follows:
$$
y(t+h) = y(t) + y'(t)h + frac{y''(t)}{2}h^2+frac{y'''(t)}{3!}h^3+frac{y''''(t)}{4!}h^4+frac{y'''''(t)}{5!}h^5+ O(h^6)
$$
Thus the last line would have been:
$$
frac{y(t+h)-y(t-h)}{2h}-y'(t) = frac{y'''(t)}{3!}h^2 + frac{y'''''(t)}{5!}h^4 + O(h^5) = O(h^4)
$$
and conclude that the error is of order $= O(h^4)$



Why is this not true?






derivatives taylor-expansion finite-differences rounding-error




derivatives taylor-expansion finite-differences rounding-error






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 9 at 6:25







ecjb

















asked Jan 8 at 13:29









ecjbecjb

1618




1618












  • $begingroup$
    You are dropping several $h^n$ factors. This explains your mistake. And the developments involves $o$, not $O$.
    $endgroup$
    – Yves Daoust
    Jan 8 at 13:34




















  • $begingroup$
    You are dropping several $h^n$ factors. This explains your mistake. And the developments involves $o$, not $O$.
    $endgroup$
    – Yves Daoust
    Jan 8 at 13:34


















$begingroup$
You are dropping several $h^n$ factors. This explains your mistake. And the developments involves $o$, not $O$.
$endgroup$
– Yves Daoust
Jan 8 at 13:34






$begingroup$
You are dropping several $h^n$ factors. This explains your mistake. And the developments involves $o$, not $O$.
$endgroup$
– Yves Daoust
Jan 8 at 13:34












2 Answers
2






active

oldest

votes


















0












$begingroup$

The correct last line is



$$
frac{y(t+h)-y(t-h)}{2h}-y'(t) = frac{y'''(t)}{3!}h^2 + frac{y'''''(t)}{5!}h^4 + o(h^4) = Theta(h^2)
$$ (unless $y'''(t)=0$).






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you for your answer @YvesDaoust. I indeed forge some $h^n$ and added them to my question. Could you elaborate a bit: why in your answer the order of order not equal to $Oh^4$? What would happen if $y'''(t) = 0?$
    $endgroup$
    – ecjb
    Jan 8 at 18:32












  • $begingroup$
    @ecjb: you confuse $O$ and $o$.
    $endgroup$
    – Yves Daoust
    Jan 8 at 21:11



















-1












$begingroup$

Because for smaller value of $h$ which is less than $1$, $h^2$ is greater than $h^5$. Hence, if the error of an algorithm is $O(h^2)$ is in $O(h^5)$ for small values of $h$ and not vice versa.






share|cite|improve this answer









$endgroup$













    Your Answer





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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    The correct last line is



    $$
    frac{y(t+h)-y(t-h)}{2h}-y'(t) = frac{y'''(t)}{3!}h^2 + frac{y'''''(t)}{5!}h^4 + o(h^4) = Theta(h^2)
    $$     (unless $y'''(t)=0$).






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thank you for your answer @YvesDaoust. I indeed forge some $h^n$ and added them to my question. Could you elaborate a bit: why in your answer the order of order not equal to $Oh^4$? What would happen if $y'''(t) = 0?$
      $endgroup$
      – ecjb
      Jan 8 at 18:32












    • $begingroup$
      @ecjb: you confuse $O$ and $o$.
      $endgroup$
      – Yves Daoust
      Jan 8 at 21:11
















    0












    $begingroup$

    The correct last line is



    $$
    frac{y(t+h)-y(t-h)}{2h}-y'(t) = frac{y'''(t)}{3!}h^2 + frac{y'''''(t)}{5!}h^4 + o(h^4) = Theta(h^2)
    $$     (unless $y'''(t)=0$).






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thank you for your answer @YvesDaoust. I indeed forge some $h^n$ and added them to my question. Could you elaborate a bit: why in your answer the order of order not equal to $Oh^4$? What would happen if $y'''(t) = 0?$
      $endgroup$
      – ecjb
      Jan 8 at 18:32












    • $begingroup$
      @ecjb: you confuse $O$ and $o$.
      $endgroup$
      – Yves Daoust
      Jan 8 at 21:11














    0












    0








    0





    $begingroup$

    The correct last line is



    $$
    frac{y(t+h)-y(t-h)}{2h}-y'(t) = frac{y'''(t)}{3!}h^2 + frac{y'''''(t)}{5!}h^4 + o(h^4) = Theta(h^2)
    $$     (unless $y'''(t)=0$).






    share|cite|improve this answer









    $endgroup$



    The correct last line is



    $$
    frac{y(t+h)-y(t-h)}{2h}-y'(t) = frac{y'''(t)}{3!}h^2 + frac{y'''''(t)}{5!}h^4 + o(h^4) = Theta(h^2)
    $$     (unless $y'''(t)=0$).







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 8 at 13:39









    Yves DaoustYves Daoust

    125k671222




    125k671222












    • $begingroup$
      Thank you for your answer @YvesDaoust. I indeed forge some $h^n$ and added them to my question. Could you elaborate a bit: why in your answer the order of order not equal to $Oh^4$? What would happen if $y'''(t) = 0?$
      $endgroup$
      – ecjb
      Jan 8 at 18:32












    • $begingroup$
      @ecjb: you confuse $O$ and $o$.
      $endgroup$
      – Yves Daoust
      Jan 8 at 21:11


















    • $begingroup$
      Thank you for your answer @YvesDaoust. I indeed forge some $h^n$ and added them to my question. Could you elaborate a bit: why in your answer the order of order not equal to $Oh^4$? What would happen if $y'''(t) = 0?$
      $endgroup$
      – ecjb
      Jan 8 at 18:32












    • $begingroup$
      @ecjb: you confuse $O$ and $o$.
      $endgroup$
      – Yves Daoust
      Jan 8 at 21:11
















    $begingroup$
    Thank you for your answer @YvesDaoust. I indeed forge some $h^n$ and added them to my question. Could you elaborate a bit: why in your answer the order of order not equal to $Oh^4$? What would happen if $y'''(t) = 0?$
    $endgroup$
    – ecjb
    Jan 8 at 18:32






    $begingroup$
    Thank you for your answer @YvesDaoust. I indeed forge some $h^n$ and added them to my question. Could you elaborate a bit: why in your answer the order of order not equal to $Oh^4$? What would happen if $y'''(t) = 0?$
    $endgroup$
    – ecjb
    Jan 8 at 18:32














    $begingroup$
    @ecjb: you confuse $O$ and $o$.
    $endgroup$
    – Yves Daoust
    Jan 8 at 21:11




    $begingroup$
    @ecjb: you confuse $O$ and $o$.
    $endgroup$
    – Yves Daoust
    Jan 8 at 21:11











    -1












    $begingroup$

    Because for smaller value of $h$ which is less than $1$, $h^2$ is greater than $h^5$. Hence, if the error of an algorithm is $O(h^2)$ is in $O(h^5)$ for small values of $h$ and not vice versa.






    share|cite|improve this answer









    $endgroup$


















      -1












      $begingroup$

      Because for smaller value of $h$ which is less than $1$, $h^2$ is greater than $h^5$. Hence, if the error of an algorithm is $O(h^2)$ is in $O(h^5)$ for small values of $h$ and not vice versa.






      share|cite|improve this answer









      $endgroup$
















        -1












        -1








        -1





        $begingroup$

        Because for smaller value of $h$ which is less than $1$, $h^2$ is greater than $h^5$. Hence, if the error of an algorithm is $O(h^2)$ is in $O(h^5)$ for small values of $h$ and not vice versa.






        share|cite|improve this answer









        $endgroup$



        Because for smaller value of $h$ which is less than $1$, $h^2$ is greater than $h^5$. Hence, if the error of an algorithm is $O(h^2)$ is in $O(h^5)$ for small values of $h$ and not vice versa.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 8 at 13:31









        OmGOmG

        2,322722




        2,322722






























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