Derive error term by using Taylor series expansions.












4












$begingroup$



Using Taylor series expansions, derive the error term for the formula
begin{equation}
f''(x)approx frac{1}{h^{2}}left [ f(x)-2f(x+h)+f(x+2h) right ].
end{equation}




I've tried it on my own way. We see that
begin{align*}
f(x+h)&=sum_{k=0}^{3}frac{h^{k}}{k!}f^{(k)}(x)+E_{n}(h)\
&=f(x)+hf'(x)+frac{h^{2}}{2}f''(x)+frac{h^{3}}{6}f'''(x)+E_{3}(h)
end{align*}
begin{align*}
f(x+2h)&=sum_{k=0}^{3}frac{(2h)^{k}}{k!}f^{(k)}(x)+E_{n}(2h)\
&=f(x)+2hf'(x)+2h^{2}f''(x)+frac{4h^{3}}{3}f'''(x)+E_{3}(2h)
end{align*}
and
begin{equation}
f(x+2h)-2f(x+h)=-f(x)+h^{2}f''(x)+h^{3}f'''(x)+E_{3}(2h)-E_{3}(h)
end{equation}
then by isolating $f''(x)$ we get
begin{equation}
f''(x)=frac{1}{h^{2}}left [ f(x+2h)-2f(x+h)+f(x) right ]-hf'''(x)-frac{1}{h^{2}}left [E_{3}(2h)-E_{3}(h) right ]
end{equation}
which isn't the right way to do since the term of "$-hf'''(x)$" is added (that's not mentioned in this problem). The rest is the error term. How do I answer it correctly?






numerical-methods







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Maybe it is a dumb question, but why don't you just stop at the $E_2$ terms?
    $endgroup$
    – Ian Mateus
    Oct 14 '13 at 0:24










  • $begingroup$
    $-(1/h^2) (E_3(2h) - E_3(h))$ isn't explicitly mentioned in this problem either.
    $endgroup$
    – Hurkyl
    Oct 14 '13 at 0:45












  • $begingroup$
    @IanMateus Thanks for your answer. It works!
    $endgroup$
    – UnknownW
    Oct 14 '13 at 1:23






  • 1




    $begingroup$
    @AjmalW: Right: but it's not the error term. The error term for the approximation you're proving $$-hf'''(x) - (1/h^2)(E_3(2h) - E_3(h))$$ (actually, I think your work is missing some factors of $2$, and that should probably be $2E_3(h)$)
    $endgroup$
    – Hurkyl
    Oct 14 '13 at 2:16








  • 1




    $begingroup$
    I suppose the confusion is over what the phrase "error term" means. In general, if we approximate $f$ with $g$, the error term to the approximation means $g - f$. $E_3(x,h)$, for example, is the error term when approximating $f(x)$ with its third order Taylor series. The error term for the approximation you're proving is not "the terms involving $E_3$", but instead "the difference between the two sides of the approximation".
    $endgroup$
    – Hurkyl
    Oct 14 '13 at 2:22


















4












$begingroup$



Using Taylor series expansions, derive the error term for the formula
begin{equation}
f''(x)approx frac{1}{h^{2}}left [ f(x)-2f(x+h)+f(x+2h) right ].
end{equation}




I've tried it on my own way. We see that
begin{align*}
f(x+h)&=sum_{k=0}^{3}frac{h^{k}}{k!}f^{(k)}(x)+E_{n}(h)\
&=f(x)+hf'(x)+frac{h^{2}}{2}f''(x)+frac{h^{3}}{6}f'''(x)+E_{3}(h)
end{align*}
begin{align*}
f(x+2h)&=sum_{k=0}^{3}frac{(2h)^{k}}{k!}f^{(k)}(x)+E_{n}(2h)\
&=f(x)+2hf'(x)+2h^{2}f''(x)+frac{4h^{3}}{3}f'''(x)+E_{3}(2h)
end{align*}
and
begin{equation}
f(x+2h)-2f(x+h)=-f(x)+h^{2}f''(x)+h^{3}f'''(x)+E_{3}(2h)-E_{3}(h)
end{equation}
then by isolating $f''(x)$ we get
begin{equation}
f''(x)=frac{1}{h^{2}}left [ f(x+2h)-2f(x+h)+f(x) right ]-hf'''(x)-frac{1}{h^{2}}left [E_{3}(2h)-E_{3}(h) right ]
end{equation}
which isn't the right way to do since the term of "$-hf'''(x)$" is added (that's not mentioned in this problem). The rest is the error term. How do I answer it correctly?






numerical-methods







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Maybe it is a dumb question, but why don't you just stop at the $E_2$ terms?
    $endgroup$
    – Ian Mateus
    Oct 14 '13 at 0:24










  • $begingroup$
    $-(1/h^2) (E_3(2h) - E_3(h))$ isn't explicitly mentioned in this problem either.
    $endgroup$
    – Hurkyl
    Oct 14 '13 at 0:45












  • $begingroup$
    @IanMateus Thanks for your answer. It works!
    $endgroup$
    – UnknownW
    Oct 14 '13 at 1:23






  • 1




    $begingroup$
    @AjmalW: Right: but it's not the error term. The error term for the approximation you're proving $$-hf'''(x) - (1/h^2)(E_3(2h) - E_3(h))$$ (actually, I think your work is missing some factors of $2$, and that should probably be $2E_3(h)$)
    $endgroup$
    – Hurkyl
    Oct 14 '13 at 2:16








  • 1




    $begingroup$
    I suppose the confusion is over what the phrase "error term" means. In general, if we approximate $f$ with $g$, the error term to the approximation means $g - f$. $E_3(x,h)$, for example, is the error term when approximating $f(x)$ with its third order Taylor series. The error term for the approximation you're proving is not "the terms involving $E_3$", but instead "the difference between the two sides of the approximation".
    $endgroup$
    – Hurkyl
    Oct 14 '13 at 2:22
















4












4








4


2



$begingroup$



Using Taylor series expansions, derive the error term for the formula
begin{equation}
f''(x)approx frac{1}{h^{2}}left [ f(x)-2f(x+h)+f(x+2h) right ].
end{equation}




I've tried it on my own way. We see that
begin{align*}
f(x+h)&=sum_{k=0}^{3}frac{h^{k}}{k!}f^{(k)}(x)+E_{n}(h)\
&=f(x)+hf'(x)+frac{h^{2}}{2}f''(x)+frac{h^{3}}{6}f'''(x)+E_{3}(h)
end{align*}
begin{align*}
f(x+2h)&=sum_{k=0}^{3}frac{(2h)^{k}}{k!}f^{(k)}(x)+E_{n}(2h)\
&=f(x)+2hf'(x)+2h^{2}f''(x)+frac{4h^{3}}{3}f'''(x)+E_{3}(2h)
end{align*}
and
begin{equation}
f(x+2h)-2f(x+h)=-f(x)+h^{2}f''(x)+h^{3}f'''(x)+E_{3}(2h)-E_{3}(h)
end{equation}
then by isolating $f''(x)$ we get
begin{equation}
f''(x)=frac{1}{h^{2}}left [ f(x+2h)-2f(x+h)+f(x) right ]-hf'''(x)-frac{1}{h^{2}}left [E_{3}(2h)-E_{3}(h) right ]
end{equation}
which isn't the right way to do since the term of "$-hf'''(x)$" is added (that's not mentioned in this problem). The rest is the error term. How do I answer it correctly?






numerical-methods







share|cite|improve this question











$endgroup$





Using Taylor series expansions, derive the error term for the formula
begin{equation}
f''(x)approx frac{1}{h^{2}}left [ f(x)-2f(x+h)+f(x+2h) right ].
end{equation}




I've tried it on my own way. We see that
begin{align*}
f(x+h)&=sum_{k=0}^{3}frac{h^{k}}{k!}f^{(k)}(x)+E_{n}(h)\
&=f(x)+hf'(x)+frac{h^{2}}{2}f''(x)+frac{h^{3}}{6}f'''(x)+E_{3}(h)
end{align*}
begin{align*}
f(x+2h)&=sum_{k=0}^{3}frac{(2h)^{k}}{k!}f^{(k)}(x)+E_{n}(2h)\
&=f(x)+2hf'(x)+2h^{2}f''(x)+frac{4h^{3}}{3}f'''(x)+E_{3}(2h)
end{align*}
and
begin{equation}
f(x+2h)-2f(x+h)=-f(x)+h^{2}f''(x)+h^{3}f'''(x)+E_{3}(2h)-E_{3}(h)
end{equation}
then by isolating $f''(x)$ we get
begin{equation}
f''(x)=frac{1}{h^{2}}left [ f(x+2h)-2f(x+h)+f(x) right ]-hf'''(x)-frac{1}{h^{2}}left [E_{3}(2h)-E_{3}(h) right ]
end{equation}
which isn't the right way to do since the term of "$-hf'''(x)$" is added (that's not mentioned in this problem). The rest is the error term. How do I answer it correctly?






numerical-methods




numerical-methods






share|cite|improve this question















share|cite|improve this question













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edited Oct 14 '13 at 0:20







UnknownW

















asked Oct 14 '13 at 0:04









UnknownWUnknownW

985822




985822








  • 1




    $begingroup$
    Maybe it is a dumb question, but why don't you just stop at the $E_2$ terms?
    $endgroup$
    – Ian Mateus
    Oct 14 '13 at 0:24










  • $begingroup$
    $-(1/h^2) (E_3(2h) - E_3(h))$ isn't explicitly mentioned in this problem either.
    $endgroup$
    – Hurkyl
    Oct 14 '13 at 0:45












  • $begingroup$
    @IanMateus Thanks for your answer. It works!
    $endgroup$
    – UnknownW
    Oct 14 '13 at 1:23






  • 1




    $begingroup$
    @AjmalW: Right: but it's not the error term. The error term for the approximation you're proving $$-hf'''(x) - (1/h^2)(E_3(2h) - E_3(h))$$ (actually, I think your work is missing some factors of $2$, and that should probably be $2E_3(h)$)
    $endgroup$
    – Hurkyl
    Oct 14 '13 at 2:16








  • 1




    $begingroup$
    I suppose the confusion is over what the phrase "error term" means. In general, if we approximate $f$ with $g$, the error term to the approximation means $g - f$. $E_3(x,h)$, for example, is the error term when approximating $f(x)$ with its third order Taylor series. The error term for the approximation you're proving is not "the terms involving $E_3$", but instead "the difference between the two sides of the approximation".
    $endgroup$
    – Hurkyl
    Oct 14 '13 at 2:22
















  • 1




    $begingroup$
    Maybe it is a dumb question, but why don't you just stop at the $E_2$ terms?
    $endgroup$
    – Ian Mateus
    Oct 14 '13 at 0:24










  • $begingroup$
    $-(1/h^2) (E_3(2h) - E_3(h))$ isn't explicitly mentioned in this problem either.
    $endgroup$
    – Hurkyl
    Oct 14 '13 at 0:45












  • $begingroup$
    @IanMateus Thanks for your answer. It works!
    $endgroup$
    – UnknownW
    Oct 14 '13 at 1:23






  • 1




    $begingroup$
    @AjmalW: Right: but it's not the error term. The error term for the approximation you're proving $$-hf'''(x) - (1/h^2)(E_3(2h) - E_3(h))$$ (actually, I think your work is missing some factors of $2$, and that should probably be $2E_3(h)$)
    $endgroup$
    – Hurkyl
    Oct 14 '13 at 2:16








  • 1




    $begingroup$
    I suppose the confusion is over what the phrase "error term" means. In general, if we approximate $f$ with $g$, the error term to the approximation means $g - f$. $E_3(x,h)$, for example, is the error term when approximating $f(x)$ with its third order Taylor series. The error term for the approximation you're proving is not "the terms involving $E_3$", but instead "the difference between the two sides of the approximation".
    $endgroup$
    – Hurkyl
    Oct 14 '13 at 2:22










1




1




$begingroup$
Maybe it is a dumb question, but why don't you just stop at the $E_2$ terms?
$endgroup$
– Ian Mateus
Oct 14 '13 at 0:24




$begingroup$
Maybe it is a dumb question, but why don't you just stop at the $E_2$ terms?
$endgroup$
– Ian Mateus
Oct 14 '13 at 0:24












$begingroup$
$-(1/h^2) (E_3(2h) - E_3(h))$ isn't explicitly mentioned in this problem either.
$endgroup$
– Hurkyl
Oct 14 '13 at 0:45






$begingroup$
$-(1/h^2) (E_3(2h) - E_3(h))$ isn't explicitly mentioned in this problem either.
$endgroup$
– Hurkyl
Oct 14 '13 at 0:45














$begingroup$
@IanMateus Thanks for your answer. It works!
$endgroup$
– UnknownW
Oct 14 '13 at 1:23




$begingroup$
@IanMateus Thanks for your answer. It works!
$endgroup$
– UnknownW
Oct 14 '13 at 1:23




1




1




$begingroup$
@AjmalW: Right: but it's not the error term. The error term for the approximation you're proving $$-hf'''(x) - (1/h^2)(E_3(2h) - E_3(h))$$ (actually, I think your work is missing some factors of $2$, and that should probably be $2E_3(h)$)
$endgroup$
– Hurkyl
Oct 14 '13 at 2:16






$begingroup$
@AjmalW: Right: but it's not the error term. The error term for the approximation you're proving $$-hf'''(x) - (1/h^2)(E_3(2h) - E_3(h))$$ (actually, I think your work is missing some factors of $2$, and that should probably be $2E_3(h)$)
$endgroup$
– Hurkyl
Oct 14 '13 at 2:16






1




1




$begingroup$
I suppose the confusion is over what the phrase "error term" means. In general, if we approximate $f$ with $g$, the error term to the approximation means $g - f$. $E_3(x,h)$, for example, is the error term when approximating $f(x)$ with its third order Taylor series. The error term for the approximation you're proving is not "the terms involving $E_3$", but instead "the difference between the two sides of the approximation".
$endgroup$
– Hurkyl
Oct 14 '13 at 2:22






$begingroup$
I suppose the confusion is over what the phrase "error term" means. In general, if we approximate $f$ with $g$, the error term to the approximation means $g - f$. $E_3(x,h)$, for example, is the error term when approximating $f(x)$ with its third order Taylor series. The error term for the approximation you're proving is not "the terms involving $E_3$", but instead "the difference between the two sides of the approximation".
$endgroup$
– Hurkyl
Oct 14 '13 at 2:22












2 Answers
2






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oldest

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$begingroup$

With the purpose of not leaving this question unanswered, this answer contains what OP already knows: stopping at $E_2$ is the cleanest way to go, since we already know how to express the error term of a Taylor series in convenient ways (Lagrange form, integral form). Some algebra quickly yields the error $h^{-2}(E_2(2h)-2E_2(h))$.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    As a central difference quotient for the position $x+h$, the Taylor expansion of the expression starts as
    $$
    frac1{h^{2}}[f(x)-2f(x+h)+f(x+2h)]=f''(x+h)+frac{h^2}{12}f^{(4)}(x+h)+O(h^4). tag{*}label{eq:1}
    $$



    By Taylor expansion of the second derivative $g(x)=f''(x)$ one gets that the deviation from $x$ to $x+h$ expands as
    begin{align}
    f''(x+h)=g(x+h)&=g(x)+g'(x)h+frac{h^2}2g''(x)+O(h^3)
    \
    &=f''(x)+hf'''(x)+frac{h^2}2f^{(4)}(x)+O(h^2). tag{**}label{eq:2}
    end{align}



    Inserting eqref{eq:2} into eqref{eq:1} gives in total
    $$
    frac1{h^{2}}[f(x)-2f(x+h)+f(x+2h)]=f''(x)+hf'''(x)+frac{7h^2}{12}f^{(4)}(x)+O(h^3),
    $$
    so that the error term from the question should be
    $$
    -hf'''(x)-frac{7h^2}{12}f^{(4)}(x)+O(h^3).
    $$
    Obviously the linear error term is the dominant one. So up to the missing $2$ in front of $E_3(h)=h^4R_3(h)$ as mentioned in the comments, the calculation is correct.






    share|cite|improve this answer











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      2 Answers
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      active

      oldest

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      2 Answers
      2






      active

      oldest

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      active

      oldest

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      active

      oldest

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      1












      $begingroup$

      With the purpose of not leaving this question unanswered, this answer contains what OP already knows: stopping at $E_2$ is the cleanest way to go, since we already know how to express the error term of a Taylor series in convenient ways (Lagrange form, integral form). Some algebra quickly yields the error $h^{-2}(E_2(2h)-2E_2(h))$.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        With the purpose of not leaving this question unanswered, this answer contains what OP already knows: stopping at $E_2$ is the cleanest way to go, since we already know how to express the error term of a Taylor series in convenient ways (Lagrange form, integral form). Some algebra quickly yields the error $h^{-2}(E_2(2h)-2E_2(h))$.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          With the purpose of not leaving this question unanswered, this answer contains what OP already knows: stopping at $E_2$ is the cleanest way to go, since we already know how to express the error term of a Taylor series in convenient ways (Lagrange form, integral form). Some algebra quickly yields the error $h^{-2}(E_2(2h)-2E_2(h))$.






          share|cite|improve this answer









          $endgroup$



          With the purpose of not leaving this question unanswered, this answer contains what OP already knows: stopping at $E_2$ is the cleanest way to go, since we already know how to express the error term of a Taylor series in convenient ways (Lagrange form, integral form). Some algebra quickly yields the error $h^{-2}(E_2(2h)-2E_2(h))$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Oct 14 '13 at 17:26









          Ian MateusIan Mateus

          4,67032452




          4,67032452























              0












              $begingroup$

              As a central difference quotient for the position $x+h$, the Taylor expansion of the expression starts as
              $$
              frac1{h^{2}}[f(x)-2f(x+h)+f(x+2h)]=f''(x+h)+frac{h^2}{12}f^{(4)}(x+h)+O(h^4). tag{*}label{eq:1}
              $$



              By Taylor expansion of the second derivative $g(x)=f''(x)$ one gets that the deviation from $x$ to $x+h$ expands as
              begin{align}
              f''(x+h)=g(x+h)&=g(x)+g'(x)h+frac{h^2}2g''(x)+O(h^3)
              \
              &=f''(x)+hf'''(x)+frac{h^2}2f^{(4)}(x)+O(h^2). tag{**}label{eq:2}
              end{align}



              Inserting eqref{eq:2} into eqref{eq:1} gives in total
              $$
              frac1{h^{2}}[f(x)-2f(x+h)+f(x+2h)]=f''(x)+hf'''(x)+frac{7h^2}{12}f^{(4)}(x)+O(h^3),
              $$
              so that the error term from the question should be
              $$
              -hf'''(x)-frac{7h^2}{12}f^{(4)}(x)+O(h^3).
              $$
              Obviously the linear error term is the dominant one. So up to the missing $2$ in front of $E_3(h)=h^4R_3(h)$ as mentioned in the comments, the calculation is correct.






              share|cite|improve this answer











              $endgroup$


















                0












                $begingroup$

                As a central difference quotient for the position $x+h$, the Taylor expansion of the expression starts as
                $$
                frac1{h^{2}}[f(x)-2f(x+h)+f(x+2h)]=f''(x+h)+frac{h^2}{12}f^{(4)}(x+h)+O(h^4). tag{*}label{eq:1}
                $$



                By Taylor expansion of the second derivative $g(x)=f''(x)$ one gets that the deviation from $x$ to $x+h$ expands as
                begin{align}
                f''(x+h)=g(x+h)&=g(x)+g'(x)h+frac{h^2}2g''(x)+O(h^3)
                \
                &=f''(x)+hf'''(x)+frac{h^2}2f^{(4)}(x)+O(h^2). tag{**}label{eq:2}
                end{align}



                Inserting eqref{eq:2} into eqref{eq:1} gives in total
                $$
                frac1{h^{2}}[f(x)-2f(x+h)+f(x+2h)]=f''(x)+hf'''(x)+frac{7h^2}{12}f^{(4)}(x)+O(h^3),
                $$
                so that the error term from the question should be
                $$
                -hf'''(x)-frac{7h^2}{12}f^{(4)}(x)+O(h^3).
                $$
                Obviously the linear error term is the dominant one. So up to the missing $2$ in front of $E_3(h)=h^4R_3(h)$ as mentioned in the comments, the calculation is correct.






                share|cite|improve this answer











                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  As a central difference quotient for the position $x+h$, the Taylor expansion of the expression starts as
                  $$
                  frac1{h^{2}}[f(x)-2f(x+h)+f(x+2h)]=f''(x+h)+frac{h^2}{12}f^{(4)}(x+h)+O(h^4). tag{*}label{eq:1}
                  $$



                  By Taylor expansion of the second derivative $g(x)=f''(x)$ one gets that the deviation from $x$ to $x+h$ expands as
                  begin{align}
                  f''(x+h)=g(x+h)&=g(x)+g'(x)h+frac{h^2}2g''(x)+O(h^3)
                  \
                  &=f''(x)+hf'''(x)+frac{h^2}2f^{(4)}(x)+O(h^2). tag{**}label{eq:2}
                  end{align}



                  Inserting eqref{eq:2} into eqref{eq:1} gives in total
                  $$
                  frac1{h^{2}}[f(x)-2f(x+h)+f(x+2h)]=f''(x)+hf'''(x)+frac{7h^2}{12}f^{(4)}(x)+O(h^3),
                  $$
                  so that the error term from the question should be
                  $$
                  -hf'''(x)-frac{7h^2}{12}f^{(4)}(x)+O(h^3).
                  $$
                  Obviously the linear error term is the dominant one. So up to the missing $2$ in front of $E_3(h)=h^4R_3(h)$ as mentioned in the comments, the calculation is correct.






                  share|cite|improve this answer











                  $endgroup$



                  As a central difference quotient for the position $x+h$, the Taylor expansion of the expression starts as
                  $$
                  frac1{h^{2}}[f(x)-2f(x+h)+f(x+2h)]=f''(x+h)+frac{h^2}{12}f^{(4)}(x+h)+O(h^4). tag{*}label{eq:1}
                  $$



                  By Taylor expansion of the second derivative $g(x)=f''(x)$ one gets that the deviation from $x$ to $x+h$ expands as
                  begin{align}
                  f''(x+h)=g(x+h)&=g(x)+g'(x)h+frac{h^2}2g''(x)+O(h^3)
                  \
                  &=f''(x)+hf'''(x)+frac{h^2}2f^{(4)}(x)+O(h^2). tag{**}label{eq:2}
                  end{align}



                  Inserting eqref{eq:2} into eqref{eq:1} gives in total
                  $$
                  frac1{h^{2}}[f(x)-2f(x+h)+f(x+2h)]=f''(x)+hf'''(x)+frac{7h^2}{12}f^{(4)}(x)+O(h^3),
                  $$
                  so that the error term from the question should be
                  $$
                  -hf'''(x)-frac{7h^2}{12}f^{(4)}(x)+O(h^3).
                  $$
                  Obviously the linear error term is the dominant one. So up to the missing $2$ in front of $E_3(h)=h^4R_3(h)$ as mentioned in the comments, the calculation is correct.







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                  edited Jan 8 at 18:57

























                  answered Aug 28 '17 at 12:58









                  LutzLLutzL

                  57.2k42054




                  57.2k42054






























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