Mixing
Derive error term by using Taylor series expansions.
$begingroup$
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
asked Oct 14 '13 at 0:04
UnknownWUnknownW
985822
$endgroup$
|
show 4 more comments
$begingroup$
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
asked Oct 14 '13 at 0:04
UnknownWUnknownW
985822
$endgroup$
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
|
show 4 more comments
$begingroup$
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
asked Oct 14 '13 at 0:04
UnknownWUnknownW
985822
$endgroup$
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
asked Oct 14 '13 at 0:04
UnknownWUnknownW
985822
asked Oct 14 '13 at 0:04
UnknownWUnknownW
985822
edited Oct 14 '13 at 0:20
UnknownW
asked Oct 14 '13 at 0:04
UnknownWUnknownW
985822
asked Oct 14 '13 at 0:04
UnknownWUnknownW
985822
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
|
show 4 more comments
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
|
show 4 more comments
2 Answers
2
active
oldest
votes
$begingroup$
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 $h^{-2}(E_2(2h)-2E_2(h))$.
answered Oct 14 '13 at 17:26
Ian MateusIan Mateus
4,67032452
$endgroup$
add a comment |
$begingroup$
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.
answered Aug 28 '17 at 12:58
LutzLLutzL
57.6k42054
$endgroup$
add a comment |
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2 Answers
2
active
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 $h^{-2}(E_2(2h)-2E_2(h))$.
answered Oct 14 '13 at 17:26
Ian MateusIan Mateus
4,67032452
$endgroup$
add a comment |
$begingroup$
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 $h^{-2}(E_2(2h)-2E_2(h))$.
answered Oct 14 '13 at 17:26
Ian MateusIan Mateus
4,67032452
$endgroup$
add a comment |
$begingroup$
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 $h^{-2}(E_2(2h)-2E_2(h))$.
answered Oct 14 '13 at 17:26
Ian MateusIan Mateus
4,67032452
$endgroup$
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 $h^{-2}(E_2(2h)-2E_2(h))$.
answered Oct 14 '13 at 17:26
Ian MateusIan Mateus
4,67032452
answered Oct 14 '13 at 17:26
Ian MateusIan Mateus
4,67032452
answered Oct 14 '13 at 17:26
Ian MateusIan Mateus
4,67032452
answered Oct 14 '13 at 17:26
Ian MateusIan Mateus
4,67032452
4,67032452
add a comment |
add a comment |
$begingroup$
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.
answered Aug 28 '17 at 12:58
LutzLLutzL
57.6k42054
$endgroup$
add a comment |
$begingroup$
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.
answered Aug 28 '17 at 12:58
LutzLLutzL
57.6k42054
$endgroup$
add a comment |
$begingroup$
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.
answered Aug 28 '17 at 12:58
LutzLLutzL
57.6k42054
$endgroup$
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.
answered Aug 28 '17 at 12:58
LutzLLutzL
57.6k42054
edited Jan 8 at 18:57
answered Aug 28 '17 at 12:58
LutzLLutzL
57.6k42054
answered Aug 28 '17 at 12:58
LutzLLutzL
57.6k42054
answered Aug 28 '17 at 12:58
LutzLLutzL
57.6k42054
57.6k42054
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1
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Maybe it is a dumb question, but why don't you just stop at the $E_2$ terms?
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– Ian Mateus
Oct 14 '13 at 0:24
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$-(1/h^2) (E_3(2h) - E_3(h))$ isn't explicitly mentioned in this problem either.
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– Hurkyl
Oct 14 '13 at 0:45
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@IanMateus Thanks for your answer. It works!
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– UnknownW
Oct 14 '13 at 1:23
1
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@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 $2E_3(h)$)
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– Hurkyl
Oct 14 '13 at 2:16
1
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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".
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– Hurkyl
Oct 14 '13 at 2:22