Proof by using Taylor's Remainder Term: Truncation Error $=frac{(b-a)h^2}{12} max{|f''(z)|}$ for Trapezoidal...
$begingroup$
I am trying to derive Truncation Error $=frac{(b-a)h^2}{12} max{|f''(z)|}$ for Trapezoidal rule of integration by using Taylor's Remainder term
My Approach:
In trapezoidal rule of integration, we divide the interval of integration $[a,b]$ into $n$ equal length sub-intervals and then approximate the function as first degree polynomial to connect the intermediate points.
Let $f(x)$ be the given function, then the Taylor series of $f(x)$ is given by:$$f(x)=f(alpha)+ (x-alpha)f'(alpha) + frac{(x-alpha)^2}{2!}f''(alpha) + frac{(x-alpha)^3}{3!}f''(alpha) + dots$$
We know:$$text{Taylor's Remainder Term}: R_{n+1}(x;c)=frac{(x-c)^{n+1}}{(n+1)!}f^{n+1}(z)$$
where $z$ is a point between $c$ and $x$
So, here
$(n+1)=2$ , $x in [a,b]$, $c=(x-h)$and $zin(x-h,x)$ ,where $h= frac{b-a}{n}$
$$implies R_2(x,x-h)=frac{(x-(x-h))^{2}}{(2)!}f''(z)$$
$$=frac{(h)^{2}}{2}f''(z) dots (i)$$
so, the largest possible error $=max|R_2(x,x-h)|$
$$=frac{(h)^{2}}{2} max|f''(z)| dots(ii)$$
But Truncation Error $=frac{(b-a)h^2}{12} max{|f''(z)|}$,
so did i missed anything?? please help anyone...
numerical-methods taylor-expansion truncation-error
asked Dec 15 '18 at 5:12
SureshSuresh
307110
$endgroup$
add a comment |
$begingroup$
I am trying to derive Truncation Error $=frac{(b-a)h^2}{12} max{|f''(z)|}$ for Trapezoidal rule of integration by using Taylor's Remainder term
My Approach:
In trapezoidal rule of integration, we divide the interval of integration $[a,b]$ into $n$ equal length sub-intervals and then approximate the function as first degree polynomial to connect the intermediate points.
Let $f(x)$ be the given function, then the Taylor series of $f(x)$ is given by:$$f(x)=f(alpha)+ (x-alpha)f'(alpha) + frac{(x-alpha)^2}{2!}f''(alpha) + frac{(x-alpha)^3}{3!}f''(alpha) + dots$$
We know:$$text{Taylor's Remainder Term}: R_{n+1}(x;c)=frac{(x-c)^{n+1}}{(n+1)!}f^{n+1}(z)$$
where $z$ is a point between $c$ and $x$
So, here
$(n+1)=2$ , $x in [a,b]$, $c=(x-h)$and $zin(x-h,x)$ ,where $h= frac{b-a}{n}$
$$implies R_2(x,x-h)=frac{(x-(x-h))^{2}}{(2)!}f''(z)$$
$$=frac{(h)^{2}}{2}f''(z) dots (i)$$
so, the largest possible error $=max|R_2(x,x-h)|$
$$=frac{(h)^{2}}{2} max|f''(z)| dots(ii)$$
But Truncation Error $=frac{(b-a)h^2}{12} max{|f''(z)|}$,
so did i missed anything?? please help anyone...
numerical-methods taylor-expansion truncation-error
asked Dec 15 '18 at 5:12
SureshSuresh
307110
$endgroup$
1
$begingroup$
Where is your $int$? You are estimating the truncation error of an integral yet no integral appears here.
$endgroup$
– xbh
Dec 15 '18 at 5:32
$begingroup$
Using the Taylor's approximation is messy. An easier way is to apply integration by parts to estimate the local error $E_n = frac{h}{2}[f(x_n) + f(x_{n+1})] - int_{x_n}^{x_{n+1}} f(x) , dx$
$endgroup$
– RRL
Dec 15 '18 at 5:38
add a comment |
$begingroup$
I am trying to derive Truncation Error $=frac{(b-a)h^2}{12} max{|f''(z)|}$ for Trapezoidal rule of integration by using Taylor's Remainder term
My Approach:
In trapezoidal rule of integration, we divide the interval of integration $[a,b]$ into $n$ equal length sub-intervals and then approximate the function as first degree polynomial to connect the intermediate points.
Let $f(x)$ be the given function, then the Taylor series of $f(x)$ is given by:$$f(x)=f(alpha)+ (x-alpha)f'(alpha) + frac{(x-alpha)^2}{2!}f''(alpha) + frac{(x-alpha)^3}{3!}f''(alpha) + dots$$
We know:$$text{Taylor's Remainder Term}: R_{n+1}(x;c)=frac{(x-c)^{n+1}}{(n+1)!}f^{n+1}(z)$$
where $z$ is a point between $c$ and $x$
So, here
$(n+1)=2$ , $x in [a,b]$, $c=(x-h)$and $zin(x-h,x)$ ,where $h= frac{b-a}{n}$
$$implies R_2(x,x-h)=frac{(x-(x-h))^{2}}{(2)!}f''(z)$$
$$=frac{(h)^{2}}{2}f''(z) dots (i)$$
so, the largest possible error $=max|R_2(x,x-h)|$
$$=frac{(h)^{2}}{2} max|f''(z)| dots(ii)$$
But Truncation Error $=frac{(b-a)h^2}{12} max{|f''(z)|}$,
so did i missed anything?? please help anyone...
numerical-methods taylor-expansion truncation-error
asked Dec 15 '18 at 5:12
SureshSuresh
307110
$endgroup$
I am trying to derive Truncation Error $=frac{(b-a)h^2}{12} max{|f''(z)|}$ for Trapezoidal rule of integration by using Taylor's Remainder term
My Approach:
In trapezoidal rule of integration, we divide the interval of integration $[a,b]$ into $n$ equal length sub-intervals and then approximate the function as first degree polynomial to connect the intermediate points.
Let $f(x)$ be the given function, then the Taylor series of $f(x)$ is given by:$$f(x)=f(alpha)+ (x-alpha)f'(alpha) + frac{(x-alpha)^2}{2!}f''(alpha) + frac{(x-alpha)^3}{3!}f''(alpha) + dots$$
We know:$$text{Taylor's Remainder Term}: R_{n+1}(x;c)=frac{(x-c)^{n+1}}{(n+1)!}f^{n+1}(z)$$
where $z$ is a point between $c$ and $x$
So, here
$(n+1)=2$ , $x in [a,b]$, $c=(x-h)$and $zin(x-h,x)$ ,where $h= frac{b-a}{n}$
$$implies R_2(x,x-h)=frac{(x-(x-h))^{2}}{(2)!}f''(z)$$
$$=frac{(h)^{2}}{2}f''(z) dots (i)$$
so, the largest possible error $=max|R_2(x,x-h)|$
$$=frac{(h)^{2}}{2} max|f''(z)| dots(ii)$$
But Truncation Error $=frac{(b-a)h^2}{12} max{|f''(z)|}$,
so did i missed anything?? please help anyone...
numerical-methods taylor-expansion truncation-error
numerical-methods taylor-expansion truncation-error
asked Dec 15 '18 at 5:12
SureshSuresh
307110
asked Dec 15 '18 at 5:12
SureshSuresh
307110
asked Dec 15 '18 at 5:12
SureshSuresh
307110
asked Dec 15 '18 at 5:12
SureshSuresh
307110
asked Dec 15 '18 at 5:12
SureshSuresh
307110
307110
1
$begingroup$
Where is your $int$? You are estimating the truncation error of an integral yet no integral appears here.
$endgroup$
– xbh
Dec 15 '18 at 5:32
$begingroup$
Using the Taylor's approximation is messy. An easier way is to apply integration by parts to estimate the local error $E_n = frac{h}{2}[f(x_n) + f(x_{n+1})] - int_{x_n}^{x_{n+1}} f(x) , dx$
$endgroup$
– RRL
Dec 15 '18 at 5:38
add a comment |
1
$begingroup$
Where is your $int$? You are estimating the truncation error of an integral yet no integral appears here.
$endgroup$
– xbh
Dec 15 '18 at 5:32
$begingroup$
Using the Taylor's approximation is messy. An easier way is to apply integration by parts to estimate the local error $E_n = frac{h}{2}[f(x_n) + f(x_{n+1})] - int_{x_n}^{x_{n+1}} f(x) , dx$
$endgroup$
– RRL
Dec 15 '18 at 5:38
1
1
$begingroup$
Where is your $int$? You are estimating the truncation error of an integral yet no integral appears here.
$endgroup$
– xbh
Dec 15 '18 at 5:32
$begingroup$
Where is your $int$? You are estimating the truncation error of an integral yet no integral appears here.
$endgroup$
– xbh
Dec 15 '18 at 5:32
$begingroup$
Using the Taylor's approximation is messy. An easier way is to apply integration by parts to estimate the local error $E_n = frac{h}{2}[f(x_n) + f(x_{n+1})] - int_{x_n}^{x_{n+1}} f(x) , dx$
$endgroup$
– RRL
Dec 15 '18 at 5:38
$begingroup$
Using the Taylor's approximation is messy. An easier way is to apply integration by parts to estimate the local error $E_n = frac{h}{2}[f(x_n) + f(x_{n+1})] - int_{x_n}^{x_{n+1}} f(x) , dx$
$endgroup$
– RRL
Dec 15 '18 at 5:38
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You need to first obtain a local approximation over a subinterval.
The error in approximating the integral on a subinterval $[x_n, x_{n+1}]$ with $h = x_{n+1} - x_n$ using the trapezoidal formula is
$$E_n = frac{h}{2}[f(x_n) + f(x_{n+1})] - int_{x_n}^{x_{n+1}} f(x) , dx.$$
Integration by parts shows this to be
$$E_n = int_{x_n}^{x_{n+1}}(x-c)f'(x) , dx,$$
where $c = (x_{n+1}+x_n)/2$ is the midpoint.
To see this, note that
$$x_{n+1} - c = c - x_n = frac{x_{n+1} - x_n}{2} = frac{h}{2},$$
and with $u = (x-c)$ and $dv = f'(x)dx$, integration by parts yields
$$int_{x_n}^{x_{n+1}}(x-c)f'(x) , dx = (x-c)f(x)|_{x_n}^{x_{n+1}} - int_{x_n}^{x_{n+1}} f(x) , dx = E_n.$$
If the second derivative is bounded with $max |f''(x)| = M$, then we can demonstrate $O(h^3)$ local accuracy. Another integration by parts yields
$$E_n = frac1{2} int_{x_n}^{x_{n+1}} [(h/2)^2 - (x-c)^2]f''(x) , dx.$$
Using the bound for $f''$ and integrating we obtain the local error bound
$$|E_n| leqslant frac{M}{12}h^3$$
Summing over $m$ subintervals where $mh = b-a$ we get the global error bound
$$ |GE| leqslant frac{Mmh^3}{12}= frac{(b-a)h^2}{12}M$$
answered Dec 15 '18 at 5:34
RRLRRL
51.6k42573
$endgroup$
1
$begingroup$
@Suresh: You're welcome. You can start by looking at this and I'll see if I can find a better reference or derive it for you.
$endgroup$
– RRL
Dec 15 '18 at 6:32
1
$begingroup$
I do however think this is the "best" approach and is worth learning.
$endgroup$
– RRL
Dec 15 '18 at 6:33
1
$begingroup$
That is correct. Now you can write the first term as $-cf(x)|_{x_n}^{x_{n+1}} = -cint_{x_n}^{x_{n+1}} f'(x) , dx$ and you will get $E_n=int_{x_n}^{x_{n+1}} (x-c)f'(x) , dx$.
$endgroup$
– RRL
Dec 15 '18 at 13:14
1
$begingroup$
+1: Not a criticism of the above, but I would like to see an error analysis that makes the error 'obvious'. The integration by parts twice certainly yields the desired result, but personally I find the (textbook) derivation a bit opaque in terms of intuition.
$endgroup$
– copper.hat
Dec 15 '18 at 20:23
1
$begingroup$
@copper.hat: Thank you. I agree. I had a more transparent way in mind showing $mathcal{O}(h^3)$ local error but could get the coefficient $1/12$. Not sure if that is important to OP.
$endgroup$
– RRL
Dec 15 '18 at 23:31
|
show 5 more comments
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$begingroup$
You need to first obtain a local approximation over a subinterval.
The error in approximating the integral on a subinterval $[x_n, x_{n+1}]$ with $h = x_{n+1} - x_n$ using the trapezoidal formula is
$$E_n = frac{h}{2}[f(x_n) + f(x_{n+1})] - int_{x_n}^{x_{n+1}} f(x) , dx.$$
Integration by parts shows this to be
$$E_n = int_{x_n}^{x_{n+1}}(x-c)f'(x) , dx,$$
where $c = (x_{n+1}+x_n)/2$ is the midpoint.
To see this, note that
$$x_{n+1} - c = c - x_n = frac{x_{n+1} - x_n}{2} = frac{h}{2},$$
and with $u = (x-c)$ and $dv = f'(x)dx$, integration by parts yields
$$int_{x_n}^{x_{n+1}}(x-c)f'(x) , dx = (x-c)f(x)|_{x_n}^{x_{n+1}} - int_{x_n}^{x_{n+1}} f(x) , dx = E_n.$$
If the second derivative is bounded with $max |f''(x)| = M$, then we can demonstrate $O(h^3)$ local accuracy. Another integration by parts yields
$$E_n = frac1{2} int_{x_n}^{x_{n+1}} [(h/2)^2 - (x-c)^2]f''(x) , dx.$$
Using the bound for $f''$ and integrating we obtain the local error bound
$$|E_n| leqslant frac{M}{12}h^3$$
Summing over $m$ subintervals where $mh = b-a$ we get the global error bound
$$ |GE| leqslant frac{Mmh^3}{12}= frac{(b-a)h^2}{12}M$$
answered Dec 15 '18 at 5:34
RRLRRL
51.6k42573
$endgroup$
1
$begingroup$
@Suresh: You're welcome. You can start by looking at this and I'll see if I can find a better reference or derive it for you.
$endgroup$
– RRL
Dec 15 '18 at 6:32
1
$begingroup$
I do however think this is the "best" approach and is worth learning.
$endgroup$
– RRL
Dec 15 '18 at 6:33
1
$begingroup$
That is correct. Now you can write the first term as $-cf(x)|_{x_n}^{x_{n+1}} = -cint_{x_n}^{x_{n+1}} f'(x) , dx$ and you will get $E_n=int_{x_n}^{x_{n+1}} (x-c)f'(x) , dx$.
$endgroup$
– RRL
Dec 15 '18 at 13:14
1
$begingroup$
+1: Not a criticism of the above, but I would like to see an error analysis that makes the error 'obvious'. The integration by parts twice certainly yields the desired result, but personally I find the (textbook) derivation a bit opaque in terms of intuition.
$endgroup$
– copper.hat
Dec 15 '18 at 20:23
1
$begingroup$
@copper.hat: Thank you. I agree. I had a more transparent way in mind showing $mathcal{O}(h^3)$ local error but could get the coefficient $1/12$. Not sure if that is important to OP.
$endgroup$
– RRL
Dec 15 '18 at 23:31
|
show 5 more comments
$begingroup$
You need to first obtain a local approximation over a subinterval.
The error in approximating the integral on a subinterval $[x_n, x_{n+1}]$ with $h = x_{n+1} - x_n$ using the trapezoidal formula is
$$E_n = frac{h}{2}[f(x_n) + f(x_{n+1})] - int_{x_n}^{x_{n+1}} f(x) , dx.$$
Integration by parts shows this to be
$$E_n = int_{x_n}^{x_{n+1}}(x-c)f'(x) , dx,$$
where $c = (x_{n+1}+x_n)/2$ is the midpoint.
To see this, note that
$$x_{n+1} - c = c - x_n = frac{x_{n+1} - x_n}{2} = frac{h}{2},$$
and with $u = (x-c)$ and $dv = f'(x)dx$, integration by parts yields
$$int_{x_n}^{x_{n+1}}(x-c)f'(x) , dx = (x-c)f(x)|_{x_n}^{x_{n+1}} - int_{x_n}^{x_{n+1}} f(x) , dx = E_n.$$
If the second derivative is bounded with $max |f''(x)| = M$, then we can demonstrate $O(h^3)$ local accuracy. Another integration by parts yields
$$E_n = frac1{2} int_{x_n}^{x_{n+1}} [(h/2)^2 - (x-c)^2]f''(x) , dx.$$
Using the bound for $f''$ and integrating we obtain the local error bound
$$|E_n| leqslant frac{M}{12}h^3$$
Summing over $m$ subintervals where $mh = b-a$ we get the global error bound
$$ |GE| leqslant frac{Mmh^3}{12}= frac{(b-a)h^2}{12}M$$
answered Dec 15 '18 at 5:34
RRLRRL
51.6k42573
$endgroup$
1
$begingroup$
@Suresh: You're welcome. You can start by looking at this and I'll see if I can find a better reference or derive it for you.
$endgroup$
– RRL
Dec 15 '18 at 6:32
1
$begingroup$
I do however think this is the "best" approach and is worth learning.
$endgroup$
– RRL
Dec 15 '18 at 6:33
1
$begingroup$
That is correct. Now you can write the first term as $-cf(x)|_{x_n}^{x_{n+1}} = -cint_{x_n}^{x_{n+1}} f'(x) , dx$ and you will get $E_n=int_{x_n}^{x_{n+1}} (x-c)f'(x) , dx$.
$endgroup$
– RRL
Dec 15 '18 at 13:14
1
$begingroup$
+1: Not a criticism of the above, but I would like to see an error analysis that makes the error 'obvious'. The integration by parts twice certainly yields the desired result, but personally I find the (textbook) derivation a bit opaque in terms of intuition.
$endgroup$
– copper.hat
Dec 15 '18 at 20:23
1
$begingroup$
@copper.hat: Thank you. I agree. I had a more transparent way in mind showing $mathcal{O}(h^3)$ local error but could get the coefficient $1/12$. Not sure if that is important to OP.
$endgroup$
– RRL
Dec 15 '18 at 23:31
|
show 5 more comments
$begingroup$
You need to first obtain a local approximation over a subinterval.
The error in approximating the integral on a subinterval $[x_n, x_{n+1}]$ with $h = x_{n+1} - x_n$ using the trapezoidal formula is
$$E_n = frac{h}{2}[f(x_n) + f(x_{n+1})] - int_{x_n}^{x_{n+1}} f(x) , dx.$$
Integration by parts shows this to be
$$E_n = int_{x_n}^{x_{n+1}}(x-c)f'(x) , dx,$$
where $c = (x_{n+1}+x_n)/2$ is the midpoint.
To see this, note that
$$x_{n+1} - c = c - x_n = frac{x_{n+1} - x_n}{2} = frac{h}{2},$$
and with $u = (x-c)$ and $dv = f'(x)dx$, integration by parts yields
$$int_{x_n}^{x_{n+1}}(x-c)f'(x) , dx = (x-c)f(x)|_{x_n}^{x_{n+1}} - int_{x_n}^{x_{n+1}} f(x) , dx = E_n.$$
If the second derivative is bounded with $max |f''(x)| = M$, then we can demonstrate $O(h^3)$ local accuracy. Another integration by parts yields
$$E_n = frac1{2} int_{x_n}^{x_{n+1}} [(h/2)^2 - (x-c)^2]f''(x) , dx.$$
Using the bound for $f''$ and integrating we obtain the local error bound
$$|E_n| leqslant frac{M}{12}h^3$$
Summing over $m$ subintervals where $mh = b-a$ we get the global error bound
$$ |GE| leqslant frac{Mmh^3}{12}= frac{(b-a)h^2}{12}M$$
answered Dec 15 '18 at 5:34
RRLRRL
51.6k42573
$endgroup$
You need to first obtain a local approximation over a subinterval.
The error in approximating the integral on a subinterval $[x_n, x_{n+1}]$ with $h = x_{n+1} - x_n$ using the trapezoidal formula is
$$E_n = frac{h}{2}[f(x_n) + f(x_{n+1})] - int_{x_n}^{x_{n+1}} f(x) , dx.$$
Integration by parts shows this to be
$$E_n = int_{x_n}^{x_{n+1}}(x-c)f'(x) , dx,$$
where $c = (x_{n+1}+x_n)/2$ is the midpoint.
To see this, note that
$$x_{n+1} - c = c - x_n = frac{x_{n+1} - x_n}{2} = frac{h}{2},$$
and with $u = (x-c)$ and $dv = f'(x)dx$, integration by parts yields
$$int_{x_n}^{x_{n+1}}(x-c)f'(x) , dx = (x-c)f(x)|_{x_n}^{x_{n+1}} - int_{x_n}^{x_{n+1}} f(x) , dx = E_n.$$
If the second derivative is bounded with $max |f''(x)| = M$, then we can demonstrate $O(h^3)$ local accuracy. Another integration by parts yields
$$E_n = frac1{2} int_{x_n}^{x_{n+1}} [(h/2)^2 - (x-c)^2]f''(x) , dx.$$
Using the bound for $f''$ and integrating we obtain the local error bound
$$|E_n| leqslant frac{M}{12}h^3$$
Summing over $m$ subintervals where $mh = b-a$ we get the global error bound
$$ |GE| leqslant frac{Mmh^3}{12}= frac{(b-a)h^2}{12}M$$
answered Dec 15 '18 at 5:34
RRLRRL
51.6k42573
edited Dec 15 '18 at 19:51
answered Dec 15 '18 at 5:34
RRLRRL
51.6k42573
answered Dec 15 '18 at 5:34
RRLRRL
51.6k42573
answered Dec 15 '18 at 5:34
RRLRRL
51.6k42573
51.6k42573
1
$begingroup$
@Suresh: You're welcome. You can start by looking at this and I'll see if I can find a better reference or derive it for you.
$endgroup$
– RRL
Dec 15 '18 at 6:32
1
$begingroup$
I do however think this is the "best" approach and is worth learning.
$endgroup$
– RRL
Dec 15 '18 at 6:33
1
$begingroup$
That is correct. Now you can write the first term as $-cf(x)|_{x_n}^{x_{n+1}} = -cint_{x_n}^{x_{n+1}} f'(x) , dx$ and you will get $E_n=int_{x_n}^{x_{n+1}} (x-c)f'(x) , dx$.
$endgroup$
– RRL
Dec 15 '18 at 13:14
1
$begingroup$
+1: Not a criticism of the above, but I would like to see an error analysis that makes the error 'obvious'. The integration by parts twice certainly yields the desired result, but personally I find the (textbook) derivation a bit opaque in terms of intuition.
$endgroup$
– copper.hat
Dec 15 '18 at 20:23
1
$begingroup$
@copper.hat: Thank you. I agree. I had a more transparent way in mind showing $mathcal{O}(h^3)$ local error but could get the coefficient $1/12$. Not sure if that is important to OP.
$endgroup$
– RRL
Dec 15 '18 at 23:31
|
show 5 more comments
1
$begingroup$
@Suresh: You're welcome. You can start by looking at this and I'll see if I can find a better reference or derive it for you.
$endgroup$
– RRL
Dec 15 '18 at 6:32
1
$begingroup$
I do however think this is the "best" approach and is worth learning.
$endgroup$
– RRL
Dec 15 '18 at 6:33
1
$begingroup$
That is correct. Now you can write the first term as $-cf(x)|_{x_n}^{x_{n+1}} = -cint_{x_n}^{x_{n+1}} f'(x) , dx$ and you will get $E_n=int_{x_n}^{x_{n+1}} (x-c)f'(x) , dx$.
$endgroup$
– RRL
Dec 15 '18 at 13:14
1
$begingroup$
+1: Not a criticism of the above, but I would like to see an error analysis that makes the error 'obvious'. The integration by parts twice certainly yields the desired result, but personally I find the (textbook) derivation a bit opaque in terms of intuition.
$endgroup$
– copper.hat
Dec 15 '18 at 20:23
1
$begingroup$
@copper.hat: Thank you. I agree. I had a more transparent way in mind showing $mathcal{O}(h^3)$ local error but could get the coefficient $1/12$. Not sure if that is important to OP.
$endgroup$
– RRL
Dec 15 '18 at 23:31
1
1
$begingroup$
@Suresh: You're welcome. You can start by looking at this and I'll see if I can find a better reference or derive it for you.
$endgroup$
– RRL
Dec 15 '18 at 6:32
$begingroup$
@Suresh: You're welcome. You can start by looking at this and I'll see if I can find a better reference or derive it for you.
$endgroup$
– RRL
Dec 15 '18 at 6:32
1
1
$begingroup$
I do however think this is the "best" approach and is worth learning.
$endgroup$
– RRL
Dec 15 '18 at 6:33
$begingroup$
I do however think this is the "best" approach and is worth learning.
$endgroup$
– RRL
Dec 15 '18 at 6:33
1
1
$begingroup$
That is correct. Now you can write the first term as $-cf(x)|_{x_n}^{x_{n+1}} = -cint_{x_n}^{x_{n+1}} f'(x) , dx$ and you will get $E_n=int_{x_n}^{x_{n+1}} (x-c)f'(x) , dx$.
$endgroup$
– RRL
Dec 15 '18 at 13:14
$begingroup$
That is correct. Now you can write the first term as $-cf(x)|_{x_n}^{x_{n+1}} = -cint_{x_n}^{x_{n+1}} f'(x) , dx$ and you will get $E_n=int_{x_n}^{x_{n+1}} (x-c)f'(x) , dx$.
$endgroup$
– RRL
Dec 15 '18 at 13:14
1
1
$begingroup$
+1: Not a criticism of the above, but I would like to see an error analysis that makes the error 'obvious'. The integration by parts twice certainly yields the desired result, but personally I find the (textbook) derivation a bit opaque in terms of intuition.
$endgroup$
– copper.hat
Dec 15 '18 at 20:23
$begingroup$
+1: Not a criticism of the above, but I would like to see an error analysis that makes the error 'obvious'. The integration by parts twice certainly yields the desired result, but personally I find the (textbook) derivation a bit opaque in terms of intuition.
$endgroup$
– copper.hat
Dec 15 '18 at 20:23
1
1
$begingroup$
@copper.hat: Thank you. I agree. I had a more transparent way in mind showing $mathcal{O}(h^3)$ local error but could get the coefficient $1/12$. Not sure if that is important to OP.
$endgroup$
– RRL
Dec 15 '18 at 23:31
$begingroup$
@copper.hat: Thank you. I agree. I had a more transparent way in mind showing $mathcal{O}(h^3)$ local error but could get the coefficient $1/12$. Not sure if that is important to OP.
$endgroup$
– RRL
Dec 15 '18 at 23:31
|
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Where is your $int$? You are estimating the truncation error of an integral yet no integral appears here.
$endgroup$
– xbh
Dec 15 '18 at 5:32
$begingroup$
Using the Taylor's approximation is messy. An easier way is to apply integration by parts to estimate the local error $E_n = frac{h}{2}[f(x_n) + f(x_{n+1})] - int_{x_n}^{x_{n+1}} f(x) , dx$
$endgroup$
– RRL
Dec 15 '18 at 5:38