Trapezoid rule error
$begingroup$
I am trying to compute the error in the trapezoid rule integration for a function $f(x)$
in the interval $[a,b]$.
I believe I have to Taylor-expand $f(x)$ around $x=a$
$f(a) + (x-a)f'(a)+ 1/2 (x-a)^2 f''(a) + ...$
and then do the integral of this, to later use it when comparing with the trapezoid rule:
$(b-a) (f(b) + f(a))/2$
However I am stuck on the integration of the Taylor expansion and on how to deduct from this the trapezoid approximation.
Can someone give me a hint on how to proceed? Thanks.
integration taylor-expansion
asked Feb 18 '16 at 21:46
EstefyEstefy
64
$endgroup$
add a comment |
$begingroup$
I am trying to compute the error in the trapezoid rule integration for a function $f(x)$
in the interval $[a,b]$.
I believe I have to Taylor-expand $f(x)$ around $x=a$
$f(a) + (x-a)f'(a)+ 1/2 (x-a)^2 f''(a) + ...$
and then do the integral of this, to later use it when comparing with the trapezoid rule:
$(b-a) (f(b) + f(a))/2$
However I am stuck on the integration of the Taylor expansion and on how to deduct from this the trapezoid approximation.
Can someone give me a hint on how to proceed? Thanks.
integration taylor-expansion
asked Feb 18 '16 at 21:46
EstefyEstefy
64
$endgroup$
add a comment |
$begingroup$
I am trying to compute the error in the trapezoid rule integration for a function $f(x)$
in the interval $[a,b]$.
I believe I have to Taylor-expand $f(x)$ around $x=a$
$f(a) + (x-a)f'(a)+ 1/2 (x-a)^2 f''(a) + ...$
and then do the integral of this, to later use it when comparing with the trapezoid rule:
$(b-a) (f(b) + f(a))/2$
However I am stuck on the integration of the Taylor expansion and on how to deduct from this the trapezoid approximation.
Can someone give me a hint on how to proceed? Thanks.
integration taylor-expansion
asked Feb 18 '16 at 21:46
EstefyEstefy
64
$endgroup$
I am trying to compute the error in the trapezoid rule integration for a function $f(x)$
in the interval $[a,b]$.
I believe I have to Taylor-expand $f(x)$ around $x=a$
$f(a) + (x-a)f'(a)+ 1/2 (x-a)^2 f''(a) + ...$
and then do the integral of this, to later use it when comparing with the trapezoid rule:
$(b-a) (f(b) + f(a))/2$
However I am stuck on the integration of the Taylor expansion and on how to deduct from this the trapezoid approximation.
Can someone give me a hint on how to proceed? Thanks.
integration taylor-expansion
integration taylor-expansion
asked Feb 18 '16 at 21:46
EstefyEstefy
64
asked Feb 18 '16 at 21:46
EstefyEstefy
64
edited Feb 18 '16 at 22:05
Estefy
asked Feb 18 '16 at 21:46
EstefyEstefy
64
asked Feb 18 '16 at 21:46
EstefyEstefy
64
asked Feb 18 '16 at 21:46
EstefyEstefy
64
64
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Estimating the error with the Taylor expansion is quite messy.
For an alternative, consider the local error on a subinterval $[x_n, x_{n+1}]$ with $h = x_{n+1} - x_n$.
The error in approximating the integral with 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 derivative is bounded as $|f'(x)| leqslant M$, we can bound the error as
$$|E_n| leqslant int_{x_n}^{x_{n+1}}|x-c||f'(x)| , dx \ leqslant Mint_{x_n}^{x_{n+1}}|x-c| , dx \ = frac{M}{4}(x_{n+1} - x_n)^2 \ = frac{M}{4}h^2.$$
Approximating the integral over $[a,b]$ with $m$ subintervals of length $h = (b-a)/m$, has a global error bound of
$$|GE| leqslant sum_{n=1}^{m} |E_n| = mfrac{M}{4}h^2 = frac{M(b-a)^2}{4m}$$
If the second derivative is bounded as $|f''(x)| leqslant M$, then we can demonstrate $O(h^3)$ local accuracy. An integration by parts of the previous integral for $E_n$ 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 and global error bounds
$$|E_n| leqslant frac{M}{12}h^3, \ |GE| leqslant frac{M(b-a)^3}{12m^2}.$$
answered Feb 18 '16 at 22:18
RRLRRL
51.5k42573
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1662083%2ftrapezoid-rule-error%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Estimating the error with the Taylor expansion is quite messy.
For an alternative, consider the local error on a subinterval $[x_n, x_{n+1}]$ with $h = x_{n+1} - x_n$.
The error in approximating the integral with 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 derivative is bounded as $|f'(x)| leqslant M$, we can bound the error as
$$|E_n| leqslant int_{x_n}^{x_{n+1}}|x-c||f'(x)| , dx \ leqslant Mint_{x_n}^{x_{n+1}}|x-c| , dx \ = frac{M}{4}(x_{n+1} - x_n)^2 \ = frac{M}{4}h^2.$$
Approximating the integral over $[a,b]$ with $m$ subintervals of length $h = (b-a)/m$, has a global error bound of
$$|GE| leqslant sum_{n=1}^{m} |E_n| = mfrac{M}{4}h^2 = frac{M(b-a)^2}{4m}$$
If the second derivative is bounded as $|f''(x)| leqslant M$, then we can demonstrate $O(h^3)$ local accuracy. An integration by parts of the previous integral for $E_n$ 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 and global error bounds
$$|E_n| leqslant frac{M}{12}h^3, \ |GE| leqslant frac{M(b-a)^3}{12m^2}.$$
answered Feb 18 '16 at 22:18
RRLRRL
51.5k42573
$endgroup$
add a comment |
$begingroup$
Estimating the error with the Taylor expansion is quite messy.
For an alternative, consider the local error on a subinterval $[x_n, x_{n+1}]$ with $h = x_{n+1} - x_n$.
The error in approximating the integral with 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 derivative is bounded as $|f'(x)| leqslant M$, we can bound the error as
$$|E_n| leqslant int_{x_n}^{x_{n+1}}|x-c||f'(x)| , dx \ leqslant Mint_{x_n}^{x_{n+1}}|x-c| , dx \ = frac{M}{4}(x_{n+1} - x_n)^2 \ = frac{M}{4}h^2.$$
Approximating the integral over $[a,b]$ with $m$ subintervals of length $h = (b-a)/m$, has a global error bound of
$$|GE| leqslant sum_{n=1}^{m} |E_n| = mfrac{M}{4}h^2 = frac{M(b-a)^2}{4m}$$
If the second derivative is bounded as $|f''(x)| leqslant M$, then we can demonstrate $O(h^3)$ local accuracy. An integration by parts of the previous integral for $E_n$ 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 and global error bounds
$$|E_n| leqslant frac{M}{12}h^3, \ |GE| leqslant frac{M(b-a)^3}{12m^2}.$$
answered Feb 18 '16 at 22:18
RRLRRL
51.5k42573
$endgroup$
add a comment |
$begingroup$
Estimating the error with the Taylor expansion is quite messy.
For an alternative, consider the local error on a subinterval $[x_n, x_{n+1}]$ with $h = x_{n+1} - x_n$.
The error in approximating the integral with 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 derivative is bounded as $|f'(x)| leqslant M$, we can bound the error as
$$|E_n| leqslant int_{x_n}^{x_{n+1}}|x-c||f'(x)| , dx \ leqslant Mint_{x_n}^{x_{n+1}}|x-c| , dx \ = frac{M}{4}(x_{n+1} - x_n)^2 \ = frac{M}{4}h^2.$$
Approximating the integral over $[a,b]$ with $m$ subintervals of length $h = (b-a)/m$, has a global error bound of
$$|GE| leqslant sum_{n=1}^{m} |E_n| = mfrac{M}{4}h^2 = frac{M(b-a)^2}{4m}$$
If the second derivative is bounded as $|f''(x)| leqslant M$, then we can demonstrate $O(h^3)$ local accuracy. An integration by parts of the previous integral for $E_n$ 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 and global error bounds
$$|E_n| leqslant frac{M}{12}h^3, \ |GE| leqslant frac{M(b-a)^3}{12m^2}.$$
answered Feb 18 '16 at 22:18
RRLRRL
51.5k42573
$endgroup$
Estimating the error with the Taylor expansion is quite messy.
For an alternative, consider the local error on a subinterval $[x_n, x_{n+1}]$ with $h = x_{n+1} - x_n$.
The error in approximating the integral with 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 derivative is bounded as $|f'(x)| leqslant M$, we can bound the error as
$$|E_n| leqslant int_{x_n}^{x_{n+1}}|x-c||f'(x)| , dx \ leqslant Mint_{x_n}^{x_{n+1}}|x-c| , dx \ = frac{M}{4}(x_{n+1} - x_n)^2 \ = frac{M}{4}h^2.$$
Approximating the integral over $[a,b]$ with $m$ subintervals of length $h = (b-a)/m$, has a global error bound of
$$|GE| leqslant sum_{n=1}^{m} |E_n| = mfrac{M}{4}h^2 = frac{M(b-a)^2}{4m}$$
If the second derivative is bounded as $|f''(x)| leqslant M$, then we can demonstrate $O(h^3)$ local accuracy. An integration by parts of the previous integral for $E_n$ 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 and global error bounds
$$|E_n| leqslant frac{M}{12}h^3, \ |GE| leqslant frac{M(b-a)^3}{12m^2}.$$
answered Feb 18 '16 at 22:18
RRLRRL
51.5k42573
edited Feb 19 '16 at 4:04
answered Feb 18 '16 at 22:18
RRLRRL
51.5k42573
answered Feb 18 '16 at 22:18
RRLRRL
51.5k42573
answered Feb 18 '16 at 22:18
RRLRRL
51.5k42573
51.5k42573
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1662083%2ftrapezoid-rule-error%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
