Approximate $int_t^{infty}e^{-delta(s-t)}omega(s)text{ds}$, with the restriction that just...
$begingroup$
Consider the following integral: $$int_t^{infty}e^{-delta(s-t)}omega(s)text{ds},$$ where $tgeq 0$ represents the current time, $delta>0$ is a fixed parameter representing the exponential decay rate, and $omega:mathbb{R}_{geq 0}rightarrow mathbb{R}$ is a given smooth bounded function, whose infinite derivatives are also bounded and with none derivative null.
I need to approximate the value of this integral with the restriction that I only have information about the actual $n$-first derivatives $omega(t)$, $dot{omega}(t)$,... ,$omega^{(n)}(t)$, with $ngeq 0$.
Recursive integration by parts leads to
$$int_t^{infty}e^{-delta(s-t)}omega(s)text{ds} = -frac{1}{delta}omega(t) + frac{1}{delta^2}dot{omega}(t)pm...+frac{(-1)^{n+1}}{delta^{n+1}}omega^{(n)}(t)+mathcal{O}_n,$$ where $$mathcal{O}_n=frac{1}{delta^{n+2}}int_t^{infty}e^{-delta(s-t)}omega^{(n+1)}(s)text{ds}.$$
What happens is the following:
Consider a fixed big $delta$ (say $deltagg1$). Then the residual $mathcal{O}_n$ is expected to be small as $1/delta^{n+2}$ is continuously decreasing with respect to $n$. This indicates that, if $deltagg1$, then the integral can be well-approximated by neglecting $mathcal{O}_n$ and, furthermore, the approximation may work better if we have more information about the actual derivatives, i.e. if we can made $n$ arbitrarily large.
However, for a fixed small $delta$ (say $deltall 1$), then the residual is continuously increasing with respect to $n$. This indicates that the approximation does not work as $mathcal{O}_n$ is expected to be big. Furthermore, the approximation may work even worse if we have more information about the actual derivatives!, i.e. making $n$ arbitrarily large.
Could anyone provide an alternative analysis explaining this issue?. Clearly, $deltagg 1$ is needed in order to approximate the integral with this expansion. But, for me is strange because it is just the exponential decay-rate... So why it should be bigger than 1 and not any other number?
calculus integration
asked Dec 21 '18 at 16:47
Alberto CastilloAlberto Castillo
212
$endgroup$
|
show 1 more comment
$begingroup$
Consider the following integral: $$int_t^{infty}e^{-delta(s-t)}omega(s)text{ds},$$ where $tgeq 0$ represents the current time, $delta>0$ is a fixed parameter representing the exponential decay rate, and $omega:mathbb{R}_{geq 0}rightarrow mathbb{R}$ is a given smooth bounded function, whose infinite derivatives are also bounded and with none derivative null.
I need to approximate the value of this integral with the restriction that I only have information about the actual $n$-first derivatives $omega(t)$, $dot{omega}(t)$,... ,$omega^{(n)}(t)$, with $ngeq 0$.
Recursive integration by parts leads to
$$int_t^{infty}e^{-delta(s-t)}omega(s)text{ds} = -frac{1}{delta}omega(t) + frac{1}{delta^2}dot{omega}(t)pm...+frac{(-1)^{n+1}}{delta^{n+1}}omega^{(n)}(t)+mathcal{O}_n,$$ where $$mathcal{O}_n=frac{1}{delta^{n+2}}int_t^{infty}e^{-delta(s-t)}omega^{(n+1)}(s)text{ds}.$$
What happens is the following:
Consider a fixed big $delta$ (say $deltagg1$). Then the residual $mathcal{O}_n$ is expected to be small as $1/delta^{n+2}$ is continuously decreasing with respect to $n$. This indicates that, if $deltagg1$, then the integral can be well-approximated by neglecting $mathcal{O}_n$ and, furthermore, the approximation may work better if we have more information about the actual derivatives, i.e. if we can made $n$ arbitrarily large.
However, for a fixed small $delta$ (say $deltall 1$), then the residual is continuously increasing with respect to $n$. This indicates that the approximation does not work as $mathcal{O}_n$ is expected to be big. Furthermore, the approximation may work even worse if we have more information about the actual derivatives!, i.e. making $n$ arbitrarily large.
Could anyone provide an alternative analysis explaining this issue?. Clearly, $deltagg 1$ is needed in order to approximate the integral with this expansion. But, for me is strange because it is just the exponential decay-rate... So why it should be bigger than 1 and not any other number?
calculus integration
asked Dec 21 '18 at 16:47
Alberto CastilloAlberto Castillo
212
$endgroup$
1
$begingroup$
I wonder what $n to infty$ is actually supposed to mean here. The way it is written it seems that $n$ depends on $omega$, so to let $n to infty$, one has to vary $omega$, but then one cannot make much non-trivial comparisons between the integrals for different $n$ and $omega$ anyway. Please clarify.
$endgroup$
– Torsten Schoeneberg
Dec 22 '18 at 6:31
$begingroup$
Well if $n$ is fixed, the dichotomy $a<1$ versus $a>1$ disappears. What you have is that the error term $mathcal {O}_n$ decreases as $a$ increases, continuously depending on $a$, without anything special happening at $a=1$; which makes perfect sense.
$endgroup$
– Torsten Schoeneberg
Dec 24 '18 at 17:34
$begingroup$
Thank you very much for your answers. You have helped me to clarify this question. I have rewritten it again in order to avoid confusions. In any case, even for the case of $n$ fixed (and $delta$ unfixed), it does not make perfect sense for me. I am asking for an alternative analysis that clarifies this behavior. Thanks!
$endgroup$
– Alberto Castillo
Jan 14 at 13:35
1
$begingroup$
I suggest finding a more descriptive title for your question.
$endgroup$
– Viktor Glombik
Jan 14 at 13:40
$begingroup$
Remark that $mathcal{O}_n$ depends not only on $delta$ and $n$, but also on $omega^{(n+1)}$ and $s$. It might very well be that that factor in the integral "cancels out" the effect of the $delta^{-n-2}$. Maybe try some easy examples (say $omega(t) = sin(t)$ and $sin^2(t)$) and see what actually happens in those error terms.
$endgroup$
– Torsten Schoeneberg
Jan 14 at 18:05
|
show 1 more comment
$begingroup$
Consider the following integral: $$int_t^{infty}e^{-delta(s-t)}omega(s)text{ds},$$ where $tgeq 0$ represents the current time, $delta>0$ is a fixed parameter representing the exponential decay rate, and $omega:mathbb{R}_{geq 0}rightarrow mathbb{R}$ is a given smooth bounded function, whose infinite derivatives are also bounded and with none derivative null.
I need to approximate the value of this integral with the restriction that I only have information about the actual $n$-first derivatives $omega(t)$, $dot{omega}(t)$,... ,$omega^{(n)}(t)$, with $ngeq 0$.
Recursive integration by parts leads to
$$int_t^{infty}e^{-delta(s-t)}omega(s)text{ds} = -frac{1}{delta}omega(t) + frac{1}{delta^2}dot{omega}(t)pm...+frac{(-1)^{n+1}}{delta^{n+1}}omega^{(n)}(t)+mathcal{O}_n,$$ where $$mathcal{O}_n=frac{1}{delta^{n+2}}int_t^{infty}e^{-delta(s-t)}omega^{(n+1)}(s)text{ds}.$$
What happens is the following:
Consider a fixed big $delta$ (say $deltagg1$). Then the residual $mathcal{O}_n$ is expected to be small as $1/delta^{n+2}$ is continuously decreasing with respect to $n$. This indicates that, if $deltagg1$, then the integral can be well-approximated by neglecting $mathcal{O}_n$ and, furthermore, the approximation may work better if we have more information about the actual derivatives, i.e. if we can made $n$ arbitrarily large.
However, for a fixed small $delta$ (say $deltall 1$), then the residual is continuously increasing with respect to $n$. This indicates that the approximation does not work as $mathcal{O}_n$ is expected to be big. Furthermore, the approximation may work even worse if we have more information about the actual derivatives!, i.e. making $n$ arbitrarily large.
Could anyone provide an alternative analysis explaining this issue?. Clearly, $deltagg 1$ is needed in order to approximate the integral with this expansion. But, for me is strange because it is just the exponential decay-rate... So why it should be bigger than 1 and not any other number?
calculus integration
asked Dec 21 '18 at 16:47
Alberto CastilloAlberto Castillo
212
$endgroup$
Consider the following integral: $$int_t^{infty}e^{-delta(s-t)}omega(s)text{ds},$$ where $tgeq 0$ represents the current time, $delta>0$ is a fixed parameter representing the exponential decay rate, and $omega:mathbb{R}_{geq 0}rightarrow mathbb{R}$ is a given smooth bounded function, whose infinite derivatives are also bounded and with none derivative null.
I need to approximate the value of this integral with the restriction that I only have information about the actual $n$-first derivatives $omega(t)$, $dot{omega}(t)$,... ,$omega^{(n)}(t)$, with $ngeq 0$.
Recursive integration by parts leads to
$$int_t^{infty}e^{-delta(s-t)}omega(s)text{ds} = -frac{1}{delta}omega(t) + frac{1}{delta^2}dot{omega}(t)pm...+frac{(-1)^{n+1}}{delta^{n+1}}omega^{(n)}(t)+mathcal{O}_n,$$ where $$mathcal{O}_n=frac{1}{delta^{n+2}}int_t^{infty}e^{-delta(s-t)}omega^{(n+1)}(s)text{ds}.$$
What happens is the following:
Consider a fixed big $delta$ (say $deltagg1$). Then the residual $mathcal{O}_n$ is expected to be small as $1/delta^{n+2}$ is continuously decreasing with respect to $n$. This indicates that, if $deltagg1$, then the integral can be well-approximated by neglecting $mathcal{O}_n$ and, furthermore, the approximation may work better if we have more information about the actual derivatives, i.e. if we can made $n$ arbitrarily large.
However, for a fixed small $delta$ (say $deltall 1$), then the residual is continuously increasing with respect to $n$. This indicates that the approximation does not work as $mathcal{O}_n$ is expected to be big. Furthermore, the approximation may work even worse if we have more information about the actual derivatives!, i.e. making $n$ arbitrarily large.
Could anyone provide an alternative analysis explaining this issue?. Clearly, $deltagg 1$ is needed in order to approximate the integral with this expansion. But, for me is strange because it is just the exponential decay-rate... So why it should be bigger than 1 and not any other number?
calculus integration
calculus integration
asked Dec 21 '18 at 16:47
Alberto CastilloAlberto Castillo
212
asked Dec 21 '18 at 16:47
Alberto CastilloAlberto Castillo
212
edited Feb 8 at 15:05
Alberto Castillo
asked Dec 21 '18 at 16:47
Alberto CastilloAlberto Castillo
212
asked Dec 21 '18 at 16:47
Alberto CastilloAlberto Castillo
212
asked Dec 21 '18 at 16:47
Alberto CastilloAlberto Castillo
212
212
1
$begingroup$
I wonder what $n to infty$ is actually supposed to mean here. The way it is written it seems that $n$ depends on $omega$, so to let $n to infty$, one has to vary $omega$, but then one cannot make much non-trivial comparisons between the integrals for different $n$ and $omega$ anyway. Please clarify.
$endgroup$
– Torsten Schoeneberg
Dec 22 '18 at 6:31
$begingroup$
Well if $n$ is fixed, the dichotomy $a<1$ versus $a>1$ disappears. What you have is that the error term $mathcal {O}_n$ decreases as $a$ increases, continuously depending on $a$, without anything special happening at $a=1$; which makes perfect sense.
$endgroup$
– Torsten Schoeneberg
Dec 24 '18 at 17:34
$begingroup$
Thank you very much for your answers. You have helped me to clarify this question. I have rewritten it again in order to avoid confusions. In any case, even for the case of $n$ fixed (and $delta$ unfixed), it does not make perfect sense for me. I am asking for an alternative analysis that clarifies this behavior. Thanks!
$endgroup$
– Alberto Castillo
Jan 14 at 13:35
1
$begingroup$
I suggest finding a more descriptive title for your question.
$endgroup$
– Viktor Glombik
Jan 14 at 13:40
$begingroup$
Remark that $mathcal{O}_n$ depends not only on $delta$ and $n$, but also on $omega^{(n+1)}$ and $s$. It might very well be that that factor in the integral "cancels out" the effect of the $delta^{-n-2}$. Maybe try some easy examples (say $omega(t) = sin(t)$ and $sin^2(t)$) and see what actually happens in those error terms.
$endgroup$
– Torsten Schoeneberg
Jan 14 at 18:05
|
show 1 more comment
1
$begingroup$
I wonder what $n to infty$ is actually supposed to mean here. The way it is written it seems that $n$ depends on $omega$, so to let $n to infty$, one has to vary $omega$, but then one cannot make much non-trivial comparisons between the integrals for different $n$ and $omega$ anyway. Please clarify.
$endgroup$
– Torsten Schoeneberg
Dec 22 '18 at 6:31
$begingroup$
Well if $n$ is fixed, the dichotomy $a<1$ versus $a>1$ disappears. What you have is that the error term $mathcal {O}_n$ decreases as $a$ increases, continuously depending on $a$, without anything special happening at $a=1$; which makes perfect sense.
$endgroup$
– Torsten Schoeneberg
Dec 24 '18 at 17:34
$begingroup$
Thank you very much for your answers. You have helped me to clarify this question. I have rewritten it again in order to avoid confusions. In any case, even for the case of $n$ fixed (and $delta$ unfixed), it does not make perfect sense for me. I am asking for an alternative analysis that clarifies this behavior. Thanks!
$endgroup$
– Alberto Castillo
Jan 14 at 13:35
1
$begingroup$
I suggest finding a more descriptive title for your question.
$endgroup$
– Viktor Glombik
Jan 14 at 13:40
$begingroup$
Remark that $mathcal{O}_n$ depends not only on $delta$ and $n$, but also on $omega^{(n+1)}$ and $s$. It might very well be that that factor in the integral "cancels out" the effect of the $delta^{-n-2}$. Maybe try some easy examples (say $omega(t) = sin(t)$ and $sin^2(t)$) and see what actually happens in those error terms.
$endgroup$
– Torsten Schoeneberg
Jan 14 at 18:05
1
1
$begingroup$
I wonder what $n to infty$ is actually supposed to mean here. The way it is written it seems that $n$ depends on $omega$, so to let $n to infty$, one has to vary $omega$, but then one cannot make much non-trivial comparisons between the integrals for different $n$ and $omega$ anyway. Please clarify.
$endgroup$
– Torsten Schoeneberg
Dec 22 '18 at 6:31
$begingroup$
I wonder what $n to infty$ is actually supposed to mean here. The way it is written it seems that $n$ depends on $omega$, so to let $n to infty$, one has to vary $omega$, but then one cannot make much non-trivial comparisons between the integrals for different $n$ and $omega$ anyway. Please clarify.
$endgroup$
– Torsten Schoeneberg
Dec 22 '18 at 6:31
$begingroup$
Well if $n$ is fixed, the dichotomy $a<1$ versus $a>1$ disappears. What you have is that the error term $mathcal {O}_n$ decreases as $a$ increases, continuously depending on $a$, without anything special happening at $a=1$; which makes perfect sense.
$endgroup$
– Torsten Schoeneberg
Dec 24 '18 at 17:34
$begingroup$
Well if $n$ is fixed, the dichotomy $a<1$ versus $a>1$ disappears. What you have is that the error term $mathcal {O}_n$ decreases as $a$ increases, continuously depending on $a$, without anything special happening at $a=1$; which makes perfect sense.
$endgroup$
– Torsten Schoeneberg
Dec 24 '18 at 17:34
$begingroup$
Thank you very much for your answers. You have helped me to clarify this question. I have rewritten it again in order to avoid confusions. In any case, even for the case of $n$ fixed (and $delta$ unfixed), it does not make perfect sense for me. I am asking for an alternative analysis that clarifies this behavior. Thanks!
$endgroup$
– Alberto Castillo
Jan 14 at 13:35
$begingroup$
Thank you very much for your answers. You have helped me to clarify this question. I have rewritten it again in order to avoid confusions. In any case, even for the case of $n$ fixed (and $delta$ unfixed), it does not make perfect sense for me. I am asking for an alternative analysis that clarifies this behavior. Thanks!
$endgroup$
– Alberto Castillo
Jan 14 at 13:35
1
1
$begingroup$
I suggest finding a more descriptive title for your question.
$endgroup$
– Viktor Glombik
Jan 14 at 13:40
$begingroup$
I suggest finding a more descriptive title for your question.
$endgroup$
– Viktor Glombik
Jan 14 at 13:40
$begingroup$
Remark that $mathcal{O}_n$ depends not only on $delta$ and $n$, but also on $omega^{(n+1)}$ and $s$. It might very well be that that factor in the integral "cancels out" the effect of the $delta^{-n-2}$. Maybe try some easy examples (say $omega(t) = sin(t)$ and $sin^2(t)$) and see what actually happens in those error terms.
$endgroup$
– Torsten Schoeneberg
Jan 14 at 18:05
$begingroup$
Remark that $mathcal{O}_n$ depends not only on $delta$ and $n$, but also on $omega^{(n+1)}$ and $s$. It might very well be that that factor in the integral "cancels out" the effect of the $delta^{-n-2}$. Maybe try some easy examples (say $omega(t) = sin(t)$ and $sin^2(t)$) and see what actually happens in those error terms.
$endgroup$
– Torsten Schoeneberg
Jan 14 at 18:05
|
show 1 more comment
0
active
oldest
votes
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%2f3048686%2fapproximate-int-t-inftye-deltas-t-omegas-textds-with-the-restri%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
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%2f3048686%2fapproximate-int-t-inftye-deltas-t-omegas-textds-with-the-restri%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
1
$begingroup$
I wonder what $n to infty$ is actually supposed to mean here. The way it is written it seems that $n$ depends on $omega$, so to let $n to infty$, one has to vary $omega$, but then one cannot make much non-trivial comparisons between the integrals for different $n$ and $omega$ anyway. Please clarify.
$endgroup$
– Torsten Schoeneberg
Dec 22 '18 at 6:31
$begingroup$
Well if $n$ is fixed, the dichotomy $a<1$ versus $a>1$ disappears. What you have is that the error term $mathcal {O}_n$ decreases as $a$ increases, continuously depending on $a$, without anything special happening at $a=1$; which makes perfect sense.
$endgroup$
– Torsten Schoeneberg
Dec 24 '18 at 17:34
$begingroup$
Thank you very much for your answers. You have helped me to clarify this question. I have rewritten it again in order to avoid confusions. In any case, even for the case of $n$ fixed (and $delta$ unfixed), it does not make perfect sense for me. I am asking for an alternative analysis that clarifies this behavior. Thanks!
$endgroup$
– Alberto Castillo
Jan 14 at 13:35
1
$begingroup$
I suggest finding a more descriptive title for your question.
$endgroup$
– Viktor Glombik
Jan 14 at 13:40
$begingroup$
Remark that $mathcal{O}_n$ depends not only on $delta$ and $n$, but also on $omega^{(n+1)}$ and $s$. It might very well be that that factor in the integral "cancels out" the effect of the $delta^{-n-2}$. Maybe try some easy examples (say $omega(t) = sin(t)$ and $sin^2(t)$) and see what actually happens in those error terms.
$endgroup$
– Torsten Schoeneberg
Jan 14 at 18:05