Approximation of $int_t^{infty}e^{-delta(s-t)}omega(s)text{ds}$, if just $omega(t)$,…,$omega^{(n)}(t)$ are...
$begingroup$
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, in this case, 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
$endgroup$
add a 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, in this case, 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
add a 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, in this case, 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, in this case, 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 Jan 17 at 8:09
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
add a 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
add a comment |
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$begingroup$
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