Proof $frac{sin(ax)}{x} rightarrow a$ as $x rightarrow 0$ in the context of a Laplace Transformation
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I'm finding it difficult to get my head around
$frac{sin(ax)}{x} rightarrow a$ as $x rightarrow 0,$
however for the Laplace transform of this function,
$mathcal{L} { frac{sin(ax)}{x} } = arctan(frac{a}{s}),$
surely, $arctan(frac{a}{s}) rightarrow 0$ as $s rightarrow infty$?
But to retain the behaviour seen in the prior limit, as
$mathcal{L}^{-1}{ frac{a}{s} } = a$,
we would require the nonsensical situation of
$arctan(frac{a}{s}) rightarrow frac{a}{s}$ as $s rightarrow infty.$
Does this create an issue for our description of this function as $t rightarrow 0$?
Is there a relationship between $s$ and $a$ that prevents us from taking this limit? I feel like I am missing something fundamental here.
limits laplace-transform
asked Nov 26 at 12:53
Tbone Willsone
64
add a comment |
up vote
-1
down vote
favorite
I'm finding it difficult to get my head around
$frac{sin(ax)}{x} rightarrow a$ as $x rightarrow 0,$
however for the Laplace transform of this function,
$mathcal{L} { frac{sin(ax)}{x} } = arctan(frac{a}{s}),$
surely, $arctan(frac{a}{s}) rightarrow 0$ as $s rightarrow infty$?
But to retain the behaviour seen in the prior limit, as
$mathcal{L}^{-1}{ frac{a}{s} } = a$,
we would require the nonsensical situation of
$arctan(frac{a}{s}) rightarrow frac{a}{s}$ as $s rightarrow infty.$
Does this create an issue for our description of this function as $t rightarrow 0$?
Is there a relationship between $s$ and $a$ that prevents us from taking this limit? I feel like I am missing something fundamental here.
limits laplace-transform
asked Nov 26 at 12:53
Tbone Willsone
64
The assertion $lim_{stoinfty}arctanleft(frac asright)=frac as$ doesn't make sense. There is no way that there is a $s$ in the value of that limit. Where did you get that from?
– José Carlos Santos
Nov 26 at 12:59
That is my problem, that there should be no $s$ dependence in the limit, and yet if we wish to recover the behaviour we see prior to the Laplace transform in the $t$ limit then we would need such a thing (which is clearly wrong). I'll edit to clear up any confusion.
– Tbone Willsone
Nov 26 at 13:02
There are other big problems. The major problem I see is that $lim_{stoinfty}F(s)$ simply has nothing to do with $lim_{tto0}f(t)$.
– David C. Ullrich
Nov 26 at 14:54
Could you please expand? I thought the final value theorem mean that it did?
– Tbone Willsone
Nov 26 at 15:08
That would imply one could truncate a series expanded like $sum ( frac{s}{a})^{n} f(s)$ after transforming a series back from Laplace space without making any assumptions on $s$ or $t$. The only assumptions you would be making would be on $f(t)$, its derivatives, and their relationship with $a$. Which is interesting if correct.
– Tbone Willsone
Nov 26 at 15:13
add a comment |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
I'm finding it difficult to get my head around
$frac{sin(ax)}{x} rightarrow a$ as $x rightarrow 0,$
however for the Laplace transform of this function,
$mathcal{L} { frac{sin(ax)}{x} } = arctan(frac{a}{s}),$
surely, $arctan(frac{a}{s}) rightarrow 0$ as $s rightarrow infty$?
But to retain the behaviour seen in the prior limit, as
$mathcal{L}^{-1}{ frac{a}{s} } = a$,
we would require the nonsensical situation of
$arctan(frac{a}{s}) rightarrow frac{a}{s}$ as $s rightarrow infty.$
Does this create an issue for our description of this function as $t rightarrow 0$?
Is there a relationship between $s$ and $a$ that prevents us from taking this limit? I feel like I am missing something fundamental here.
limits laplace-transform
asked Nov 26 at 12:53
Tbone Willsone
64
I'm finding it difficult to get my head around
$frac{sin(ax)}{x} rightarrow a$ as $x rightarrow 0,$
however for the Laplace transform of this function,
$mathcal{L} { frac{sin(ax)}{x} } = arctan(frac{a}{s}),$
surely, $arctan(frac{a}{s}) rightarrow 0$ as $s rightarrow infty$?
But to retain the behaviour seen in the prior limit, as
$mathcal{L}^{-1}{ frac{a}{s} } = a$,
we would require the nonsensical situation of
$arctan(frac{a}{s}) rightarrow frac{a}{s}$ as $s rightarrow infty.$
Does this create an issue for our description of this function as $t rightarrow 0$?
Is there a relationship between $s$ and $a$ that prevents us from taking this limit? I feel like I am missing something fundamental here.
limits laplace-transform
limits laplace-transform
asked Nov 26 at 12:53
Tbone Willsone
64
asked Nov 26 at 12:53
Tbone Willsone
64
edited Nov 26 at 13:07
asked Nov 26 at 12:53
Tbone Willsone
64
asked Nov 26 at 12:53
Tbone Willsone
64
asked Nov 26 at 12:53
Tbone Willsone
64
64
The assertion $lim_{stoinfty}arctanleft(frac asright)=frac as$ doesn't make sense. There is no way that there is a $s$ in the value of that limit. Where did you get that from?
– José Carlos Santos
Nov 26 at 12:59
That is my problem, that there should be no $s$ dependence in the limit, and yet if we wish to recover the behaviour we see prior to the Laplace transform in the $t$ limit then we would need such a thing (which is clearly wrong). I'll edit to clear up any confusion.
– Tbone Willsone
Nov 26 at 13:02
There are other big problems. The major problem I see is that $lim_{stoinfty}F(s)$ simply has nothing to do with $lim_{tto0}f(t)$.
– David C. Ullrich
Nov 26 at 14:54
Could you please expand? I thought the final value theorem mean that it did?
– Tbone Willsone
Nov 26 at 15:08
That would imply one could truncate a series expanded like $sum ( frac{s}{a})^{n} f(s)$ after transforming a series back from Laplace space without making any assumptions on $s$ or $t$. The only assumptions you would be making would be on $f(t)$, its derivatives, and their relationship with $a$. Which is interesting if correct.
– Tbone Willsone
Nov 26 at 15:13
add a comment |
The assertion $lim_{stoinfty}arctanleft(frac asright)=frac as$ doesn't make sense. There is no way that there is a $s$ in the value of that limit. Where did you get that from?
– José Carlos Santos
Nov 26 at 12:59
That is my problem, that there should be no $s$ dependence in the limit, and yet if we wish to recover the behaviour we see prior to the Laplace transform in the $t$ limit then we would need such a thing (which is clearly wrong). I'll edit to clear up any confusion.
– Tbone Willsone
Nov 26 at 13:02
There are other big problems. The major problem I see is that $lim_{stoinfty}F(s)$ simply has nothing to do with $lim_{tto0}f(t)$.
– David C. Ullrich
Nov 26 at 14:54
Could you please expand? I thought the final value theorem mean that it did?
– Tbone Willsone
Nov 26 at 15:08
That would imply one could truncate a series expanded like $sum ( frac{s}{a})^{n} f(s)$ after transforming a series back from Laplace space without making any assumptions on $s$ or $t$. The only assumptions you would be making would be on $f(t)$, its derivatives, and their relationship with $a$. Which is interesting if correct.
– Tbone Willsone
Nov 26 at 15:13
The assertion $lim_{stoinfty}arctanleft(frac asright)=frac as$ doesn't make sense. There is no way that there is a $s$ in the value of that limit. Where did you get that from?
– José Carlos Santos
Nov 26 at 12:59
The assertion $lim_{stoinfty}arctanleft(frac asright)=frac as$ doesn't make sense. There is no way that there is a $s$ in the value of that limit. Where did you get that from?
– José Carlos Santos
Nov 26 at 12:59
That is my problem, that there should be no $s$ dependence in the limit, and yet if we wish to recover the behaviour we see prior to the Laplace transform in the $t$ limit then we would need such a thing (which is clearly wrong). I'll edit to clear up any confusion.
– Tbone Willsone
Nov 26 at 13:02
That is my problem, that there should be no $s$ dependence in the limit, and yet if we wish to recover the behaviour we see prior to the Laplace transform in the $t$ limit then we would need such a thing (which is clearly wrong). I'll edit to clear up any confusion.
– Tbone Willsone
Nov 26 at 13:02
There are other big problems. The major problem I see is that $lim_{stoinfty}F(s)$ simply has nothing to do with $lim_{tto0}f(t)$.
– David C. Ullrich
Nov 26 at 14:54
There are other big problems. The major problem I see is that $lim_{stoinfty}F(s)$ simply has nothing to do with $lim_{tto0}f(t)$.
– David C. Ullrich
Nov 26 at 14:54
Could you please expand? I thought the final value theorem mean that it did?
– Tbone Willsone
Nov 26 at 15:08
Could you please expand? I thought the final value theorem mean that it did?
– Tbone Willsone
Nov 26 at 15:08
That would imply one could truncate a series expanded like $sum ( frac{s}{a})^{n} f(s)$ after transforming a series back from Laplace space without making any assumptions on $s$ or $t$. The only assumptions you would be making would be on $f(t)$, its derivatives, and their relationship with $a$. Which is interesting if correct.
– Tbone Willsone
Nov 26 at 15:13
That would imply one could truncate a series expanded like $sum ( frac{s}{a})^{n} f(s)$ after transforming a series back from Laplace space without making any assumptions on $s$ or $t$. The only assumptions you would be making would be on $f(t)$, its derivatives, and their relationship with $a$. Which is interesting if correct.
– Tbone Willsone
Nov 26 at 15:13
add a comment |
1 Answer
1
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oldest
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up vote
-1
down vote
accepted
The final value theorem for Laplace's transform actually states:
$$
lim_{sto infty} sF(s) = lim_{tto 0^+} f(t)
$$
Thus:
$$
lim_{sto infty} s left[text{sgn}(a) frac{pi}{2}-arctan left( frac{s}{a} right) right] = lim_{tto 0} frac{sin(at)}{t}
$$
Check the result for the Laplace transform with Wolfram Alpha.
The limit above for $s$ can be computed with L'Hopital's rule and yields $a$, but that I guess is possibly pointless when the limit for the sine is being put into question.
answered Nov 26 at 13:06
Mefitico
920117
Brilliant, thank you!
– Tbone Willsone
Nov 26 at 13:10
Out of interest, does this mean that if we were to expand this function into a series, it would converge on $a$ faster than it would on $s$ or $t$?
– Tbone Willsone
Nov 26 at 13:13
If you are talking about $sin(at)/t$ ... You would also need to specify the notion to assess the velocity of convergence. Also, I would think the series would converge faster depending on the values chosen for either $t$ and $a$.
– Mefitico
Nov 26 at 13:20
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
-1
down vote
accepted
The final value theorem for Laplace's transform actually states:
$$
lim_{sto infty} sF(s) = lim_{tto 0^+} f(t)
$$
Thus:
$$
lim_{sto infty} s left[text{sgn}(a) frac{pi}{2}-arctan left( frac{s}{a} right) right] = lim_{tto 0} frac{sin(at)}{t}
$$
Check the result for the Laplace transform with Wolfram Alpha.
The limit above for $s$ can be computed with L'Hopital's rule and yields $a$, but that I guess is possibly pointless when the limit for the sine is being put into question.
answered Nov 26 at 13:06
Mefitico
920117
Brilliant, thank you!
– Tbone Willsone
Nov 26 at 13:10
Out of interest, does this mean that if we were to expand this function into a series, it would converge on $a$ faster than it would on $s$ or $t$?
– Tbone Willsone
Nov 26 at 13:13
If you are talking about $sin(at)/t$ ... You would also need to specify the notion to assess the velocity of convergence. Also, I would think the series would converge faster depending on the values chosen for either $t$ and $a$.
– Mefitico
Nov 26 at 13:20
add a comment |
up vote
-1
down vote
accepted
The final value theorem for Laplace's transform actually states:
$$
lim_{sto infty} sF(s) = lim_{tto 0^+} f(t)
$$
Thus:
$$
lim_{sto infty} s left[text{sgn}(a) frac{pi}{2}-arctan left( frac{s}{a} right) right] = lim_{tto 0} frac{sin(at)}{t}
$$
Check the result for the Laplace transform with Wolfram Alpha.
The limit above for $s$ can be computed with L'Hopital's rule and yields $a$, but that I guess is possibly pointless when the limit for the sine is being put into question.
answered Nov 26 at 13:06
Mefitico
920117
Brilliant, thank you!
– Tbone Willsone
Nov 26 at 13:10
Out of interest, does this mean that if we were to expand this function into a series, it would converge on $a$ faster than it would on $s$ or $t$?
– Tbone Willsone
Nov 26 at 13:13
If you are talking about $sin(at)/t$ ... You would also need to specify the notion to assess the velocity of convergence. Also, I would think the series would converge faster depending on the values chosen for either $t$ and $a$.
– Mefitico
Nov 26 at 13:20
add a comment |
up vote
-1
down vote
accepted
up vote
-1
down vote
accepted
The final value theorem for Laplace's transform actually states:
$$
lim_{sto infty} sF(s) = lim_{tto 0^+} f(t)
$$
Thus:
$$
lim_{sto infty} s left[text{sgn}(a) frac{pi}{2}-arctan left( frac{s}{a} right) right] = lim_{tto 0} frac{sin(at)}{t}
$$
Check the result for the Laplace transform with Wolfram Alpha.
The limit above for $s$ can be computed with L'Hopital's rule and yields $a$, but that I guess is possibly pointless when the limit for the sine is being put into question.
answered Nov 26 at 13:06
Mefitico
920117
The final value theorem for Laplace's transform actually states:
$$
lim_{sto infty} sF(s) = lim_{tto 0^+} f(t)
$$
Thus:
$$
lim_{sto infty} s left[text{sgn}(a) frac{pi}{2}-arctan left( frac{s}{a} right) right] = lim_{tto 0} frac{sin(at)}{t}
$$
Check the result for the Laplace transform with Wolfram Alpha.
The limit above for $s$ can be computed with L'Hopital's rule and yields $a$, but that I guess is possibly pointless when the limit for the sine is being put into question.
answered Nov 26 at 13:06
Mefitico
920117
edited Nov 26 at 16:48
answered Nov 26 at 13:06
Mefitico
920117
answered Nov 26 at 13:06
Mefitico
920117
answered Nov 26 at 13:06
Mefitico
920117
920117
Brilliant, thank you!
– Tbone Willsone
Nov 26 at 13:10
Out of interest, does this mean that if we were to expand this function into a series, it would converge on $a$ faster than it would on $s$ or $t$?
– Tbone Willsone
Nov 26 at 13:13
If you are talking about $sin(at)/t$ ... You would also need to specify the notion to assess the velocity of convergence. Also, I would think the series would converge faster depending on the values chosen for either $t$ and $a$.
– Mefitico
Nov 26 at 13:20
add a comment |
Brilliant, thank you!
– Tbone Willsone
Nov 26 at 13:10
Out of interest, does this mean that if we were to expand this function into a series, it would converge on $a$ faster than it would on $s$ or $t$?
– Tbone Willsone
Nov 26 at 13:13
If you are talking about $sin(at)/t$ ... You would also need to specify the notion to assess the velocity of convergence. Also, I would think the series would converge faster depending on the values chosen for either $t$ and $a$.
– Mefitico
Nov 26 at 13:20
Brilliant, thank you!
– Tbone Willsone
Nov 26 at 13:10
Brilliant, thank you!
– Tbone Willsone
Nov 26 at 13:10
Out of interest, does this mean that if we were to expand this function into a series, it would converge on $a$ faster than it would on $s$ or $t$?
– Tbone Willsone
Nov 26 at 13:13
Out of interest, does this mean that if we were to expand this function into a series, it would converge on $a$ faster than it would on $s$ or $t$?
– Tbone Willsone
Nov 26 at 13:13
If you are talking about $sin(at)/t$ ... You would also need to specify the notion to assess the velocity of convergence. Also, I would think the series would converge faster depending on the values chosen for either $t$ and $a$.
– Mefitico
Nov 26 at 13:20
If you are talking about $sin(at)/t$ ... You would also need to specify the notion to assess the velocity of convergence. Also, I would think the series would converge faster depending on the values chosen for either $t$ and $a$.
– Mefitico
Nov 26 at 13:20
add a comment |
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The assertion $lim_{stoinfty}arctanleft(frac asright)=frac as$ doesn't make sense. There is no way that there is a $s$ in the value of that limit. Where did you get that from?
– José Carlos Santos
Nov 26 at 12:59
That is my problem, that there should be no $s$ dependence in the limit, and yet if we wish to recover the behaviour we see prior to the Laplace transform in the $t$ limit then we would need such a thing (which is clearly wrong). I'll edit to clear up any confusion.
– Tbone Willsone
Nov 26 at 13:02
There are other big problems. The major problem I see is that $lim_{stoinfty}F(s)$ simply has nothing to do with $lim_{tto0}f(t)$.
– David C. Ullrich
Nov 26 at 14:54
Could you please expand? I thought the final value theorem mean that it did?
– Tbone Willsone
Nov 26 at 15:08
That would imply one could truncate a series expanded like $sum ( frac{s}{a})^{n} f(s)$ after transforming a series back from Laplace space without making any assumptions on $s$ or $t$. The only assumptions you would be making would be on $f(t)$, its derivatives, and their relationship with $a$. Which is interesting if correct.
– Tbone Willsone
Nov 26 at 15:13