Functions without Laplace transform?
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We have just started working with Laplace transformations at our university course. One of the I came across as following:
Provide three examples of functions for which the Laplace transform does not exist.
We use the book written by Kreyszig "Advanced Engineering Mathematics".
I would have answered where $S > 0$ and $S > K$, I am not sure what the third kind of function would be. In addition I am unsure if this is even the answer they are looking for, I am hoping someone more experienced could provide me with some suggestions. Worst case scenario I get my answer in two weeks time when I hand in the assignment :)
examples-counterexamples laplace-transform
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add a comment |
$begingroup$
We have just started working with Laplace transformations at our university course. One of the I came across as following:
Provide three examples of functions for which the Laplace transform does not exist.
We use the book written by Kreyszig "Advanced Engineering Mathematics".
I would have answered where $S > 0$ and $S > K$, I am not sure what the third kind of function would be. In addition I am unsure if this is even the answer they are looking for, I am hoping someone more experienced could provide me with some suggestions. Worst case scenario I get my answer in two weeks time when I hand in the assignment :)
examples-counterexamples laplace-transform
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3
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Laplace transform for $f(t)$ works if $int_0^infty e^{-st} f(t) dt$ converges. So if your function grows so fast that no decaying exponential can stop it then the integral diverges. Think about fast growing functions, even faster than any $e^{at}$ to find an example.
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– Maesumi
Feb 24 '13 at 17:07
add a comment |
$begingroup$
We have just started working with Laplace transformations at our university course. One of the I came across as following:
Provide three examples of functions for which the Laplace transform does not exist.
We use the book written by Kreyszig "Advanced Engineering Mathematics".
I would have answered where $S > 0$ and $S > K$, I am not sure what the third kind of function would be. In addition I am unsure if this is even the answer they are looking for, I am hoping someone more experienced could provide me with some suggestions. Worst case scenario I get my answer in two weeks time when I hand in the assignment :)
examples-counterexamples laplace-transform
$endgroup$
We have just started working with Laplace transformations at our university course. One of the I came across as following:
Provide three examples of functions for which the Laplace transform does not exist.
We use the book written by Kreyszig "Advanced Engineering Mathematics".
I would have answered where $S > 0$ and $S > K$, I am not sure what the third kind of function would be. In addition I am unsure if this is even the answer they are looking for, I am hoping someone more experienced could provide me with some suggestions. Worst case scenario I get my answer in two weeks time when I hand in the assignment :)
examples-counterexamples laplace-transform
examples-counterexamples laplace-transform
edited Apr 30 '18 at 10:30
GNUSupporter 8964民主女神 地下教會
12.7k72445
12.7k72445
asked Feb 24 '13 at 17:00
WobbleyWobbley
12314
12314
3
$begingroup$
Laplace transform for $f(t)$ works if $int_0^infty e^{-st} f(t) dt$ converges. So if your function grows so fast that no decaying exponential can stop it then the integral diverges. Think about fast growing functions, even faster than any $e^{at}$ to find an example.
$endgroup$
– Maesumi
Feb 24 '13 at 17:07
add a comment |
3
$begingroup$
Laplace transform for $f(t)$ works if $int_0^infty e^{-st} f(t) dt$ converges. So if your function grows so fast that no decaying exponential can stop it then the integral diverges. Think about fast growing functions, even faster than any $e^{at}$ to find an example.
$endgroup$
– Maesumi
Feb 24 '13 at 17:07
3
3
$begingroup$
Laplace transform for $f(t)$ works if $int_0^infty e^{-st} f(t) dt$ converges. So if your function grows so fast that no decaying exponential can stop it then the integral diverges. Think about fast growing functions, even faster than any $e^{at}$ to find an example.
$endgroup$
– Maesumi
Feb 24 '13 at 17:07
$begingroup$
Laplace transform for $f(t)$ works if $int_0^infty e^{-st} f(t) dt$ converges. So if your function grows so fast that no decaying exponential can stop it then the integral diverges. Think about fast growing functions, even faster than any $e^{at}$ to find an example.
$endgroup$
– Maesumi
Feb 24 '13 at 17:07
add a comment |
4 Answers
4
active
oldest
votes
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Hint: The function $e^{e^t}$ could be an answer?
Aside: Bilateral Laplace transform of $e^{-e^{-t}}$ is Gamma function?
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add a comment |
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Find something that grows too fast for the exponentially-decaying damping factor $e^{-st}$ (where $t$ is the variable of integration) in the defining integral of the transform to cancel. What do you know of that grows more quickly than an exponential function?
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add a comment |
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Find some functions that exist $limlimits_{tto a}f(t)$ diverges for some positive real numbers $a$ , e.g. $tan t$ .
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add a comment |
$begingroup$
$$F(t)=frac{1}{t}$$
Laplace transformation of this function does not exists
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add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: The function $e^{e^t}$ could be an answer?
Aside: Bilateral Laplace transform of $e^{-e^{-t}}$ is Gamma function?
$endgroup$
add a comment |
$begingroup$
Hint: The function $e^{e^t}$ could be an answer?
Aside: Bilateral Laplace transform of $e^{-e^{-t}}$ is Gamma function?
$endgroup$
add a comment |
$begingroup$
Hint: The function $e^{e^t}$ could be an answer?
Aside: Bilateral Laplace transform of $e^{-e^{-t}}$ is Gamma function?
$endgroup$
Hint: The function $e^{e^t}$ could be an answer?
Aside: Bilateral Laplace transform of $e^{-e^{-t}}$ is Gamma function?
edited Apr 8 '13 at 7:36
robjohn♦
266k27304626
266k27304626
answered Apr 8 '13 at 6:17
Kannan NambiarKannan Nambiar
111
111
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$begingroup$
Find something that grows too fast for the exponentially-decaying damping factor $e^{-st}$ (where $t$ is the variable of integration) in the defining integral of the transform to cancel. What do you know of that grows more quickly than an exponential function?
$endgroup$
add a comment |
$begingroup$
Find something that grows too fast for the exponentially-decaying damping factor $e^{-st}$ (where $t$ is the variable of integration) in the defining integral of the transform to cancel. What do you know of that grows more quickly than an exponential function?
$endgroup$
add a comment |
$begingroup$
Find something that grows too fast for the exponentially-decaying damping factor $e^{-st}$ (where $t$ is the variable of integration) in the defining integral of the transform to cancel. What do you know of that grows more quickly than an exponential function?
$endgroup$
Find something that grows too fast for the exponentially-decaying damping factor $e^{-st}$ (where $t$ is the variable of integration) in the defining integral of the transform to cancel. What do you know of that grows more quickly than an exponential function?
answered Apr 8 '13 at 7:39
The_SympathizerThe_Sympathizer
7,4852245
7,4852245
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add a comment |
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Find some functions that exist $limlimits_{tto a}f(t)$ diverges for some positive real numbers $a$ , e.g. $tan t$ .
$endgroup$
add a comment |
$begingroup$
Find some functions that exist $limlimits_{tto a}f(t)$ diverges for some positive real numbers $a$ , e.g. $tan t$ .
$endgroup$
add a comment |
$begingroup$
Find some functions that exist $limlimits_{tto a}f(t)$ diverges for some positive real numbers $a$ , e.g. $tan t$ .
$endgroup$
Find some functions that exist $limlimits_{tto a}f(t)$ diverges for some positive real numbers $a$ , e.g. $tan t$ .
answered Apr 8 '13 at 8:33
doraemonpauldoraemonpaul
12.5k31660
12.5k31660
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$$F(t)=frac{1}{t}$$
Laplace transformation of this function does not exists
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add a comment |
$begingroup$
$$F(t)=frac{1}{t}$$
Laplace transformation of this function does not exists
$endgroup$
add a comment |
$begingroup$
$$F(t)=frac{1}{t}$$
Laplace transformation of this function does not exists
$endgroup$
$$F(t)=frac{1}{t}$$
Laplace transformation of this function does not exists
edited Apr 30 '18 at 9:53
Alex Vong
1,284819
1,284819
answered Apr 30 '18 at 9:25
RKMVRKMV
1
1
add a comment |
add a comment |
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$begingroup$
Laplace transform for $f(t)$ works if $int_0^infty e^{-st} f(t) dt$ converges. So if your function grows so fast that no decaying exponential can stop it then the integral diverges. Think about fast growing functions, even faster than any $e^{at}$ to find an example.
$endgroup$
– Maesumi
Feb 24 '13 at 17:07