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Series Approximation Before and After Inverse Laplace Transformation
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We have the following series involving some Laplace transformed function $F(s)$,
$$sum^{infty}_{n=0} left( frac{s}{a} right)^{n} F(s).$$
Using the following formula,
$$left( frac{1}{a} right)^{n} mathcal{L} { sum^{infty}_{n=0} f^{(n)}(t) } = s^{n}F(s) + left( frac{1}{a} right)^{n} left( sum^{n-1}_{m=0} f^{(m)}(0) s^{(n-1)-m}right) ,$$
if $a rightarrow infty$ we truncate the series and take the following leading terms,
$$mathcal{L}^{-1}{F(s)+frac{s}{a}F(s)}=f(t)+frac{1}{a}f^{prime}(t)+frac{1}{a}delta(t)f(0).$$
Then we must make assumptions regarding $s$ and its relationship to $a$ which prevents us from using this simplified formula when $t rightarrow 0$. Yet if we inverse Laplace transform the entire series and take the leading term afterwards then we are no longer making these assumptions. Instead this depends on the value of the function, $f$, at $t=0$, its derivative, and the Dirac Delta function. Are these equivalent or is there a discrepancy between these two truncations? Are neither, one, or both of these descriptions suitable when describing $t rightarrow 0$?
laplace-transform inverselaplace
asked 20 hours ago
Tbone Willsone
62
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up vote
0
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We have the following series involving some Laplace transformed function $F(s)$,
$$sum^{infty}_{n=0} left( frac{s}{a} right)^{n} F(s).$$
Using the following formula,
$$left( frac{1}{a} right)^{n} mathcal{L} { sum^{infty}_{n=0} f^{(n)}(t) } = s^{n}F(s) + left( frac{1}{a} right)^{n} left( sum^{n-1}_{m=0} f^{(m)}(0) s^{(n-1)-m}right) ,$$
if $a rightarrow infty$ we truncate the series and take the following leading terms,
$$mathcal{L}^{-1}{F(s)+frac{s}{a}F(s)}=f(t)+frac{1}{a}f^{prime}(t)+frac{1}{a}delta(t)f(0).$$
Then we must make assumptions regarding $s$ and its relationship to $a$ which prevents us from using this simplified formula when $t rightarrow 0$. Yet if we inverse Laplace transform the entire series and take the leading term afterwards then we are no longer making these assumptions. Instead this depends on the value of the function, $f$, at $t=0$, its derivative, and the Dirac Delta function. Are these equivalent or is there a discrepancy between these two truncations? Are neither, one, or both of these descriptions suitable when describing $t rightarrow 0$?
laplace-transform inverselaplace
asked 20 hours ago
Tbone Willsone
62
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
We have the following series involving some Laplace transformed function $F(s)$,
$$sum^{infty}_{n=0} left( frac{s}{a} right)^{n} F(s).$$
Using the following formula,
$$left( frac{1}{a} right)^{n} mathcal{L} { sum^{infty}_{n=0} f^{(n)}(t) } = s^{n}F(s) + left( frac{1}{a} right)^{n} left( sum^{n-1}_{m=0} f^{(m)}(0) s^{(n-1)-m}right) ,$$
if $a rightarrow infty$ we truncate the series and take the following leading terms,
$$mathcal{L}^{-1}{F(s)+frac{s}{a}F(s)}=f(t)+frac{1}{a}f^{prime}(t)+frac{1}{a}delta(t)f(0).$$
Then we must make assumptions regarding $s$ and its relationship to $a$ which prevents us from using this simplified formula when $t rightarrow 0$. Yet if we inverse Laplace transform the entire series and take the leading term afterwards then we are no longer making these assumptions. Instead this depends on the value of the function, $f$, at $t=0$, its derivative, and the Dirac Delta function. Are these equivalent or is there a discrepancy between these two truncations? Are neither, one, or both of these descriptions suitable when describing $t rightarrow 0$?
laplace-transform inverselaplace
asked 20 hours ago
Tbone Willsone
62
We have the following series involving some Laplace transformed function $F(s)$,
$$sum^{infty}_{n=0} left( frac{s}{a} right)^{n} F(s).$$
Using the following formula,
$$left( frac{1}{a} right)^{n} mathcal{L} { sum^{infty}_{n=0} f^{(n)}(t) } = s^{n}F(s) + left( frac{1}{a} right)^{n} left( sum^{n-1}_{m=0} f^{(m)}(0) s^{(n-1)-m}right) ,$$
if $a rightarrow infty$ we truncate the series and take the following leading terms,
$$mathcal{L}^{-1}{F(s)+frac{s}{a}F(s)}=f(t)+frac{1}{a}f^{prime}(t)+frac{1}{a}delta(t)f(0).$$
Then we must make assumptions regarding $s$ and its relationship to $a$ which prevents us from using this simplified formula when $t rightarrow 0$. Yet if we inverse Laplace transform the entire series and take the leading term afterwards then we are no longer making these assumptions. Instead this depends on the value of the function, $f$, at $t=0$, its derivative, and the Dirac Delta function. Are these equivalent or is there a discrepancy between these two truncations? Are neither, one, or both of these descriptions suitable when describing $t rightarrow 0$?
laplace-transform inverselaplace
laplace-transform inverselaplace
asked 20 hours ago
Tbone Willsone
62
asked 20 hours ago
Tbone Willsone
62
asked 20 hours ago
Tbone Willsone
62
asked 20 hours ago
Tbone Willsone
62
asked 20 hours ago
Tbone Willsone
62
62
add a comment |
add a comment |
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