Mixing
Laplace transform $mathcal L(f^{(k)})(z)=z^k mathcal L (f)(z)-sum_{j=0}^{k-1}f^{(j)}(0)z^{k-1-j}$
$begingroup$
Let $f in C^k[0,+infty)$ and $ sigma(f):=inf{s in mathbb{R}:exp(-st)f(t) in L^1[0,+infty)}$.
I want to show that for $Re z>max_{j=0,ldots,k}sigma(f^{(j)})$
$$mathcal L(f^{(k)})(z)=z^k mathcal L (f)(z)-sum_{j=0}^{k-1}f^{(j)}(0)z^{k-1-j}$$
I tried the following, using induction and partial integration. I used the induction hypothesis in the third step.
my attempt
Can this be done like this? And does $e^{-zt}f^{(k-1)}(t)$ equal $0$ when plugging for $t=infty$?
real-analysis complex-analysis laplace-transform
asked Jan 27 at 14:42
user636610user636610
82
$endgroup$
add a comment |
$begingroup$
Let $f in C^k[0,+infty)$ and $ sigma(f):=inf{s in mathbb{R}:exp(-st)f(t) in L^1[0,+infty)}$.
I want to show that for $Re z>max_{j=0,ldots,k}sigma(f^{(j)})$
$$mathcal L(f^{(k)})(z)=z^k mathcal L (f)(z)-sum_{j=0}^{k-1}f^{(j)}(0)z^{k-1-j}$$
I tried the following, using induction and partial integration. I used the induction hypothesis in the third step.
my attempt
Can this be done like this? And does $e^{-zt}f^{(k-1)}(t)$ equal $0$ when plugging for $t=infty$?
real-analysis complex-analysis laplace-transform
asked Jan 27 at 14:42
user636610user636610
82
$endgroup$
add a comment |
$begingroup$
Let $f in C^k[0,+infty)$ and $ sigma(f):=inf{s in mathbb{R}:exp(-st)f(t) in L^1[0,+infty)}$.
I want to show that for $Re z>max_{j=0,ldots,k}sigma(f^{(j)})$
$$mathcal L(f^{(k)})(z)=z^k mathcal L (f)(z)-sum_{j=0}^{k-1}f^{(j)}(0)z^{k-1-j}$$
I tried the following, using induction and partial integration. I used the induction hypothesis in the third step.
my attempt
Can this be done like this? And does $e^{-zt}f^{(k-1)}(t)$ equal $0$ when plugging for $t=infty$?
real-analysis complex-analysis laplace-transform
asked Jan 27 at 14:42
user636610user636610
82
$endgroup$
Let $f in C^k[0,+infty)$ and $ sigma(f):=inf{s in mathbb{R}:exp(-st)f(t) in L^1[0,+infty)}$.
I want to show that for $Re z>max_{j=0,ldots,k}sigma(f^{(j)})$
$$mathcal L(f^{(k)})(z)=z^k mathcal L (f)(z)-sum_{j=0}^{k-1}f^{(j)}(0)z^{k-1-j}$$
I tried the following, using induction and partial integration. I used the induction hypothesis in the third step.
my attempt
Can this be done like this? And does $e^{-zt}f^{(k-1)}(t)$ equal $0$ when plugging for $t=infty$?
real-analysis complex-analysis laplace-transform
real-analysis complex-analysis laplace-transform
asked Jan 27 at 14:42
user636610user636610
82
asked Jan 27 at 14:42
user636610user636610
82
asked Jan 27 at 14:42
user636610user636610
82
asked Jan 27 at 14:42
user636610user636610
82
asked Jan 27 at 14:42
user636610user636610
82
82
add a comment |
add a comment |
1 Answer
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$begingroup$
Well, first of all if we define
$$
F(z) = mathcal{L}(f)(z)
$$
The basic case for $k = 0$ essentially follows from the definition above. Now if you assume
$$
mathcal L(f^{(k-1)})(z)=z^{k-1} F(z) -sum_{j=0}^{k-2}f^{(j)}(0)z^{k-1-j}
$$
you can use integration by parts for case $k$ since
$$
mathcal{L}(f^{(k)})(z) = mathcal{L}(Df^{(k-1)})(z) = left. f^{(k-1)}(t)e^{-zt} right|_{0}^{+infty} + z mathcal{L}(f^{(k-1)})(z) = -f^{(k-1)}(0) + z mathcal{L}(f^{(k-1)})(z),
$$
here $D$ denotes the differential operator. If you substitute to $mathcal{L}(f^{(k-1)})(z)$ the expression given by your inductive hypothesis you'll get your results.
Of course to make the integrals valid you need to use your other assumptions, I just focused on the calculations.
answered Jan 27 at 14:56
user8469759user8469759
1,5681618
$endgroup$
$begingroup$
What you do can be formulated as "generalized integration by parts" proofwiki.org/wiki/Generalized_Integration_by_Parts
$endgroup$
– Jean Marie
Feb 12 at 11:52
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Well, first of all if we define
$$
F(z) = mathcal{L}(f)(z)
$$
The basic case for $k = 0$ essentially follows from the definition above. Now if you assume
$$
mathcal L(f^{(k-1)})(z)=z^{k-1} F(z) -sum_{j=0}^{k-2}f^{(j)}(0)z^{k-1-j}
$$
you can use integration by parts for case $k$ since
$$
mathcal{L}(f^{(k)})(z) = mathcal{L}(Df^{(k-1)})(z) = left. f^{(k-1)}(t)e^{-zt} right|_{0}^{+infty} + z mathcal{L}(f^{(k-1)})(z) = -f^{(k-1)}(0) + z mathcal{L}(f^{(k-1)})(z),
$$
here $D$ denotes the differential operator. If you substitute to $mathcal{L}(f^{(k-1)})(z)$ the expression given by your inductive hypothesis you'll get your results.
Of course to make the integrals valid you need to use your other assumptions, I just focused on the calculations.
answered Jan 27 at 14:56
user8469759user8469759
1,5681618
$endgroup$
$begingroup$
What you do can be formulated as "generalized integration by parts" proofwiki.org/wiki/Generalized_Integration_by_Parts
$endgroup$
– Jean Marie
Feb 12 at 11:52
add a comment |
$begingroup$
Well, first of all if we define
$$
F(z) = mathcal{L}(f)(z)
$$
The basic case for $k = 0$ essentially follows from the definition above. Now if you assume
$$
mathcal L(f^{(k-1)})(z)=z^{k-1} F(z) -sum_{j=0}^{k-2}f^{(j)}(0)z^{k-1-j}
$$
you can use integration by parts for case $k$ since
$$
mathcal{L}(f^{(k)})(z) = mathcal{L}(Df^{(k-1)})(z) = left. f^{(k-1)}(t)e^{-zt} right|_{0}^{+infty} + z mathcal{L}(f^{(k-1)})(z) = -f^{(k-1)}(0) + z mathcal{L}(f^{(k-1)})(z),
$$
here $D$ denotes the differential operator. If you substitute to $mathcal{L}(f^{(k-1)})(z)$ the expression given by your inductive hypothesis you'll get your results.
Of course to make the integrals valid you need to use your other assumptions, I just focused on the calculations.
answered Jan 27 at 14:56
user8469759user8469759
1,5681618
$endgroup$
$begingroup$
What you do can be formulated as "generalized integration by parts" proofwiki.org/wiki/Generalized_Integration_by_Parts
$endgroup$
– Jean Marie
Feb 12 at 11:52
add a comment |
$begingroup$
Well, first of all if we define
$$
F(z) = mathcal{L}(f)(z)
$$
The basic case for $k = 0$ essentially follows from the definition above. Now if you assume
$$
mathcal L(f^{(k-1)})(z)=z^{k-1} F(z) -sum_{j=0}^{k-2}f^{(j)}(0)z^{k-1-j}
$$
you can use integration by parts for case $k$ since
$$
mathcal{L}(f^{(k)})(z) = mathcal{L}(Df^{(k-1)})(z) = left. f^{(k-1)}(t)e^{-zt} right|_{0}^{+infty} + z mathcal{L}(f^{(k-1)})(z) = -f^{(k-1)}(0) + z mathcal{L}(f^{(k-1)})(z),
$$
here $D$ denotes the differential operator. If you substitute to $mathcal{L}(f^{(k-1)})(z)$ the expression given by your inductive hypothesis you'll get your results.
Of course to make the integrals valid you need to use your other assumptions, I just focused on the calculations.
answered Jan 27 at 14:56
user8469759user8469759
1,5681618
$endgroup$
Well, first of all if we define
$$
F(z) = mathcal{L}(f)(z)
$$
The basic case for $k = 0$ essentially follows from the definition above. Now if you assume
$$
mathcal L(f^{(k-1)})(z)=z^{k-1} F(z) -sum_{j=0}^{k-2}f^{(j)}(0)z^{k-1-j}
$$
you can use integration by parts for case $k$ since
$$
mathcal{L}(f^{(k)})(z) = mathcal{L}(Df^{(k-1)})(z) = left. f^{(k-1)}(t)e^{-zt} right|_{0}^{+infty} + z mathcal{L}(f^{(k-1)})(z) = -f^{(k-1)}(0) + z mathcal{L}(f^{(k-1)})(z),
$$
here $D$ denotes the differential operator. If you substitute to $mathcal{L}(f^{(k-1)})(z)$ the expression given by your inductive hypothesis you'll get your results.
Of course to make the integrals valid you need to use your other assumptions, I just focused on the calculations.
answered Jan 27 at 14:56
user8469759user8469759
1,5681618
answered Jan 27 at 14:56
user8469759user8469759
1,5681618
answered Jan 27 at 14:56
user8469759user8469759
1,5681618
answered Jan 27 at 14:56
user8469759user8469759
1,5681618
1,5681618
$begingroup$
What you do can be formulated as "generalized integration by parts" proofwiki.org/wiki/Generalized_Integration_by_Parts
$endgroup$
– Jean Marie
Feb 12 at 11:52
add a comment |
$begingroup$
What you do can be formulated as "generalized integration by parts" proofwiki.org/wiki/Generalized_Integration_by_Parts
$endgroup$
– Jean Marie
Feb 12 at 11:52
$begingroup$
What you do can be formulated as "generalized integration by parts" proofwiki.org/wiki/Generalized_Integration_by_Parts
$endgroup$
– Jean Marie
Feb 12 at 11:52
$begingroup$
What you do can be formulated as "generalized integration by parts" proofwiki.org/wiki/Generalized_Integration_by_Parts
$endgroup$
– Jean Marie
Feb 12 at 11:52
add a comment |
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