Laplace Transform of $y^n$
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When the Laplace transform of $y$ is denoted $Y(s)$, we have formulas for the derivatives of $y$ without actually knowing what $y$ is. Is there an explicit formula for $y^2$? More generally, $y^n$?
ordinary-differential-equations laplace-transform
edited Jan 8 at 3:02
El borito
668216
asked Jan 8 at 1:50
Ace DavenportAce Davenport
394
$endgroup$
add a comment |
$begingroup$
When the Laplace transform of $y$ is denoted $Y(s)$, we have formulas for the derivatives of $y$ without actually knowing what $y$ is. Is there an explicit formula for $y^2$? More generally, $y^n$?
ordinary-differential-equations laplace-transform
edited Jan 8 at 3:02
El borito
668216
asked Jan 8 at 1:50
Ace DavenportAce Davenport
394
$endgroup$
$begingroup$
Do you just want the laplace transform of a basic power function, or a something like $y(x)^2$, where it's an entire function taken to a power?
$endgroup$
– Calvin Godfrey
Jan 8 at 3:04
$begingroup$
The entire function taken to a power - sorry if I was unclear.
$endgroup$
– Ace Davenport
Jan 8 at 3:07
1
$begingroup$
Generally, no..
$endgroup$
– Dylan
Jan 8 at 3:30
add a comment |
$begingroup$
When the Laplace transform of $y$ is denoted $Y(s)$, we have formulas for the derivatives of $y$ without actually knowing what $y$ is. Is there an explicit formula for $y^2$? More generally, $y^n$?
ordinary-differential-equations laplace-transform
edited Jan 8 at 3:02
El borito
668216
asked Jan 8 at 1:50
Ace DavenportAce Davenport
394
$endgroup$
When the Laplace transform of $y$ is denoted $Y(s)$, we have formulas for the derivatives of $y$ without actually knowing what $y$ is. Is there an explicit formula for $y^2$? More generally, $y^n$?
ordinary-differential-equations laplace-transform
ordinary-differential-equations laplace-transform
edited Jan 8 at 3:02
El borito
668216
asked Jan 8 at 1:50
Ace DavenportAce Davenport
394
edited Jan 8 at 3:02
El borito
668216
asked Jan 8 at 1:50
Ace DavenportAce Davenport
394
edited Jan 8 at 3:02
El borito
668216
edited Jan 8 at 3:02
El borito
668216
edited Jan 8 at 3:02
El borito
668216
668216
asked Jan 8 at 1:50
Ace DavenportAce Davenport
394
asked Jan 8 at 1:50
Ace DavenportAce Davenport
394
asked Jan 8 at 1:50
Ace DavenportAce Davenport
394
394
$begingroup$
Do you just want the laplace transform of a basic power function, or a something like $y(x)^2$, where it's an entire function taken to a power?
$endgroup$
– Calvin Godfrey
Jan 8 at 3:04
$begingroup$
The entire function taken to a power - sorry if I was unclear.
$endgroup$
– Ace Davenport
Jan 8 at 3:07
1
$begingroup$
Generally, no..
$endgroup$
– Dylan
Jan 8 at 3:30
add a comment |
$begingroup$
Do you just want the laplace transform of a basic power function, or a something like $y(x)^2$, where it's an entire function taken to a power?
$endgroup$
– Calvin Godfrey
Jan 8 at 3:04
$begingroup$
The entire function taken to a power - sorry if I was unclear.
$endgroup$
– Ace Davenport
Jan 8 at 3:07
1
$begingroup$
Generally, no..
$endgroup$
– Dylan
Jan 8 at 3:30
$begingroup$
Do you just want the laplace transform of a basic power function, or a something like $y(x)^2$, where it's an entire function taken to a power?
$endgroup$
– Calvin Godfrey
Jan 8 at 3:04
$begingroup$
Do you just want the laplace transform of a basic power function, or a something like $y(x)^2$, where it's an entire function taken to a power?
$endgroup$
– Calvin Godfrey
Jan 8 at 3:04
$begingroup$
The entire function taken to a power - sorry if I was unclear.
$endgroup$
– Ace Davenport
Jan 8 at 3:07
$begingroup$
The entire function taken to a power - sorry if I was unclear.
$endgroup$
– Ace Davenport
Jan 8 at 3:07
1
1
$begingroup$
Generally, no..
$endgroup$
– Dylan
Jan 8 at 3:30
$begingroup$
Generally, no..
$endgroup$
– Dylan
Jan 8 at 3:30
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
From this math stackexchange link, it seems there's no quick and easy way to do it. From the Wikipedia page on Laplace transform, it lists how to calculate $mathcal{L}(f(t)g(t))$, in which case you could simply say $f(t)=g(t)$. But the formula deals with the convolution of two functions, namely
$$mathcal{L}(f(t)g(t))=frac{1}{2pi i}lim_{Ttoinfty}int_{c-iT}^{c+iT}F(sigma)G(s-sigma)d,sigma$$
And this integration is done along the vertical line $textrm{Re}(sigma)=c$, where $F(sigma)$ and $G(s-sigma)$ are the Laplace transform of $f(t)$ and $g(t)$, respectively. So it's clear that for an arbitrary function, taking the laplace of $f(t)^n$ is not an easy task.
answered Jan 8 at 3:17
Calvin GodfreyCalvin Godfrey
633311
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add a comment |
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1 Answer
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active
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1 Answer
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active
oldest
votes
active
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active
oldest
votes
$begingroup$
From this math stackexchange link, it seems there's no quick and easy way to do it. From the Wikipedia page on Laplace transform, it lists how to calculate $mathcal{L}(f(t)g(t))$, in which case you could simply say $f(t)=g(t)$. But the formula deals with the convolution of two functions, namely
$$mathcal{L}(f(t)g(t))=frac{1}{2pi i}lim_{Ttoinfty}int_{c-iT}^{c+iT}F(sigma)G(s-sigma)d,sigma$$
And this integration is done along the vertical line $textrm{Re}(sigma)=c$, where $F(sigma)$ and $G(s-sigma)$ are the Laplace transform of $f(t)$ and $g(t)$, respectively. So it's clear that for an arbitrary function, taking the laplace of $f(t)^n$ is not an easy task.
answered Jan 8 at 3:17
Calvin GodfreyCalvin Godfrey
633311
$endgroup$
add a comment |
$begingroup$
From this math stackexchange link, it seems there's no quick and easy way to do it. From the Wikipedia page on Laplace transform, it lists how to calculate $mathcal{L}(f(t)g(t))$, in which case you could simply say $f(t)=g(t)$. But the formula deals with the convolution of two functions, namely
$$mathcal{L}(f(t)g(t))=frac{1}{2pi i}lim_{Ttoinfty}int_{c-iT}^{c+iT}F(sigma)G(s-sigma)d,sigma$$
And this integration is done along the vertical line $textrm{Re}(sigma)=c$, where $F(sigma)$ and $G(s-sigma)$ are the Laplace transform of $f(t)$ and $g(t)$, respectively. So it's clear that for an arbitrary function, taking the laplace of $f(t)^n$ is not an easy task.
answered Jan 8 at 3:17
Calvin GodfreyCalvin Godfrey
633311
$endgroup$
add a comment |
$begingroup$
From this math stackexchange link, it seems there's no quick and easy way to do it. From the Wikipedia page on Laplace transform, it lists how to calculate $mathcal{L}(f(t)g(t))$, in which case you could simply say $f(t)=g(t)$. But the formula deals with the convolution of two functions, namely
$$mathcal{L}(f(t)g(t))=frac{1}{2pi i}lim_{Ttoinfty}int_{c-iT}^{c+iT}F(sigma)G(s-sigma)d,sigma$$
And this integration is done along the vertical line $textrm{Re}(sigma)=c$, where $F(sigma)$ and $G(s-sigma)$ are the Laplace transform of $f(t)$ and $g(t)$, respectively. So it's clear that for an arbitrary function, taking the laplace of $f(t)^n$ is not an easy task.
answered Jan 8 at 3:17
Calvin GodfreyCalvin Godfrey
633311
$endgroup$
From this math stackexchange link, it seems there's no quick and easy way to do it. From the Wikipedia page on Laplace transform, it lists how to calculate $mathcal{L}(f(t)g(t))$, in which case you could simply say $f(t)=g(t)$. But the formula deals with the convolution of two functions, namely
$$mathcal{L}(f(t)g(t))=frac{1}{2pi i}lim_{Ttoinfty}int_{c-iT}^{c+iT}F(sigma)G(s-sigma)d,sigma$$
And this integration is done along the vertical line $textrm{Re}(sigma)=c$, where $F(sigma)$ and $G(s-sigma)$ are the Laplace transform of $f(t)$ and $g(t)$, respectively. So it's clear that for an arbitrary function, taking the laplace of $f(t)^n$ is not an easy task.
answered Jan 8 at 3:17
Calvin GodfreyCalvin Godfrey
633311
answered Jan 8 at 3:17
Calvin GodfreyCalvin Godfrey
633311
answered Jan 8 at 3:17
Calvin GodfreyCalvin Godfrey
633311
answered Jan 8 at 3:17
Calvin GodfreyCalvin Godfrey
633311
633311
add a comment |
add a comment |
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$begingroup$
Do you just want the laplace transform of a basic power function, or a something like $y(x)^2$, where it's an entire function taken to a power?
$endgroup$
– Calvin Godfrey
Jan 8 at 3:04
$begingroup$
The entire function taken to a power - sorry if I was unclear.
$endgroup$
– Ace Davenport
Jan 8 at 3:07
1
$begingroup$
Generally, no..
$endgroup$
– Dylan
Jan 8 at 3:30