Laplace transform of complementary error function $operatornameerfc(1/sqrtt)$- using infinite series Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Finding Laplace Transform of a FunctionSolve transport equations by using Laplace transformLaplace transform of complementary error function using infinite seriesInverse Laplace Transform to arrive at Error FunctionLaplace transforms and error functionLaplace Transform $sinh(sqrtt)$Laplace transform $mathcalL_xleft[frac1(x+9) sqrtx+8right](s)$How to get the Laplace transformation of function $fraca2sqrtπtoperatornameerf(frac-a4t)$Formula for complementary cumulative distribution function (CCDF) using Laplace and inverse Laplace transformLaplace Transform of Complementary Error Function
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Laplace transform of complementary error function $operatornameerfc(1/sqrtt)$- using infinite series
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Finding Laplace Transform of a FunctionSolve transport equations by using Laplace transformLaplace transform of complementary error function using infinite seriesInverse Laplace Transform to arrive at Error FunctionLaplace transforms and error functionLaplace Transform $sinh(sqrtt)$Laplace transform $mathcalL_xleft[frac1(x+9) sqrtx+8right](s)$How to get the Laplace transformation of function $fraca2sqrtπtoperatornameerf(frac-a4t)$Formula for complementary cumulative distribution function (CCDF) using Laplace and inverse Laplace transformLaplace Transform of Complementary Error Function
$begingroup$
The Laplace transform of the complementary error function $operatornameerfcleft(frac1sqrttright)$ is
$$Lleftoperatornameerfcleft(frac1sqrttright)right=Lleft1-operatornameerfleft(frac1sqrttright) right$$
That is
$$beginaligned
Lleftoperatornameerfcleft(frac1sqrttright)right
&=Lleft1-frac2sqrtpi int_0^1/sqrtt e^-x^2 dx right\
&=Lleft1-frac2sqrtpi sum_n=0^infty frac(-1)^nn! int_0^1/sqrtt x^2n dx right \
&=Lleft1-frac2sqrtpi sum_n=0^infty frac(-1)^nn!(2n+1) frac1t^n+1/2 right \
&=frac1p - frac2sqrtpi sum_n=0^infty frac(-1)^nn!(2n+1) Lleftfrac1t^n+1/2 right \
endaligned$$
After this step, I do not know how to proceed. Please help, if you know the procedure or if there is any mistake please point out. Thank you!
laplace-transform
edited Aug 22 '16 at 1:29
Bungo
13.7k22148
asked Aug 22 '16 at 1:08
Venkatesan MurugesanVenkatesan Murugesan
1413
$endgroup$
add a comment |
$begingroup$
The Laplace transform of the complementary error function $operatornameerfcleft(frac1sqrttright)$ is
$$Lleftoperatornameerfcleft(frac1sqrttright)right=Lleft1-operatornameerfleft(frac1sqrttright) right$$
That is
$$beginaligned
Lleftoperatornameerfcleft(frac1sqrttright)right
&=Lleft1-frac2sqrtpi int_0^1/sqrtt e^-x^2 dx right\
&=Lleft1-frac2sqrtpi sum_n=0^infty frac(-1)^nn! int_0^1/sqrtt x^2n dx right \
&=Lleft1-frac2sqrtpi sum_n=0^infty frac(-1)^nn!(2n+1) frac1t^n+1/2 right \
&=frac1p - frac2sqrtpi sum_n=0^infty frac(-1)^nn!(2n+1) Lleftfrac1t^n+1/2 right \
endaligned$$
After this step, I do not know how to proceed. Please help, if you know the procedure or if there is any mistake please point out. Thank you!
laplace-transform
edited Aug 22 '16 at 1:29
Bungo
13.7k22148
asked Aug 22 '16 at 1:08
Venkatesan MurugesanVenkatesan Murugesan
1413
$endgroup$
$begingroup$
There is a problem because $displaystyle mathcalLleft(frac1t^n+1/2right)$ doesn't exist when $n>1/2$ !
$endgroup$
– Aron OME
Feb 28 at 16:07
add a comment |
$begingroup$
The Laplace transform of the complementary error function $operatornameerfcleft(frac1sqrttright)$ is
$$Lleftoperatornameerfcleft(frac1sqrttright)right=Lleft1-operatornameerfleft(frac1sqrttright) right$$
That is
$$beginaligned
Lleftoperatornameerfcleft(frac1sqrttright)right
&=Lleft1-frac2sqrtpi int_0^1/sqrtt e^-x^2 dx right\
&=Lleft1-frac2sqrtpi sum_n=0^infty frac(-1)^nn! int_0^1/sqrtt x^2n dx right \
&=Lleft1-frac2sqrtpi sum_n=0^infty frac(-1)^nn!(2n+1) frac1t^n+1/2 right \
&=frac1p - frac2sqrtpi sum_n=0^infty frac(-1)^nn!(2n+1) Lleftfrac1t^n+1/2 right \
endaligned$$
After this step, I do not know how to proceed. Please help, if you know the procedure or if there is any mistake please point out. Thank you!
laplace-transform
edited Aug 22 '16 at 1:29
Bungo
13.7k22148
asked Aug 22 '16 at 1:08
Venkatesan MurugesanVenkatesan Murugesan
1413
$endgroup$
The Laplace transform of the complementary error function $operatornameerfcleft(frac1sqrttright)$ is
$$Lleftoperatornameerfcleft(frac1sqrttright)right=Lleft1-operatornameerfleft(frac1sqrttright) right$$
That is
$$beginaligned
Lleftoperatornameerfcleft(frac1sqrttright)right
&=Lleft1-frac2sqrtpi int_0^1/sqrtt e^-x^2 dx right\
&=Lleft1-frac2sqrtpi sum_n=0^infty frac(-1)^nn! int_0^1/sqrtt x^2n dx right \
&=Lleft1-frac2sqrtpi sum_n=0^infty frac(-1)^nn!(2n+1) frac1t^n+1/2 right \
&=frac1p - frac2sqrtpi sum_n=0^infty frac(-1)^nn!(2n+1) Lleftfrac1t^n+1/2 right \
endaligned$$
After this step, I do not know how to proceed. Please help, if you know the procedure or if there is any mistake please point out. Thank you!
laplace-transform
laplace-transform
edited Aug 22 '16 at 1:29
Bungo
13.7k22148
asked Aug 22 '16 at 1:08
Venkatesan MurugesanVenkatesan Murugesan
1413
edited Aug 22 '16 at 1:29
Bungo
13.7k22148
asked Aug 22 '16 at 1:08
Venkatesan MurugesanVenkatesan Murugesan
1413
edited Aug 22 '16 at 1:29
Bungo
13.7k22148
edited Aug 22 '16 at 1:29
Bungo
13.7k22148
edited Aug 22 '16 at 1:29
Bungo
13.7k22148
13.7k22148
asked Aug 22 '16 at 1:08
Venkatesan MurugesanVenkatesan Murugesan
1413
asked Aug 22 '16 at 1:08
Venkatesan MurugesanVenkatesan Murugesan
1413
asked Aug 22 '16 at 1:08
Venkatesan MurugesanVenkatesan Murugesan
1413
1413
$begingroup$
There is a problem because $displaystyle mathcalLleft(frac1t^n+1/2right)$ doesn't exist when $n>1/2$ !
$endgroup$
– Aron OME
Feb 28 at 16:07
add a comment |
$begingroup$
There is a problem because $displaystyle mathcalLleft(frac1t^n+1/2right)$ doesn't exist when $n>1/2$ !
$endgroup$
– Aron OME
Feb 28 at 16:07
$begingroup$
There is a problem because $displaystyle mathcalLleft(frac1t^n+1/2right)$ doesn't exist when $n>1/2$ !
$endgroup$
– Aron OME
Feb 28 at 16:07
$begingroup$
There is a problem because $displaystyle mathcalLleft(frac1t^n+1/2right)$ doesn't exist when $n>1/2$ !
$endgroup$
– Aron OME
Feb 28 at 16:07
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It looks fine. Now you just have to exploit
$$mathcalL(1)=frac1s,qquad mathcalLleft(frac1t^n+1/2right) = s^n-frac12Gammaleft(frac12-nright)=s^n-frac12(-1)^nfrac2^nsqrtpi(2n-1)!!. $$
Anyway, It would have been faster to exploit the properties of the Laplace transform. $textErfc$ is defined by an integral, and through a change of variable it is not difficult to check that
$$ mathcalLleft(textErfcleft(frac1sqrttright)right) = colorredfrace^-2sqrtss. $$
answered Aug 22 '16 at 1:26
Jack D'AurizioJack D'Aurizio
292k33284674
$endgroup$
1
$begingroup$
Sir, I have obtained the result by exploiting the properties of the Laplace transform. However, my interest is to get the Laplace transform using infinite series of $operatornameerfcleft(frac1sqrttright)$. I have used the result as suggested by you. But I didn't get the required result. Please help if possible. –
$endgroup$
– Venkatesan Murugesan
Aug 22 '16 at 15:52
add a comment |
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1 Answer
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1 Answer
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$begingroup$
It looks fine. Now you just have to exploit
$$mathcalL(1)=frac1s,qquad mathcalLleft(frac1t^n+1/2right) = s^n-frac12Gammaleft(frac12-nright)=s^n-frac12(-1)^nfrac2^nsqrtpi(2n-1)!!. $$
Anyway, It would have been faster to exploit the properties of the Laplace transform. $textErfc$ is defined by an integral, and through a change of variable it is not difficult to check that
$$ mathcalLleft(textErfcleft(frac1sqrttright)right) = colorredfrace^-2sqrtss. $$
answered Aug 22 '16 at 1:26
Jack D'AurizioJack D'Aurizio
292k33284674
$endgroup$
1
$begingroup$
Sir, I have obtained the result by exploiting the properties of the Laplace transform. However, my interest is to get the Laplace transform using infinite series of $operatornameerfcleft(frac1sqrttright)$. I have used the result as suggested by you. But I didn't get the required result. Please help if possible. –
$endgroup$
– Venkatesan Murugesan
Aug 22 '16 at 15:52
add a comment |
$begingroup$
It looks fine. Now you just have to exploit
$$mathcalL(1)=frac1s,qquad mathcalLleft(frac1t^n+1/2right) = s^n-frac12Gammaleft(frac12-nright)=s^n-frac12(-1)^nfrac2^nsqrtpi(2n-1)!!. $$
Anyway, It would have been faster to exploit the properties of the Laplace transform. $textErfc$ is defined by an integral, and through a change of variable it is not difficult to check that
$$ mathcalLleft(textErfcleft(frac1sqrttright)right) = colorredfrace^-2sqrtss. $$
answered Aug 22 '16 at 1:26
Jack D'AurizioJack D'Aurizio
292k33284674
$endgroup$
1
$begingroup$
Sir, I have obtained the result by exploiting the properties of the Laplace transform. However, my interest is to get the Laplace transform using infinite series of $operatornameerfcleft(frac1sqrttright)$. I have used the result as suggested by you. But I didn't get the required result. Please help if possible. –
$endgroup$
– Venkatesan Murugesan
Aug 22 '16 at 15:52
add a comment |
$begingroup$
It looks fine. Now you just have to exploit
$$mathcalL(1)=frac1s,qquad mathcalLleft(frac1t^n+1/2right) = s^n-frac12Gammaleft(frac12-nright)=s^n-frac12(-1)^nfrac2^nsqrtpi(2n-1)!!. $$
Anyway, It would have been faster to exploit the properties of the Laplace transform. $textErfc$ is defined by an integral, and through a change of variable it is not difficult to check that
$$ mathcalLleft(textErfcleft(frac1sqrttright)right) = colorredfrace^-2sqrtss. $$
answered Aug 22 '16 at 1:26
Jack D'AurizioJack D'Aurizio
292k33284674
$endgroup$
It looks fine. Now you just have to exploit
$$mathcalL(1)=frac1s,qquad mathcalLleft(frac1t^n+1/2right) = s^n-frac12Gammaleft(frac12-nright)=s^n-frac12(-1)^nfrac2^nsqrtpi(2n-1)!!. $$
Anyway, It would have been faster to exploit the properties of the Laplace transform. $textErfc$ is defined by an integral, and through a change of variable it is not difficult to check that
$$ mathcalLleft(textErfcleft(frac1sqrttright)right) = colorredfrace^-2sqrtss. $$
answered Aug 22 '16 at 1:26
Jack D'AurizioJack D'Aurizio
292k33284674
answered Aug 22 '16 at 1:26
Jack D'AurizioJack D'Aurizio
292k33284674
answered Aug 22 '16 at 1:26
Jack D'AurizioJack D'Aurizio
292k33284674
answered Aug 22 '16 at 1:26
Jack D'AurizioJack D'Aurizio
292k33284674
292k33284674
1
$begingroup$
Sir, I have obtained the result by exploiting the properties of the Laplace transform. However, my interest is to get the Laplace transform using infinite series of $operatornameerfcleft(frac1sqrttright)$. I have used the result as suggested by you. But I didn't get the required result. Please help if possible. –
$endgroup$
– Venkatesan Murugesan
Aug 22 '16 at 15:52
add a comment |
1
$begingroup$
Sir, I have obtained the result by exploiting the properties of the Laplace transform. However, my interest is to get the Laplace transform using infinite series of $operatornameerfcleft(frac1sqrttright)$. I have used the result as suggested by you. But I didn't get the required result. Please help if possible. –
$endgroup$
– Venkatesan Murugesan
Aug 22 '16 at 15:52
1
1
$begingroup$
Sir, I have obtained the result by exploiting the properties of the Laplace transform. However, my interest is to get the Laplace transform using infinite series of $operatornameerfcleft(frac1sqrttright)$. I have used the result as suggested by you. But I didn't get the required result. Please help if possible. –
$endgroup$
– Venkatesan Murugesan
Aug 22 '16 at 15:52
$begingroup$
Sir, I have obtained the result by exploiting the properties of the Laplace transform. However, my interest is to get the Laplace transform using infinite series of $operatornameerfcleft(frac1sqrttright)$. I have used the result as suggested by you. But I didn't get the required result. Please help if possible. –
$endgroup$
– Venkatesan Murugesan
Aug 22 '16 at 15:52
add a comment |
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$begingroup$
There is a problem because $displaystyle mathcalLleft(frac1t^n+1/2right)$ doesn't exist when $n>1/2$ !
$endgroup$
– Aron OME
Feb 28 at 16:07