Divergency of the asymptotic expansion of $operatornameEi(x)$
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Consider function $f(z)=e^-zint^zfrace^ttdt$, it satisfies the differential equation $w'+w=frac1z$. Let $(w')$'s asymptotic expansion as $ztoinfty$ be $w'sim Ce^-z+sumlimits_n=1^inftyfraca_nz^n$. Substitute it into $w'+w=frac1z$ gives $wsim Ce^-z+sumlimits_n=1^inftyfrac(n-1)!z^n$.
My questions
(i)If I let $w'=Ce^-z+sumlimits_n=1^inftyfraca_nz^n$, I won't get a result because of the divergency of the series. Why does the series diverge?
(ii)Is there a way to determine whether the asymptotic expansion of $int f(z)dz$ or a differential equation's solution converge or diverge?
Ideas trying to answer (i)
Let $t=frac1z$, $tto0$. $fracdwdt-frac1t^2w=-frac1t$. $P(t)=-frac1t^2$ and $Q(t)=-frac1t$. They have poles at $t=0$. Probably that is why the asymptotic expansion diverges?
calculus integration differential-equations asymptotics
asked Sep 4 at 9:04
Kemono Chen
68416
add a comment |Â
up vote
0
down vote
favorite
Consider function $f(z)=e^-zint^zfrace^ttdt$, it satisfies the differential equation $w'+w=frac1z$. Let $(w')$'s asymptotic expansion as $ztoinfty$ be $w'sim Ce^-z+sumlimits_n=1^inftyfraca_nz^n$. Substitute it into $w'+w=frac1z$ gives $wsim Ce^-z+sumlimits_n=1^inftyfrac(n-1)!z^n$.
My questions
(i)If I let $w'=Ce^-z+sumlimits_n=1^inftyfraca_nz^n$, I won't get a result because of the divergency of the series. Why does the series diverge?
(ii)Is there a way to determine whether the asymptotic expansion of $int f(z)dz$ or a differential equation's solution converge or diverge?
Ideas trying to answer (i)
Let $t=frac1z$, $tto0$. $fracdwdt-frac1t^2w=-frac1t$. $P(t)=-frac1t^2$ and $Q(t)=-frac1t$. They have poles at $t=0$. Probably that is why the asymptotic expansion diverges?
calculus integration differential-equations asymptotics
asked Sep 4 at 9:04
Kemono Chen
68416
It diverges because $frac(n-1)!z^n$ does not go to zero for any $z$ if $nrightarrow infty$. The summands are essentially the inverse of the terms of the Taylor series for $e^x$.
– gammatester
Sep 4 at 9:14
@gammatester Yes, I am trying to find an explanation from the integral and differential equation themselves.
– Kemono Chen
Sep 4 at 9:22
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Consider function $f(z)=e^-zint^zfrace^ttdt$, it satisfies the differential equation $w'+w=frac1z$. Let $(w')$'s asymptotic expansion as $ztoinfty$ be $w'sim Ce^-z+sumlimits_n=1^inftyfraca_nz^n$. Substitute it into $w'+w=frac1z$ gives $wsim Ce^-z+sumlimits_n=1^inftyfrac(n-1)!z^n$.
My questions
(i)If I let $w'=Ce^-z+sumlimits_n=1^inftyfraca_nz^n$, I won't get a result because of the divergency of the series. Why does the series diverge?
(ii)Is there a way to determine whether the asymptotic expansion of $int f(z)dz$ or a differential equation's solution converge or diverge?
Ideas trying to answer (i)
Let $t=frac1z$, $tto0$. $fracdwdt-frac1t^2w=-frac1t$. $P(t)=-frac1t^2$ and $Q(t)=-frac1t$. They have poles at $t=0$. Probably that is why the asymptotic expansion diverges?
calculus integration differential-equations asymptotics
asked Sep 4 at 9:04
Kemono Chen
68416
Consider function $f(z)=e^-zint^zfrace^ttdt$, it satisfies the differential equation $w'+w=frac1z$. Let $(w')$'s asymptotic expansion as $ztoinfty$ be $w'sim Ce^-z+sumlimits_n=1^inftyfraca_nz^n$. Substitute it into $w'+w=frac1z$ gives $wsim Ce^-z+sumlimits_n=1^inftyfrac(n-1)!z^n$.
My questions
(i)If I let $w'=Ce^-z+sumlimits_n=1^inftyfraca_nz^n$, I won't get a result because of the divergency of the series. Why does the series diverge?
(ii)Is there a way to determine whether the asymptotic expansion of $int f(z)dz$ or a differential equation's solution converge or diverge?
Ideas trying to answer (i)
Let $t=frac1z$, $tto0$. $fracdwdt-frac1t^2w=-frac1t$. $P(t)=-frac1t^2$ and $Q(t)=-frac1t$. They have poles at $t=0$. Probably that is why the asymptotic expansion diverges?
calculus integration differential-equations asymptotics
calculus integration differential-equations asymptotics
asked Sep 4 at 9:04
Kemono Chen
68416
asked Sep 4 at 9:04
Kemono Chen
68416
asked Sep 4 at 9:04
Kemono Chen
68416
asked Sep 4 at 9:04
Kemono Chen
68416
asked Sep 4 at 9:04
Kemono Chen
68416
68416
It diverges because $frac(n-1)!z^n$ does not go to zero for any $z$ if $nrightarrow infty$. The summands are essentially the inverse of the terms of the Taylor series for $e^x$.
– gammatester
Sep 4 at 9:14
@gammatester Yes, I am trying to find an explanation from the integral and differential equation themselves.
– Kemono Chen
Sep 4 at 9:22
add a comment |Â
It diverges because $frac(n-1)!z^n$ does not go to zero for any $z$ if $nrightarrow infty$. The summands are essentially the inverse of the terms of the Taylor series for $e^x$.
– gammatester
Sep 4 at 9:14
@gammatester Yes, I am trying to find an explanation from the integral and differential equation themselves.
– Kemono Chen
Sep 4 at 9:22
It diverges because $frac(n-1)!z^n$ does not go to zero for any $z$ if $nrightarrow infty$. The summands are essentially the inverse of the terms of the Taylor series for $e^x$.
– gammatester
Sep 4 at 9:14
It diverges because $frac(n-1)!z^n$ does not go to zero for any $z$ if $nrightarrow infty$. The summands are essentially the inverse of the terms of the Taylor series for $e^x$.
– gammatester
Sep 4 at 9:14
@gammatester Yes, I am trying to find an explanation from the integral and differential equation themselves.
– Kemono Chen
Sep 4 at 9:22
@gammatester Yes, I am trying to find an explanation from the integral and differential equation themselves.
– Kemono Chen
Sep 4 at 9:22
add a comment |Â
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It diverges because $frac(n-1)!z^n$ does not go to zero for any $z$ if $nrightarrow infty$. The summands are essentially the inverse of the terms of the Taylor series for $e^x$.
– gammatester
Sep 4 at 9:14
@gammatester Yes, I am trying to find an explanation from the integral and differential equation themselves.
– Kemono Chen
Sep 4 at 9:22