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$frac{c^t e^t}{t^{t+1/2}}$ and $e^{-kt^2}$ which decay faster as $t rightarrow infty$？ where $c$ is...
$frac{c^t e^t}{t^{t+1/2}}$ and $e^{-kt^2}$ which decay faster as $t rightarrow infty$？ where $c,k$ are constants . how to see that?
calculus asymptotics
edited Nov 21 '18 at 21:30
J.G.
23.3k22137
asked Nov 21 '18 at 21:26
ShaoyuPeiShaoyuPei
1698
add a comment |
$frac{c^t e^t}{t^{t+1/2}}$ and $e^{-kt^2}$ which decay faster as $t rightarrow infty$？ where $c,k$ are constants . how to see that?
calculus asymptotics
edited Nov 21 '18 at 21:30
J.G.
23.3k22137
asked Nov 21 '18 at 21:26
ShaoyuPeiShaoyuPei
1698
add a comment |
$frac{c^t e^t}{t^{t+1/2}}$ and $e^{-kt^2}$ which decay faster as $t rightarrow infty$？ where $c,k$ are constants . how to see that?
calculus asymptotics
edited Nov 21 '18 at 21:30
J.G.
23.3k22137
asked Nov 21 '18 at 21:26
ShaoyuPeiShaoyuPei
1698
$frac{c^t e^t}{t^{t+1/2}}$ and $e^{-kt^2}$ which decay faster as $t rightarrow infty$？ where $c,k$ are constants . how to see that?
calculus asymptotics
calculus asymptotics
edited Nov 21 '18 at 21:30
J.G.
23.3k22137
asked Nov 21 '18 at 21:26
ShaoyuPeiShaoyuPei
1698
edited Nov 21 '18 at 21:30
J.G.
23.3k22137
asked Nov 21 '18 at 21:26
ShaoyuPeiShaoyuPei
1698
edited Nov 21 '18 at 21:30
J.G.
23.3k22137
edited Nov 21 '18 at 21:30
J.G.
23.3k22137
edited Nov 21 '18 at 21:30
J.G.
23.3k22137
23.3k22137
asked Nov 21 '18 at 21:26
ShaoyuPeiShaoyuPei
1698
asked Nov 21 '18 at 21:26
ShaoyuPeiShaoyuPei
1698
asked Nov 21 '18 at 21:26
ShaoyuPeiShaoyuPei
1698
1698
add a comment |
add a comment |
1 Answer
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Consider the respective logarithms $t(ln c+1-ln t)-frac{1}{2}ln tsim -tln t,,-kt^2$. The latter heads to $-infty$ faster as $ttoinfty$, so $e^{-kt^2}$ is the faster-decaying function.
answered Nov 21 '18 at 21:29
J.G.J.G.
23.3k22137
thx ,that's helpful
– ShaoyuPei
Nov 21 '18 at 21:30
add a comment |
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1 Answer
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1 Answer
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oldest
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votes
Consider the respective logarithms $t(ln c+1-ln t)-frac{1}{2}ln tsim -tln t,,-kt^2$. The latter heads to $-infty$ faster as $ttoinfty$, so $e^{-kt^2}$ is the faster-decaying function.
answered Nov 21 '18 at 21:29
J.G.J.G.
23.3k22137
thx ,that's helpful
– ShaoyuPei
Nov 21 '18 at 21:30
add a comment |
Consider the respective logarithms $t(ln c+1-ln t)-frac{1}{2}ln tsim -tln t,,-kt^2$. The latter heads to $-infty$ faster as $ttoinfty$, so $e^{-kt^2}$ is the faster-decaying function.
answered Nov 21 '18 at 21:29
J.G.J.G.
23.3k22137
thx ,that's helpful
– ShaoyuPei
Nov 21 '18 at 21:30
add a comment |
Consider the respective logarithms $t(ln c+1-ln t)-frac{1}{2}ln tsim -tln t,,-kt^2$. The latter heads to $-infty$ faster as $ttoinfty$, so $e^{-kt^2}$ is the faster-decaying function.
answered Nov 21 '18 at 21:29
J.G.J.G.
23.3k22137
Consider the respective logarithms $t(ln c+1-ln t)-frac{1}{2}ln tsim -tln t,,-kt^2$. The latter heads to $-infty$ faster as $ttoinfty$, so $e^{-kt^2}$ is the faster-decaying function.
answered Nov 21 '18 at 21:29
J.G.J.G.
23.3k22137
answered Nov 21 '18 at 21:29
J.G.J.G.
23.3k22137
answered Nov 21 '18 at 21:29
J.G.J.G.
23.3k22137
answered Nov 21 '18 at 21:29
J.G.J.G.
23.3k22137
23.3k22137
thx ,that's helpful
– ShaoyuPei
Nov 21 '18 at 21:30
add a comment |
thx ,that's helpful
– ShaoyuPei
Nov 21 '18 at 21:30
thx ,that's helpful
– ShaoyuPei
Nov 21 '18 at 21:30
thx ,that's helpful
– ShaoyuPei
Nov 21 '18 at 21:30
add a comment |
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