$frac{c^t e^t}{t^{t+1/2}}$ and $e^{-kt^2}$ which decay faster as $t rightarrow infty$？ where $c$ is...
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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 $c,k$ are constants . how to see that?
calculus asymptotics
edited Nov 21 at 21:30
J.G.
19.2k21932
asked Nov 21 at 21:26
ShaoyuPei
1577
add a comment |
up vote
0
down vote
favorite
$frac{c^t e^t}{t^{t+1/2}}$ 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 at 21:30
J.G.
19.2k21932
asked Nov 21 at 21:26
ShaoyuPei
1577
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
$frac{c^t e^t}{t^{t+1/2}}$ 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 at 21:30
J.G.
19.2k21932
asked Nov 21 at 21:26
ShaoyuPei
1577
$frac{c^t e^t}{t^{t+1/2}}$ 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 at 21:30
J.G.
19.2k21932
asked Nov 21 at 21:26
ShaoyuPei
1577
edited Nov 21 at 21:30
J.G.
19.2k21932
asked Nov 21 at 21:26
ShaoyuPei
1577
edited Nov 21 at 21:30
J.G.
19.2k21932
edited Nov 21 at 21:30
J.G.
19.2k21932
edited Nov 21 at 21:30
J.G.
19.2k21932
19.2k21932
asked Nov 21 at 21:26
ShaoyuPei
1577
asked Nov 21 at 21:26
ShaoyuPei
1577
asked Nov 21 at 21:26
ShaoyuPei
1577
1577
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
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 at 21:29
J.G.
19.2k21932
thx ,that's helpful
– ShaoyuPei
Nov 21 at 21:30
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
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 at 21:29
J.G.
19.2k21932
thx ,that's helpful
– ShaoyuPei
Nov 21 at 21:30
add a comment |
up vote
2
down vote
accepted
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 at 21:29
J.G.
19.2k21932
thx ,that's helpful
– ShaoyuPei
Nov 21 at 21:30
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
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 at 21:29
J.G.
19.2k21932
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 at 21:29
J.G.
19.2k21932
answered Nov 21 at 21:29
J.G.
19.2k21932
answered Nov 21 at 21:29
J.G.
19.2k21932
answered Nov 21 at 21:29
J.G.
19.2k21932
19.2k21932
thx ,that's helpful
– ShaoyuPei
Nov 21 at 21:30
add a comment |
thx ,that's helpful
– ShaoyuPei
Nov 21 at 21:30
thx ,that's helpful
– ShaoyuPei
Nov 21 at 21:30
thx ,that's helpful
– ShaoyuPei
Nov 21 at 21:30
add a comment |
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