$frac{c^t e^t}{t^{t+1/2}}$ and $e^{-kt^2}$ which decay faster as $t rightarrow infty$？ where $c$ is...
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.
18.8k21932
asked Nov 21 at 21:26
ShaoyuPei
1427
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.
18.8k21932
asked Nov 21 at 21:26
ShaoyuPei
1427
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.
18.8k21932
asked Nov 21 at 21:26
ShaoyuPei
1427
$frac{c^t e^t}{t^{t+1/2}}$ 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.
18.8k21932
asked Nov 21 at 21:26
ShaoyuPei
1427
edited Nov 21 at 21:30
J.G.
18.8k21932
asked Nov 21 at 21:26
ShaoyuPei
1427
edited Nov 21 at 21:30
J.G.
18.8k21932
edited Nov 21 at 21:30
J.G.
18.8k21932
edited Nov 21 at 21:30
J.G.
18.8k21932
18.8k21932
asked Nov 21 at 21:26
ShaoyuPei
1427
asked Nov 21 at 21:26
ShaoyuPei
1427
asked Nov 21 at 21:26
ShaoyuPei
1427
1427
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.
18.8k21932
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.
18.8k21932
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.
18.8k21932
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.
18.8k21932
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.
18.8k21932
answered Nov 21 at 21:29
J.G.
18.8k21932
answered Nov 21 at 21:29
J.G.
18.8k21932
answered Nov 21 at 21:29
J.G.
18.8k21932
18.8k21932
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 |
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3008391%2ffracct-ettt1-2-and-e-kt2-which-decay-faster-as-t-rightarrow%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
