Mixing
Question about asymptotic integral expansion. Reference?
$begingroup$
I need to compute a series expansion in $epsilon$ (at least to second order) of integrals like
$$g(epsilon):=int_0^epsilon f(epsilon,v), dv,$$ as $epsilon to 0^+$, where the functions in question are very complicated. (Too complicated, in my opinion, to express as Taylor series themselves.)
Does anyone have a reference for such things? I have tried naively expanding around $0$, but some of the derivatives blow up there, so I haven't been able to reconcile terms like $g''(0)$ which don't exist.
(The derivatives at $epsilon=0$ are functions of $v$. But, they are not integrable on $[0,epsilon]$)
Thanks.
EDIT: per request, here is one such $f$:
Note that we have $|v|<epsilon$ and $epsilon>0$ small.
EDIT #2: Would still like a reference for the theory of integral expansions if someone has one they like.
reference-request asymptotics taylor-expansion
asked Jan 21 at 21:49
MathIsArtMathIsArt
1278
$endgroup$
|
show 3 more comments
$begingroup$
I need to compute a series expansion in $epsilon$ (at least to second order) of integrals like
$$g(epsilon):=int_0^epsilon f(epsilon,v), dv,$$ as $epsilon to 0^+$, where the functions in question are very complicated. (Too complicated, in my opinion, to express as Taylor series themselves.)
Does anyone have a reference for such things? I have tried naively expanding around $0$, but some of the derivatives blow up there, so I haven't been able to reconcile terms like $g''(0)$ which don't exist.
(The derivatives at $epsilon=0$ are functions of $v$. But, they are not integrable on $[0,epsilon]$)
Thanks.
EDIT: per request, here is one such $f$:
Note that we have $|v|<epsilon$ and $epsilon>0$ small.
EDIT #2: Would still like a reference for the theory of integral expansions if someone has one they like.
reference-request asymptotics taylor-expansion
asked Jan 21 at 21:49
MathIsArtMathIsArt
1278
$endgroup$
$begingroup$
Remember that under some assumptions, $$g'(epsilon) = f(epsilon, epsilon) + int_0^{epsilon} frac{d}{d epsilon} f(epsilon, v) dv$$ So you could calculate $g'(0)$ yourself and do the same thing for the second derivative. Is this of any help?
$endgroup$
– Harnak
Jan 21 at 22:17
$begingroup$
Thanks for your comment. Yes, you are right, and that is what I have been doing so far. My issue is that the second derivative of $f$ wrt $epsilon$ goes to infinity as $epsilon to 0^+$, so I don't know if it's valid for me to write something like $$g(epsilon) = f(0,0)epsilon + (epsilon^2/2)(f_v(0,0) + 2f_epsilon(0,0)) + O(epsilon^3)$$
$endgroup$
– MathIsArt
Jan 21 at 22:38
$begingroup$
Then the desired expansion may not exist. Have you tried to numerically compute the derivatives?
$endgroup$
– marty cohen
Jan 21 at 22:53
$begingroup$
Can I see $f$'s expression?
$endgroup$
– Harnak
Jan 21 at 22:55
$begingroup$
Well, that's horrible lol. Anyway, I'll tell you what I was thinking: even if $f''$ goes to infinity, you may find a function g which is integrable in a right neighbourhood of 0 and such that $f(epsilon, v) leq g(v)$ for every $epsilon$ in that neighbourhood. In that case, $int_0^{epsilon} frac{partial^2}{partial epsilon^2} f(epsilon, v) , dv to 0$. I don't have time to try this right now.
$endgroup$
– Harnak
Jan 21 at 23:26
|
show 3 more comments
$begingroup$
I need to compute a series expansion in $epsilon$ (at least to second order) of integrals like
$$g(epsilon):=int_0^epsilon f(epsilon,v), dv,$$ as $epsilon to 0^+$, where the functions in question are very complicated. (Too complicated, in my opinion, to express as Taylor series themselves.)
Does anyone have a reference for such things? I have tried naively expanding around $0$, but some of the derivatives blow up there, so I haven't been able to reconcile terms like $g''(0)$ which don't exist.
(The derivatives at $epsilon=0$ are functions of $v$. But, they are not integrable on $[0,epsilon]$)
Thanks.
EDIT: per request, here is one such $f$:
Note that we have $|v|<epsilon$ and $epsilon>0$ small.
EDIT #2: Would still like a reference for the theory of integral expansions if someone has one they like.
reference-request asymptotics taylor-expansion
asked Jan 21 at 21:49
MathIsArtMathIsArt
1278
$endgroup$
I need to compute a series expansion in $epsilon$ (at least to second order) of integrals like
$$g(epsilon):=int_0^epsilon f(epsilon,v), dv,$$ as $epsilon to 0^+$, where the functions in question are very complicated. (Too complicated, in my opinion, to express as Taylor series themselves.)
Does anyone have a reference for such things? I have tried naively expanding around $0$, but some of the derivatives blow up there, so I haven't been able to reconcile terms like $g''(0)$ which don't exist.
(The derivatives at $epsilon=0$ are functions of $v$. But, they are not integrable on $[0,epsilon]$)
Thanks.
EDIT: per request, here is one such $f$:
Note that we have $|v|<epsilon$ and $epsilon>0$ small.
EDIT #2: Would still like a reference for the theory of integral expansions if someone has one they like.
reference-request asymptotics taylor-expansion
reference-request asymptotics taylor-expansion
asked Jan 21 at 21:49
MathIsArtMathIsArt
1278
asked Jan 21 at 21:49
MathIsArtMathIsArt
1278
edited Jan 22 at 0:09
MathIsArt
asked Jan 21 at 21:49
MathIsArtMathIsArt
1278
asked Jan 21 at 21:49
MathIsArtMathIsArt
1278
asked Jan 21 at 21:49
MathIsArtMathIsArt
1278
1278
$begingroup$
Remember that under some assumptions, $$g'(epsilon) = f(epsilon, epsilon) + int_0^{epsilon} frac{d}{d epsilon} f(epsilon, v) dv$$ So you could calculate $g'(0)$ yourself and do the same thing for the second derivative. Is this of any help?
$endgroup$
– Harnak
Jan 21 at 22:17
$begingroup$
Thanks for your comment. Yes, you are right, and that is what I have been doing so far. My issue is that the second derivative of $f$ wrt $epsilon$ goes to infinity as $epsilon to 0^+$, so I don't know if it's valid for me to write something like $$g(epsilon) = f(0,0)epsilon + (epsilon^2/2)(f_v(0,0) + 2f_epsilon(0,0)) + O(epsilon^3)$$
$endgroup$
– MathIsArt
Jan 21 at 22:38
$begingroup$
Then the desired expansion may not exist. Have you tried to numerically compute the derivatives?
$endgroup$
– marty cohen
Jan 21 at 22:53
$begingroup$
Can I see $f$'s expression?
$endgroup$
– Harnak
Jan 21 at 22:55
$begingroup$
Well, that's horrible lol. Anyway, I'll tell you what I was thinking: even if $f''$ goes to infinity, you may find a function g which is integrable in a right neighbourhood of 0 and such that $f(epsilon, v) leq g(v)$ for every $epsilon$ in that neighbourhood. In that case, $int_0^{epsilon} frac{partial^2}{partial epsilon^2} f(epsilon, v) , dv to 0$. I don't have time to try this right now.
$endgroup$
– Harnak
Jan 21 at 23:26
|
show 3 more comments
$begingroup$
Remember that under some assumptions, $$g'(epsilon) = f(epsilon, epsilon) + int_0^{epsilon} frac{d}{d epsilon} f(epsilon, v) dv$$ So you could calculate $g'(0)$ yourself and do the same thing for the second derivative. Is this of any help?
$endgroup$
– Harnak
Jan 21 at 22:17
$begingroup$
Thanks for your comment. Yes, you are right, and that is what I have been doing so far. My issue is that the second derivative of $f$ wrt $epsilon$ goes to infinity as $epsilon to 0^+$, so I don't know if it's valid for me to write something like $$g(epsilon) = f(0,0)epsilon + (epsilon^2/2)(f_v(0,0) + 2f_epsilon(0,0)) + O(epsilon^3)$$
$endgroup$
– MathIsArt
Jan 21 at 22:38
$begingroup$
Then the desired expansion may not exist. Have you tried to numerically compute the derivatives?
$endgroup$
– marty cohen
Jan 21 at 22:53
$begingroup$
Can I see $f$'s expression?
$endgroup$
– Harnak
Jan 21 at 22:55
$begingroup$
Well, that's horrible lol. Anyway, I'll tell you what I was thinking: even if $f''$ goes to infinity, you may find a function g which is integrable in a right neighbourhood of 0 and such that $f(epsilon, v) leq g(v)$ for every $epsilon$ in that neighbourhood. In that case, $int_0^{epsilon} frac{partial^2}{partial epsilon^2} f(epsilon, v) , dv to 0$. I don't have time to try this right now.
$endgroup$
– Harnak
Jan 21 at 23:26
$begingroup$
Remember that under some assumptions, $$g'(epsilon) = f(epsilon, epsilon) + int_0^{epsilon} frac{d}{d epsilon} f(epsilon, v) dv$$ So you could calculate $g'(0)$ yourself and do the same thing for the second derivative. Is this of any help?
$endgroup$
– Harnak
Jan 21 at 22:17
$begingroup$
Remember that under some assumptions, $$g'(epsilon) = f(epsilon, epsilon) + int_0^{epsilon} frac{d}{d epsilon} f(epsilon, v) dv$$ So you could calculate $g'(0)$ yourself and do the same thing for the second derivative. Is this of any help?
$endgroup$
– Harnak
Jan 21 at 22:17
$begingroup$
Thanks for your comment. Yes, you are right, and that is what I have been doing so far. My issue is that the second derivative of $f$ wrt $epsilon$ goes to infinity as $epsilon to 0^+$, so I don't know if it's valid for me to write something like $$g(epsilon) = f(0,0)epsilon + (epsilon^2/2)(f_v(0,0) + 2f_epsilon(0,0)) + O(epsilon^3)$$
$endgroup$
– MathIsArt
Jan 21 at 22:38
$begingroup$
Thanks for your comment. Yes, you are right, and that is what I have been doing so far. My issue is that the second derivative of $f$ wrt $epsilon$ goes to infinity as $epsilon to 0^+$, so I don't know if it's valid for me to write something like $$g(epsilon) = f(0,0)epsilon + (epsilon^2/2)(f_v(0,0) + 2f_epsilon(0,0)) + O(epsilon^3)$$
$endgroup$
– MathIsArt
Jan 21 at 22:38
$begingroup$
Then the desired expansion may not exist. Have you tried to numerically compute the derivatives?
$endgroup$
– marty cohen
Jan 21 at 22:53
$begingroup$
Then the desired expansion may not exist. Have you tried to numerically compute the derivatives?
$endgroup$
– marty cohen
Jan 21 at 22:53
$begingroup$
Can I see $f$'s expression?
$endgroup$
– Harnak
Jan 21 at 22:55
$begingroup$
Can I see $f$'s expression?
$endgroup$
– Harnak
Jan 21 at 22:55
$begingroup$
Well, that's horrible lol. Anyway, I'll tell you what I was thinking: even if $f''$ goes to infinity, you may find a function g which is integrable in a right neighbourhood of 0 and such that $f(epsilon, v) leq g(v)$ for every $epsilon$ in that neighbourhood. In that case, $int_0^{epsilon} frac{partial^2}{partial epsilon^2} f(epsilon, v) , dv to 0$. I don't have time to try this right now.
$endgroup$
– Harnak
Jan 21 at 23:26
$begingroup$
Well, that's horrible lol. Anyway, I'll tell you what I was thinking: even if $f''$ goes to infinity, you may find a function g which is integrable in a right neighbourhood of 0 and such that $f(epsilon, v) leq g(v)$ for every $epsilon$ in that neighbourhood. In that case, $int_0^{epsilon} frac{partial^2}{partial epsilon^2} f(epsilon, v) , dv to 0$. I don't have time to try this right now.
$endgroup$
– Harnak
Jan 21 at 23:26
|
show 3 more comments
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$begingroup$
Remember that under some assumptions, $$g'(epsilon) = f(epsilon, epsilon) + int_0^{epsilon} frac{d}{d epsilon} f(epsilon, v) dv$$ So you could calculate $g'(0)$ yourself and do the same thing for the second derivative. Is this of any help?
$endgroup$
– Harnak
Jan 21 at 22:17
$begingroup$
Thanks for your comment. Yes, you are right, and that is what I have been doing so far. My issue is that the second derivative of $f$ wrt $epsilon$ goes to infinity as $epsilon to 0^+$, so I don't know if it's valid for me to write something like $$g(epsilon) = f(0,0)epsilon + (epsilon^2/2)(f_v(0,0) + 2f_epsilon(0,0)) + O(epsilon^3)$$
$endgroup$
– MathIsArt
Jan 21 at 22:38
$begingroup$
Then the desired expansion may not exist. Have you tried to numerically compute the derivatives?
$endgroup$
– marty cohen
Jan 21 at 22:53
$begingroup$
Can I see $f$'s expression?
$endgroup$
– Harnak
Jan 21 at 22:55
$begingroup$
Well, that's horrible lol. Anyway, I'll tell you what I was thinking: even if $f''$ goes to infinity, you may find a function g which is integrable in a right neighbourhood of 0 and such that $f(epsilon, v) leq g(v)$ for every $epsilon$ in that neighbourhood. In that case, $int_0^{epsilon} frac{partial^2}{partial epsilon^2} f(epsilon, v) , dv to 0$. I don't have time to try this right now.
$endgroup$
– Harnak
Jan 21 at 23:26