Oscillatory Cancellation of Integral
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Under which conditions for a real function $f(x)$ is
$$
left| int_1^x left fleft(fracxwright) - fleft(leftlfloor fracxw rightrfloorright) right cosleft(wright) , rm dw right|
$$
finite, as $xrightarrow infty$.
The point here is that the $cos $ function is present for cancellation of most parts. I was thinking to expand
$$
fleft(fracxwright) = fleft(leftlfloorfracxwrightrfloorright) + f'left(leftlfloorfracxwrightrfloorright) underbraceleft(fracxw - leftlfloorfracxwrightrfloor right)_cal O(1) + cal Oleft( left(fracxw - leftlfloorfracxwrightrfloor right)^2 right)
$$
so that $|f'(x)|$ must at least be bounded as $xrightarrow infty$.
But is it true that for any $cal O(1)$ function $g(x)$
$$
left| int_1^x g(w) cosleft(wright) , rm dw right|
$$
is finite as $xrightarrow infty$? I mean it is obviously not true if I choose $g(x)=cos(x)$, so there must be some restrictions as in $cos(x)$ and $g(x)$ are not correlated.
I hope it is clear what I want. I was trying to make it somewhat rigorous.
real-analysis oscillatory-integral
asked Sep 10 at 21:08
Diger
757310
add a comment |Â
up vote
0
down vote
favorite
Under which conditions for a real function $f(x)$ is
$$
left| int_1^x left fleft(fracxwright) - fleft(leftlfloor fracxw rightrfloorright) right cosleft(wright) , rm dw right|
$$
finite, as $xrightarrow infty$.
The point here is that the $cos $ function is present for cancellation of most parts. I was thinking to expand
$$
fleft(fracxwright) = fleft(leftlfloorfracxwrightrfloorright) + f'left(leftlfloorfracxwrightrfloorright) underbraceleft(fracxw - leftlfloorfracxwrightrfloor right)_cal O(1) + cal Oleft( left(fracxw - leftlfloorfracxwrightrfloor right)^2 right)
$$
so that $|f'(x)|$ must at least be bounded as $xrightarrow infty$.
But is it true that for any $cal O(1)$ function $g(x)$
$$
left| int_1^x g(w) cosleft(wright) , rm dw right|
$$
is finite as $xrightarrow infty$? I mean it is obviously not true if I choose $g(x)=cos(x)$, so there must be some restrictions as in $cos(x)$ and $g(x)$ are not correlated.
I hope it is clear what I want. I was trying to make it somewhat rigorous.
real-analysis oscillatory-integral
asked Sep 10 at 21:08
Diger
757310
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Under which conditions for a real function $f(x)$ is
$$
left| int_1^x left fleft(fracxwright) - fleft(leftlfloor fracxw rightrfloorright) right cosleft(wright) , rm dw right|
$$
finite, as $xrightarrow infty$.
The point here is that the $cos $ function is present for cancellation of most parts. I was thinking to expand
$$
fleft(fracxwright) = fleft(leftlfloorfracxwrightrfloorright) + f'left(leftlfloorfracxwrightrfloorright) underbraceleft(fracxw - leftlfloorfracxwrightrfloor right)_cal O(1) + cal Oleft( left(fracxw - leftlfloorfracxwrightrfloor right)^2 right)
$$
so that $|f'(x)|$ must at least be bounded as $xrightarrow infty$.
But is it true that for any $cal O(1)$ function $g(x)$
$$
left| int_1^x g(w) cosleft(wright) , rm dw right|
$$
is finite as $xrightarrow infty$? I mean it is obviously not true if I choose $g(x)=cos(x)$, so there must be some restrictions as in $cos(x)$ and $g(x)$ are not correlated.
I hope it is clear what I want. I was trying to make it somewhat rigorous.
real-analysis oscillatory-integral
asked Sep 10 at 21:08
Diger
757310
Under which conditions for a real function $f(x)$ is
$$
left| int_1^x left fleft(fracxwright) - fleft(leftlfloor fracxw rightrfloorright) right cosleft(wright) , rm dw right|
$$
finite, as $xrightarrow infty$.
The point here is that the $cos $ function is present for cancellation of most parts. I was thinking to expand
$$
fleft(fracxwright) = fleft(leftlfloorfracxwrightrfloorright) + f'left(leftlfloorfracxwrightrfloorright) underbraceleft(fracxw - leftlfloorfracxwrightrfloor right)_cal O(1) + cal Oleft( left(fracxw - leftlfloorfracxwrightrfloor right)^2 right)
$$
so that $|f'(x)|$ must at least be bounded as $xrightarrow infty$.
But is it true that for any $cal O(1)$ function $g(x)$
$$
left| int_1^x g(w) cosleft(wright) , rm dw right|
$$
is finite as $xrightarrow infty$? I mean it is obviously not true if I choose $g(x)=cos(x)$, so there must be some restrictions as in $cos(x)$ and $g(x)$ are not correlated.
I hope it is clear what I want. I was trying to make it somewhat rigorous.
real-analysis oscillatory-integral
real-analysis oscillatory-integral
asked Sep 10 at 21:08
Diger
757310
asked Sep 10 at 21:08
Diger
757310
edited Sep 10 at 21:20
asked Sep 10 at 21:08
Diger
757310
asked Sep 10 at 21:08
Diger
757310
asked Sep 10 at 21:08
Diger
757310
757310
add a comment |Â
add a comment |Â
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