Proving $1-exp(-4x^2/ pi) ge text{erf}(x)^2$
$begingroup$
This is a follow-up to this question. I came across a tighter potential bound and checked it numerically for $0le xle 5$. I think it holds for all positive $x$, can anyone see a proof?
$$1-exp(-4x^2/ pi) ge text{erf}(x)^2$$
Note: using analysis for previous question you can show that $1-exp(-k x^2)$ is an upper bound on $text{erf}(x)^2$ for $k=2$ and a lower bound when $k=1$. The factor $k=4/pi$ comes out when numerically searching for tightest upper bound. Not only does it seem to give an upper bound, but it also tracks $text{erf}(x)^2$ very closely.
Dashed graph below is $text{erf}(x)^2$, red is $1-exp(-k x^2)$ for $k=4/pi$, other two graphs are for $k=1$ and $k=2$
inequality
edited Dec 27 '18 at 21:49
Glorfindel
3,41381930
asked Oct 15 '10 at 22:00
Yaroslav BulatovYaroslav Bulatov
1,88411526
$endgroup$
add a comment |
$begingroup$
This is a follow-up to this question. I came across a tighter potential bound and checked it numerically for $0le xle 5$. I think it holds for all positive $x$, can anyone see a proof?
$$1-exp(-4x^2/ pi) ge text{erf}(x)^2$$
Note: using analysis for previous question you can show that $1-exp(-k x^2)$ is an upper bound on $text{erf}(x)^2$ for $k=2$ and a lower bound when $k=1$. The factor $k=4/pi$ comes out when numerically searching for tightest upper bound. Not only does it seem to give an upper bound, but it also tracks $text{erf}(x)^2$ very closely.
Dashed graph below is $text{erf}(x)^2$, red is $1-exp(-k x^2)$ for $k=4/pi$, other two graphs are for $k=1$ and $k=2$
inequality
edited Dec 27 '18 at 21:49
Glorfindel
3,41381930
asked Oct 15 '10 at 22:00
Yaroslav BulatovYaroslav Bulatov
1,88411526
$endgroup$
1
$begingroup$
The inequality is correct and I think it can be proved by some differentiation and Taylor series manipulations. The Taylor series expansion about$x=0$also tells us why$k=4/pi$is the best possible. But I don't know if there is a simple proof of the inequality. Would love to see one.
$endgroup$
– Dinesh
Oct 16 '10 at 0:30
add a comment |
$begingroup$
This is a follow-up to this question. I came across a tighter potential bound and checked it numerically for $0le xle 5$. I think it holds for all positive $x$, can anyone see a proof?
$$1-exp(-4x^2/ pi) ge text{erf}(x)^2$$
Note: using analysis for previous question you can show that $1-exp(-k x^2)$ is an upper bound on $text{erf}(x)^2$ for $k=2$ and a lower bound when $k=1$. The factor $k=4/pi$ comes out when numerically searching for tightest upper bound. Not only does it seem to give an upper bound, but it also tracks $text{erf}(x)^2$ very closely.
Dashed graph below is $text{erf}(x)^2$, red is $1-exp(-k x^2)$ for $k=4/pi$, other two graphs are for $k=1$ and $k=2$
inequality
edited Dec 27 '18 at 21:49
Glorfindel
3,41381930
asked Oct 15 '10 at 22:00
Yaroslav BulatovYaroslav Bulatov
1,88411526
$endgroup$
This is a follow-up to this question. I came across a tighter potential bound and checked it numerically for $0le xle 5$. I think it holds for all positive $x$, can anyone see a proof?
$$1-exp(-4x^2/ pi) ge text{erf}(x)^2$$
Note: using analysis for previous question you can show that $1-exp(-k x^2)$ is an upper bound on $text{erf}(x)^2$ for $k=2$ and a lower bound when $k=1$. The factor $k=4/pi$ comes out when numerically searching for tightest upper bound. Not only does it seem to give an upper bound, but it also tracks $text{erf}(x)^2$ very closely.
Dashed graph below is $text{erf}(x)^2$, red is $1-exp(-k x^2)$ for $k=4/pi$, other two graphs are for $k=1$ and $k=2$
inequality
inequality
edited Dec 27 '18 at 21:49
Glorfindel
3,41381930
asked Oct 15 '10 at 22:00
Yaroslav BulatovYaroslav Bulatov
1,88411526
edited Dec 27 '18 at 21:49
Glorfindel
3,41381930
asked Oct 15 '10 at 22:00
Yaroslav BulatovYaroslav Bulatov
1,88411526
edited Dec 27 '18 at 21:49
Glorfindel
3,41381930
edited Dec 27 '18 at 21:49
Glorfindel
3,41381930
edited Dec 27 '18 at 21:49
Glorfindel
3,41381930
3,41381930
asked Oct 15 '10 at 22:00
Yaroslav BulatovYaroslav Bulatov
1,88411526
asked Oct 15 '10 at 22:00
Yaroslav BulatovYaroslav Bulatov
1,88411526
asked Oct 15 '10 at 22:00
Yaroslav BulatovYaroslav Bulatov
1,88411526
1,88411526
1
$begingroup$
The inequality is correct and I think it can be proved by some differentiation and Taylor series manipulations. The Taylor series expansion about$x=0$also tells us why$k=4/pi$is the best possible. But I don't know if there is a simple proof of the inequality. Would love to see one.
$endgroup$
– Dinesh
Oct 16 '10 at 0:30
add a comment |
1
$begingroup$
The inequality is correct and I think it can be proved by some differentiation and Taylor series manipulations. The Taylor series expansion about$x=0$also tells us why$k=4/pi$is the best possible. But I don't know if there is a simple proof of the inequality. Would love to see one.
$endgroup$
– Dinesh
Oct 16 '10 at 0:30
1
1
$begingroup$
The inequality is correct and I think it can be proved by some differentiation and Taylor series manipulations. The Taylor series expansion about
$x=0$ also tells us why $k=4/pi$ is the best possible. But I don't know if there is a simple proof of the inequality. Would love to see one.$endgroup$
– Dinesh
Oct 16 '10 at 0:30
$begingroup$
The inequality is correct and I think it can be proved by some differentiation and Taylor series manipulations. The Taylor series expansion about
$x=0$ also tells us why $k=4/pi$ is the best possible. But I don't know if there is a simple proof of the inequality. Would love to see one.$endgroup$
– Dinesh
Oct 16 '10 at 0:30
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
As before we consider
$$erf(x)^2={4over pi}int_0^xint_0^x exp{-(s^2+t^2)} ds dt$$
Now compare this with the same over the area which is given by the quarter of a circle of radius $displaystyle frac{2x}{sqrt{pi}}$. The area of this is same as the area of the square of side $x$.
Since $displaystyle e^{-(s^2 + t^2)}$ decreases as $displaystyle s^2 + t^2$ increases, we are done!
The integral over the non-common area for the quarter circle is greater than the integral over the non-common area of the square (which lies outside the circle and so $displaystyle s^2 + t^2$ is higher in that region).
answered Oct 16 '10 at 2:20
AryabhataAryabhata
70.3k6157247
$endgroup$
3
$begingroup$
Yep. This is another illustration of the rearrangement inequality: for an arbitrary radial function $f(x) = f(|x|) geq 0$, such that $partial_r f leq 0$, we can ask: of all the sets $Omega$ of area 1, when is the integral $int_Omega f(x) dx$ the greatest? The answer would be the ball centered at the origin with total area 1.
$endgroup$
– Willie Wong
Oct 16 '10 at 2:26
$begingroup$
Thanks for pointing out the rearrangement inequality, it seems to make the proof very easy!
$endgroup$
– Yaroslav Bulatov
Oct 16 '10 at 3:32
$begingroup$
Ah!!! This is such a lovely proof.
$endgroup$
– Dinesh
Oct 16 '10 at 3:43
$begingroup$
@Willie: Rearrangement Inequality? I thought that was something completely different. Wikipedia seems to agree: en.wikipedia.org/wiki/Rearrangement_inequality. Can you please provide a reference which uses this name for the problem you state?
$endgroup$
– Aryabhata
Oct 16 '10 at 15:03
$begingroup$
@Moron: Beautifully done! I will delete my answer.
$endgroup$
– user940
Oct 16 '10 at 15:31
|
show 4 more comments
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$begingroup$
As before we consider
$$erf(x)^2={4over pi}int_0^xint_0^x exp{-(s^2+t^2)} ds dt$$
Now compare this with the same over the area which is given by the quarter of a circle of radius $displaystyle frac{2x}{sqrt{pi}}$. The area of this is same as the area of the square of side $x$.
Since $displaystyle e^{-(s^2 + t^2)}$ decreases as $displaystyle s^2 + t^2$ increases, we are done!
The integral over the non-common area for the quarter circle is greater than the integral over the non-common area of the square (which lies outside the circle and so $displaystyle s^2 + t^2$ is higher in that region).
answered Oct 16 '10 at 2:20
AryabhataAryabhata
70.3k6157247
$endgroup$
3
$begingroup$
Yep. This is another illustration of the rearrangement inequality: for an arbitrary radial function $f(x) = f(|x|) geq 0$, such that $partial_r f leq 0$, we can ask: of all the sets $Omega$ of area 1, when is the integral $int_Omega f(x) dx$ the greatest? The answer would be the ball centered at the origin with total area 1.
$endgroup$
– Willie Wong
Oct 16 '10 at 2:26
$begingroup$
Thanks for pointing out the rearrangement inequality, it seems to make the proof very easy!
$endgroup$
– Yaroslav Bulatov
Oct 16 '10 at 3:32
$begingroup$
Ah!!! This is such a lovely proof.
$endgroup$
– Dinesh
Oct 16 '10 at 3:43
$begingroup$
@Willie: Rearrangement Inequality? I thought that was something completely different. Wikipedia seems to agree: en.wikipedia.org/wiki/Rearrangement_inequality. Can you please provide a reference which uses this name for the problem you state?
$endgroup$
– Aryabhata
Oct 16 '10 at 15:03
$begingroup$
@Moron: Beautifully done! I will delete my answer.
$endgroup$
– user940
Oct 16 '10 at 15:31
|
show 4 more comments
$begingroup$
As before we consider
$$erf(x)^2={4over pi}int_0^xint_0^x exp{-(s^2+t^2)} ds dt$$
Now compare this with the same over the area which is given by the quarter of a circle of radius $displaystyle frac{2x}{sqrt{pi}}$. The area of this is same as the area of the square of side $x$.
Since $displaystyle e^{-(s^2 + t^2)}$ decreases as $displaystyle s^2 + t^2$ increases, we are done!
The integral over the non-common area for the quarter circle is greater than the integral over the non-common area of the square (which lies outside the circle and so $displaystyle s^2 + t^2$ is higher in that region).
answered Oct 16 '10 at 2:20
AryabhataAryabhata
70.3k6157247
$endgroup$
3
$begingroup$
Yep. This is another illustration of the rearrangement inequality: for an arbitrary radial function $f(x) = f(|x|) geq 0$, such that $partial_r f leq 0$, we can ask: of all the sets $Omega$ of area 1, when is the integral $int_Omega f(x) dx$ the greatest? The answer would be the ball centered at the origin with total area 1.
$endgroup$
– Willie Wong
Oct 16 '10 at 2:26
$begingroup$
Thanks for pointing out the rearrangement inequality, it seems to make the proof very easy!
$endgroup$
– Yaroslav Bulatov
Oct 16 '10 at 3:32
$begingroup$
Ah!!! This is such a lovely proof.
$endgroup$
– Dinesh
Oct 16 '10 at 3:43
$begingroup$
@Willie: Rearrangement Inequality? I thought that was something completely different. Wikipedia seems to agree: en.wikipedia.org/wiki/Rearrangement_inequality. Can you please provide a reference which uses this name for the problem you state?
$endgroup$
– Aryabhata
Oct 16 '10 at 15:03
$begingroup$
@Moron: Beautifully done! I will delete my answer.
$endgroup$
– user940
Oct 16 '10 at 15:31
|
show 4 more comments
$begingroup$
As before we consider
$$erf(x)^2={4over pi}int_0^xint_0^x exp{-(s^2+t^2)} ds dt$$
Now compare this with the same over the area which is given by the quarter of a circle of radius $displaystyle frac{2x}{sqrt{pi}}$. The area of this is same as the area of the square of side $x$.
Since $displaystyle e^{-(s^2 + t^2)}$ decreases as $displaystyle s^2 + t^2$ increases, we are done!
The integral over the non-common area for the quarter circle is greater than the integral over the non-common area of the square (which lies outside the circle and so $displaystyle s^2 + t^2$ is higher in that region).
answered Oct 16 '10 at 2:20
AryabhataAryabhata
70.3k6157247
$endgroup$
As before we consider
$$erf(x)^2={4over pi}int_0^xint_0^x exp{-(s^2+t^2)} ds dt$$
Now compare this with the same over the area which is given by the quarter of a circle of radius $displaystyle frac{2x}{sqrt{pi}}$. The area of this is same as the area of the square of side $x$.
Since $displaystyle e^{-(s^2 + t^2)}$ decreases as $displaystyle s^2 + t^2$ increases, we are done!
The integral over the non-common area for the quarter circle is greater than the integral over the non-common area of the square (which lies outside the circle and so $displaystyle s^2 + t^2$ is higher in that region).
answered Oct 16 '10 at 2:20
AryabhataAryabhata
70.3k6157247
edited Oct 16 '10 at 2:26
answered Oct 16 '10 at 2:20
AryabhataAryabhata
70.3k6157247
answered Oct 16 '10 at 2:20
AryabhataAryabhata
70.3k6157247
answered Oct 16 '10 at 2:20
AryabhataAryabhata
70.3k6157247
70.3k6157247
3
$begingroup$
Yep. This is another illustration of the rearrangement inequality: for an arbitrary radial function $f(x) = f(|x|) geq 0$, such that $partial_r f leq 0$, we can ask: of all the sets $Omega$ of area 1, when is the integral $int_Omega f(x) dx$ the greatest? The answer would be the ball centered at the origin with total area 1.
$endgroup$
– Willie Wong
Oct 16 '10 at 2:26
$begingroup$
Thanks for pointing out the rearrangement inequality, it seems to make the proof very easy!
$endgroup$
– Yaroslav Bulatov
Oct 16 '10 at 3:32
$begingroup$
Ah!!! This is such a lovely proof.
$endgroup$
– Dinesh
Oct 16 '10 at 3:43
$begingroup$
@Willie: Rearrangement Inequality? I thought that was something completely different. Wikipedia seems to agree: en.wikipedia.org/wiki/Rearrangement_inequality. Can you please provide a reference which uses this name for the problem you state?
$endgroup$
– Aryabhata
Oct 16 '10 at 15:03
$begingroup$
@Moron: Beautifully done! I will delete my answer.
$endgroup$
– user940
Oct 16 '10 at 15:31
|
show 4 more comments
3
$begingroup$
Yep. This is another illustration of the rearrangement inequality: for an arbitrary radial function $f(x) = f(|x|) geq 0$, such that $partial_r f leq 0$, we can ask: of all the sets $Omega$ of area 1, when is the integral $int_Omega f(x) dx$ the greatest? The answer would be the ball centered at the origin with total area 1.
$endgroup$
– Willie Wong
Oct 16 '10 at 2:26
$begingroup$
Thanks for pointing out the rearrangement inequality, it seems to make the proof very easy!
$endgroup$
– Yaroslav Bulatov
Oct 16 '10 at 3:32
$begingroup$
Ah!!! This is such a lovely proof.
$endgroup$
– Dinesh
Oct 16 '10 at 3:43
$begingroup$
@Willie: Rearrangement Inequality? I thought that was something completely different. Wikipedia seems to agree: en.wikipedia.org/wiki/Rearrangement_inequality. Can you please provide a reference which uses this name for the problem you state?
$endgroup$
– Aryabhata
Oct 16 '10 at 15:03
$begingroup$
@Moron: Beautifully done! I will delete my answer.
$endgroup$
– user940
Oct 16 '10 at 15:31
3
3
$begingroup$
Yep. This is another illustration of the rearrangement inequality: for an arbitrary radial function $f(x) = f(|x|) geq 0$, such that $partial_r f leq 0$, we can ask: of all the sets $Omega$ of area 1, when is the integral $int_Omega f(x) dx$ the greatest? The answer would be the ball centered at the origin with total area 1.
$endgroup$
– Willie Wong
Oct 16 '10 at 2:26
$begingroup$
Yep. This is another illustration of the rearrangement inequality: for an arbitrary radial function $f(x) = f(|x|) geq 0$, such that $partial_r f leq 0$, we can ask: of all the sets $Omega$ of area 1, when is the integral $int_Omega f(x) dx$ the greatest? The answer would be the ball centered at the origin with total area 1.
$endgroup$
– Willie Wong
Oct 16 '10 at 2:26
$begingroup$
Thanks for pointing out the rearrangement inequality, it seems to make the proof very easy!
$endgroup$
– Yaroslav Bulatov
Oct 16 '10 at 3:32
$begingroup$
Thanks for pointing out the rearrangement inequality, it seems to make the proof very easy!
$endgroup$
– Yaroslav Bulatov
Oct 16 '10 at 3:32
$begingroup$
Ah!!! This is such a lovely proof.
$endgroup$
– Dinesh
Oct 16 '10 at 3:43
$begingroup$
Ah!!! This is such a lovely proof.
$endgroup$
– Dinesh
Oct 16 '10 at 3:43
$begingroup$
@Willie: Rearrangement Inequality? I thought that was something completely different. Wikipedia seems to agree: en.wikipedia.org/wiki/Rearrangement_inequality. Can you please provide a reference which uses this name for the problem you state?
$endgroup$
– Aryabhata
Oct 16 '10 at 15:03
$begingroup$
@Willie: Rearrangement Inequality? I thought that was something completely different. Wikipedia seems to agree: en.wikipedia.org/wiki/Rearrangement_inequality. Can you please provide a reference which uses this name for the problem you state?
$endgroup$
– Aryabhata
Oct 16 '10 at 15:03
$begingroup$
@Moron: Beautifully done! I will delete my answer.
$endgroup$
– user940
Oct 16 '10 at 15:31
$begingroup$
@Moron: Beautifully done! I will delete my answer.
$endgroup$
– user940
Oct 16 '10 at 15:31
|
show 4 more comments
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$begingroup$
The inequality is correct and I think it can be proved by some differentiation and Taylor series manipulations. The Taylor series expansion about
$x=0$also tells us why$k=4/pi$is the best possible. But I don't know if there is a simple proof of the inequality. Would love to see one.$endgroup$
– Dinesh
Oct 16 '10 at 0:30