Proving $1-exp(-4x^2/ pi) ge text{erf}(x)^2$












7












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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


















7












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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
















7












7








7


1



$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 27 '18 at 21:49









Glorfindel

3,41381930




3,41381930










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
















  • 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












1 Answer
1






active

oldest

votes


















10












$begingroup$

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).






share|cite|improve this answer











$endgroup$









  • 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












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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









10












$begingroup$

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).






share|cite|improve this answer











$endgroup$









  • 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
















10












$begingroup$

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).






share|cite|improve this answer











$endgroup$









  • 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














10












10








10





$begingroup$

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).






share|cite|improve this answer











$endgroup$



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).







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Oct 16 '10 at 2:26

























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














  • 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


















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