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Simplify or upper bound this summation
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I am not a mathematician, so sorry for this trivial question. Is there a way to simplify or to upper bound the following summation: $$sum_i=1^n e^frac-i^2sigma^2 $$
Can I apply the geometric series?
EDIT: I have a difficulty because of the power $2$, i.e if the summation is in this way $sum_i=1^n e^frac-isigma^2 $ then it is easy to apply the geometric series!!
sequences-and-series geometric-series
asked 3 hours ago
user8003788
213
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up vote
4
down vote
favorite
I am not a mathematician, so sorry for this trivial question. Is there a way to simplify or to upper bound the following summation: $$sum_i=1^n e^frac-i^2sigma^2 $$
Can I apply the geometric series?
EDIT: I have a difficulty because of the power $2$, i.e if the summation is in this way $sum_i=1^n e^frac-isigma^2 $ then it is easy to apply the geometric series!!
sequences-and-series geometric-series
asked 3 hours ago
user8003788
213
1
Don't be sorry for the question! The only way to get better as mathematician is to ask questions. (See also the first annotation to this post)
– Jacob Manaker
3 hours ago
1
Seems that you could try the integral $int_0^infty exp(-x^2),mathrm dx$ to bound that.
– xbh
3 hours ago
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
I am not a mathematician, so sorry for this trivial question. Is there a way to simplify or to upper bound the following summation: $$sum_i=1^n e^frac-i^2sigma^2 $$
Can I apply the geometric series?
EDIT: I have a difficulty because of the power $2$, i.e if the summation is in this way $sum_i=1^n e^frac-isigma^2 $ then it is easy to apply the geometric series!!
sequences-and-series geometric-series
asked 3 hours ago
user8003788
213
I am not a mathematician, so sorry for this trivial question. Is there a way to simplify or to upper bound the following summation: $$sum_i=1^n e^frac-i^2sigma^2 $$
Can I apply the geometric series?
EDIT: I have a difficulty because of the power $2$, i.e if the summation is in this way $sum_i=1^n e^frac-isigma^2 $ then it is easy to apply the geometric series!!
sequences-and-series geometric-series
sequences-and-series geometric-series
asked 3 hours ago
user8003788
213
asked 3 hours ago
user8003788
213
edited 3 hours ago
asked 3 hours ago
user8003788
213
asked 3 hours ago
user8003788
213
asked 3 hours ago
user8003788
213
213
1
Don't be sorry for the question! The only way to get better as mathematician is to ask questions. (See also the first annotation to this post)
– Jacob Manaker
3 hours ago
1
Seems that you could try the integral $int_0^infty exp(-x^2),mathrm dx$ to bound that.
– xbh
3 hours ago
add a comment |Â
1
Don't be sorry for the question! The only way to get better as mathematician is to ask questions. (See also the first annotation to this post)
– Jacob Manaker
3 hours ago
1
Seems that you could try the integral $int_0^infty exp(-x^2),mathrm dx$ to bound that.
– xbh
3 hours ago
1
1
Don't be sorry for the question! The only way to get better as mathematician is to ask questions. (See also the first annotation to this post)
– Jacob Manaker
3 hours ago
Don't be sorry for the question! The only way to get better as mathematician is to ask questions. (See also the first annotation to this post)
– Jacob Manaker
3 hours ago
1
1
Seems that you could try the integral $int_0^infty exp(-x^2),mathrm dx$ to bound that.
– xbh
3 hours ago
Seems that you could try the integral $int_0^infty exp(-x^2),mathrm dx$ to bound that.
– xbh
3 hours ago
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
3
down vote
Alternative:
Since $f(x) = exp(-x^2/sigma^2) searrow 0$, we can write
$
DeclareMathOperatordiff,d!
$
beginalign*
&sum_1^n expleft(-frac j^2sigma^2right) \
&= sum_1^n int_j-1^j expleft(-frac j^2sigma^2right)diff x \
&leqslant sum_1^n int_j-1^j expleft(-frac x^2sigma^2right) diff x \
&=sigma int_0^n expleft(-frac x^2sigma^2right) diff left(frac x sigma right)\
&= sigma int_0^n/sigma exp(-x^2)diff x\
&leqslant sigma int_0^+inftyexp(-x^2)diff x\
&= frac sigma 2 sqrt pi
endalign*
answered 2 hours ago
xbh
3,575320
1
Nicely done! Sometimes simplest is best.
– Jacob Manaker
1 hour ago
add a comment |Â
up vote
3
down vote
TL;DR: three relatively easy bounds are the numbered equations below.
You cannot directly apply the formula for the geometric series for the reason mentioned in your edit. But note that $igeq1$, so we have $$sum_i=1^nexpleft(-fraci^2sigma^2right)leqsum_i=1^nexpleft(-fracicdot1sigma^2right)$$ The latter, of course, is a geometric sum. Taking the sum over all $i$ (including $i=0$), we get $$(1-e^-sigma^-2)^-1 tag1 labeleqn:first$$ The calculation for finitely many terms isn't much harder, and only differs by an exponentially decreasing factor.
If this isn't a strong enough bound, there are other techniques. If $n<sigma$, then we can get very far elementarily. Note that $e^xgeq x+1$; dividing each side, we get $$e^-xleq(1+x)^-1=sum_k=0^infty(-x)^k$$ if $|x|<1$. Taking $x=left(fracisigmaright)^2$, we thus obtain beginalign*
sum_i=1^ne^-fraci^2sigma^2&leqsum_i=1^nsum_k=0^inftyleft(-left(fracisigmaright)^2right)^k \
&=sum_k=0^infty(-1)^ksum_i=1^nleft(fracisigmaright)^2k tag* labeleqn:star
endalign*
(We can interchange sums because one is finite.) Now, for all $k$, the function $left(fraccdotsigmaright)^2k$ is increasing on $[0,infty)$; we thus have $$int_0^nleft(fracisigmaright)^2k,dileqsum_i=1^nleft(fracisigmaright)^2kleqleft(fracnsigmaright)^2k+int_1^nleft(fracisigmaright)^2k,di$$ Evaluating the integrals and simplifying, we have $$0leqsum_i=1^nleft(fracisigmaright)^2k-fracn2k+1left(fracnsigmaright)^2kleqleft(fracnsigmaright)^2kleft(1-frac1(2k+1)n^2kright)$$
Substituting into $eqrefeqn:star$, we get beginalign*
sum_i=1^ne^-fraci^2sigma^2&leqsum_k=0^inftyfrac(-1)^kn2k+1left(fracnsigmaright)^2k-sum_j=0^inftyleft(fracnsigmaright)^4j+2left(1-frac1(4j+3)n^4j+2right) \
&leqsum_k=0^inftyfrac(-1)^kn2k+1left(fracnsigmaright)^2k-sum_j=0^inftyleft(fracnsigmaright)^4j+2 \
&=sigmatan^-1left(fracnsigmaright)-fracleft(fracnsigmaright)^21-left(fracnsigmaright)^4hspace4em(n<sigma) tag2
endalign*
Finally, for the general case we can achieve a slight improvement on $eqrefeqn:first$ via the theory of majorization. $x_i_i=1^nmapstosum_i=1^nexpleft(-fracx_isigma^2right)$ is convex and symmetric in its arguments, hence Schur-convex. Let $b_i=i^2$ and $a_i=left(frac2n-13right)i$. Clearly, for all $mleq n$, we have $$sum_i=1^ma_i=fracm(m-1)2cdotfrac2n-13geqfracm(m-1)(2m-1)6=sum_i=1^mb_i$$ with equality if $m=n$. Thus $veca$ majorizes $vecb$, so beginalign*
sum_i=1^nexpleft(-fraci^2sigma^2right)&=sum_i=1^nexpleft(-fracb_isigma^2right) \
&leqsum_i=1^nexpleft(-fraca_isigma^2right) \
&=sum_i=1^nexpleft(-frac(2n-1)i3sigma^2right) \
&leqsum_i=0^inftyexpleft(-frac(2n-1)i3sigma^2right) \
&leqleft(1-expleft(frac2n-13sigma^2right)right)^-1 tag3
endalign*
answered 3 hours ago
Jacob Manaker
982413
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
Alternative:
Since $f(x) = exp(-x^2/sigma^2) searrow 0$, we can write
$
DeclareMathOperatordiff,d!
$
beginalign*
&sum_1^n expleft(-frac j^2sigma^2right) \
&= sum_1^n int_j-1^j expleft(-frac j^2sigma^2right)diff x \
&leqslant sum_1^n int_j-1^j expleft(-frac x^2sigma^2right) diff x \
&=sigma int_0^n expleft(-frac x^2sigma^2right) diff left(frac x sigma right)\
&= sigma int_0^n/sigma exp(-x^2)diff x\
&leqslant sigma int_0^+inftyexp(-x^2)diff x\
&= frac sigma 2 sqrt pi
endalign*
answered 2 hours ago
xbh
3,575320
1
Nicely done! Sometimes simplest is best.
– Jacob Manaker
1 hour ago
add a comment |Â
up vote
3
down vote
Alternative:
Since $f(x) = exp(-x^2/sigma^2) searrow 0$, we can write
$
DeclareMathOperatordiff,d!
$
beginalign*
&sum_1^n expleft(-frac j^2sigma^2right) \
&= sum_1^n int_j-1^j expleft(-frac j^2sigma^2right)diff x \
&leqslant sum_1^n int_j-1^j expleft(-frac x^2sigma^2right) diff x \
&=sigma int_0^n expleft(-frac x^2sigma^2right) diff left(frac x sigma right)\
&= sigma int_0^n/sigma exp(-x^2)diff x\
&leqslant sigma int_0^+inftyexp(-x^2)diff x\
&= frac sigma 2 sqrt pi
endalign*
answered 2 hours ago
xbh
3,575320
1
Nicely done! Sometimes simplest is best.
– Jacob Manaker
1 hour ago
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Alternative:
Since $f(x) = exp(-x^2/sigma^2) searrow 0$, we can write
$
DeclareMathOperatordiff,d!
$
beginalign*
&sum_1^n expleft(-frac j^2sigma^2right) \
&= sum_1^n int_j-1^j expleft(-frac j^2sigma^2right)diff x \
&leqslant sum_1^n int_j-1^j expleft(-frac x^2sigma^2right) diff x \
&=sigma int_0^n expleft(-frac x^2sigma^2right) diff left(frac x sigma right)\
&= sigma int_0^n/sigma exp(-x^2)diff x\
&leqslant sigma int_0^+inftyexp(-x^2)diff x\
&= frac sigma 2 sqrt pi
endalign*
answered 2 hours ago
xbh
3,575320
Alternative:
Since $f(x) = exp(-x^2/sigma^2) searrow 0$, we can write
$
DeclareMathOperatordiff,d!
$
beginalign*
&sum_1^n expleft(-frac j^2sigma^2right) \
&= sum_1^n int_j-1^j expleft(-frac j^2sigma^2right)diff x \
&leqslant sum_1^n int_j-1^j expleft(-frac x^2sigma^2right) diff x \
&=sigma int_0^n expleft(-frac x^2sigma^2right) diff left(frac x sigma right)\
&= sigma int_0^n/sigma exp(-x^2)diff x\
&leqslant sigma int_0^+inftyexp(-x^2)diff x\
&= frac sigma 2 sqrt pi
endalign*
answered 2 hours ago
xbh
3,575320
answered 2 hours ago
xbh
3,575320
answered 2 hours ago
xbh
3,575320
answered 2 hours ago
xbh
3,575320
3,575320
1
Nicely done! Sometimes simplest is best.
– Jacob Manaker
1 hour ago
add a comment |Â
1
Nicely done! Sometimes simplest is best.
– Jacob Manaker
1 hour ago
1
1
Nicely done! Sometimes simplest is best.
– Jacob Manaker
1 hour ago
Nicely done! Sometimes simplest is best.
– Jacob Manaker
1 hour ago
add a comment |Â
up vote
3
down vote
TL;DR: three relatively easy bounds are the numbered equations below.
You cannot directly apply the formula for the geometric series for the reason mentioned in your edit. But note that $igeq1$, so we have $$sum_i=1^nexpleft(-fraci^2sigma^2right)leqsum_i=1^nexpleft(-fracicdot1sigma^2right)$$ The latter, of course, is a geometric sum. Taking the sum over all $i$ (including $i=0$), we get $$(1-e^-sigma^-2)^-1 tag1 labeleqn:first$$ The calculation for finitely many terms isn't much harder, and only differs by an exponentially decreasing factor.
If this isn't a strong enough bound, there are other techniques. If $n<sigma$, then we can get very far elementarily. Note that $e^xgeq x+1$; dividing each side, we get $$e^-xleq(1+x)^-1=sum_k=0^infty(-x)^k$$ if $|x|<1$. Taking $x=left(fracisigmaright)^2$, we thus obtain beginalign*
sum_i=1^ne^-fraci^2sigma^2&leqsum_i=1^nsum_k=0^inftyleft(-left(fracisigmaright)^2right)^k \
&=sum_k=0^infty(-1)^ksum_i=1^nleft(fracisigmaright)^2k tag* labeleqn:star
endalign*
(We can interchange sums because one is finite.) Now, for all $k$, the function $left(fraccdotsigmaright)^2k$ is increasing on $[0,infty)$; we thus have $$int_0^nleft(fracisigmaright)^2k,dileqsum_i=1^nleft(fracisigmaright)^2kleqleft(fracnsigmaright)^2k+int_1^nleft(fracisigmaright)^2k,di$$ Evaluating the integrals and simplifying, we have $$0leqsum_i=1^nleft(fracisigmaright)^2k-fracn2k+1left(fracnsigmaright)^2kleqleft(fracnsigmaright)^2kleft(1-frac1(2k+1)n^2kright)$$
Substituting into $eqrefeqn:star$, we get beginalign*
sum_i=1^ne^-fraci^2sigma^2&leqsum_k=0^inftyfrac(-1)^kn2k+1left(fracnsigmaright)^2k-sum_j=0^inftyleft(fracnsigmaright)^4j+2left(1-frac1(4j+3)n^4j+2right) \
&leqsum_k=0^inftyfrac(-1)^kn2k+1left(fracnsigmaright)^2k-sum_j=0^inftyleft(fracnsigmaright)^4j+2 \
&=sigmatan^-1left(fracnsigmaright)-fracleft(fracnsigmaright)^21-left(fracnsigmaright)^4hspace4em(n<sigma) tag2
endalign*
Finally, for the general case we can achieve a slight improvement on $eqrefeqn:first$ via the theory of majorization. $x_i_i=1^nmapstosum_i=1^nexpleft(-fracx_isigma^2right)$ is convex and symmetric in its arguments, hence Schur-convex. Let $b_i=i^2$ and $a_i=left(frac2n-13right)i$. Clearly, for all $mleq n$, we have $$sum_i=1^ma_i=fracm(m-1)2cdotfrac2n-13geqfracm(m-1)(2m-1)6=sum_i=1^mb_i$$ with equality if $m=n$. Thus $veca$ majorizes $vecb$, so beginalign*
sum_i=1^nexpleft(-fraci^2sigma^2right)&=sum_i=1^nexpleft(-fracb_isigma^2right) \
&leqsum_i=1^nexpleft(-fraca_isigma^2right) \
&=sum_i=1^nexpleft(-frac(2n-1)i3sigma^2right) \
&leqsum_i=0^inftyexpleft(-frac(2n-1)i3sigma^2right) \
&leqleft(1-expleft(frac2n-13sigma^2right)right)^-1 tag3
endalign*
answered 3 hours ago
Jacob Manaker
982413
add a comment |Â
up vote
3
down vote
TL;DR: three relatively easy bounds are the numbered equations below.
You cannot directly apply the formula for the geometric series for the reason mentioned in your edit. But note that $igeq1$, so we have $$sum_i=1^nexpleft(-fraci^2sigma^2right)leqsum_i=1^nexpleft(-fracicdot1sigma^2right)$$ The latter, of course, is a geometric sum. Taking the sum over all $i$ (including $i=0$), we get $$(1-e^-sigma^-2)^-1 tag1 labeleqn:first$$ The calculation for finitely many terms isn't much harder, and only differs by an exponentially decreasing factor.
If this isn't a strong enough bound, there are other techniques. If $n<sigma$, then we can get very far elementarily. Note that $e^xgeq x+1$; dividing each side, we get $$e^-xleq(1+x)^-1=sum_k=0^infty(-x)^k$$ if $|x|<1$. Taking $x=left(fracisigmaright)^2$, we thus obtain beginalign*
sum_i=1^ne^-fraci^2sigma^2&leqsum_i=1^nsum_k=0^inftyleft(-left(fracisigmaright)^2right)^k \
&=sum_k=0^infty(-1)^ksum_i=1^nleft(fracisigmaright)^2k tag* labeleqn:star
endalign*
(We can interchange sums because one is finite.) Now, for all $k$, the function $left(fraccdotsigmaright)^2k$ is increasing on $[0,infty)$; we thus have $$int_0^nleft(fracisigmaright)^2k,dileqsum_i=1^nleft(fracisigmaright)^2kleqleft(fracnsigmaright)^2k+int_1^nleft(fracisigmaright)^2k,di$$ Evaluating the integrals and simplifying, we have $$0leqsum_i=1^nleft(fracisigmaright)^2k-fracn2k+1left(fracnsigmaright)^2kleqleft(fracnsigmaright)^2kleft(1-frac1(2k+1)n^2kright)$$
Substituting into $eqrefeqn:star$, we get beginalign*
sum_i=1^ne^-fraci^2sigma^2&leqsum_k=0^inftyfrac(-1)^kn2k+1left(fracnsigmaright)^2k-sum_j=0^inftyleft(fracnsigmaright)^4j+2left(1-frac1(4j+3)n^4j+2right) \
&leqsum_k=0^inftyfrac(-1)^kn2k+1left(fracnsigmaright)^2k-sum_j=0^inftyleft(fracnsigmaright)^4j+2 \
&=sigmatan^-1left(fracnsigmaright)-fracleft(fracnsigmaright)^21-left(fracnsigmaright)^4hspace4em(n<sigma) tag2
endalign*
Finally, for the general case we can achieve a slight improvement on $eqrefeqn:first$ via the theory of majorization. $x_i_i=1^nmapstosum_i=1^nexpleft(-fracx_isigma^2right)$ is convex and symmetric in its arguments, hence Schur-convex. Let $b_i=i^2$ and $a_i=left(frac2n-13right)i$. Clearly, for all $mleq n$, we have $$sum_i=1^ma_i=fracm(m-1)2cdotfrac2n-13geqfracm(m-1)(2m-1)6=sum_i=1^mb_i$$ with equality if $m=n$. Thus $veca$ majorizes $vecb$, so beginalign*
sum_i=1^nexpleft(-fraci^2sigma^2right)&=sum_i=1^nexpleft(-fracb_isigma^2right) \
&leqsum_i=1^nexpleft(-fraca_isigma^2right) \
&=sum_i=1^nexpleft(-frac(2n-1)i3sigma^2right) \
&leqsum_i=0^inftyexpleft(-frac(2n-1)i3sigma^2right) \
&leqleft(1-expleft(frac2n-13sigma^2right)right)^-1 tag3
endalign*
answered 3 hours ago
Jacob Manaker
982413
add a comment |Â
up vote
3
down vote
up vote
3
down vote
TL;DR: three relatively easy bounds are the numbered equations below.
You cannot directly apply the formula for the geometric series for the reason mentioned in your edit. But note that $igeq1$, so we have $$sum_i=1^nexpleft(-fraci^2sigma^2right)leqsum_i=1^nexpleft(-fracicdot1sigma^2right)$$ The latter, of course, is a geometric sum. Taking the sum over all $i$ (including $i=0$), we get $$(1-e^-sigma^-2)^-1 tag1 labeleqn:first$$ The calculation for finitely many terms isn't much harder, and only differs by an exponentially decreasing factor.
If this isn't a strong enough bound, there are other techniques. If $n<sigma$, then we can get very far elementarily. Note that $e^xgeq x+1$; dividing each side, we get $$e^-xleq(1+x)^-1=sum_k=0^infty(-x)^k$$ if $|x|<1$. Taking $x=left(fracisigmaright)^2$, we thus obtain beginalign*
sum_i=1^ne^-fraci^2sigma^2&leqsum_i=1^nsum_k=0^inftyleft(-left(fracisigmaright)^2right)^k \
&=sum_k=0^infty(-1)^ksum_i=1^nleft(fracisigmaright)^2k tag* labeleqn:star
endalign*
(We can interchange sums because one is finite.) Now, for all $k$, the function $left(fraccdotsigmaright)^2k$ is increasing on $[0,infty)$; we thus have $$int_0^nleft(fracisigmaright)^2k,dileqsum_i=1^nleft(fracisigmaright)^2kleqleft(fracnsigmaright)^2k+int_1^nleft(fracisigmaright)^2k,di$$ Evaluating the integrals and simplifying, we have $$0leqsum_i=1^nleft(fracisigmaright)^2k-fracn2k+1left(fracnsigmaright)^2kleqleft(fracnsigmaright)^2kleft(1-frac1(2k+1)n^2kright)$$
Substituting into $eqrefeqn:star$, we get beginalign*
sum_i=1^ne^-fraci^2sigma^2&leqsum_k=0^inftyfrac(-1)^kn2k+1left(fracnsigmaright)^2k-sum_j=0^inftyleft(fracnsigmaright)^4j+2left(1-frac1(4j+3)n^4j+2right) \
&leqsum_k=0^inftyfrac(-1)^kn2k+1left(fracnsigmaright)^2k-sum_j=0^inftyleft(fracnsigmaright)^4j+2 \
&=sigmatan^-1left(fracnsigmaright)-fracleft(fracnsigmaright)^21-left(fracnsigmaright)^4hspace4em(n<sigma) tag2
endalign*
Finally, for the general case we can achieve a slight improvement on $eqrefeqn:first$ via the theory of majorization. $x_i_i=1^nmapstosum_i=1^nexpleft(-fracx_isigma^2right)$ is convex and symmetric in its arguments, hence Schur-convex. Let $b_i=i^2$ and $a_i=left(frac2n-13right)i$. Clearly, for all $mleq n$, we have $$sum_i=1^ma_i=fracm(m-1)2cdotfrac2n-13geqfracm(m-1)(2m-1)6=sum_i=1^mb_i$$ with equality if $m=n$. Thus $veca$ majorizes $vecb$, so beginalign*
sum_i=1^nexpleft(-fraci^2sigma^2right)&=sum_i=1^nexpleft(-fracb_isigma^2right) \
&leqsum_i=1^nexpleft(-fraca_isigma^2right) \
&=sum_i=1^nexpleft(-frac(2n-1)i3sigma^2right) \
&leqsum_i=0^inftyexpleft(-frac(2n-1)i3sigma^2right) \
&leqleft(1-expleft(frac2n-13sigma^2right)right)^-1 tag3
endalign*
answered 3 hours ago
Jacob Manaker
982413
TL;DR: three relatively easy bounds are the numbered equations below.
You cannot directly apply the formula for the geometric series for the reason mentioned in your edit. But note that $igeq1$, so we have $$sum_i=1^nexpleft(-fraci^2sigma^2right)leqsum_i=1^nexpleft(-fracicdot1sigma^2right)$$ The latter, of course, is a geometric sum. Taking the sum over all $i$ (including $i=0$), we get $$(1-e^-sigma^-2)^-1 tag1 labeleqn:first$$ The calculation for finitely many terms isn't much harder, and only differs by an exponentially decreasing factor.
If this isn't a strong enough bound, there are other techniques. If $n<sigma$, then we can get very far elementarily. Note that $e^xgeq x+1$; dividing each side, we get $$e^-xleq(1+x)^-1=sum_k=0^infty(-x)^k$$ if $|x|<1$. Taking $x=left(fracisigmaright)^2$, we thus obtain beginalign*
sum_i=1^ne^-fraci^2sigma^2&leqsum_i=1^nsum_k=0^inftyleft(-left(fracisigmaright)^2right)^k \
&=sum_k=0^infty(-1)^ksum_i=1^nleft(fracisigmaright)^2k tag* labeleqn:star
endalign*
(We can interchange sums because one is finite.) Now, for all $k$, the function $left(fraccdotsigmaright)^2k$ is increasing on $[0,infty)$; we thus have $$int_0^nleft(fracisigmaright)^2k,dileqsum_i=1^nleft(fracisigmaright)^2kleqleft(fracnsigmaright)^2k+int_1^nleft(fracisigmaright)^2k,di$$ Evaluating the integrals and simplifying, we have $$0leqsum_i=1^nleft(fracisigmaright)^2k-fracn2k+1left(fracnsigmaright)^2kleqleft(fracnsigmaright)^2kleft(1-frac1(2k+1)n^2kright)$$
Substituting into $eqrefeqn:star$, we get beginalign*
sum_i=1^ne^-fraci^2sigma^2&leqsum_k=0^inftyfrac(-1)^kn2k+1left(fracnsigmaright)^2k-sum_j=0^inftyleft(fracnsigmaright)^4j+2left(1-frac1(4j+3)n^4j+2right) \
&leqsum_k=0^inftyfrac(-1)^kn2k+1left(fracnsigmaright)^2k-sum_j=0^inftyleft(fracnsigmaright)^4j+2 \
&=sigmatan^-1left(fracnsigmaright)-fracleft(fracnsigmaright)^21-left(fracnsigmaright)^4hspace4em(n<sigma) tag2
endalign*
Finally, for the general case we can achieve a slight improvement on $eqrefeqn:first$ via the theory of majorization. $x_i_i=1^nmapstosum_i=1^nexpleft(-fracx_isigma^2right)$ is convex and symmetric in its arguments, hence Schur-convex. Let $b_i=i^2$ and $a_i=left(frac2n-13right)i$. Clearly, for all $mleq n$, we have $$sum_i=1^ma_i=fracm(m-1)2cdotfrac2n-13geqfracm(m-1)(2m-1)6=sum_i=1^mb_i$$ with equality if $m=n$. Thus $veca$ majorizes $vecb$, so beginalign*
sum_i=1^nexpleft(-fraci^2sigma^2right)&=sum_i=1^nexpleft(-fracb_isigma^2right) \
&leqsum_i=1^nexpleft(-fraca_isigma^2right) \
&=sum_i=1^nexpleft(-frac(2n-1)i3sigma^2right) \
&leqsum_i=0^inftyexpleft(-frac(2n-1)i3sigma^2right) \
&leqleft(1-expleft(frac2n-13sigma^2right)right)^-1 tag3
endalign*
answered 3 hours ago
Jacob Manaker
982413
edited 1 hour ago
answered 3 hours ago
Jacob Manaker
982413
answered 3 hours ago
Jacob Manaker
982413
answered 3 hours ago
Jacob Manaker
982413
982413
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1
Don't be sorry for the question! The only way to get better as mathematician is to ask questions. (See also the first annotation to this post)
– Jacob Manaker
3 hours ago
1
Seems that you could try the integral $int_0^infty exp(-x^2),mathrm dx$ to bound that.
– xbh
3 hours ago