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How to simplify or upperbound 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 upperbound the following summation:
$$ sum_i=1^nexpleft(-fraci^2sigma^2right).$$
Can I use geometric series?
EDIT: I have difficulty because of the power $2$, i.e if the summation would be $ sumlimits_i=1^nexpleft(-fracisigma^2right) $ then it would be easy to apply geometric series!
sequences-and-series geometric-series
edited 7 mins ago
Jim
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asked 5 hours ago
user8003788
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up vote
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I am not a mathematician, so sorry for this trivial question. Is there a way to simplify or to upperbound the following summation:
$$ sum_i=1^nexpleft(-fraci^2sigma^2right).$$
Can I use geometric series?
EDIT: I have difficulty because of the power $2$, i.e if the summation would be $ sumlimits_i=1^nexpleft(-fracisigma^2right) $ then it would be easy to apply geometric series!
sequences-and-series geometric-series
edited 7 mins ago
Jim
1147
asked 5 hours ago
user8003788
263
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
5 hours ago
1
Seems that you could try the integral $int_0^infty exp(-x^2),mathrm dx$ to bound that.
– xbh
5 hours ago
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
I am not a mathematician, so sorry for this trivial question. Is there a way to simplify or to upperbound the following summation:
$$ sum_i=1^nexpleft(-fraci^2sigma^2right).$$
Can I use geometric series?
EDIT: I have difficulty because of the power $2$, i.e if the summation would be $ sumlimits_i=1^nexpleft(-fracisigma^2right) $ then it would be easy to apply geometric series!
sequences-and-series geometric-series
edited 7 mins ago
Jim
1147
asked 5 hours ago
user8003788
263
I am not a mathematician, so sorry for this trivial question. Is there a way to simplify or to upperbound the following summation:
$$ sum_i=1^nexpleft(-fraci^2sigma^2right).$$
Can I use geometric series?
EDIT: I have difficulty because of the power $2$, i.e if the summation would be $ sumlimits_i=1^nexpleft(-fracisigma^2right) $ then it would be easy to apply geometric series!
sequences-and-series geometric-series
sequences-and-series geometric-series
edited 7 mins ago
Jim
1147
asked 5 hours ago
user8003788
263
edited 7 mins ago
Jim
1147
asked 5 hours ago
user8003788
263
edited 7 mins ago
Jim
1147
edited 7 mins ago
Jim
1147
edited 7 mins ago
Jim
1147
1147
asked 5 hours ago
user8003788
263
asked 5 hours ago
user8003788
263
asked 5 hours ago
user8003788
263
263
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
5 hours ago
1
Seems that you could try the integral $int_0^infty exp(-x^2),mathrm dx$ to bound that.
– xbh
5 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
5 hours ago
1
Seems that you could try the integral $int_0^infty exp(-x^2),mathrm dx$ to bound that.
– xbh
5 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
5 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
5 hours ago
1
1
Seems that you could try the integral $int_0^infty exp(-x^2),mathrm dx$ to bound that.
– xbh
5 hours ago
Seems that you could try the integral $int_0^infty exp(-x^2),mathrm dx$ to bound that.
– xbh
5 hours ago
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
7
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 4 hours ago
xbh
3,625320
1
Nicely done! Sometimes simplest is best.
– Jacob Manaker
3 hours ago
add a comment |Â
up vote
5
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 5 hours ago
Jacob Manaker
1,002413
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up vote
0
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There's a rather trivial upper bound that $frac-i^2sigma^2$ is negative, so exponentiating it results in a number less than 1, so the sum is at most $n$. If you want a constant upper bound, you can upper bound it with the geometric series.
answered 21 mins ago
Acccumulation
5,8342616
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
7
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 4 hours ago
xbh
3,625320
1
Nicely done! Sometimes simplest is best.
– Jacob Manaker
3 hours ago
add a comment |Â
up vote
7
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 4 hours ago
xbh
3,625320
1
Nicely done! Sometimes simplest is best.
– Jacob Manaker
3 hours ago
add a comment |Â
up vote
7
down vote
up vote
7
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 4 hours ago
xbh
3,625320
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 4 hours ago
xbh
3,625320
answered 4 hours ago
xbh
3,625320
answered 4 hours ago
xbh
3,625320
answered 4 hours ago
xbh
3,625320
3,625320
1
Nicely done! Sometimes simplest is best.
– Jacob Manaker
3 hours ago
add a comment |Â
1
Nicely done! Sometimes simplest is best.
– Jacob Manaker
3 hours ago
1
1
Nicely done! Sometimes simplest is best.
– Jacob Manaker
3 hours ago
Nicely done! Sometimes simplest is best.
– Jacob Manaker
3 hours ago
add a comment |Â
up vote
5
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 5 hours ago
Jacob Manaker
1,002413
add a comment |Â
up vote
5
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 5 hours ago
Jacob Manaker
1,002413
add a comment |Â
up vote
5
down vote
up vote
5
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 5 hours ago
Jacob Manaker
1,002413
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 5 hours ago
Jacob Manaker
1,002413
edited 3 hours ago
answered 5 hours ago
Jacob Manaker
1,002413
answered 5 hours ago
Jacob Manaker
1,002413
answered 5 hours ago
Jacob Manaker
1,002413
1,002413
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add a comment |Â
up vote
0
down vote
There's a rather trivial upper bound that $frac-i^2sigma^2$ is negative, so exponentiating it results in a number less than 1, so the sum is at most $n$. If you want a constant upper bound, you can upper bound it with the geometric series.
answered 21 mins ago
Acccumulation
5,8342616
add a comment |Â
up vote
0
down vote
There's a rather trivial upper bound that $frac-i^2sigma^2$ is negative, so exponentiating it results in a number less than 1, so the sum is at most $n$. If you want a constant upper bound, you can upper bound it with the geometric series.
answered 21 mins ago
Acccumulation
5,8342616
add a comment |Â
up vote
0
down vote
up vote
0
down vote
There's a rather trivial upper bound that $frac-i^2sigma^2$ is negative, so exponentiating it results in a number less than 1, so the sum is at most $n$. If you want a constant upper bound, you can upper bound it with the geometric series.
answered 21 mins ago
Acccumulation
5,8342616
There's a rather trivial upper bound that $frac-i^2sigma^2$ is negative, so exponentiating it results in a number less than 1, so the sum is at most $n$. If you want a constant upper bound, you can upper bound it with the geometric series.
answered 21 mins ago
Acccumulation
5,8342616
answered 21 mins ago
Acccumulation
5,8342616
answered 21 mins ago
Acccumulation
5,8342616
answered 21 mins ago
Acccumulation
5,8342616
5,8342616
add a comment |Â
add a comment |Â
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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
5 hours ago
1
Seems that you could try the integral $int_0^infty exp(-x^2),mathrm dx$ to bound that.
– xbh
5 hours ago