Inequality involving sum of exponentials

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Let



$lambda = fracsumlimits_n=1^infty lambda_n B_n e^-lambda_n t_0sumlimits_n=1^infty B_n e^-lambda_n t_0,,$
with $lambda_n$, $B_n$, $t_0$ positive real numbers.



Does the following always hold for $t_1 > t_0$



$fracsumlimits_n=1^infty B_n e^-lambda_n t_1sumlimits_n=1^infty B_n e^-lambda_n t_0 ge
frace^-lambda t_1e^-lambda t_0,?
$



[Edit: inequality sign corrected from '<' to $ge$ after Kavi Rama Murthy's answer and random's comment.]




inequality exponential-function




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  • 1




    With all $lambda_n$ equal and $Sigma B_n$ convergent there is equality.
    – random
    Aug 2 at 12:07














up vote
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Let



$lambda = fracsumlimits_n=1^infty lambda_n B_n e^-lambda_n t_0sumlimits_n=1^infty B_n e^-lambda_n t_0,,$
with $lambda_n$, $B_n$, $t_0$ positive real numbers.



Does the following always hold for $t_1 > t_0$



$fracsumlimits_n=1^infty B_n e^-lambda_n t_1sumlimits_n=1^infty B_n e^-lambda_n t_0 ge
frace^-lambda t_1e^-lambda t_0,?
$



[Edit: inequality sign corrected from '<' to $ge$ after Kavi Rama Murthy's answer and random's comment.]




inequality exponential-function




share|cite|improve this question

















  • 1




    With all $lambda_n$ equal and $Sigma B_n$ convergent there is equality.
    – random
    Aug 2 at 12:07












up vote
1
down vote

favorite
1









up vote
1
down vote

favorite
1






1





Let



$lambda = fracsumlimits_n=1^infty lambda_n B_n e^-lambda_n t_0sumlimits_n=1^infty B_n e^-lambda_n t_0,,$
with $lambda_n$, $B_n$, $t_0$ positive real numbers.



Does the following always hold for $t_1 > t_0$



$fracsumlimits_n=1^infty B_n e^-lambda_n t_1sumlimits_n=1^infty B_n e^-lambda_n t_0 ge
frace^-lambda t_1e^-lambda t_0,?
$



[Edit: inequality sign corrected from '<' to $ge$ after Kavi Rama Murthy's answer and random's comment.]




inequality exponential-function




share|cite|improve this question













Let



$lambda = fracsumlimits_n=1^infty lambda_n B_n e^-lambda_n t_0sumlimits_n=1^infty B_n e^-lambda_n t_0,,$
with $lambda_n$, $B_n$, $t_0$ positive real numbers.



Does the following always hold for $t_1 > t_0$



$fracsumlimits_n=1^infty B_n e^-lambda_n t_1sumlimits_n=1^infty B_n e^-lambda_n t_0 ge
frace^-lambda t_1e^-lambda t_0,?
$



[Edit: inequality sign corrected from '<' to $ge$ after Kavi Rama Murthy's answer and random's comment.]





inequality exponential-function





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edited Aug 2 at 14:38
























asked Aug 2 at 10:09









toliveira

597312




597312







  • 1




    With all $lambda_n$ equal and $Sigma B_n$ convergent there is equality.
    – random
    Aug 2 at 12:07












  • 1




    With all $lambda_n$ equal and $Sigma B_n$ convergent there is equality.
    – random
    Aug 2 at 12:07







1




1




With all $lambda_n$ equal and $Sigma B_n$ convergent there is equality.
– random
Aug 2 at 12:07




With all $lambda_n$ equal and $Sigma B_n$ convergent there is equality.
– random
Aug 2 at 12:07










2 Answers
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Yes, let $f(t)=sumlimits_n=1^infty B_n e^-lambda_n t$ and note that $sumlimits_n=1^infty lambda_n B_n e^-lambda_n t=-sumlimits_n=1^infty frac ddt(B_n e^-lambda_n t)$.



The problem statement implies that both series converge for $t=t_0$ and therefore uniformly converge for $tge t_0$, so $sumlimits_n=1^infty frac ddt(B_n e^-lambda_n t)=frac ddtsumlimits_n=1^infty (B_n e^-lambda_n t)$ and one has $$lambda(t) = fracsumlimits_n=1^infty lambda_n B_n e^-lambda_n tsumlimits_n=1^infty B_n e^-lambda_n t=-fracd log fdt$$



The expression for $lambda(t)$ can also be interpreted as a weighted arithmetic mean of the individual $lambda_n$ and since with growing $t$ the normalized weights for smaller $lambda_n$ increase at the expense of those of larger $lambda_n$, $lambda$ must be a non-increasing function of $t$.



By the mean value theorem there is a $t_i in (t_0,t)$ for which $log f(t)-log f(t_0)=-lambda(t_i)(t-t_0)ge -lambda(t_0)(t-t_0)$, which is equivalent with $fracf(t)f(t_0)ge e^-lambda(t_0)(t-t_0)$.






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  • I found it brilliant! Thank you.
    – toliveira
    Aug 3 at 18:34

















up vote
1
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You cannot have such an inequality. RHS $<1$. If such an inequality holds for all choices of $B_n$'s we can get $1$ as a limting value of LHS leading to a contradiction. For example you can let $B_n$ and $lambda_n$ approach $0$ for all $n>1$, $lambda_1=1$ and $B_1$ approach $infty$.






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  • Thank you very much for pointing it out. I corrected the inequality sign so that the question now makes more sense.
    – toliveira
    Aug 2 at 10:36










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2 Answers
2






active

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2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
1
down vote



accepted










Yes, let $f(t)=sumlimits_n=1^infty B_n e^-lambda_n t$ and note that $sumlimits_n=1^infty lambda_n B_n e^-lambda_n t=-sumlimits_n=1^infty frac ddt(B_n e^-lambda_n t)$.



The problem statement implies that both series converge for $t=t_0$ and therefore uniformly converge for $tge t_0$, so $sumlimits_n=1^infty frac ddt(B_n e^-lambda_n t)=frac ddtsumlimits_n=1^infty (B_n e^-lambda_n t)$ and one has $$lambda(t) = fracsumlimits_n=1^infty lambda_n B_n e^-lambda_n tsumlimits_n=1^infty B_n e^-lambda_n t=-fracd log fdt$$



The expression for $lambda(t)$ can also be interpreted as a weighted arithmetic mean of the individual $lambda_n$ and since with growing $t$ the normalized weights for smaller $lambda_n$ increase at the expense of those of larger $lambda_n$, $lambda$ must be a non-increasing function of $t$.



By the mean value theorem there is a $t_i in (t_0,t)$ for which $log f(t)-log f(t_0)=-lambda(t_i)(t-t_0)ge -lambda(t_0)(t-t_0)$, which is equivalent with $fracf(t)f(t_0)ge e^-lambda(t_0)(t-t_0)$.






share|cite|improve this answer





















  • I found it brilliant! Thank you.
    – toliveira
    Aug 3 at 18:34














up vote
1
down vote



accepted










Yes, let $f(t)=sumlimits_n=1^infty B_n e^-lambda_n t$ and note that $sumlimits_n=1^infty lambda_n B_n e^-lambda_n t=-sumlimits_n=1^infty frac ddt(B_n e^-lambda_n t)$.



The problem statement implies that both series converge for $t=t_0$ and therefore uniformly converge for $tge t_0$, so $sumlimits_n=1^infty frac ddt(B_n e^-lambda_n t)=frac ddtsumlimits_n=1^infty (B_n e^-lambda_n t)$ and one has $$lambda(t) = fracsumlimits_n=1^infty lambda_n B_n e^-lambda_n tsumlimits_n=1^infty B_n e^-lambda_n t=-fracd log fdt$$



The expression for $lambda(t)$ can also be interpreted as a weighted arithmetic mean of the individual $lambda_n$ and since with growing $t$ the normalized weights for smaller $lambda_n$ increase at the expense of those of larger $lambda_n$, $lambda$ must be a non-increasing function of $t$.



By the mean value theorem there is a $t_i in (t_0,t)$ for which $log f(t)-log f(t_0)=-lambda(t_i)(t-t_0)ge -lambda(t_0)(t-t_0)$, which is equivalent with $fracf(t)f(t_0)ge e^-lambda(t_0)(t-t_0)$.






share|cite|improve this answer





















  • I found it brilliant! Thank you.
    – toliveira
    Aug 3 at 18:34












up vote
1
down vote



accepted







up vote
1
down vote



accepted






Yes, let $f(t)=sumlimits_n=1^infty B_n e^-lambda_n t$ and note that $sumlimits_n=1^infty lambda_n B_n e^-lambda_n t=-sumlimits_n=1^infty frac ddt(B_n e^-lambda_n t)$.



The problem statement implies that both series converge for $t=t_0$ and therefore uniformly converge for $tge t_0$, so $sumlimits_n=1^infty frac ddt(B_n e^-lambda_n t)=frac ddtsumlimits_n=1^infty (B_n e^-lambda_n t)$ and one has $$lambda(t) = fracsumlimits_n=1^infty lambda_n B_n e^-lambda_n tsumlimits_n=1^infty B_n e^-lambda_n t=-fracd log fdt$$



The expression for $lambda(t)$ can also be interpreted as a weighted arithmetic mean of the individual $lambda_n$ and since with growing $t$ the normalized weights for smaller $lambda_n$ increase at the expense of those of larger $lambda_n$, $lambda$ must be a non-increasing function of $t$.



By the mean value theorem there is a $t_i in (t_0,t)$ for which $log f(t)-log f(t_0)=-lambda(t_i)(t-t_0)ge -lambda(t_0)(t-t_0)$, which is equivalent with $fracf(t)f(t_0)ge e^-lambda(t_0)(t-t_0)$.






share|cite|improve this answer













Yes, let $f(t)=sumlimits_n=1^infty B_n e^-lambda_n t$ and note that $sumlimits_n=1^infty lambda_n B_n e^-lambda_n t=-sumlimits_n=1^infty frac ddt(B_n e^-lambda_n t)$.



The problem statement implies that both series converge for $t=t_0$ and therefore uniformly converge for $tge t_0$, so $sumlimits_n=1^infty frac ddt(B_n e^-lambda_n t)=frac ddtsumlimits_n=1^infty (B_n e^-lambda_n t)$ and one has $$lambda(t) = fracsumlimits_n=1^infty lambda_n B_n e^-lambda_n tsumlimits_n=1^infty B_n e^-lambda_n t=-fracd log fdt$$



The expression for $lambda(t)$ can also be interpreted as a weighted arithmetic mean of the individual $lambda_n$ and since with growing $t$ the normalized weights for smaller $lambda_n$ increase at the expense of those of larger $lambda_n$, $lambda$ must be a non-increasing function of $t$.



By the mean value theorem there is a $t_i in (t_0,t)$ for which $log f(t)-log f(t_0)=-lambda(t_i)(t-t_0)ge -lambda(t_0)(t-t_0)$, which is equivalent with $fracf(t)f(t_0)ge e^-lambda(t_0)(t-t_0)$.







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered Aug 3 at 13:44









random

23815




23815











  • I found it brilliant! Thank you.
    – toliveira
    Aug 3 at 18:34
















  • I found it brilliant! Thank you.
    – toliveira
    Aug 3 at 18:34















I found it brilliant! Thank you.
– toliveira
Aug 3 at 18:34




I found it brilliant! Thank you.
– toliveira
Aug 3 at 18:34










up vote
1
down vote













You cannot have such an inequality. RHS $<1$. If such an inequality holds for all choices of $B_n$'s we can get $1$ as a limting value of LHS leading to a contradiction. For example you can let $B_n$ and $lambda_n$ approach $0$ for all $n>1$, $lambda_1=1$ and $B_1$ approach $infty$.






share|cite|improve this answer





















  • Thank you very much for pointing it out. I corrected the inequality sign so that the question now makes more sense.
    – toliveira
    Aug 2 at 10:36














up vote
1
down vote













You cannot have such an inequality. RHS $<1$. If such an inequality holds for all choices of $B_n$'s we can get $1$ as a limting value of LHS leading to a contradiction. For example you can let $B_n$ and $lambda_n$ approach $0$ for all $n>1$, $lambda_1=1$ and $B_1$ approach $infty$.






share|cite|improve this answer





















  • Thank you very much for pointing it out. I corrected the inequality sign so that the question now makes more sense.
    – toliveira
    Aug 2 at 10:36












up vote
1
down vote










up vote
1
down vote









You cannot have such an inequality. RHS $<1$. If such an inequality holds for all choices of $B_n$'s we can get $1$ as a limting value of LHS leading to a contradiction. For example you can let $B_n$ and $lambda_n$ approach $0$ for all $n>1$, $lambda_1=1$ and $B_1$ approach $infty$.






share|cite|improve this answer













You cannot have such an inequality. RHS $<1$. If such an inequality holds for all choices of $B_n$'s we can get $1$ as a limting value of LHS leading to a contradiction. For example you can let $B_n$ and $lambda_n$ approach $0$ for all $n>1$, $lambda_1=1$ and $B_1$ approach $infty$.







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered Aug 2 at 10:22









Kavi Rama Murthy

19.2k2829




19.2k2829











  • Thank you very much for pointing it out. I corrected the inequality sign so that the question now makes more sense.
    – toliveira
    Aug 2 at 10:36
















  • Thank you very much for pointing it out. I corrected the inequality sign so that the question now makes more sense.
    – toliveira
    Aug 2 at 10:36















Thank you very much for pointing it out. I corrected the inequality sign so that the question now makes more sense.
– toliveira
Aug 2 at 10:36




Thank you very much for pointing it out. I corrected the inequality sign so that the question now makes more sense.
– toliveira
Aug 2 at 10:36












 

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