Bound for $e^{-alpha x}$












1












$begingroup$


For a part of my proof I need to establish that $e^{-alpha x} lt h(x)$, where $alpha,x gt 0, $ and $x,alpha inmathbb{R}$. I thought for a while and couldn't find a function independent of $alpha$ that fulfills my criteria. Any ideas?






exponential-function functional-inequalities







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Are you asking for an example of a function $h$ for which the inequality is true?
    $endgroup$
    – kimchi lover
    Jan 15 at 14:21






  • 2




    $begingroup$
    $h(x)=1$ fits the bill.
    $endgroup$
    – Adrian Keister
    Jan 15 at 14:31






  • 1




    $begingroup$
    If you don't want any specific $h$, there are a lot (and basically useless) functions that bound $e^{-ax}$ from above. For example given any strictly positive integer $n$, $h_n(x)=n$ is always strictly larger than $e^{-ax}$ and independent of $a$.
    $endgroup$
    – James
    Jan 15 at 14:36












  • $begingroup$
    I'm voting to close the question for lack of context. Without context, we cannot know what kinda of upper bound you want, and the question becomes too open-ended.
    $endgroup$
    – Simply Beautiful Art
    Jan 15 at 14:37






  • 1




    $begingroup$
    $$ alpha,x>0implies e^{-alpha x}<1.$$
    $endgroup$
    – Julián Aguirre
    Jan 15 at 15:15
















1












$begingroup$


For a part of my proof I need to establish that $e^{-alpha x} lt h(x)$, where $alpha,x gt 0, $ and $x,alpha inmathbb{R}$. I thought for a while and couldn't find a function independent of $alpha$ that fulfills my criteria. Any ideas?






exponential-function functional-inequalities







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Are you asking for an example of a function $h$ for which the inequality is true?
    $endgroup$
    – kimchi lover
    Jan 15 at 14:21






  • 2




    $begingroup$
    $h(x)=1$ fits the bill.
    $endgroup$
    – Adrian Keister
    Jan 15 at 14:31






  • 1




    $begingroup$
    If you don't want any specific $h$, there are a lot (and basically useless) functions that bound $e^{-ax}$ from above. For example given any strictly positive integer $n$, $h_n(x)=n$ is always strictly larger than $e^{-ax}$ and independent of $a$.
    $endgroup$
    – James
    Jan 15 at 14:36












  • $begingroup$
    I'm voting to close the question for lack of context. Without context, we cannot know what kinda of upper bound you want, and the question becomes too open-ended.
    $endgroup$
    – Simply Beautiful Art
    Jan 15 at 14:37






  • 1




    $begingroup$
    $$ alpha,x>0implies e^{-alpha x}<1.$$
    $endgroup$
    – Julián Aguirre
    Jan 15 at 15:15














1












1








1


0



$begingroup$


For a part of my proof I need to establish that $e^{-alpha x} lt h(x)$, where $alpha,x gt 0, $ and $x,alpha inmathbb{R}$. I thought for a while and couldn't find a function independent of $alpha$ that fulfills my criteria. Any ideas?






exponential-function functional-inequalities







share|cite|improve this question











$endgroup$




For a part of my proof I need to establish that $e^{-alpha x} lt h(x)$, where $alpha,x gt 0, $ and $x,alpha inmathbb{R}$. I thought for a while and couldn't find a function independent of $alpha$ that fulfills my criteria. Any ideas?






exponential-function functional-inequalities




exponential-function functional-inequalities






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 15 at 14:32









Adrian Keister

5,28171933




5,28171933










asked Jan 15 at 14:14









Michael MaierMichael Maier

859




859








  • 1




    $begingroup$
    Are you asking for an example of a function $h$ for which the inequality is true?
    $endgroup$
    – kimchi lover
    Jan 15 at 14:21






  • 2




    $begingroup$
    $h(x)=1$ fits the bill.
    $endgroup$
    – Adrian Keister
    Jan 15 at 14:31






  • 1




    $begingroup$
    If you don't want any specific $h$, there are a lot (and basically useless) functions that bound $e^{-ax}$ from above. For example given any strictly positive integer $n$, $h_n(x)=n$ is always strictly larger than $e^{-ax}$ and independent of $a$.
    $endgroup$
    – James
    Jan 15 at 14:36












  • $begingroup$
    I'm voting to close the question for lack of context. Without context, we cannot know what kinda of upper bound you want, and the question becomes too open-ended.
    $endgroup$
    – Simply Beautiful Art
    Jan 15 at 14:37






  • 1




    $begingroup$
    $$ alpha,x>0implies e^{-alpha x}<1.$$
    $endgroup$
    – Julián Aguirre
    Jan 15 at 15:15














  • 1




    $begingroup$
    Are you asking for an example of a function $h$ for which the inequality is true?
    $endgroup$
    – kimchi lover
    Jan 15 at 14:21






  • 2




    $begingroup$
    $h(x)=1$ fits the bill.
    $endgroup$
    – Adrian Keister
    Jan 15 at 14:31






  • 1




    $begingroup$
    If you don't want any specific $h$, there are a lot (and basically useless) functions that bound $e^{-ax}$ from above. For example given any strictly positive integer $n$, $h_n(x)=n$ is always strictly larger than $e^{-ax}$ and independent of $a$.
    $endgroup$
    – James
    Jan 15 at 14:36












  • $begingroup$
    I'm voting to close the question for lack of context. Without context, we cannot know what kinda of upper bound you want, and the question becomes too open-ended.
    $endgroup$
    – Simply Beautiful Art
    Jan 15 at 14:37






  • 1




    $begingroup$
    $$ alpha,x>0implies e^{-alpha x}<1.$$
    $endgroup$
    – Julián Aguirre
    Jan 15 at 15:15








1




1




$begingroup$
Are you asking for an example of a function $h$ for which the inequality is true?
$endgroup$
– kimchi lover
Jan 15 at 14:21




$begingroup$
Are you asking for an example of a function $h$ for which the inequality is true?
$endgroup$
– kimchi lover
Jan 15 at 14:21




2




2




$begingroup$
$h(x)=1$ fits the bill.
$endgroup$
– Adrian Keister
Jan 15 at 14:31




$begingroup$
$h(x)=1$ fits the bill.
$endgroup$
– Adrian Keister
Jan 15 at 14:31




1




1




$begingroup$
If you don't want any specific $h$, there are a lot (and basically useless) functions that bound $e^{-ax}$ from above. For example given any strictly positive integer $n$, $h_n(x)=n$ is always strictly larger than $e^{-ax}$ and independent of $a$.
$endgroup$
– James
Jan 15 at 14:36






$begingroup$
If you don't want any specific $h$, there are a lot (and basically useless) functions that bound $e^{-ax}$ from above. For example given any strictly positive integer $n$, $h_n(x)=n$ is always strictly larger than $e^{-ax}$ and independent of $a$.
$endgroup$
– James
Jan 15 at 14:36














$begingroup$
I'm voting to close the question for lack of context. Without context, we cannot know what kinda of upper bound you want, and the question becomes too open-ended.
$endgroup$
– Simply Beautiful Art
Jan 15 at 14:37




$begingroup$
I'm voting to close the question for lack of context. Without context, we cannot know what kinda of upper bound you want, and the question becomes too open-ended.
$endgroup$
– Simply Beautiful Art
Jan 15 at 14:37




1




1




$begingroup$
$$ alpha,x>0implies e^{-alpha x}<1.$$
$endgroup$
– Julián Aguirre
Jan 15 at 15:15




$begingroup$
$$ alpha,x>0implies e^{-alpha x}<1.$$
$endgroup$
– Julián Aguirre
Jan 15 at 15:15










1 Answer
1






active

oldest

votes


















-1












$begingroup$

Also note that:
$$e^{-alpha x}=sum_{n=0}^inftyfrac{(-1)^nalpha^nx^n}{n!}$$
A polynomial of this expansion could be used, although take care to see that it is consistently greater as this is a series with alternating sign






share|cite|improve this answer









$endgroup$













  • $begingroup$
    But a polynomial (of positive degree) gives a worse bound than the obvious $h(x)=1$.
    $endgroup$
    – David C. Ullrich
    Jan 15 at 17:41










  • $begingroup$
    The higher order used the closer this approximation is surely, especially at smaller values of $x$, since $h(x)=1$ is most accurate for $xtoinfty$
    $endgroup$
    – Henry Lee
    Jan 15 at 18:53












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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









-1












$begingroup$

Also note that:
$$e^{-alpha x}=sum_{n=0}^inftyfrac{(-1)^nalpha^nx^n}{n!}$$
A polynomial of this expansion could be used, although take care to see that it is consistently greater as this is a series with alternating sign






share|cite|improve this answer









$endgroup$













  • $begingroup$
    But a polynomial (of positive degree) gives a worse bound than the obvious $h(x)=1$.
    $endgroup$
    – David C. Ullrich
    Jan 15 at 17:41










  • $begingroup$
    The higher order used the closer this approximation is surely, especially at smaller values of $x$, since $h(x)=1$ is most accurate for $xtoinfty$
    $endgroup$
    – Henry Lee
    Jan 15 at 18:53
















-1












$begingroup$

Also note that:
$$e^{-alpha x}=sum_{n=0}^inftyfrac{(-1)^nalpha^nx^n}{n!}$$
A polynomial of this expansion could be used, although take care to see that it is consistently greater as this is a series with alternating sign






share|cite|improve this answer









$endgroup$













  • $begingroup$
    But a polynomial (of positive degree) gives a worse bound than the obvious $h(x)=1$.
    $endgroup$
    – David C. Ullrich
    Jan 15 at 17:41










  • $begingroup$
    The higher order used the closer this approximation is surely, especially at smaller values of $x$, since $h(x)=1$ is most accurate for $xtoinfty$
    $endgroup$
    – Henry Lee
    Jan 15 at 18:53














-1












-1








-1





$begingroup$

Also note that:
$$e^{-alpha x}=sum_{n=0}^inftyfrac{(-1)^nalpha^nx^n}{n!}$$
A polynomial of this expansion could be used, although take care to see that it is consistently greater as this is a series with alternating sign






share|cite|improve this answer









$endgroup$



Also note that:
$$e^{-alpha x}=sum_{n=0}^inftyfrac{(-1)^nalpha^nx^n}{n!}$$
A polynomial of this expansion could be used, although take care to see that it is consistently greater as this is a series with alternating sign







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 15 at 16:40









Henry LeeHenry Lee

2,181319




2,181319












  • $begingroup$
    But a polynomial (of positive degree) gives a worse bound than the obvious $h(x)=1$.
    $endgroup$
    – David C. Ullrich
    Jan 15 at 17:41










  • $begingroup$
    The higher order used the closer this approximation is surely, especially at smaller values of $x$, since $h(x)=1$ is most accurate for $xtoinfty$
    $endgroup$
    – Henry Lee
    Jan 15 at 18:53


















  • $begingroup$
    But a polynomial (of positive degree) gives a worse bound than the obvious $h(x)=1$.
    $endgroup$
    – David C. Ullrich
    Jan 15 at 17:41










  • $begingroup$
    The higher order used the closer this approximation is surely, especially at smaller values of $x$, since $h(x)=1$ is most accurate for $xtoinfty$
    $endgroup$
    – Henry Lee
    Jan 15 at 18:53
















$begingroup$
But a polynomial (of positive degree) gives a worse bound than the obvious $h(x)=1$.
$endgroup$
– David C. Ullrich
Jan 15 at 17:41




$begingroup$
But a polynomial (of positive degree) gives a worse bound than the obvious $h(x)=1$.
$endgroup$
– David C. Ullrich
Jan 15 at 17:41












$begingroup$
The higher order used the closer this approximation is surely, especially at smaller values of $x$, since $h(x)=1$ is most accurate for $xtoinfty$
$endgroup$
– Henry Lee
Jan 15 at 18:53




$begingroup$
The higher order used the closer this approximation is surely, especially at smaller values of $x$, since $h(x)=1$ is most accurate for $xtoinfty$
$endgroup$
– Henry Lee
Jan 15 at 18:53


















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