Mixing
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
edited Jan 15 at 14:32
Adrian Keister
5,23271933
asked Jan 15 at 14:14
Michael MaierMichael Maier
859
$endgroup$
|
show 4 more comments
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
edited Jan 15 at 14:32
Adrian Keister
5,23271933
asked Jan 15 at 14:14
Michael MaierMichael Maier
859
$endgroup$
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
|
show 4 more comments
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
edited Jan 15 at 14:32
Adrian Keister
5,23271933
asked Jan 15 at 14:14
Michael MaierMichael Maier
859
$endgroup$
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
edited Jan 15 at 14:32
Adrian Keister
5,23271933
asked Jan 15 at 14:14
Michael MaierMichael Maier
859
edited Jan 15 at 14:32
Adrian Keister
5,23271933
asked Jan 15 at 14:14
Michael MaierMichael Maier
859
edited Jan 15 at 14:32
Adrian Keister
5,23271933
edited Jan 15 at 14:32
Adrian Keister
5,23271933
edited Jan 15 at 14:32
Adrian Keister
5,23271933
5,23271933
asked Jan 15 at 14:14
Michael MaierMichael Maier
859
asked Jan 15 at 14:14
Michael MaierMichael Maier
859
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
|
show 4 more comments
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
|
show 4 more comments
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
answered Jan 15 at 16:40
Henry LeeHenry Lee
2,042219
$endgroup$
$begingroup$
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
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
draft saved
draft discarded
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3074479%2fbound-for-e-alpha-x%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
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
answered Jan 15 at 16:40
Henry LeeHenry Lee
2,042219
$endgroup$
$begingroup$
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
add a comment |
-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
answered Jan 15 at 16:40
Henry LeeHenry Lee
2,042219
$endgroup$
$begingroup$
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
add a comment |
-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
answered Jan 15 at 16:40
Henry LeeHenry Lee
2,042219
$endgroup$
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
answered Jan 15 at 16:40
Henry LeeHenry Lee
2,042219
answered Jan 15 at 16:40
Henry LeeHenry Lee
2,042219
answered Jan 15 at 16:40
Henry LeeHenry Lee
2,042219
answered Jan 15 at 16:40
Henry LeeHenry Lee
2,042219
2,042219
$begingroup$
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
add a comment |
$begingroup$
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
add a comment |
draft saved
draft discarded
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
draft saved
draft discarded
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3074479%2fbound-for-e-alpha-x%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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