$0<e^-sqrt x<frac1x^a$ inequality The 2019 Stack Overflow Developer Survey Results Are InProve $sqrt[n]nleq 1+frac2sqrtn$Calculate the following: $lim limits_n to infty sqrt[n]e^n+(1+frac1n)^n^2$Prove or disprove the inequality $sqrt1+sin b-sqrt1+sin aleqfracb-a2$How to prove this inequality $ left|fracsin xx - fracsin yyright| le sqrt 2left$Inequality $log xle frac2e , sqrtx$Inequality $|ln(fracx+sqrt1+x^2y+sqrt1+y^2)| leq |x-y|$Prove inequality $arccos left( fracsin 1-sin x1-x right) leq sqrtfrac1+x+x^23$Compute $lim_ntoinftyint_0^fracpi2 fracsin^nxsqrt1+x, dx$Prove $int_0^inftyfrac1e^sxsqrt1+s^2ds < arctanleft(frac1xright),quadforall xge1$.Proving an inequality involving logarithm
Getting crown tickets for Statue of Liberty
Match Roman Numerals
How can I define good in a religion that claims no moral authority?
If I can cast sorceries at instant speed, can I use sorcery-speed activated abilities at instant speed?
What's the name of these plastic connectors
Why didn't the Event Horizon Telescope team mention Sagittarius A*?
How to translate "being like"?
How to charge AirPods to keep battery healthy?
How can I add encounters in the Lost Mine of Phandelver campaign without giving PCs too much XP?
Deal with toxic manager when you can't quit
Worn-tile Scrabble
What is this sharp, curved notch on my knife for?
Why does the nucleus not repel itself?
Is Cinnamon a desktop environment or a window manager? (Or both?)
What is the motivation for a law requiring 2 parties to consent for recording a conversation
What is this business jet?
Merge two greps into single one
How to notate time signature switching consistently every measure
Mathematics of imaging the black hole
Inverse Relationship Between Precision and Recall
Get name of standard action overriden in Visualforce contorller
writing variables above the numbers in tikz picture
Cooking pasta in a water boiler
What does Linus Torvalds mean when he says that Git "never ever" tracks a file?
$0
The 2019 Stack Overflow Developer Survey Results Are InProve $sqrt[n]nleq 1+frac2sqrtn$Calculate the following: $lim limits_n to infty sqrt[n]e^n+(1+frac1n)^n^2$Prove or disprove the inequality $sqrt1+sin b-sqrt1+sin aleqfracb-a2$How to prove this inequality $ left|fracsin xx - fracsin yyright| le sqrt frac1x - frac1yright$Inequality $log xle frac2e , sqrtx$Inequality $|ln(fracx+sqrt1+x^2y+sqrt1+y^2)| leq |x-y|$Prove inequality $arccos left( fracsin 1-sin x1-x right) leq sqrtfrac1+x+x^23$Compute $lim_ntoinftyint_0^fracpi2 fracsin^nxsqrt1+x, dx$Prove $int_0^inftyfrac1e^sxsqrt1+s^2ds < arctanleft(frac1xright),quadforall xge1$.Proving an inequality involving logarithm
$begingroup$
How can we prove that $0<e^-sqrt x<frac1x^a$ is true for all $x>B(a)$ for some $B(a)$ and some fixed $a>0$. By the mean theorem I can only receive $frac11+sqrt x>e^-sqrt x$ and after that $frac1sqrt x>e^-sqrt x$. Then I have next inequality for $x>1$ because $frac1x^b>frac1sqrt x>e^-sqrt x$ but it I have only for $b<1/2$. Thanks for your help.
calculus logarithms exponential-function
edited Apr 7 at 20:50
Michael Rozenberg
110k1896201
asked Apr 5 at 21:15
CtSm0x1CtSm0x1
203
New contributor
CtSm0x1 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
How can we prove that $0<e^-sqrt x<frac1x^a$ is true for all $x>B(a)$ for some $B(a)$ and some fixed $a>0$. By the mean theorem I can only receive $frac11+sqrt x>e^-sqrt x$ and after that $frac1sqrt x>e^-sqrt x$. Then I have next inequality for $x>1$ because $frac1x^b>frac1sqrt x>e^-sqrt x$ but it I have only for $b<1/2$. Thanks for your help.
calculus logarithms exponential-function
edited Apr 7 at 20:50
Michael Rozenberg
110k1896201
asked Apr 5 at 21:15
CtSm0x1CtSm0x1
203
New contributor
CtSm0x1 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
The square root doesn't really matter: if you put $y=sqrtx$, you can focus on prooving that $e^-y<frac1y^2a$ for $y$ large enough. To show that, try to calculate the limit $lim_yrightarrowinfty frace^yy^2a$. Suggestion: use l'Hopital rule.
$endgroup$
– Adam Latosiński
Apr 5 at 21:31
add a comment |
$begingroup$
How can we prove that $0<e^-sqrt x<frac1x^a$ is true for all $x>B(a)$ for some $B(a)$ and some fixed $a>0$. By the mean theorem I can only receive $frac11+sqrt x>e^-sqrt x$ and after that $frac1sqrt x>e^-sqrt x$. Then I have next inequality for $x>1$ because $frac1x^b>frac1sqrt x>e^-sqrt x$ but it I have only for $b<1/2$. Thanks for your help.
calculus logarithms exponential-function
edited Apr 7 at 20:50
Michael Rozenberg
110k1896201
asked Apr 5 at 21:15
CtSm0x1CtSm0x1
203
New contributor
CtSm0x1 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
How can we prove that $0<e^-sqrt x<frac1x^a$ is true for all $x>B(a)$ for some $B(a)$ and some fixed $a>0$. By the mean theorem I can only receive $frac11+sqrt x>e^-sqrt x$ and after that $frac1sqrt x>e^-sqrt x$. Then I have next inequality for $x>1$ because $frac1x^b>frac1sqrt x>e^-sqrt x$ but it I have only for $b<1/2$. Thanks for your help.
calculus logarithms exponential-function
calculus logarithms exponential-function
edited Apr 7 at 20:50
Michael Rozenberg
110k1896201
asked Apr 5 at 21:15
CtSm0x1CtSm0x1
203
New contributor
CtSm0x1 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Apr 7 at 20:50
Michael Rozenberg
110k1896201
asked Apr 5 at 21:15
CtSm0x1CtSm0x1
203
New contributor
CtSm0x1 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Apr 7 at 20:50
Michael Rozenberg
110k1896201
edited Apr 7 at 20:50
Michael Rozenberg
110k1896201
edited Apr 7 at 20:50
Michael Rozenberg
110k1896201
110k1896201
asked Apr 5 at 21:15
CtSm0x1CtSm0x1
203
New contributor
CtSm0x1 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Apr 5 at 21:15
CtSm0x1CtSm0x1
203
asked Apr 5 at 21:15
CtSm0x1CtSm0x1
203
203
New contributor
CtSm0x1 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
CtSm0x1 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
CtSm0x1 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
$begingroup$
The square root doesn't really matter: if you put $y=sqrtx$, you can focus on prooving that $e^-y<frac1y^2a$ for $y$ large enough. To show that, try to calculate the limit $lim_yrightarrowinfty frace^yy^2a$. Suggestion: use l'Hopital rule.
$endgroup$
– Adam Latosiński
Apr 5 at 21:31
add a comment |
1
$begingroup$
The square root doesn't really matter: if you put $y=sqrtx$, you can focus on prooving that $e^-y<frac1y^2a$ for $y$ large enough. To show that, try to calculate the limit $lim_yrightarrowinfty frace^yy^2a$. Suggestion: use l'Hopital rule.
$endgroup$
– Adam Latosiński
Apr 5 at 21:31
1
1
$begingroup$
The square root doesn't really matter: if you put $y=sqrtx$, you can focus on prooving that $e^-y<frac1y^2a$ for $y$ large enough. To show that, try to calculate the limit $lim_yrightarrowinfty frace^yy^2a$. Suggestion: use l'Hopital rule.
$endgroup$
– Adam Latosiński
Apr 5 at 21:31
$begingroup$
The square root doesn't really matter: if you put $y=sqrtx$, you can focus on prooving that $e^-y<frac1y^2a$ for $y$ large enough. To show that, try to calculate the limit $lim_yrightarrowinfty frace^yy^2a$. Suggestion: use l'Hopital rule.
$endgroup$
– Adam Latosiński
Apr 5 at 21:31
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The inequality is equivalent to $-sqrtx<-alog x$ or $sqrtx>2alogsqrtx$. Set $t=sqrtx$ and consider the function
$$
f(t)=t-2alog t
$$
Since $lim_ttoinftyf(t)=infty$, there is some number $C(a)$ such that $f(t)>0$ for $t>C(a)$. Now set $B(a)=C(a)^2$.
answered Apr 5 at 22:41
egregegreg
186k1486208
$endgroup$
add a comment |
$begingroup$
Rewrite our inequality in the following form:
$$fracsqrtxlnx>a.$$
Now, prove that $fracsqrtxlnx>lnx$ for all $x>5510$ and we can take
$$B(a)=maxleft5510,e^aright+1$$
answered Apr 5 at 21:20
Michael RozenbergMichael Rozenberg
110k1896201
$endgroup$
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
);
);
CtSm0x1 is a new contributor. Be nice, and check out our Code of Conduct.
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%2f3176421%2f0e-sqrt-x-frac1xa-inequality%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The inequality is equivalent to $-sqrtx<-alog x$ or $sqrtx>2alogsqrtx$. Set $t=sqrtx$ and consider the function
$$
f(t)=t-2alog t
$$
Since $lim_ttoinftyf(t)=infty$, there is some number $C(a)$ such that $f(t)>0$ for $t>C(a)$. Now set $B(a)=C(a)^2$.
answered Apr 5 at 22:41
egregegreg
186k1486208
$endgroup$
add a comment |
$begingroup$
The inequality is equivalent to $-sqrtx<-alog x$ or $sqrtx>2alogsqrtx$. Set $t=sqrtx$ and consider the function
$$
f(t)=t-2alog t
$$
Since $lim_ttoinftyf(t)=infty$, there is some number $C(a)$ such that $f(t)>0$ for $t>C(a)$. Now set $B(a)=C(a)^2$.
answered Apr 5 at 22:41
egregegreg
186k1486208
$endgroup$
add a comment |
$begingroup$
The inequality is equivalent to $-sqrtx<-alog x$ or $sqrtx>2alogsqrtx$. Set $t=sqrtx$ and consider the function
$$
f(t)=t-2alog t
$$
Since $lim_ttoinftyf(t)=infty$, there is some number $C(a)$ such that $f(t)>0$ for $t>C(a)$. Now set $B(a)=C(a)^2$.
answered Apr 5 at 22:41
egregegreg
186k1486208
$endgroup$
The inequality is equivalent to $-sqrtx<-alog x$ or $sqrtx>2alogsqrtx$. Set $t=sqrtx$ and consider the function
$$
f(t)=t-2alog t
$$
Since $lim_ttoinftyf(t)=infty$, there is some number $C(a)$ such that $f(t)>0$ for $t>C(a)$. Now set $B(a)=C(a)^2$.
answered Apr 5 at 22:41
egregegreg
186k1486208
answered Apr 5 at 22:41
egregegreg
186k1486208
answered Apr 5 at 22:41
egregegreg
186k1486208
answered Apr 5 at 22:41
egregegreg
186k1486208
186k1486208
add a comment |
add a comment |
$begingroup$
Rewrite our inequality in the following form:
$$fracsqrtxlnx>a.$$
Now, prove that $fracsqrtxlnx>lnx$ for all $x>5510$ and we can take
$$B(a)=maxleft5510,e^aright+1$$
answered Apr 5 at 21:20
Michael RozenbergMichael Rozenberg
110k1896201
$endgroup$
add a comment |
$begingroup$
Rewrite our inequality in the following form:
$$fracsqrtxlnx>a.$$
Now, prove that $fracsqrtxlnx>lnx$ for all $x>5510$ and we can take
$$B(a)=maxleft5510,e^aright+1$$
answered Apr 5 at 21:20
Michael RozenbergMichael Rozenberg
110k1896201
$endgroup$
add a comment |
$begingroup$
Rewrite our inequality in the following form:
$$fracsqrtxlnx>a.$$
Now, prove that $fracsqrtxlnx>lnx$ for all $x>5510$ and we can take
$$B(a)=maxleft5510,e^aright+1$$
answered Apr 5 at 21:20
Michael RozenbergMichael Rozenberg
110k1896201
$endgroup$
Rewrite our inequality in the following form:
$$fracsqrtxlnx>a.$$
Now, prove that $fracsqrtxlnx>lnx$ for all $x>5510$ and we can take
$$B(a)=maxleft5510,e^aright+1$$
answered Apr 5 at 21:20
Michael RozenbergMichael Rozenberg
110k1896201
edited Apr 6 at 4:05
answered Apr 5 at 21:20
Michael RozenbergMichael Rozenberg
110k1896201
answered Apr 5 at 21:20
Michael RozenbergMichael Rozenberg
110k1896201
answered Apr 5 at 21:20
Michael RozenbergMichael Rozenberg
110k1896201
110k1896201
add a comment |
add a comment |
CtSm0x1 is a new contributor. Be nice, and check out our Code of Conduct.
CtSm0x1 is a new contributor. Be nice, and check out our Code of Conduct.
CtSm0x1 is a new contributor. Be nice, and check out our Code of Conduct.
CtSm0x1 is a new contributor. Be nice, and check out our Code of Conduct.
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.
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%2f3176421%2f0e-sqrt-x-frac1xa-inequality%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
$begingroup$
The square root doesn't really matter: if you put $y=sqrtx$, you can focus on prooving that $e^-y<frac1y^2a$ for $y$ large enough. To show that, try to calculate the limit $lim_yrightarrowinfty frace^yy^2a$. Suggestion: use l'Hopital rule.
$endgroup$
– Adam Latosiński
Apr 5 at 21:31