Proving the sequence $(x_n)$ ($n in mathbb N$) with $x_0$ $in$ ($0$,$frac{1}{2}$) and...












1















Let $f:mathbb R^*tomathbb R$ be $f(x)=dfrac{e^frac1x}{x^2}$. Prove that the sequence $(x_n)$ ($ninmathbb N$) with $$x_0inleft(0,frac12right), x_{n+1}=fleft(frac1{x_n}right)$$ is bounded and $limlimits_{ntoinfty}x_n=0$.




I think the problem can be solved working from the inequality $0lt x_0 lt frac{1}{2}$ and doing that I realize that the upper bound of the terms shrinks and that should be "clear" but what would be a mathematically accepted proof?



Just to make sure I am no getting anything wrong here is my "calculations":
$0lt x_0 lt 1/2$

Raising terms to the power of -$1$:
$2lt frac{1}{x_0}ltinfty$

Applying f to every term:
$0lt x_1lt f(2)$

Then doing the same thing again results:
$0lt x_2lt fbiggl(frac{4}{e^{frac{1}{2}}}biggr)$
$frac{4}{e^{frac{1}{2}}} gt 2$ and given that $f$ is decreasing for any $x gt -1/2$ then ultimately the value of the upper bound is actually shrinking.



I am guessing this is not enough to prove the convergence of the sequence, right?






real-analysis limits







share|cite|improve this question




















  • 1




    So $x_{n+1}=g(x_n)$ where $g(x)=x^2e^x$. For what values of $x$, $g(x)<x$ holds?
    – Song
    6 hours ago
















1















Let $f:mathbb R^*tomathbb R$ be $f(x)=dfrac{e^frac1x}{x^2}$. Prove that the sequence $(x_n)$ ($ninmathbb N$) with $$x_0inleft(0,frac12right), x_{n+1}=fleft(frac1{x_n}right)$$ is bounded and $limlimits_{ntoinfty}x_n=0$.




I think the problem can be solved working from the inequality $0lt x_0 lt frac{1}{2}$ and doing that I realize that the upper bound of the terms shrinks and that should be "clear" but what would be a mathematically accepted proof?



Just to make sure I am no getting anything wrong here is my "calculations":
$0lt x_0 lt 1/2$

Raising terms to the power of -$1$:
$2lt frac{1}{x_0}ltinfty$

Applying f to every term:
$0lt x_1lt f(2)$

Then doing the same thing again results:
$0lt x_2lt fbiggl(frac{4}{e^{frac{1}{2}}}biggr)$
$frac{4}{e^{frac{1}{2}}} gt 2$ and given that $f$ is decreasing for any $x gt -1/2$ then ultimately the value of the upper bound is actually shrinking.



I am guessing this is not enough to prove the convergence of the sequence, right?






real-analysis limits







share|cite|improve this question




















  • 1




    So $x_{n+1}=g(x_n)$ where $g(x)=x^2e^x$. For what values of $x$, $g(x)<x$ holds?
    – Song
    6 hours ago














1












1








1








Let $f:mathbb R^*tomathbb R$ be $f(x)=dfrac{e^frac1x}{x^2}$. Prove that the sequence $(x_n)$ ($ninmathbb N$) with $$x_0inleft(0,frac12right), x_{n+1}=fleft(frac1{x_n}right)$$ is bounded and $limlimits_{ntoinfty}x_n=0$.




I think the problem can be solved working from the inequality $0lt x_0 lt frac{1}{2}$ and doing that I realize that the upper bound of the terms shrinks and that should be "clear" but what would be a mathematically accepted proof?



Just to make sure I am no getting anything wrong here is my "calculations":
$0lt x_0 lt 1/2$

Raising terms to the power of -$1$:
$2lt frac{1}{x_0}ltinfty$

Applying f to every term:
$0lt x_1lt f(2)$

Then doing the same thing again results:
$0lt x_2lt fbiggl(frac{4}{e^{frac{1}{2}}}biggr)$
$frac{4}{e^{frac{1}{2}}} gt 2$ and given that $f$ is decreasing for any $x gt -1/2$ then ultimately the value of the upper bound is actually shrinking.



I am guessing this is not enough to prove the convergence of the sequence, right?






real-analysis limits







share|cite|improve this question
















Let $f:mathbb R^*tomathbb R$ be $f(x)=dfrac{e^frac1x}{x^2}$. Prove that the sequence $(x_n)$ ($ninmathbb N$) with $$x_0inleft(0,frac12right), x_{n+1}=fleft(frac1{x_n}right)$$ is bounded and $limlimits_{ntoinfty}x_n=0$.




I think the problem can be solved working from the inequality $0lt x_0 lt frac{1}{2}$ and doing that I realize that the upper bound of the terms shrinks and that should be "clear" but what would be a mathematically accepted proof?



Just to make sure I am no getting anything wrong here is my "calculations":
$0lt x_0 lt 1/2$

Raising terms to the power of -$1$:
$2lt frac{1}{x_0}ltinfty$

Applying f to every term:
$0lt x_1lt f(2)$

Then doing the same thing again results:
$0lt x_2lt fbiggl(frac{4}{e^{frac{1}{2}}}biggr)$
$frac{4}{e^{frac{1}{2}}} gt 2$ and given that $f$ is decreasing for any $x gt -1/2$ then ultimately the value of the upper bound is actually shrinking.



I am guessing this is not enough to prove the convergence of the sequence, right?






real-analysis limits




real-analysis limits






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 6 hours ago









Saad

19.7k92252




19.7k92252










asked 6 hours ago









Radu Gabriel

263




263








  • 1




    So $x_{n+1}=g(x_n)$ where $g(x)=x^2e^x$. For what values of $x$, $g(x)<x$ holds?
    – Song
    6 hours ago














  • 1




    So $x_{n+1}=g(x_n)$ where $g(x)=x^2e^x$. For what values of $x$, $g(x)<x$ holds?
    – Song
    6 hours ago








1




1




So $x_{n+1}=g(x_n)$ where $g(x)=x^2e^x$. For what values of $x$, $g(x)<x$ holds?
– Song
6 hours ago




So $x_{n+1}=g(x_n)$ where $g(x)=x^2e^x$. For what values of $x$, $g(x)<x$ holds?
– Song
6 hours ago










1 Answer
1






active

oldest

votes


















2














Let's denote $g(x)=fleft(frac{1}{x}right)=x^2 e^x$.



We have to study the sequence
$$x_{n+1} = g(x_n)$$



where $x_0 in (0,1/2)$.



You can prove that $g$ is strictly increasing on $[0,1/2)$, that $g(x) <x$ on $(0,1/2)$ and that the equation $g(x) = x$ has $0$ for unique solution on the interval $[0,1/2)$.



Hence for $x_0 in (0,1/2)$, the sequence $(x_n)$ is strictly decreasing and bounded below by $0$. Therefore it converges. As $g$ is continuous, $(x_n)$ converges to a fix point of $g$, that is towards $0$.






share|cite|improve this answer























  • You've helped me a lot. Just to be sure though, my logic was right but the answer wouldn't have been enough, right?
    – Radu Gabriel
    5 hours ago






  • 1




    @RaduGabriel Yes, you started well but you're missing precise arguments to prove convergence.
    – mathcounterexamples.net
    5 hours ago











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


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3060343%2fproving-the-sequence-x-n-n-in-mathbb-n-with-x-0-in-0-frac1%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









2














Let's denote $g(x)=fleft(frac{1}{x}right)=x^2 e^x$.



We have to study the sequence
$$x_{n+1} = g(x_n)$$



where $x_0 in (0,1/2)$.



You can prove that $g$ is strictly increasing on $[0,1/2)$, that $g(x) <x$ on $(0,1/2)$ and that the equation $g(x) = x$ has $0$ for unique solution on the interval $[0,1/2)$.



Hence for $x_0 in (0,1/2)$, the sequence $(x_n)$ is strictly decreasing and bounded below by $0$. Therefore it converges. As $g$ is continuous, $(x_n)$ converges to a fix point of $g$, that is towards $0$.






share|cite|improve this answer























  • You've helped me a lot. Just to be sure though, my logic was right but the answer wouldn't have been enough, right?
    – Radu Gabriel
    5 hours ago






  • 1




    @RaduGabriel Yes, you started well but you're missing precise arguments to prove convergence.
    – mathcounterexamples.net
    5 hours ago
















2














Let's denote $g(x)=fleft(frac{1}{x}right)=x^2 e^x$.



We have to study the sequence
$$x_{n+1} = g(x_n)$$



where $x_0 in (0,1/2)$.



You can prove that $g$ is strictly increasing on $[0,1/2)$, that $g(x) <x$ on $(0,1/2)$ and that the equation $g(x) = x$ has $0$ for unique solution on the interval $[0,1/2)$.



Hence for $x_0 in (0,1/2)$, the sequence $(x_n)$ is strictly decreasing and bounded below by $0$. Therefore it converges. As $g$ is continuous, $(x_n)$ converges to a fix point of $g$, that is towards $0$.






share|cite|improve this answer























  • You've helped me a lot. Just to be sure though, my logic was right but the answer wouldn't have been enough, right?
    – Radu Gabriel
    5 hours ago






  • 1




    @RaduGabriel Yes, you started well but you're missing precise arguments to prove convergence.
    – mathcounterexamples.net
    5 hours ago














2












2








2






Let's denote $g(x)=fleft(frac{1}{x}right)=x^2 e^x$.



We have to study the sequence
$$x_{n+1} = g(x_n)$$



where $x_0 in (0,1/2)$.



You can prove that $g$ is strictly increasing on $[0,1/2)$, that $g(x) <x$ on $(0,1/2)$ and that the equation $g(x) = x$ has $0$ for unique solution on the interval $[0,1/2)$.



Hence for $x_0 in (0,1/2)$, the sequence $(x_n)$ is strictly decreasing and bounded below by $0$. Therefore it converges. As $g$ is continuous, $(x_n)$ converges to a fix point of $g$, that is towards $0$.






share|cite|improve this answer














Let's denote $g(x)=fleft(frac{1}{x}right)=x^2 e^x$.



We have to study the sequence
$$x_{n+1} = g(x_n)$$



where $x_0 in (0,1/2)$.



You can prove that $g$ is strictly increasing on $[0,1/2)$, that $g(x) <x$ on $(0,1/2)$ and that the equation $g(x) = x$ has $0$ for unique solution on the interval $[0,1/2)$.



Hence for $x_0 in (0,1/2)$, the sequence $(x_n)$ is strictly decreasing and bounded below by $0$. Therefore it converges. As $g$ is continuous, $(x_n)$ converges to a fix point of $g$, that is towards $0$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 5 hours ago

























answered 5 hours ago









mathcounterexamples.net

24.6k21753




24.6k21753












  • You've helped me a lot. Just to be sure though, my logic was right but the answer wouldn't have been enough, right?
    – Radu Gabriel
    5 hours ago






  • 1




    @RaduGabriel Yes, you started well but you're missing precise arguments to prove convergence.
    – mathcounterexamples.net
    5 hours ago


















  • You've helped me a lot. Just to be sure though, my logic was right but the answer wouldn't have been enough, right?
    – Radu Gabriel
    5 hours ago






  • 1




    @RaduGabriel Yes, you started well but you're missing precise arguments to prove convergence.
    – mathcounterexamples.net
    5 hours ago
















You've helped me a lot. Just to be sure though, my logic was right but the answer wouldn't have been enough, right?
– Radu Gabriel
5 hours ago




You've helped me a lot. Just to be sure though, my logic was right but the answer wouldn't have been enough, right?
– Radu Gabriel
5 hours ago




1




1




@RaduGabriel Yes, you started well but you're missing precise arguments to prove convergence.
– mathcounterexamples.net
5 hours ago




@RaduGabriel Yes, you started well but you're missing precise arguments to prove convergence.
– mathcounterexamples.net
5 hours ago


















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.





Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


Please pay close attention to the following guidance:


  • 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.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3060343%2fproving-the-sequence-x-n-n-in-mathbb-n-with-x-0-in-0-frac1%23new-answer', 'question_page');
}
);

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







-

Popular posts from this blog

1300-talet

Image
Den här artikeln handlar om hundraårsperioden 1300–1399. För decenniet 1300–1309, se 1300-talet (decennium). .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output
Read more

1300-talet

Image
Den här artikeln handlar om hundraårsperioden 1300–1399. För decenniet 1300–1309, se 1300-talet (decennium). .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output
Read more

Has there ever been an instance of an active nuclear power plant within or near a war zone?

Image
36 3 The question Has there ever been a military combat going on around an active (running or in outage but not decommissioned) commercial nuclear power plant? Or dangerously near it? If yes, how did the plant and its personnel fare in such situation? Why I ask Let's assume for the scope of this question modern commercial nuclear power plants (CANDU, PWR, BWR of generation II plus all of gen III and newer) are usually reasonably safe to operate given that the engineering has been done right and there is a whole cohort of very well trained personnel on-site at all times, well rested and with considerable resources on their hands. Things hard to get with an armed conflict like the recent Syrian war raging around. I can't seem to find any reference or a comprehensive article about this, hence this question.
Read more