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












1












$begingroup$



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







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  • 1




    $begingroup$
    So $x_{n+1}=g(x_n)$ where $g(x)=x^2e^x$. For what values of $x$, $g(x)<x$ holds?
    $endgroup$
    – Song
    Jan 3 at 8:05
















1












$begingroup$



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











$endgroup$








  • 1




    $begingroup$
    So $x_{n+1}=g(x_n)$ where $g(x)=x^2e^x$. For what values of $x$, $g(x)<x$ holds?
    $endgroup$
    – Song
    Jan 3 at 8:05














1












1








1





$begingroup$



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











$endgroup$





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 Jan 3 at 7:59









Saad

19.7k92352




19.7k92352










asked Jan 3 at 7:50









Radu GabrielRadu Gabriel

463




463








  • 1




    $begingroup$
    So $x_{n+1}=g(x_n)$ where $g(x)=x^2e^x$. For what values of $x$, $g(x)<x$ holds?
    $endgroup$
    – Song
    Jan 3 at 8:05














  • 1




    $begingroup$
    So $x_{n+1}=g(x_n)$ where $g(x)=x^2e^x$. For what values of $x$, $g(x)<x$ holds?
    $endgroup$
    – Song
    Jan 3 at 8:05








1




1




$begingroup$
So $x_{n+1}=g(x_n)$ where $g(x)=x^2e^x$. For what values of $x$, $g(x)<x$ holds?
$endgroup$
– Song
Jan 3 at 8:05




$begingroup$
So $x_{n+1}=g(x_n)$ where $g(x)=x^2e^x$. For what values of $x$, $g(x)<x$ holds?
$endgroup$
– Song
Jan 3 at 8:05










1 Answer
1






active

oldest

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2












$begingroup$

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











$endgroup$













  • $begingroup$
    You've helped me a lot. Just to be sure though, my logic was right but the answer wouldn't have been enough, right?
    $endgroup$
    – Radu Gabriel
    Jan 3 at 8:26






  • 1




    $begingroup$
    @RaduGabriel Yes, you started well but you're missing precise arguments to prove convergence.
    $endgroup$
    – mathcounterexamples.net
    Jan 3 at 9:00











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1 Answer
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active

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2












$begingroup$

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











$endgroup$













  • $begingroup$
    You've helped me a lot. Just to be sure though, my logic was right but the answer wouldn't have been enough, right?
    $endgroup$
    – Radu Gabriel
    Jan 3 at 8:26






  • 1




    $begingroup$
    @RaduGabriel Yes, you started well but you're missing precise arguments to prove convergence.
    $endgroup$
    – mathcounterexamples.net
    Jan 3 at 9:00
















2












$begingroup$

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











$endgroup$













  • $begingroup$
    You've helped me a lot. Just to be sure though, my logic was right but the answer wouldn't have been enough, right?
    $endgroup$
    – Radu Gabriel
    Jan 3 at 8:26






  • 1




    $begingroup$
    @RaduGabriel Yes, you started well but you're missing precise arguments to prove convergence.
    $endgroup$
    – mathcounterexamples.net
    Jan 3 at 9:00














2












2








2





$begingroup$

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











$endgroup$



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 Jan 3 at 8:26

























answered Jan 3 at 8:18









mathcounterexamples.netmathcounterexamples.net

25.8k21953




25.8k21953












  • $begingroup$
    You've helped me a lot. Just to be sure though, my logic was right but the answer wouldn't have been enough, right?
    $endgroup$
    – Radu Gabriel
    Jan 3 at 8:26






  • 1




    $begingroup$
    @RaduGabriel Yes, you started well but you're missing precise arguments to prove convergence.
    $endgroup$
    – mathcounterexamples.net
    Jan 3 at 9:00


















  • $begingroup$
    You've helped me a lot. Just to be sure though, my logic was right but the answer wouldn't have been enough, right?
    $endgroup$
    – Radu Gabriel
    Jan 3 at 8:26






  • 1




    $begingroup$
    @RaduGabriel Yes, you started well but you're missing precise arguments to prove convergence.
    $endgroup$
    – mathcounterexamples.net
    Jan 3 at 9:00
















$begingroup$
You've helped me a lot. Just to be sure though, my logic was right but the answer wouldn't have been enough, right?
$endgroup$
– Radu Gabriel
Jan 3 at 8:26




$begingroup$
You've helped me a lot. Just to be sure though, my logic was right but the answer wouldn't have been enough, right?
$endgroup$
– Radu Gabriel
Jan 3 at 8:26




1




1




$begingroup$
@RaduGabriel Yes, you started well but you're missing precise arguments to prove convergence.
$endgroup$
– mathcounterexamples.net
Jan 3 at 9:00




$begingroup$
@RaduGabriel Yes, you started well but you're missing precise arguments to prove convergence.
$endgroup$
– mathcounterexamples.net
Jan 3 at 9:00


















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