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







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  • 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
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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











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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


















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