Mixing
Proving the sequence $(x_n)$ ($n in mathbb N$) with $x_0$ $in$ ($0$,$frac{1}{2}$) and...
$begingroup$
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
edited Jan 3 at 7:59
Saad
19.7k92352
asked Jan 3 at 7:50
Radu GabrielRadu Gabriel
463
$endgroup$
add a comment |
$begingroup$
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
edited Jan 3 at 7:59
Saad
19.7k92352
asked Jan 3 at 7:50
Radu GabrielRadu Gabriel
463
$endgroup$
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
add a comment |
$begingroup$
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
edited Jan 3 at 7:59
Saad
19.7k92352
asked Jan 3 at 7:50
Radu GabrielRadu Gabriel
463
$endgroup$
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
edited Jan 3 at 7:59
Saad
19.7k92352
asked Jan 3 at 7:50
Radu GabrielRadu Gabriel
463
edited Jan 3 at 7:59
Saad
19.7k92352
asked Jan 3 at 7:50
Radu GabrielRadu Gabriel
463
edited Jan 3 at 7:59
Saad
19.7k92352
edited Jan 3 at 7:59
Saad
19.7k92352
edited Jan 3 at 7:59
Saad
19.7k92352
19.7k92352
asked Jan 3 at 7:50
Radu GabrielRadu Gabriel
463
asked Jan 3 at 7:50
Radu GabrielRadu Gabriel
463
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
add a comment |
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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 $0$.
answered Jan 3 at 8:18
mathcounterexamples.netmathcounterexamples.net
25.8k21953
$endgroup$
$begingroup$
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
add a comment |
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$begingroup$
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 $0$.
answered Jan 3 at 8:18
mathcounterexamples.netmathcounterexamples.net
25.8k21953
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $0$.
answered Jan 3 at 8:18
mathcounterexamples.netmathcounterexamples.net
25.8k21953
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $0$.
answered Jan 3 at 8:18
mathcounterexamples.netmathcounterexamples.net
25.8k21953
$endgroup$
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 $0$.
answered Jan 3 at 8:18
mathcounterexamples.netmathcounterexamples.net
25.8k21953
edited Jan 3 at 8:26
answered Jan 3 at 8:18
mathcounterexamples.netmathcounterexamples.net
25.8k21953
answered Jan 3 at 8:18
mathcounterexamples.netmathcounterexamples.net
25.8k21953
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
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