My Math Forum Monotone Convergence Theorem and Cauchy Convergence Criterion
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 October 25th, 2018, 08:00 PM #1 Newbie   Joined: Oct 2018 From: Indonesia Posts: 4 Thanks: 0 Math Focus: Nonlinear Waves and Differential Equations Monotone Convergence Theorem and Cauchy Convergence Criterion Hello. I would be grateful for anyone who can solve this problem for me : Show that the sequence $\left(z_n\right)$ given by $z_{n+1}=\sqrt{a+z_n}$ for $a>0, z_1>0$ converges by using Monotone Convergence Theorem (or Cauchy Convergence Criterion)!
 October 25th, 2018, 09:14 PM #2 Senior Member   Joined: Sep 2016 From: USA Posts: 669 Thanks: 440 Math Focus: Dynamical systems, analytic function theory, numerics Here is a hint: Apply the mean value theorem to successive terms $|z_{n+1} - z_{n}| = |\sqrt{a + z_n} - \sqrt{a + z_{n-1}}| \leq \frac{ |z_n - z_{n+1}|}{2\sqrt{a}}$ Thanks from gusrianputra
 October 25th, 2018, 09:15 PM #3 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,690 Thanks: 2669 Math Focus: Mainly analysis and algebra \left.\begin{aligned}z_{n+1} &= \sqrt{a+z_n} \\ z_1 &\gt 0 \end{aligned} \right\} \implies z_{n} \gt 0 \; \forall n \ge 1 This means that we can compare the magnitude of $z_{n+1}^2$ and $z_n^2$ as proxies for the magnitude of $z_{n+1}$ and $z_n$. \begin{align} z_{n+1}^2 &= a+z_n \\ z_{n+1}^2 - z_n^2 &= a + z_n - z_n^2 \end{align} The quadratic expression on the right factorises as $\big(b - z_n\big)\big((b-1) + z_n\big)$ where $b(b-1)=a$ which has a unique solution with $b \gt 1$ because $a \gt 0$. Thus we have three cases:$z_n < b$: in which case $\big(b - z_n\big)\big((b-1) + z_n\big) \gt 0$ and thus $z_{n+1}^2 > z_n^2 \implies z_{n+1} > z_n$ $z_n > b$: in which case $\big(b - z_n\big)\big((b-1) + z_n\big) \lt 0$ and thus $z_{n+1}^2 < z_n^2 \implies z_{n+1} < z_n$ $z_n = b$: in which case $\big(b - z_n\big)\big((b-1) + z_n\big) = 0$ and thus $z_{n+1}^2 = z_n^2 \implies z_{n+1} = z_n$ Now, we compare the magnitude of $z_{n+1}^2$ and $b^2$ as proxies for $z_{n+1}$ and $b$. \begin{align}z_{n+1}^2 - b^2 &= a + z_n - b^2 \\ &= b(b-1) + z_n -b^2 \\ &= z_n -b\end{align} Here,if $z_n < b$ we have $z_{n+1}^2 < b^2$ and so $z_{n+1} < b$ too. if $z_n > b$ we have $z_{n+1}^2 > b^2$ and so $z_{n+1} > b$ too. and if $z_n = b$ we have $z_{n+1}^2 = b^2$ and so $z_{n+1} = b$ too. Therefore:If $z_n < b$ we have $z_n < z_{n+1} < b$ If $z_n > b$ we have $z_n > z_{n+1} > b$ If $z_n = b$ we have $z_n = z_{n+1} = b$ Thus:If $z_1 < b$, the sequence is increasing and bounded above by $b$. If $z_1 > b$, the sequence is decreasing and bounded below by $b$. If $z_1 = b$, the sequence is constant. Thanks from gusrianputra Last edited by v8archie; October 25th, 2018 at 09:18 PM.
 October 25th, 2018, 10:06 PM #4 Newbie   Joined: Oct 2018 From: Indonesia Posts: 4 Thanks: 0 Math Focus: Nonlinear Waves and Differential Equations Thanks for such a great hint, SDK. I don't think I could use the hint for now, but of course it would be useful in my next course of advanced real analysis.
 October 25th, 2018, 10:09 PM #5 Newbie   Joined: Oct 2018 From: Indonesia Posts: 4 Thanks: 0 Math Focus: Nonlinear Waves and Differential Equations Thank you very much, v8archie. This is what I am looking for.

 Tags cauchy, convergence, criterion, monotone, theorem

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