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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 Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Johanovegas Calculus 2 January 23rd, 2016 01:34 PM Chemist@ Calculus 7 November 1st, 2014 02:16 PM MageKnight Real Analysis 16 May 18th, 2013 01:33 PM ali Real Analysis 1 April 3rd, 2010 09:28 PM problem Real Analysis 1 April 3rd, 2010 02:41 AM

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