My Math Forum sum of infinite series

 Algebra Pre-Algebra and Basic Algebra Math Forum

 February 18th, 2016, 01:38 PM #1 Member   Joined: Dec 2015 From: Europe Posts: 42 Thanks: 1 sum of infinite series $\displaystyle \sqrt{i+\sqrt{i+\sqrt{i+...}}}$ infinite line. What is this expression equal to? Last edited by skipjack; February 18th, 2016 at 04:52 PM.
 February 18th, 2016, 01:52 PM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,655 Thanks: 2633 Math Focus: Mainly analysis and algebra Assuming that the series converges to a value $s$, we have \newcommand{\i}{\mathrm i}\begin{aligned} s&=\sqrt{\i +\sqrt{\i +\sqrt{\i +\ldots}}} \\ s^2-\i&=\sqrt{\i +\sqrt{\i +\ldots}}= s\end{aligned} Now solve the quadratic for $s$. Thanks from 123qwerty and topsquark
 February 18th, 2016, 04:13 PM #3 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,655 Thanks: 2633 Math Focus: Mainly analysis and algebra It's worth highlighting that the assumption made at the top of my previous post is non-trivial. Every one of those square root signs can take either of two values, given a value for the expression underneath it. Thanks from topsquark
February 19th, 2016, 06:47 AM   #4
Banned Camp

Joined: Jun 2014
From: Earth

Posts: 945
Thanks: 191

Quote:
 Originally Posted by v8archie Every one of those square root signs can take either of two values, given a value for the expression underneath it.
The square root function of an imaginary number returns one value.

Examples:

$\displaystyle \sqrt{i} \ = \ \dfrac{ i + 1}{\sqrt{2}}$

$\displaystyle -\sqrt{i} \ = \ \dfrac{- i - 1}{\sqrt{2}}$

 February 19th, 2016, 07:55 AM #5 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,655 Thanks: 2633 Math Focus: Mainly analysis and algebra Actually, the principal square root function has only one value. But you are correct that it is the function referenced in the given expression. Last edited by skipjack; February 19th, 2016 at 10:09 PM.
 February 19th, 2016, 08:57 AM #6 Member   Joined: Dec 2015 From: Europe Posts: 42 Thanks: 1 Thanks so for any infinite series written by this expression. We can make it into a quadratic equation which only has on real solution (because the other is negative) ?
 February 19th, 2016, 09:42 AM #7 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,655 Thanks: 2633 Math Focus: Mainly analysis and algebra This doesn't have any real solutions.
February 19th, 2016, 05:09 PM   #8
Senior Member

Joined: Oct 2013
From: New York, USA

Posts: 639
Thanks: 85

Quote:
 Originally Posted by Inflekx12 Thanks so for any infinite series written by this expression. We can make it into a quadratic equation which only has on real solution (because the other is negative) ?
A quadratic equation can have both solutions real or both solutions complex, but not one of each.

 February 19th, 2016, 06:01 PM #9 Math Team     Joined: May 2013 From: The Astral plane Posts: 2,156 Thanks: 876 Math Focus: Wibbly wobbly timey-wimey stuff. Now here's the bonus question: If we have two complex solutions how do we know which one is correct? They both can't be. Yes, I know we need to check each solution, but is there a way to check without calculating the series itself? -Dan
 February 19th, 2016, 06:18 PM #10 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,655 Thanks: 2633 Math Focus: Mainly analysis and algebra The principal root of $z= r\mathrm e^{\mathrm i \theta}$ where $-\pi \lt \theta \le \pi$ is $\sqrt z = \sqrt r \mathrm e^{\mathrm i \theta \over 2}$. Of course, the secondary root should also work as a solution, but I have my doubts that you'd be able to switch between principal and secondary roots and still get convergent expression. Not that we are guaranteed a convergent expression when we select one root consistently, but it does seem more likely. You might also note that calculating the expression itself is non-trivial. It's not like a convergent infinite series where we can select how many terms to calculate, but those terms are complete, and the remainder get smaller and smaller in magnitude. In this expression, you need the infinite part to calculate any of it.

 Tags infinite, series, sum

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post Dragonkiller Complex Analysis 4 August 2nd, 2012 02:30 PM oddlogic Calculus 7 April 1st, 2011 04:49 PM oddlogic Calculus 3 March 31st, 2011 06:46 PM parmar Real Analysis 2 December 2nd, 2010 03:30 PM Salient Calculus 1 May 5th, 2009 04:48 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top