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February 18th, 2016, 01:38 PM   #1
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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.
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February 18th, 2016, 01:52 PM   #2
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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 $s$.
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February 18th, 2016, 04:13 PM   #3
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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.
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February 19th, 2016, 06:47 AM   #4
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Quote:
Originally Posted by v8archie View Post
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:


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


$\displaystyle -\sqrt{i} \ = \ \dfrac{- i - 1}{\sqrt{2}}$
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February 19th, 2016, 07:55 AM   #5
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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.
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February 19th, 2016, 08:57 AM   #6
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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) ?
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February 19th, 2016, 09:42 AM   #7
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This doesn't have any real solutions.
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February 19th, 2016, 05:09 PM   #8
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Originally Posted by Inflekx12 View Post
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.
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February 19th, 2016, 06:01 PM   #9
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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
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February 19th, 2016, 06:18 PM   #10
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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.
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