Radicals of 2

$\sqrt { 2\sqrt { 2\sqrt { 2\sqrt { 2 } } } } =\quad ?\\ \qquad \qquad .\\ \qquad \qquad .\\ \qquad \qquad .$

#Algebra

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

Here is how we could write out the results of each step:

$2^{\frac{1}{2}}$

$(2^{\frac{1}{2}})(2^{\frac{1}{4}})$

$(2^{\frac{1}{2}})(2^{\frac{1}{4}})(2^{\frac{1}{8}})$

and so on, each time adding another factor of $2$ , raised to half of the preceding factor's power.

If we were to continue this process to infinity, we could combine all the powers like this:

$2^{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+~ ...}$

As it turns out, the infinite sum $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+~ ...$ is a simple geometric series which can be shown to converge to $1$ . Thus, this expression simplifies to

$2^{1}$ , or just $\boxed{2}$

David Stiff - 1 year, 3 months ago

Wow! Thanks for the solution!

I changed my name. - 1 year, 3 months ago

Glad you found it helpful!

David Stiff - 1 year, 3 months ago

Nikola Alfredi - 1 year, 3 months ago

We do it by taking it x. And square it .... but we see after squaring the product on the right is 2x.... .

But, I do not like this approachas not all infinities are equal... You can see this in real life, compare the number of points of two lines of varying length, boyh have infinite points but aren't equal.

Nikola Alfredi - 1 year, 3 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$