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$$ \sqrt{ a \sqrt{ a \sqrt{a \cdots}}}=\text{ ?} $$ options were given as

  1. $0$
  2. $-a$
  3. $a$
  4. $1$

i did not know how to solve it or what it was related to. Could anyone please explain the concept and/or provide helpful references or links?

Apple
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    Did you mean $\sqrt{a\sqrt{a\sqrt{a\ldots}}}$? – Ángel Mario Gallegos Jun 28 '15 at 17:18
  • Well, I don't like the dots there. They make everything very ambiguous. – Zardo Jun 28 '15 at 17:18
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    BTW, possible answers are 1 and 3. – Zardo Jun 28 '15 at 17:19
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    Welcome to Stackexchange. A few things. First, use a descriptive title and leave off personal information. Second, you must include your work and thoughts. Third, use proper grammar. This question will likely not be received well and is likely to be closed. –  Jun 28 '15 at 17:20
  • Hint: Rewrite it as an infinite product of powers of $a$ and take logarithms. – A.P. Jun 28 '15 at 17:26
  • If $x=\sqrt{ a \sqrt{ a \sqrt{a \cdots}}}$ then $x^2=a\sqrt{ a \sqrt{ a \sqrt{a \cdots}}}=ax$ so you have $x^2=ax$, so $x(x-a)=0$. That implies $x=0$ or $x=a$. ${}\qquad{}$ – Michael Hardy Jun 28 '15 at 17:31

3 Answers3

6

That's just $a^{1/2}\cdot a^{1/4} \cdots = a^{1/2+1/4+1/8+\cdots}$.

You can use the summation $\sum_{i=1}^{\infty} \frac{1}{2^i} = 1$ to get the answer.

Martín Forsberg Conde
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    I understand that this could be written better, but why the downvote? – A.P. Jun 28 '15 at 17:30
  • @A.P. Perhaps the downvoter thought this answer was claiming $1$ was the correct choice (I did initially). – pjs36 Jun 28 '15 at 17:34
4

Let $x_1=\sqrt{a}$, and $x_n=\sqrt{ax_{n-1}}$, for $n\ge 2$ if the limit $$x:=\lim_{n\to \infty}x_n$$ If that limit exist. We have \begin{align*} x^2&=a\lim_{n\to\infty}x_n\\ &=ax \end{align*} So, $x=0$ or $x=a$.

Ángel Mario Gallegos
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2

Supposing that $x=\sqrt{a\sqrt{a\sqrt{a\sqrt{\cdots}}}}$, then squaring both sides, we have $x^2 = a\sqrt{a\sqrt{a\sqrt{a\sqrt{\cdots}}}} = ax$. If we assume $x$ is nonzero, we may divide both sides by $x$ yielding $x=a$. Otherwise, $x=0$; this occurs when $a=0\;(=x)$, so both possible cases are encapsulated in the response $x=a$.

Kite
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  • If $x=0$ and $a=17$ then $x^2 = ax,$ so it's not clear to me why we should conclude that if $x^2=ax$ and $x=0$ then $a=0,$ as you seem to be doing. – David K Jun 29 '15 at 06:47