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The task is to find the the value of this expression:
$$\sqrt{7\sqrt{7\sqrt{7\sqrt{7\,\cdots}}}}$$

We can assume
$$\sqrt{7\sqrt{7\sqrt{7\sqrt{7....}}}} = x$$ Then we replace the nested part with $x$ $$\sqrt{7x} = x$$ Then square both sides and solve the equation
$$\begin{align} 7x &= x^2 \\[4pt] x^2 - 7x &= 0 \\[4pt] x(x - 7) &= 0 \\[1em] x_1 = 7,\quad& x_2 = 0 \end{align}$$

The problem is I assume $0$ is not a valid solution, but I can't explain why. It almost makes sense it could be $0$ as the numbers get progressively smaller.

Can someone explain how can we reject the solution $0$?

Edit: I think I understand it better now. But what I'm most confused about is why we get the invalid solutions here in the first place. Could it be that the problem is in the beginning where we could write the equation in different ways, like $\sqrt{7\sqrt{7x}} = x ?$ This way we would get 4 different solutions.

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    $\begingroup$ Sometimes, where the sequence converges depends greatly on what you consider to be a partial sum $\endgroup$
    – FShrike
    Aug 1 '21 at 11:37
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    $\begingroup$ You could show it can't be zero by showing that the sequence $a_{n+1}=\sqrt{7a_n}, a_1=\sqrt{7}$ is increasing $\endgroup$
    – A. Goodier
    Aug 1 '21 at 11:43
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    $\begingroup$ Or write as $7^{\frac12+\frac14+\frac18+\ldots}=7^{\sum_{k=1}^{\infty} (\frac12)^k}$ to see that it's $7$ $\endgroup$
    – A. Goodier
    Aug 1 '21 at 11:43
  • $\begingroup$ Also see math.stackexchange.com/questions/1136748/… $\endgroup$
    – A. Goodier
    Aug 1 '21 at 11:54
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    $\begingroup$ As A.Goodier noted, as soon as we realize that the expression is $7$ in some non-negative power, we can conclude that it can't be zero $\endgroup$
    – Vasya
    Aug 1 '21 at 12:14
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Your substitution has no sense. It is just a formal one.

Unless you define properly what

$$ \sqrt{7\sqrt{7\sqrt{7\sqrt{7\sqrt{7...}}}}}$$

means, you cannot make your substitution and expect to get any answer, let alone the correct one. The one that you got is possibly true, but not necessarily true.

There are many infinite expressions that do not have single possible interpretation. Some of them are infinite, some of them are undefined etc.

So what does

$$ \sqrt{7\sqrt{7\sqrt{7\sqrt{7\sqrt{7...}}}}}$$

mean?

  1. Suppose that you want to say, it is

$$ \sqrt{7},\sqrt{7\sqrt{7}},\sqrt{7\sqrt{7\sqrt{7}}},...$$

taken to infinity. Now since:

$$ \sqrt{7}<\sqrt{7\sqrt{7}}<\sqrt{7\sqrt{7\sqrt{7}}}<...$$

the solution cannot be $0$. It has to be a number that is greater than $\sqrt{7}$ if this series converges at all.

  1. Another way of defining your expression is

$$ \sqrt{a},\sqrt{7\sqrt{a}},\sqrt{7\sqrt{7\sqrt{a}}},...$$

and now arguing what effect $a$ has when we take this to infinity. Since:

$$\sqrt{a}=a^{\frac1{2}},\sqrt{7\sqrt{a}}=\sqrt{7}a^{\frac1{2^2}},\sqrt{7\sqrt{7\sqrt{a}}}=\sqrt{7\sqrt{7}}a^{\frac1{2^3}},...$$

we can see that both values 1. and 2., if they converge at all, are the same except when $a=0$.

So, what you are talking about is the rule that if a series: $f_1, f_2, f_3...$ has an accumulation point then it is one of the solutions of $f_n=f_{n+1}$ taken at infinity. This is exactly what you are doing. But then it is the 2. definition that you are actually using. You are saying if $f_n=x$ then $f_{n+1}=\sqrt{7x}$, right?

And indeed you got that one of the possible solutions is $7$ and another is $0$. Notice that $\sqrt{7x}=x$ does not exclude having any $a$ on its own. The same equation is valid for whatever $a$ you take, hence it can be $a=0$. Not to forget that even after you've got $7$ by this method, you have to prove that the series converges.

If you take the version number 1., you cannot use $\sqrt{7x}=x$ since that is not how you create the next element. You create the next element by multiplying the previous one by $7^{\frac1{2^{n}}}$, as this is how you nest in the next $7$ deeper inside. So it is:

$$f_1=7^{\frac1{2}}$$ $$f_2=7^{\frac1{2}}7^{\frac1{2^2}}$$ $$f_{n+1}=f_n 7^{\frac1{2^{n+1}}}$$

In this version, it does not help if you use $f_{n+1}=f_n$ at infinity since it is giving $$x=x 7^{\frac1{2^{\infty}}}$$

or

$$x=x$$

but if you use another method of unwrapping you have:

$$f_{n+1}=\prod\limits_{k=1}^{n+1}7^{\frac1{2^{k}}}$$

$$f_{n+1}=7^{\sum\limits_{k=1}^{n+1}\frac1{2^{k}}}$$

or at infinity

$$7^{\sum\limits_{k=1}^{\infty}\frac1{2^{k}}}=7$$

So you are using the correct method that is giving you the solution based on initial conditions, i.e. $a$, while you are not aware that you need this initial condition in the first place.

You do. Even then not all solutions of $f_{n+1}=f_{n}$ are necessarily the true accumulation point. The rule says if you have an accumulation point then it is one of the solutions. On top of that, in order to prove that this accumulation point is the solution, you need to prove that the series converges. Proving that the series converges is done separately from finding the solution in virtually all cases. Handling 2. is giving the obvious clue of how this is to be done. It is just reducing it all to $\sum\limits_{k=1}^{\infty}\frac1{2^{k}}$ which we did already. You just need to add the argument about how $a$ behaves at infinity, and the job done.

So your obvious definition may have two interpretations, one of which, indeed, may have $0$ as a solution.

Edit

To understand the complexity, take that you have actually meant:

$$\sqrt{7}, \sqrt{7\sqrt{7\sqrt{7}}}, \sqrt{7\sqrt{7\sqrt{7\sqrt{7\sqrt{7}}}}},...$$

expanding in both directions. It looks "the same" at infinity, right? And where would this go to? Using your rule it is:

$$x=\sqrt{7\sqrt{x\sqrt{7}}}$$

and this has (surprise, surprise,) two possible solutions: $0$ and $7^{\frac{5}{6}}$.

Pick some other arrangement over how this is extended, and do not be surprised with the solution that you may get. You can easily construct the one that is not going to converge at all.

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    $\begingroup$ Excellent answer! I particularly like your last example, as it is a very clear illustration of the kind of meaningless consequences we can get from ill-defined expressions! $\endgroup$
    – user21820
    Aug 1 '21 at 17:25
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The infinite expression in question is generally interpreted as meaning the limit of the recursive sequence $$ x_{0} = \sqrt{7},\qquad x_{n+1} = \sqrt{7x_{n}}\quad\text{for $n \geq 0$.} $$ Since

  • $x_{0} \geq 1$, and
  • For each $n \geq 0$, $x_{n} \geq 1$ implies $x_{n+1} = \sqrt{7x_{n}} \geq \sqrt{7} \geq 1$,

induction on $n$ implies $x_{n} \geq 1$ for all $n$, so the limit is no smaller than $1$. That rules out $0$ as a limit.

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Another solution would be to notice that $nth$ term of this sequence is given by $$7^{x_{n}}$$ where $$x_n = 1/2 +1/2^{2} + ... + 1/2^{n}$$ Since power is a continuous function and $x_n$ tends to 1 we have the claim.

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    $\begingroup$ You're not answering the actual question: "Can someone explain how can we reject the solution 0?" $\endgroup$
    – user21820
    Aug 1 '21 at 12:50
  • $\begingroup$ Read the solution carefully please - power function is continuous and $x_n$ tends to $1$. $\endgroup$
    – Salcio
    Aug 1 '21 at 14:29
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    $\begingroup$ The question is "how can we reject 0 in my attempted solution", not Salcio's solution. $\endgroup$
    – user21820
    Aug 1 '21 at 14:32
  • $\begingroup$ If it is your concern then one simple reason is that each term of the sequence is bigger than $sqrt(7)$ . The proof is by induction. Once you establish this you should conclude that zero can not be the limit. $\endgroup$
    – Salcio
    Aug 1 '21 at 15:02
  • $\begingroup$ Sure, I know it's easy, but you should include that in your answer. By the way, I'm not the downvoter haha.. $\endgroup$
    – user21820
    Aug 1 '21 at 15:04
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Extraneous solutions are usually a result of an algebraic manipulation of an equation. In this case, by squaring the equation, you are forcing it to have “two” solutions. This is the reason why it pays to check if solutions make sense.

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    $\begingroup$ The squaring did not force any solutions in this case. Already at $\sqrt{7x} = x$ are both $x= 0$ and $x= 7$ solutions. $\endgroup$ Aug 1 '21 at 16:28
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let $(x_n)_{n\in\mathbb N}$ be a sequence defined in the following form :

$\forall n\in \mathbb N : x_n=a^{\frac{2^n-1}{2^n}} . a\in\mathbb N^*$

it's easy to prove this sequence is convergent and its limit is $a$ as $n\longrightarrow \infty$

$\lim_{n\to\infty} x_n=a^{\lim_{n\to\infty}\frac{2^n-1}{2^n}}=a$

so we put $a=7$ we fin $\lim_{n\to\infty}x_n=\lim_{n\to\infty}(7^{\frac{2^n-1}{2^n}})=\sqrt{7\sqrt{7\sqrt{7\sqrt.....}}}=7$

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