Let the following expression, with $n \in \mathbb{N}$
$$ T_n = \underbrace{\sqrt{x \sqrt{x \sqrt{x \dots \sqrt{x}}}}}_{\text{n times}} $$
It's easy to see that
$$\lim_{n \to \infty} T_n = x$$
Find all x where that limit is true
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1What do you mean by 'assigning a numerical set for $x$'? – Winther Mar 01 '15 at 23:50
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1I do not know what you mean, by assign a numerical set, do you mean find all x where that limit is true? What exactly are you asking for? – Bill Trok Mar 01 '15 at 23:51
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yes Bill Trok, sorry i'm not used to write math in english. – Gustavo Viana Mar 01 '15 at 23:58
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What i'm trying to say is: is this valid for N, R, or C? – Gustavo Viana Mar 02 '15 at 00:00
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The relation $T_{n+1} = \sqrt{xT_n}$ together with induction can be useful to determine this. – Winther Mar 02 '15 at 00:00
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You may observe that $$ T_n = \underbrace{\sqrt{x \sqrt{x \sqrt{x \dots \sqrt{x}}}}}_{\text{n times}}=x^{\large \frac12+\frac1{2^2}+...+\frac{1}{2^n}}=x^{\large 1-\frac{1}{2^n}}.$$ Hence for any complex number with non negative real part $x$ you have $$\lim_{n \to \infty} T_n = x.$$
Olivier Oloa
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$x=0$ too ... and (if we use the principal square roots) any complex number with nonnegative real part. – hmakholm left over Monica Mar 02 '15 at 00:06
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@kobe For me "positive" doesn't mean "strictly positive". Does it? – Olivier Oloa Mar 02 '15 at 00:07
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1In English, "positive" means $>0$. This is distinct from the French word "positif" which means $\ge 0$. – hmakholm left over Monica Mar 02 '15 at 00:10
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1The "hence" doesn't quite follow in the case of complex $x$; there a more ad-hoc argument is needed. – hmakholm left over Monica Mar 02 '15 at 00:13
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@HenningMakholm I don't see the point since in the complex case you have $\displaystyle x^{\large 1-\frac{1}{2^n}}=exp\left((1-\frac{1}{2^n})\log x\right)$ with $\Re x > 0$ giving the conclusion (for $x=0$, the result being direct). – Olivier Oloa Mar 02 '15 at 00:20
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1x @Olivier: Not so fast. Significantly more than that is needed to argue that the multi-valued nature of the complex square roots and logarithms will not create problems. As one example, the straightforward derivation of $x^{1-2^{-n}}$ that works in the real case depends on the rule $\sqrt{ab}=\sqrt a \sqrt b$ -- but that isn't true in general for complex numbers. – hmakholm left over Monica Mar 02 '15 at 00:36
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@HenningMakholm Of course it is not true in general, but here we have considered the principal value of the logarithm. – Olivier Oloa Mar 02 '15 at 00:45
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1x @Olivier: Just because it's true doesn't mean that you have presented a convincing argument that it is. – hmakholm left over Monica Mar 02 '15 at 00:49