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How do we show that there is only one solution to,$$\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+x}}}}=\sqrt[3]{6+\sqrt[3]{6+\sqrt[3]{6+\sqrt[3]{6+x}}}}$$

I guess it is only $x=2$. Please help.

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    $\begingroup$ So no one else has to type it in: W|A only finds $x=2$ $\endgroup$
    – apnorton
    Jan 24, 2014 at 4:22
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    $\begingroup$ The four who voted to close chose the option: "This question does not appear to be about math within the scope defined in the help center." Really?! $\endgroup$ Jan 24, 2014 at 5:48
  • $\begingroup$ @Tito Piezas III: Thanks for edit. $\endgroup$
    – mathkiss
    Jan 24, 2014 at 7:14

3 Answers 3

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Let $\;f(x) = \sqrt{2+x}\;$ and $\;g(x) = \sqrt[3]{6+x}$, they are strictly increasing function in $x$ when $x \ge -2$. Since $(x+2)^3 - (x+6)^2 = (x-2)(x^2 + 7x + 14)$ and $x^2 + 7x + 14 > 0$ for all $x$, we have $$\begin{cases} f(x) > g(x) > 2,& x > 2\\f(x) = g(x) = 2,& x = 2\\f(x) < g(x) < 2, & x <2\end{cases}$$ So for any $x > 2$, we have $$\begin{align} & f(x) > g(x) > 2\\ \implies & f(f(x)) > f(g(x)) > g(g(x)) > 2\\ \implies & f(f(f(x))) > f(g(g(x)) > g(g(g(x)) > 2\\ \implies & f(f(f(f(x))) > f(g(g(g(x))) > g(g(g(g(x)))) > 2\\ \implies & f(f(f(f(x))) \ne g(g(g(g(x)))) \end{align}$$ Please note that in above deductions, we are using following reasoning repeatedly. $$\underbrace{g\circ\cdots\circ g(x)}_{k \text{ terms}} > 2 \implies f(\underbrace{g\circ\cdots\circ g(x)}_{k \text{ terms}}) > \underbrace{g\circ\cdots\circ g(x)}_{k+1 \text{ terms}} > 2.$$

Similar logic shows that $f(f(f(f(x)))) \ne g(g(g(g(x))))$ for $x < 2$. As a result, $x = 2$ is the only solution for the equation $f(f(f(f(x)))) = g(g(g(g(x))))$.

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    $\begingroup$ Should we vote to have this re-opened? $\endgroup$ Jan 24, 2014 at 5:53
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    $\begingroup$ @TitoPiezasIII sure, why not. the question itself is interesting. $\endgroup$ Jan 24, 2014 at 5:56
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A proof by induction. Let: $$f_n(x) = \sqrt[3]{6+\sqrt[3]{6+\ldots+\sqrt[3]{6+x}}},\ g_n(x) = \sqrt{2+\sqrt{2+\ldots+\sqrt{2+x}}}$$ With $n$ terms. Then for $n=1$ you can easily solve the cubic equation to show that $f_1=g_1 $ only at $x=2$ (over the reals).

Now assume our claim is true for $n$, i.e. that $f_n(x)=g_n(x)$ iff for $x=2$. Then for $n+1$, raise to the sixth power to get that: $$(6+f_n(x))^2=(2+g_n(x))^3$$ Clearly, this equality is true for $x=2$ since $f_n(2)=g_n(2)$. Now, if our claim is false and this equality holds for some $x_0\neq 2$, then: $$g_n(x_0) = (6+f_n(x_0))^{2/3}-2$$ But since $dg_n/df_n < 1$ for all $x>0$, by the mean value theorem we have a contradiction. $$$$ Thus we have proven that for any number of $n$ the only (real) solution of $f_n(x)=g_n(x)$ is $x=2$.

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  • $\begingroup$ Wait, I'm not sure how you are using the MVT here. $\endgroup$
    – Braindead
    Jan 24, 2014 at 5:39
  • $\begingroup$ @Braindead - just to prove that if $g(x)=h(x)$ at some point, and the derivative of one is is always greater then the other, they will never coincide. $\endgroup$ Jan 24, 2014 at 5:41
  • $\begingroup$ Okay, so you assume that $f_{n+1}(x_0) = g_{n+1}(x_0)$ for some $x_0 \ne2$. Then you get that equation. But how / where do you get $g'_n < f'_n$ for all $x>0$? $\endgroup$
    – Braindead
    Jan 24, 2014 at 5:49
  • $\begingroup$ @Braindead - do the math :) take the derivative $dg/df$ and show it is smaller than one. $\endgroup$ Jan 24, 2014 at 5:52
  • $\begingroup$ @TitoPiezasIII - sure, it's a nice question :) $\endgroup$ Jan 24, 2014 at 5:54
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Hint: raise both sides to the sixth power.

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  • $\begingroup$ Doesn't that make things worse? $\endgroup$
    – Braindead
    Jan 24, 2014 at 5:29
  • $\begingroup$ Oh, ok. But there really should be a policy that if more than one person has already given a substantial answer, then that question should not be closed. $\endgroup$ Jan 24, 2014 at 6:17
  • $\begingroup$ I wouldn't do it that way. I would just let x equal to a number that's 2, less than 2, and greater than 2. Through trial and error, you can see that x = 2 is the only acceptable answer for this problem. If the question requires a proof, you could do it by induction. I'm just starting out, so I can't really do it yet...only through what I could see at the moment. $\endgroup$
    – usukidoll
    Jan 24, 2014 at 7:36

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