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Suggest me a hint to solve:$$\psi=\lim_{x\to0}{\frac{\sqrt{1-\cos x+\sqrt{1-\cos x+\sqrt{1-\cos x+\cdots}}}-1}{x^2}}$$

My try, enter image description here

RE60K
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2 Answers2

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Hint

If you use Taylor series, you can start from lab bhattacharjee's answer

$$y=\frac{1\pm\sqrt{5-4\cos x}}2$$ and use the fact that around $x=0$, $\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{24}+O\left(x^5\right)$. So,$$5-4\cos (x)=1+2 x^2-\frac{x^4}{6}+O\left(x^5\right)$$ Now, use the fact that $$\sqrt{1+y}=1+\frac{y}{2}-\frac{y^2}{8}+O\left(y^3\right)$$ and replace $y$ by $(2 x^2-\frac{x^4}{6})$. So,$$\sqrt{5-4\cos (x)}=1+x^2-\frac{7 x^4}{12}+O\left(x^5\right)$$

I am sure that you can take from here and easily find lab bhattacharjee's results.

Claude Leibovici
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Let $\displaystyle y=\sqrt{1-\cos x+\sqrt{1-\cos x+\sqrt{1-\cos x+\cdots}}}$

$$\implies y^2=1-\cos x+y\iff y^2-y+\cos x-1=0$$

$$y=\frac{1\pm\sqrt{1-4(\cos x-1)}}2=\frac{1\pm\sqrt{5-4\cos x}}2$$

For $y\ge0$ and $\displaystyle x\to0,x\ne0\cos x<1,\sqrt{5-4\cos x}>1\implies 1-\sqrt{5-4\cos x}<0$ hence can be safely discarded

So, $$\lim_{x\to0}{\frac{\sqrt{1-\cos x+\sqrt{1-\cos x+\sqrt{1-\cos x+\cdots}}}-1}{x^2}}$$

$$=\lim_{x\to0}\frac{\sqrt{5-4\cos x}-1}{2x^2}$$

$$=\lim_{x\to0}\frac{5-4\cos x-1}{2x^2}\cdot\frac1{\lim_{x\to0}\sqrt{5-4\cos x}+1}$$

Now $$\lim_{x\to0}\frac{5-4\cos x-1}{2x^2}=\frac42\lim_{x\to0}\frac{1-\cos x}{x^2}$$

$x=2y\implies$ $$\lim_{x\to0}\frac{1-\cos x}{x^2}=\lim_{y\to0}\frac{1-\cos2y}{(2y)^2}=\frac24\left(\lim_{y\to0}\frac{\sin y}y\right)^2=\cdots$$

Paramanand Singh
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lab bhattacharjee
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  • Aditya was already there,,, – chubakueno Jul 05 '14 at 05:49
  • how can you eliminate - sign even though it gives 0 as satisfied by $y\ge0$ – RE60K Jul 05 '14 at 05:51
  • sorry, but the thing i want to ask is: $$y=\frac{1\pm\sqrt{5-4\cos x}}{2}$$ is correct but if you put $x\to0$ or $\cos x\to1 $ , you get $$y\to\frac{1\pm\sqrt{5-4}}{2}=\frac{1\pm1}{2}=0,1$$, both of which are $y\ge0$ – RE60K Jul 05 '14 at 06:05
  • @Aditya, Thanks for your observation. But, I think the other limit is undefined – lab bhattacharjee Jul 05 '14 at 06:12
  • that's what my problem is, even if it comes infinite, it doesn't prove/suggest to take the one limit which exists;; or is there any other means to prove which one satisfies the given condition? – RE60K Jul 05 '14 at 06:15
  • btw your answer is surely useful ,save this "y-problem" – RE60K Jul 05 '14 at 06:15
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    @Aditya, For $$x\to0,x\ne0\cos x<1,\sqrt{5-4\cos x}>1\implies 1-\sqrt{5-4\cos x}<0$$ hence can be safely discarded – lab bhattacharjee Jul 05 '14 at 06:22