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Let $f(x)=\lim_{n\to\infty}\ln\left(\sqrt{e^{\cos x}\sqrt{e^{3\cos x}\sqrt{e^{5\cos x}...\sqrt{e^{(2n+1)\cos x}}}}}\right)$ and if $g(x)=\left[\frac{f(x)}3\right]$, then the number of points in $[0,2\pi]$ where $g(x)$ is discontinuous, is/are (where $[*]$ represents greatest integer function)...

$\cos x$ varies from $-1$ to $1$.

If $\cos x=-1$ then as $n\to\infty$, $e^{(2n+1)\cos x}=0$

If $\cos x=0$ then as $n\to\infty$, $e^{(2n+1)\cos x}=1$

If $\cos x=\infty$ then as $n\to\infty$, $e^{(2n+1)\cos x}=\infty$

Also, since $\cos x$ is changing these values in continuous manner, can we use that to comment about the continuity of $f(x)$?

But before that, I would need to solve the limit. Not able to do so. Can you give any pointers? Thanks.

EDIT: $f(x)=\lim_{n\to\infty}\ln\left(e^{\left(\frac12+\frac3{2^2}+\frac5{2^3}+...+\frac{(2n+1)}{2^{n+1}}\right)\cos x}\right)$

Let $$S=\frac12+\frac3{2^2}+\frac5{2^3}+...+\frac{(2n+1)}{2^{n+1}}\\\frac S2=\frac1{2^2}+\frac3{2^3}+\frac5{2^4}+...+\frac{(2n+1)}{2^{n+2}}$$

Subtracting, $$\frac S2=\frac12+\frac12+...\text{(n+1) times}-\frac{2n+1}{2^{n+2}}=\frac {n+1}2-\frac{2n+1}{2^{n+2}}\\\implies S=n+1-\frac{2n+1}{2^{n+1}}$$

I think my $S$ is coming out to be $\infty$. Can you point out the mistake? Thanks.

EDIT$2$: After reading the comments, I reworked and found $S=3$.

Thus, $f(x)=3\cos x\implies\left[\frac{f(x)}3\right]=[\cos x]$

Thus, there are $4$ points of discontinuity $\{0,\frac\pi2,\frac{3\pi}2,2\pi\}$.

aarbee
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  • $\sqrt{e^y} = e^{\frac y2}$. Start using this from the innermost square root : you'll get $e^{...}$ once all the square roots simplify, and that $...$ , you can think about using some trigonometric identities to evaluate it. Of course, you are taking an $\ln$ so the $e$ will vanish, leaving you with some trigonometric expression. – Sarvesh Ravichandran Iyer Sep 10 '21 at 07:33
  • @TeresaLisbon Thanks for the hint. I have edited my post. Can you check? Thanks. – aarbee Sep 10 '21 at 08:05
  • The third term in the difference is $2/8$, not $1/2$ – Empy2 Sep 10 '21 at 08:31
  • @Empy2 Thanks. I'll edit in a while. – aarbee Sep 10 '21 at 08:34

1 Answers1

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Hint:

Let $r = e^{\cos(x)}$. Then verify by induction or otherwise that $$ \sqrt{r \sqrt{r^3 \sqrt{r^5 \ldots \sqrt{r ^{2n+1} }}} } = r^{\frac12+ \frac{3}{2^2} + \ldots+ \frac{2n+1}{2^{n+1}} }$$

Now evaluate $S(n) = \sum_{k=0}^{n}\frac{2k+1}{2^{k+1}}$ using this and geometric series.

Then evaluate $f(x) = \lim_{n\to \infty} \ln( e^{S(n)\cos(x)}) = \lim_{n\to \infty} S(n)\cos (x)$ then proceed to find the answer.

Update

I think there is a problem in calculating $S(n)$. To evaluate it use the formula given here, then we see that $$ \sum_{k=0}^{n}\frac{k}{2^k} = \frac{2^{n + 1} - n - 2}{2^{n}}$$

So $$S(n) = \frac{{3 \cdot 2^{n + 1} - 2 \, n - 5}} {2^{n + 1}}$$

Now can you find the $\lim_{n\to \infty}S(n)$?

Infinity_hunter
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