finding limit with $\cos$ function occur $n$ times

Question

Finding $\displaystyle \lim_{x\rightarrow 0}\frac{1-\cos(1-\cos(1-\cos(1-\cdots \cdots (1-\cos x))))}{x^{2^n}}$

where number of $\cos$ is $n$ times

when $x\rightarrow 0$ then $\displaystyle 1-\cos x = 2\sin^2 \frac{x}{2} \rightarrow 2\frac{x}{2} = x$

so $1-\cos (1-\cos x) = 1-\cos x$

some help me., thanks

My only thought is L'Hospitals. . – Simply Beautiful Art Feb 01 '17 at 13:34 — Simply Beautiful Art, Feb 01 '17 at 13:34

score 2 · Answer 1 · edited Apr 13 '17 at 12:21

2

Let $f^{n} $ denote the composition of $f$ with itself $n$ times and let $f^{0}(x)=x$. Then the numerator of the given expression (whose limit is to be evaluated) is equal to $f^{n} (x) $ where $f(x) = 1-\cos x$. Note that as $x\to 0$ each of the functions $f^{n} (x) \to 0$ and also note that we have $$\lim_{x\to 0}\frac{1 - \cos x} {x^{2}}=\frac{1}{2}$$ Replacing $x$ by $f^{n-1}(x)$ in the above equation we get $$\lim_{x\to 0}\frac{f^{n} (x)} {(f^{n-1}(x))^{2}} = \frac{1} {2} $$ Now we need to replace $n$ by $n-1$ in above equation and square it and continue like this and multiply resulting equations to get $$\lim_{x\to 0}\frac{f^{n}(x)}{(f^{0}(x))^{2^{n}}}= \frac{1}{2}\cdot\frac{1}{2^{2}}\cdots\frac{1}{2^{2^{n-1}}}$$ and thus the desired limit is $1/2^{1+2+2^{2}+\cdots +2^{n-1}}$ or $1/2^{2^{n} - 1}$.

See a similar answer to a related question.

edited Apr 13 '17 at 12:21

Community

1

answered Feb 02 '17 at 07:46

Paramanand Singh

87,309

This is exactly the same as my answer. – Rene Schipperus Feb 02 '17 at 12:24
@ReneSchipperus: I don't think there is substantial difference between any of the answers here and yet each one is independently arrived at. – Paramanand Singh Feb 02 '17 at 12:56
@ReneSchipperus: this answer is inspired by another answer from past which I have linked here. – Paramanand Singh Feb 02 '17 at 12:57
Well there is a substantal difference between the other answers, ( I discussed it with the other posters at the time), and yours comes 18 hours after the others. – Rene Schipperus Feb 02 '17 at 12:59
@ReneSchipperus: everyone is entitled to his or her own opinion here and this is what is best about such sites like MSE. – Paramanand Singh Feb 02 '17 at 13:01
You are a Parasite. – Rene Schipperus Feb 02 '17 at 13:03
@ReneSchipperus : thanks for undeserved downvote and feedback. I hope this closes the matter. – Paramanand Singh Feb 02 '17 at 13:09

Rene Schipperus · Answer 2 · 2017-02-01T14:29:43.060

Take the simple case $$\frac{1-\cos(1-\cos x)}{x^4}$$

$$=\frac{1-\cos(1-\cos x)}{(1-\cos x)^2}\left(\frac{1-\cos x}{x^2}\right)^2$$ $$\rightarrow \frac{1}{2}\left(\frac{1}{2}\right)^2$$

Can you see how to do the induction ?

So you have in general if $f(n)=1-\cos(1-\cos(1-\cos(1-\cdots \cdots (1-\cos x))))$

$$\frac{f(n)}{x^{2^n}}=\frac{f(n)}{f(n-1)^2}\left(\frac{f(n-1)}{x^{2^{n-1}}}\right)^2\rightarrow \frac{1}{2}L_{n-1}^2$$

And if we write $L_n=\left(\frac{1}{2}\right)^{e_n}$

we get the recurrence relation

$$e_n=2e_{n-1}+1$$ which means $$e_n=2^n-1$$

Nosrati · Answer 3 · 2017-02-01T14:33:11.187

1

$$\lim_{x\rightarrow 0}\frac{1-\cos x}{x^2}=\frac12$$ thus $1-\cos x\to\dfrac12x^2$ with substantiation $$\displaystyle \lim_{x\rightarrow 0}\frac{1-\cos(1-\cos(1-\cos(1-\cdots \cdots (1-\cos x))))}{x^{2^n}}$$ $$=\displaystyle \lim_{x\rightarrow 0}\frac{1-\cos(1-\cos(1-\cos(1-\cdots \cdots (\dfrac12x^2))))}{x^{2^n}}$$ $$=\displaystyle \lim_{x\rightarrow 0}\frac{1-\cos(1-\cos(1-\cos(1-\cdots \cdots \dfrac12(\dfrac12x^2)^2)))}{x^{2^n}}$$ $$=\displaystyle \lim_{x\rightarrow 0}\frac{\left(\dfrac12\right)^{2^n-1}x^{2^n}}{x^{2^n}}$$ $$=\left(\dfrac12\right)^{2^n-1}$$

edited Feb 01 '17 at 14:33

answered Feb 01 '17 at 13:39

Nosrati

29,995

The exponent of $\frac{1}{2}$ is $2^n-1$. – Rene Schipperus Feb 01 '17 at 14:20
See my edited answer. – Rene Schipperus Feb 01 '17 at 14:30
@ReneSchipperus Thanks a lot – Nosrati Feb 01 '17 at 14:32

finding limit with $\cos$ function occur $n$ times

3 Answers3