Integral Bee Preparation -- Trouble with this beautiful integral

In summary: This is the approach that I would follow too, but there is one more needed step: to explain why the limit ##L(x)## exists in the first place. I think the easiest argument would be to notice that the map ##t\mapsto cos(1-t)## is a contraction and apply contraction-mapping. Usually you need the contraction to decrease the distance by a factor bounded away from 1 but since you can take the domain to be compact it's enough that it strictly decreases distance between distinct points. So the sequence ##f_n## is a Cauchy sequence and hence converges to something.In summary, the conversation discusses various approaches to solving the integral and proving the convergence of the integrand. One approach involves using
  • #1
YAYA12345
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0
TL;DR Summary
While I was preparing for an integrals contest, I had a doubt about the following integral, I tried several substitutions but nothing worked.I would appreciate your support for this beautiful integral.
$$ \int\limits_{0}^{1/2} \cos(1-\cos(1-\cos(...(1-\cos(x))...) \ \mathrm{d}x$$
While I was preparing for an integrals contest, I had a doubt about the following integral, I tried several substitutions but nothing worked.I would appreciate your support for this beautiful integral.
$$ \int\limits_{0}^{1/2} \cos(1-\cos(1-\cos(...(1-\cos(x))...) \ \mathrm{d}x$$
 
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  • #2
YAYA12345 said:
TL;DR Summary: While I was preparing for an integrals contest, I had a doubt about the following integral, I tried several substitutions but nothing worked.I would appreciate your support for this beautiful integral.
$$ \int\limits_{0}^{1/2} \cos(1-\cos(1-\cos(...(1-\cos(x))...) \ \mathrm{d}x$$

While I was preparing for an integrals contest, I had a doubt about the following integral, I tried several substitutions but nothing worked.I would appreciate your support for this beautiful integral.
$$ \int\limits_{0}^{1/2} \cos(1-\cos(1-\cos(...(1-\cos(x))...) \ \mathrm{d}x$$
Looks as if your integrand converges to identical ##1.##
 
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  • #3
¿Cómo lo sabes?
Translation by mentor: How do you know that?

@YAYA12345 -- per the forum rules, posts must be in English.
 
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  • #6
YAYA12345 said:
¿Y cómo podría probarlo? ¿Puedes apoyarme con una pista?
I suggest we continue in English.

I would start with the Taylor expansion:
\begin{align*}
\cos(x_0) &= 1- \dfrac{x_0^2}{2!}+\dfrac{x_0^4}{4!}-\dfrac{x_0^6}{6!}\pm \ldots \\
1-\cos(x_0) &= \dfrac{x_0^2}{2!}-\dfrac{x_0^4}{4!}+\dfrac{x_0^6}{6!}\mp \ldots =:x_1\\
\cos(x_1) &= 1 - \dfrac{x_1^2}{2!}+\dfrac{x_1^4}{4!}-\dfrac{x_1^6}{6!}\pm \ldots \\
&=1- O(x_0^4) \\
& etc.
\end{align*}
The second term is already small on ##I:=[0\, , \,0.5].## Iteration looks as if
$$
cos(1-cos(1-cos(1-cos(1-cos(x))))) = 1 + O\left(\left(\dfrac{x}{2}\right)^{2^5}\right)
$$
or similar. The big-O term quickly becomes negligible on ##I.##
 
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  • #7
My approach would be to calculate a few integrals of increasing complexity; i.e., ##\int \cos(1 - \cos(x))dx, \int \cos(1 - \
cos(1 - \cos(x))))dx##, and maybe one more to see if a pattern emerges. If so, I would then try a proof by induction. I can't guarantee this would work, but it's the direction I would start with.
 
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  • #8
YAYA12345 said:
¿Y cómo podría probarlo? ¿Puedes apoyarme con una pista?

(Approximate) Translation by mentor -- And how could it be proven? Can you lead me to a track?

Another idea:

Let's define ##f_0(x)=\cos(x)## and ##f_n(x)=\cos(1-f_{n-1}(x)).## We want to show that ##L(x):=\lim_{n \to \infty}f_n(x) \equiv 1.##

We have
\begin{align*}
L(x)&=\lim_{n \to \infty}f_n(x)\\&=\lim_{n \to \infty}(\cos(1-f_{n-1}(x)))\\
&=\cos(1-\lim_{n \to \infty}f_{n-1}(x))\\&=\cos(1-L(x))
\end{align*}
and so ##L(x)=\cos(1-L(x)).##

So you only have to show that ##x=\cos(1-x)## has only the solution ##x=1.##
 
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  • #9
fresh_42 said:
So you only have to show that ##x=\cos(1-x)## has only the solution ##x=1.##

As long as we're doing calculus: x and ##\cos(1-x)## are both increasing on ##[0,1]## (it can't be a negative x because cosine is always positive on ##[-1,1]##) and the slope of cosine is strictly less than 1 on the interval, so they can only intersect at most once.

I don't know what any of this has to do with an integration bee though.
 
  • #10
Office_Shredder said:
As long as we're doing calculus: x and ##\cos(1-x)## are both increasing on ##[0,1]## (it can't be a negative x because cosine is always positive on ##[-1,1]##) and the slope of cosine is strictly less than 1 on the interval, so they can only intersect at most once.

I don't know what any of this has to do with an integration bee though.
I tried to leave at least a bit for the OP to do.
 
  • #11
fresh_42 said:
So you only have to show that ##x=\cos(1-x)## has only the solution ##x=1.##

This is the approach that I would follow too, but there is one more needed step: to explain why the limit ##L(x)## exists in the first place. I think the easiest argument would be to notice that the map ##t\mapsto cos(1-t)## is a contraction and apply contraction-mapping. Usually you need the contraction to decrease the distance by a factor bounded away from 1 but since you can take the domain to be compact it's enough that it strictly decreases distance between distinct points.
 
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