# Integral Bee Preparation -- Trouble with this beautiful integral

• I
• YAYA12345
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
YAYA12345
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.
$$\int\limits_{0}^{1/2} \cos(1-\cos(1-\cos(...(1-\cos(x))...) \ \mathrm{d}x$$

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.##

YAYA12345
¿Cómo lo sabes?
Translation by mentor: How do you know that?

@YAYA12345 -- per the forum rules, posts must be in English.

Last edited by a moderator:
fresh_42 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?

Last edited by a moderator:
YAYA12345 said:
¿Y cómo podría probarlo? ¿Puedes apoyarme con una pista?
I suggest we continue in English.

\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.##

Last edited:
malawi_glenn
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.

malawi_glenn
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.##

Infrared, dextercioby and malawi_glenn
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.

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.

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.

Office_Shredder and fresh_42

## What is Integral Bee Preparation?

Integral Bee Preparation is a method used by beekeepers to prepare their hives for the winter season. It involves creating a nutrient-rich food source for the bees to consume during the colder months, as well as insulating the hive to protect the bees from harsh weather conditions.

## Why is Integral Bee Preparation important?

Integral Bee Preparation is important because it helps ensure the survival of the bee colony during the winter season. By providing the bees with enough food and insulation, they are able to maintain their body temperature and stay active, which is crucial for their survival.

## What are the steps involved in Integral Bee Preparation?

The first step in Integral Bee Preparation is to assess the health of the hive and make sure the bees have enough food stores. Then, the beekeeper will add supplemental feedings, such as sugar syrup or pollen patties, to help the bees build up their food reserves. Finally, the hive is insulated with materials such as straw or foam to protect the bees from the cold.

## What are some common problems encountered with Integral Bee Preparation?

One common problem with Integral Bee Preparation is that the bees may not consume all of the supplemental feedings provided, which can lead to mold growth and other issues. Another problem is that the insulation may not be enough to protect the bees from extreme weather conditions.

## How can I troubleshoot issues with Integral Bee Preparation?

If you are experiencing issues with Integral Bee Preparation, it is best to consult with an experienced beekeeper or local beekeeping association. They can provide guidance on the best methods for your specific hive and climate, as well as offer solutions for any problems that may arise.

• Calculus
Replies
8
Views
120
• Calculus
Replies
8
Views
382
• Calculus
Replies
3
Views
997
• Calculus
Replies
16
Views
1K
• Calculus
Replies
6
Views
1K
• Calculus
Replies
2
Views
248
• Calculus
Replies
3
Views
604
• Calculus
Replies
3
Views
286
• Calculus
Replies
29
Views
673
• Calculus
Replies
4
Views
298