# Area enclosed by a function involving 2 power towers

• I
• Saracen Rue
In summary, the conversation discusses a function involving up-arrow notation and its behavior for increasing values of k. The area under the graph of the function was found to approach a constant value as k approaches infinity. To find this value, it is necessary to evaluate an integral. To prove that the integral converges, it is shown that the function is bounded between 0 and 1 and that it is strictly increasing. This leads to the conclusion that the function approaches a constant value as k increases.
Saracen Rue
TL;DR Summary
Evaluate the following integral ##\int_0^1 ((−ln(x↑↑(2k)))↑↑(2k+1))dx## as ##k \to \infty##
I've been playing around with Up-Arrow notation quite a lot lately and have come up with the following "thought experiment" so to speak. Consider the following function: $$f(x)=(−ln(x↑↑(2k)))↑↑(2k+1)$$ $$\text{Where }k∈\mathbb{Z} ^+$$

In the image below we can see some examples of what this function looks like for the first ##5## values of ##k##:

It can be visually seen from the graphs that while the area under each graph does become larger for increasing values of ##k##, the rate at which the area is increases decrease. However, I went ahead and took the liberty of doing the calculations anyway. I found the difference in area between ## f_{k=2}(x)## and ## f_{k=1}(x)##, and then between ## f_{k=3}(x)## and ## f_{k=2}(x)##, and so on. I then took these values and plotted them as such:

(Note: The graphs of ##f_{k=10}(x)##and ##f_{k=11}(x)## weren't included previously, they are only being used here now to help with the accuracy of the curve fitting)

Each point was plotted in the following manor: ##(1, \int_0^1(f_{k=2}(x)-f_{k=1}(x))dx), (2, \int_0^1(f_{k=3}(x)-f_{k=2}(x))dx), \text{etc}.##

The points were found to fit the general relationship of ##y_A \approx m_1 x_A^{a_1x_A+b_1}+c_1## extremely well, have an ##R^2## value very close to ##1##. This leads me to believe that this relationship appropriately represents the difference in area between different values of ##k## in ##f(x)##. Therefore, because this regression curve approaches ##0## as ##x## approaches ##\infty##, the area enclosed by the graph of ##f(x)=(−ln⁡(x↑↑(2k)))↑↑(2k+1)## must approach a constant value as ##k \to \infty##.

So, we know it does approach a constant value, we just need to find out what said value is. In other words, we need to evaluate $$\int_0^1 ((−ln(x↑↑(2k)))↑↑(2k+1))dx \text{ as } k \to \infty$$

The differences going to zero is not sufficient. ##\displaystyle \sum \frac 1 n## has the differences between partial sums going to zero, too, but clearly the sum diverges.

I think you can prove that 1 > fk > fk-1 for all x between 0 and 1, that is sufficient for convergence.

mfb said:
The differences going to zero is not sufficient. ##\displaystyle \sum \frac 1 n## has the differences between partial sums going to zero, too, but clearly the sum diverges.

I think you can prove that 1 > fk > fk-1 for all x between 0 and 1, that is sufficient for convergence.
Thank you for letting me know. I was never taught about convergent/divergent series in school, so I'm still not very confident with this area of maths. I would very much appreciate if you could help with how to go about proving ##1## ##>## ##f_k## ##>## ##f_{k-1}## for all ##x## between ##0## and ##1##.

This should work (details to be worked out):
• Show that ##1/e \leq x↑↑(2k) \leq 1##. It follows that ##0 \leq -\ln( x↑↑(2k)) \leq 1##, which means your integrand is between 0 and 1. That gives the upper bound on the integral.
• Show that ##x↑↑(2k) \leq x↑↑(2k-2)##. It follows that ##-\ln( x↑↑(2k)) \geq -\ln( x↑↑(2k))##.
• Show that ##x ↑↑ (2k+1) > x ↑↑ (2k-1)## and that ##x ↑↑ (2k+1) > y ↑↑ (2k+1)## for x>y.
• Combine step 2 and 3 to show that ##f_k > f_{k-1}##.

## 1. What is a power tower?

A power tower is a mathematical concept in which an exponent is raised to another exponent, and this process is repeated multiple times. For example, a power tower of 2 with a height of 3 would be written as 2^(2^2), which is equivalent to 2^4 or 16.

## 2. How do you calculate the area enclosed by a function involving 2 power towers?

To calculate the area enclosed by a function involving 2 power towers, you would need to first graph the function and determine the boundaries of the enclosed region. Then, you can use integration techniques to find the area under the curve within those boundaries.

## 3. What is the significance of finding the area enclosed by a function involving 2 power towers?

Finding the area enclosed by a function involving 2 power towers can help in understanding the behavior and properties of the function. It can also be useful in solving real-world problems that involve exponential growth or decay.

## 4. Are there any specific methods or techniques for finding the area enclosed by a function involving 2 power towers?

Yes, there are various integration techniques that can be used to find the area enclosed by a function involving 2 power towers, such as the substitution method or the integration by parts method. The choice of method may depend on the complexity of the function.

## 5. Can the area enclosed by a function involving 2 power towers be negative?

No, the area enclosed by a function involving 2 power towers cannot be negative. This is because the area under a curve is always considered to be positive, and the power tower function involves only positive values.

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