When You Have Too Many Ex's Part 1

Calculus Level 4

What is the slope of the line tangent to the graph of

$\huge{y=x^{x^{x^{x^{x^{.^{.^{.}}}}}}}}$

at the point $x=\sqrt{2}$ ?

If your answer can be expressed as

$p\sqrt{q}\displaystyle \sum_{k=0}^{\infty}\ln^{k}(r)$

for positive, prime integers $(p,q,r)$ , find $p+q+r$ .

Note: the x's continue to infinity.

The answer is 6.

1 solution

Trevor Arashiro
Jan 4, 2016

please forgive me for not including dx and dy
Define $\Bbb{D}(g(x))=g'(x)$ for typing sake

$y=x^{x^{x^{x^{x^{.^{.^{.}}}}}}}$

$\ln(y)=\ln(x)x^{x^{x^{x^{x^{.^{.^{.}}}}}}}$

$\large{\dfrac{y'}{y}=\ln{x}\cdot \Bbb{D}\left(x^{x^{x^{x^{x^{.^{.^{.}}}}}}}\right)+ x^{\left(x^{x^{x^{x^{.^{.^{.}}}}}}-1\right)}}$

$\large{y'=\ln{x}\cdot \Bbb{D}\left(x^{x^{x^{x^{x^{.^{.^{.}}}}}}}\right)\cdot x^{x^{x^{x^{x^{.^{.^{.}}}}}}}+x^{x^{x^{x^{x^{.^{.^{.}}}}}}}\cdot x^{\left(x^{x^{x^{x^{.^{.^{.}}}}}}-1\right)}}$

Now, for visual simplicity, let $y=x^{x^{x^{x^{x^{.^{.^{.}}}}}}}=f(x)$ and $y'=\Bbb{D}\left(x^{x^{x^{x^{x^{.^{.^{.}}}}}}}\right)=f'(x)$

$\boxed{f'(x)=f'(x) \left(\ln(x)f(x)\right)+\dfrac{f^2(x)}{x}}$

Substituting for itself on the right side

$f'(x)=\left(f'(x)(\ln(x)f(x))+\dfrac{f^2(x)}{x}\right) \left(\ln(x)f(x)\right)+\dfrac{f^2(x)}{x}$

$f'(x)=f'(x)\ln^2(x)f^2(x)+\left(f(x)\ln(x)\right)\dfrac{f^2(x)}{x}+\dfrac{f^2(x)}{x}$

One more time

$f'(x)=f'(x)\ln^3(x)f^3(x)+\left(f(x)\ln(x)\right)^2\dfrac{f^2(x)}{x}+\left(f(x)\ln(x)\right)\dfrac{f^2(x)}{x}+\dfrac{f^2(x)}{x}$

This turns into the infinite summation

$\large{f'(x)=\displaystyle\lim_{a\rightarrow \infty}f'(x) \left(\ln(x)f(x)\right)^a+ \dfrac{f^2(x)}{x}\displaystyle \sum_{k=0}^{a} \left(\ln(x)f(x)\right)^k}$

$\large{f'(x)=\displaystyle\lim_{a\rightarrow \infty} \dfrac{\cfrac{f^2(x)}{x}\displaystyle \sum_{k=0}^{a} \left(\ln(x)f(x)\right)^k}{1-\left(\ln(x)f(x)\right)^a}}$

Plugging in for $x$

$\large{f'(x)=\displaystyle\lim_{a\rightarrow \infty} \dfrac{\cfrac{f^2(\sqrt{2})}{\sqrt{2}}\displaystyle \sum_{k=0}^{a} \left(\ln(\sqrt{2})f(\sqrt{2})\right)^k}{1-\left(\ln(\sqrt{2})f(\sqrt{2})\right)^a}}$

Quick proof $\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{.^{.^{.}}}}}}}=2$

$x^{x^{x^{x^{x^{.^{.^{.}}}}}}}=2$

$x^2=2$

$x=\sqrt{2}$

Plugging in $f\left(\sqrt{2}\right)=2$

$\large{f'(x)=\displaystyle\lim_{a\rightarrow \infty} \dfrac{\cfrac{2^2}{\sqrt{2}}\displaystyle \sum_{k=0}^{a} \left(\ln(\sqrt{2})\left(2\right)\right)^k}{1-\left(\ln(\sqrt{2})\left(2\right)\right)^a}}$

Since $\ln(2)<0\Longrightarrow \left(\ln(2)\right)^{\infty}=0$

$\large{f'(x)= 2\sqrt{2}\displaystyle \sum_{k=0}^{\infty} \ln^k(2)}$

$\therefore p=q=r=2\longrightarrow p+q+r=\boxed{6}$

Moderator note:

As a side note, you should be careful with existence and uniqueness. For example, in your simple proof that the successive powers of $\sqrt{2}$ equals to 2, you should provide an explanation of why such a value exists. For example, it is obvious that the successive powers of 2 do not give a finite value.

You need to show that this is true:

This turns into the infinite summation

$\large{f'(x)=\displaystyle\lim_{a\rightarrow \infty}f'(x) \left(\ln(x)f(x)\right)^a+ \dfrac{f^2(x)}{x}\displaystyle \sum_{k=0}^{a} \left(\ln(x)f(x)\right)^k}$

Plus, there's a simpler way of solving it. We have $y = x^y$ , take log and differentiate with respect to $x$ , we get $\dfrac{dy}{dx} \left(\dfrac1y - \ln x\right) = \dfrac yx$ , set $(x,y) = (\sqrt2,2)$ gives $\dfrac{dy}{dx} = \dfrac{2\sqrt2}{1-\ln2}$ , and you can recognize that this is a sum of a geometric progression.

Pi Han Goh - 5 years, 5 months ago

I think it is more obvious as to why it turns into the sum now, do I need to prove it more indepth?

Also, could you please post a slightly more detailed version of your method? I'm not sure how your method works.

Trevor Arashiro - 5 years, 5 months ago

It is not obvious at all, it may appear true from a glance, but you still need to list down the steps to get to the answer.

Brian Charlesworth already did that.

By the way, the summation in question starts from k=0, not k=1. and you shouldn't write $(\ln2)^\infty$ , but you should write it as $\displaystyle \lim_{n\to\infty} (\ln 2)^n = 0$ . or $(\ln2)^n{ \rightarrow} 0$ . And you need to show that x^x^x.... converges to 2 when x = sqrt2

Pi Han Goh - 5 years, 5 months ago

Hey Trevor. Are you back at school yet? This is quite the detailed solution, (although I took the same approach as Pi Han Goh). In the text of your question I think that you'll need to have the sum starting at $k = 0$ rather than $k = 1$ , (which is indeed what you have in your solution).

Brian Charlesworth - 5 years, 5 months ago

Yes, I'm back in school but things are going pretty smoothly now.

I took a look at Pi Han's solution and I'm not sure I understand it. I posted this problem mainly because I was messing around with these infinite stacks of x and practicing differentiation for the AP exam (and yes, I ordered an AP calc textbook for Christmas :P. Oh the nostalgia).

And thanks for fixing my problem... again XD

Trevor Arashiro - 5 years, 5 months ago

Pi Han's method uses a combination of logarithmic and implicit differentiation. Starting with $y = x^{y}$ we take the natural log of both sides, (with the condition that $x \gt 0$ ), to end up with $\ln(y) = y\ln(x)$ . Implicit differentiation (along with the product rule) then gives us that

$\dfrac{1}{y} \dfrac{dy}{dx} = \dfrac{dy}{dx} \ln(x) + y*\dfrac{1}{x} \Longrightarrow \dfrac{dy}{dx} = \dfrac{y^{2}}{x(1 - y\ln(x))}$ .

Then as $y(\sqrt{2}) = 2$ we have that at $x = \sqrt{2}$

$\dfrac{dy}{dx} = \dfrac{4}{\sqrt{2}(1 - 2\ln(2^{\frac{1}{2}}))} = \dfrac{2\sqrt{2}}{1 - \ln(2)} = 2\sqrt{2}\displaystyle\sum_{k=0}^{\infty} (\ln(2))^{k}.$

Brian Charlesworth - 5 years, 5 months ago

When You Have Too Many Ex's Part 1

The answer is 6.

1 solution

Moderator note:

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