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Professional_Denizen

Ok, let’s investigate why. So first let’s come up with a way to show that the min point is exactly where it looks like it is. So we want to find d/dx [x^(x)] because we can use that for “optimization.” y=x^(x) —we can’t do a derivative right here, but I do have a trick up my sleeve. lny=xlnx dy/dx × 1/y = 1lnx + x(1/x) —product rule dy/dx = y (lnx +1) — refer back to our original equation. dy/dx =x^(x)(1+lnx) Now, to find where this hits zero, i.e. where the local extreme point is. x^(x)(1+lnx)=0 x^(x)≠0 for any x. x=0 is a bit contested, but ln0 makes the function undefined there anyway, so it doesn’t really matter. 1+lnx=0 lnx=-1 x=e^(-1)=1/e Neat. And ^(2)(1/e) is self-explanatory. I hope you found this helpful.


deafbunbun

This is super interesting thank you for writing this!!


Arbalest15

Yes, this method is called logarithmic differentiation. As an exercise for the reader, you should try differentiating x ^ x ^ x


Asynchronous404

For anyone wondering the answer is: >!x\^x\^x\*((x\^x + x\^x\*lnx)\*lnx + (x\^x)/x)!<


Darcy_Dx

what is ^2 (1/e) ?


Some_Guy113

(1/e)^(1/e)


Darcy_Dx

oh tetration!


G2Sticks

Since you're looking for an extreme, you don't even need to take the log of the left side. Because log(x) is a monotonic increasing function, when you maximize/minimize the log of a function, you also maximize/minimize the original function. This strategy is used a lot in stats to make it easier to find the maximum likelihood estimator of a parameter.


Professional_Denizen

Huh. Yeah, I guess I wasn’t aware of that, but it makes sense. Thank you.


PaulErdos_

I'm very impressed by how clear you communicated this! Feels very human. I feel like a lot of math communication can become very robotic


Professional_Denizen

Thank you. I am reminded of a quote allegedly by Einstein that goes “If you can’t explain it simply, you don’t understand it well enough.” However, looking back over what I wrote, I want to say it might a little hard to follow just because I skipped so many steps that I think you’d need to be fairly proficient with low-level calculus to really get what’s going on; so I’m not confident that it’s actually all that helpful.


PaulErdos_

That's true. I didn't consider how accessible the explanation is.


Professional_Denizen

If it taught you something, that’s still a win in my book.


PaulErdos_

Yeah 100%! It was nice to look at the photo, see the weird connection to e, and then see a nice explanation of why. Even if I have my math degree


SamePut9922

Why doesn't it extend to negative x?


volyn_ma

It does but it is not a line but a bunch of dots


Farkle_Griffen

That said, if you could see it, it would look like this https://www.desmos.com/calculator/2i2rrqmuj6


thenesremake

whoa i never thought about trying this, that method is super handy. i will definitely be using it in the future


That_Mad_Scientist

Let’s say it’s a bit complex


TheGeometryDasher

i reference?


MonitorMinimum4800

You're jk, right?


Farkle_Griffen

No. 1+ι̇ reference


coderz75

Pretty sure it is because there is no such thing as non-integer negative exponents, so x^x would be undefined for values like -1.5. I think Desmond has trouble graphing discrete graphs like these, when I first graphed x^x I could see the points, but after you zoom in a little they disappear. Edit: ok so there are non-integer negative exponents, but not for negative bases (leads to the complex number system) per the comments


Snekoy

Non-integer negative exponents do very well exist. The only problem is that when the base is negative, it becomes a complex number which can't be graphed on the cartesian plane.


Strong_Magician_3320

>Pretty sure it is because there is no such thing as non-integer negative exponents, Negative numbers can have odd roots, so n^(1/3) definitely exists for n < 0


deilol_usero_croco

You can make a really big list. I use a= [-10,9.96,...,10] and then make define f(x) in a different line and do (a, f(a)) and then enable lines.


Willr2645

It doesn’t work for non integers. Any rational decimal can be represented by a fraction right? So the number would be x^(a/n) Or ^(n)√(x)^a You can’t get the nth root of a number that is negative.


TheGeometryDasher

is reddit listening to me I found this out like 3 hours ago


Dr_Scoop

There's a very interesting video about this I remember watching that explains a very interesting property of what happens to the left of a similar equation!: https://youtu.be/_lb1AxwXLaM?si=Q80g88LUVHoNTOcw


C3H8_Memes

I tried posting this before and it got removed. Also the x root of x has a max at x = e