Is x^x=0 solvable?

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Let's discuss if the power tower equation x^x=0 is solvable or not? Of course, we will discuss the real cases and the complex cases. Then we talk about the limit.

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If x is a complex number other than 0 we can write x=r*e^iθ and then we see that x^x=(re^iθ)^x=r^x e^ixθ
In order for this expression to be 0 we need either r^x=0 or e^ixθ=0. But since r>0 hence r^x != 0. Likewise e^ixθ != for any x. Thus x^x=0 has no solution in the complex numbers

bartekabuz
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If u want to know about the complex solution, u can find bprp's old video with title "the tetration of (1+i) and the formula (a+bi)^(c+di)"

khoozu
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You say "not my best video", but I really enjoyed it!

OverLordGoldDragon
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Must be the first vid without "of course, otherwise, how could I make this video?" Love the subversion of expectations

peterciccone
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For me 0^0 = 1/2, take the average value of 0 and 1, im sure everyone will be happy 😊

silverblade
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Naturally, let’s be rational for a moment. This is too complex to be real.

crsmtl
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Hey there BPRP, been watching for some 6 years now. Good stuff man

kindafool
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For the limit x^x you can do:
lim x^x = lim exp(ln(x)*x) = exp(lim ln(x)*x) because exp is a continuous function. Then lim ln(x) * x = lim ln(x) / (1/x) = lim (1/x) / (-1/x^2) = lim 1 / (-1 / x) = lim -x = 0 (using de l'Hopital) and therefore lim exp(ln(x)*x) = lim x^x = 1.

christianstieger
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This is an example of a pedagogically responsible explanation.

meccamiles
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I understand that mathematician say : "It has no solution, because negative Infinity is not a real or complex number"... But, it is obvious that assymptoticly -inf (in integer) converges to solution. It could be really interesting to make some physical experiment, where the nature say if the -inf is the solution of this equation.... For me, from technicial - engineering aspect, it is solid solution (but not ideal); The only question for me is, if the nature supports the infinity .. :)

petersagitarius
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What do you mean not your best video? I enjoyed every single second. Good job

alvinoceanohorsky
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My high school teacher once declared 0^0 is undefined because powers can be written as fractions, so 0^0 equals 0/0. I never questioned it until now.

hbfrts
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1:55 both arguments basically portray when an unstoppable force meets an immovable object

shadismith
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I believe as long as x is real, case 6 with the limit works. lim x->-inf (x^x) = lim x->inf ((-x)^(-x)).

(-x)^(-x)=(-1)^(-x) * x^(-x)

The latter factor always approaches zero as x approaches infinity. The former factor is cyclical and the absolute value is always equal to one, so (-x)^(-x) always approaches 0 as x approaches infinity and thus x^x approaches zero as x approaches negative infinity.

I think there’s an extension of this argument if you allow x to approach negative infinity from paths in the complex plane, but I haven’t fully formulated it yet.

i_am_anxious
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To me, it has to be undefined. The reason why anything to the power of 0 is 1 is because a number divided by itself is always equal to 1, but since division by 0, even on itself, is always undefined, it has to be undefined.

ToufG
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Hello sir, I just wanted to thank you for making these videos because they've helped me out a lot.

phantoniex
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This is where new math is to be found. Idk why no one looks here or ever talks about numbers like these. Juat as we thouught i couldnt exist, it does. So this probably does too. Same with |x|=-1

Snakeyes
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I think the lim as x-> negative infinity is valid. You can see that if you try x= -10, -100, -1000 then x^x surely is approaching zero. It could be argued that the function is discontinuous in the negatives, but for going to -infinity at every point the function exists x^x approaches zero as x gets more negative.

adamrussell
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Here comments is more difficult than videos. Please get this man a medal 🏅🏅. Salute professors.

tsheringdorjigamming
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I think 💭 the first video I watched from was pi^e vs e^pi.

Back then I think was entering masters in math/stats and today I’m a year away from PhD in statistics.

Always inspiring!

Btw did you ever did a video on Riemann–Stieltjes integration????

I recall it but can’t find it

rick