Is This Even Possible?

🙏Support me by becoming a channel member!

Disclaimer: This video is for entertainment purposes only and should not be considered academic. Though all information is provided in good faith, no warranty of any kind, expressed or implied, is made with regards to the accuracy, validity, reliability, consistency, adequacy, or completeness of this information.

#math #brithemathguy #algebra

[BriTheMathGuy]

🎓Become a Math Master With My Intro To Proofs Course! (FREE ON YOUTUBE)

[knutthompson]

The interesting thing to do now is solving 4=x^x^x^x... . Using the simple trick from the beginning of your video, the answer comes out to sqrt(2) again, which doesn't make sense. How can sqrt(2)^sqrt(2)^sqrt(2)^... equal both 2 and 4? The issue is the problem actually does NOT have an answer in the case of 4, so we were wrong to assume it does. Going through the induction proof you can figure out where it fails.

[CatInABaseballCap]

*School be like:*
In 5–6 paragraphs, explain why (√𝑥)² =𝑥

[danielrhouck]

2:34: “If it’s bounded above by some number, in this case 2, and it’s always increasing towards 2, well it must converge at 2.”
You’re missing a step. This sequence is bounded above by 17 as well. This shows it converges, not that it converges *at 17*. You did already say *if* it converges it at 2, but that’s a different argument.

[Inspirator_AG]

Exploiting infinity. That is how it is possible.

[cheeseburgermonkey]

love the videos you make, they're super entertaining and educational as well

[kensmusic]

5:50 here of course, the continuity of the exponential function is used to drag the limit into the exponent

[oitthegroit]

I have an interesting problem for you: is the continued fraction actually equal to 1/(sqrt(e)-1) (approximately 1.54149...)? Or is it just a coincidence? Is there any way to prove it is, or isn't equal to that value?

[koendos]

I solved for x (sqrt(2)). And then checked if sqrt(2) is less than exp(1/e). if so, it will converge.

[Inspirator_AG]

Equations that are infinite end up containing _themselves._

[amoledzeppelin]

(haven't watched yet)
Log both parts by base 2: 1 = (x^x^..) log2 x
Since x^x^x... = 2, we get 1 = 2log2 x
log2 x = 0.5, so x = 2^0.5 = √2

[pinedelgado]

Awesome as a possum with a blossom math!!! I may not understand all of it at this time--BUT, I love this!!!

[cmilkau]

"..."-expressions are usually meant to be read in the way that the "..." can be left out to make an element of the sequence. This means that this sequence needs to start with 1, as leaving out an exponent corresponds to an exponent of 1. In this case, the second element works out to be 1 anyway, so no harm done, but this can matter when the recursion has multiple fixed points.

[marble]

And the x in the thumbnail is just 2¹, how nice

[ilias-]

1:20 When taking the log of x^y and seperating u have to take absolute value : log(x^y)= y log(abs(x)) which is obvious since log(x^2)=2log(abs(x)). So your "proof" that x can't be negative is false.

[poutineausyropderable]

Yes, but if your function is 1.5 - 1/n

It's bounded above by 2.

And always increase

[cmilkau]

This is a maths channel, so please don't argue that negative solutions don't exist because the logarithm is not defined for negative numbers. The opposite is true, you are only allowed to use the (real) logarithm when there are no negative solutions, or you are ready to lose those negative solutions (perhaps recovering them later).

[Immadeus]

Holy shit someone made math fun to understand

[sujanrajakannan]

No, logarithms are definite for negative numbers like this
log__b(-a)=ln(-a)/lnb
=(lna+ln-1)/lnb
=(lna+πi)/lnb

[cmilkau]

I really love the "domino" metaphor for complete induction. Will use this from now on to explain it!