What does an irrational exponent mean?

Рет қаралды 136,604

5 жыл бұрын

We will go over how we define a number raised to an irrational exponent. We will be using 2^sqrt(2) as an example. This is actually related to calculus limits!
Read more on, Gelfond-Schneider constant 2^sqrt(2) en.wikipedia.org/wiki/Gelfond...
Check out Newton's Method vs. Euler's Method here: • Square Root of 2, Newt... 0
👉 Ultimate Integrals On Your Wall: teespring.com/calc-2-integral...

Пікірлер: 283

@blackpenredpen 5 жыл бұрын

Note: 2^sqrt(2) is actually transcendental (thus irrational), using Gelfond-Schneider Theorem.

@VJZ-YT 5 жыл бұрын

the only person who is not infected by tik tok

@paulovictorfagundescampos7008 5 жыл бұрын

solve 2^(√x)=i and 2^x=i

@m.guypirate6900 2 жыл бұрын

Cmon math! Cant you come up with something better than this?

@toshb1384 5 жыл бұрын

how is this any different than saying 2^(√2) = lim (n->√2) 2^n

@darkdelphin834 4 жыл бұрын

EXACTLY what I was looking for-for a while! Even asked my teacher and had no answer... Thank you for making this!

@tommasoferrari6144 4 жыл бұрын

Why 2^sqrt(2) = 2^[2^1/2] = 2^[2*1/2] = 2^[1] = 2 is wrong?

@mtaur4113 3 жыл бұрын

exp and log are more convenient tools here, but the approach in the video is more intuitive. If you use exp and log, those need definitions and it is important to verify that exp(n ln x) = x^n for positive x and integer n.

@theophonchana5025 2 жыл бұрын

2^(Square root of 2) = irrational

@foreachepsilon 5 жыл бұрын

More importantly, is sqrt(2)^sqrt(2) rational or irrational?

@olanmills64 3 жыл бұрын

I still didn't get a sense of what it "means" to take an irrational exponent of an integer

@WarpRulez 5 жыл бұрын

My first instinct, from having watched so many of your videos, was to start with

@harshsrivastava9570 5 жыл бұрын

Very cool! I am not familiar with the Gelfond-Schneider Theorem so could you explain the proof of transcendence of 2^√2

@Calilasseia 3 жыл бұрын

I would approach this by treating it as an infinite product. Of course, we have the issue of convergence to worry about, but you can deal with that by noting that 2^1.4 is finite, 2^1.5 is finite, and since 2^sqrt(2) lies between these values, it too converges, and therefore the infinite product also converges.

@kaiiverson1769 5 жыл бұрын

The way that I always though it was calculated was to write it in terms of e. You could then use the equation lim(n->infinity) (1+x/n)^n where x is the power of e.

@btdpro752 5 жыл бұрын

Very interesting, keep the good work up.

@aaab6054 4 жыл бұрын

Nice Video, but I just wanted to point out that for irrationals the continued fraction sequence will always converge fastest.(ie. the decimal approximation sequence or any other can only at best converge on an irrational as fast as the continued fraction)