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

My first instinct, from having watched so many of your videos, was to start with exp(ln(2^sqrt(2))) and go from there.

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

solve 2^(√x)=i and 2^x=i

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

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.

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.

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.

Very interesting, keep the good work up.

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)

I've been wondering about this for a very long time. thank you so much bprp.

Even on Thanksgiving we get a math lecture. Happy Thanksgiving.

plz do newton's method for those I've never learned the technique

Why it is not as such? 2^sqrt2 = 2^2^1/2 = 2^1 = 2. I assume I am badly mistaken somewhere, but I cannot figure out where.

How would the calculator know how to calculate it without a definition? I just define it to be lim n->sqrt2 of 2^n.

You’re an amazing mathematician. Did you have those values memorized or are you able to do those calculations in your head?

“Weo weo” - Blackpenredpen. Love ur videos ❤️

For this one could you alternatively use Newton's method from calculus?

I would maybe use a sequence that makes it obvious how to compute each subsequent term. For example a_0 = 2, a_n+1 = 1/2 a_n + 1/a_n converges to sqrt(2) (this is just Newton's method of course).

Could you explain what taking a number to the power of i? I know it has something to do with rotating around the unit circle but why?

I thought you were going to use the newton’s method sequence, which might make this definition more satisfying, since you don’t already have to know the decimal expansion of sqrt(2)

Ok, if a sequence rn coverges to sqrt(2), then 2^rn converges to 2^sqrt(2). But if there is another sequence pn that converges to sqrt(2), does 2^pn converges to 2^sqrt(2) too? What ensures convergence? (I love your channel)

Great video! I have watched 3blue1brown video on this topic where he said that exponents can be regarded as infinite polynomial e^x = 1 + x/1 + (x^2) /2! + (x^3) / 3! +... (for instance 2^(2^(1/2)) = e ^(2^(1/2) * ln2) and then instead of x plug ln2 * 2^(1/2) in the infinite polynomial)

a lot of people in the comments are talking about definitions in terms of the exponential function, and they raise a very good point! the best definition is relative, and both are useful in their own way. in fact, the two definitions basically parallel the two common definitions of the real numbers. the real numbers are defined as the unique complete ordered field up to isomorphism, and completeness can be phrased in two ways: • a complete metric space is one in which every cauchy sequence converges. intuitively, this means that any sequence where the terms get really close to each other has some limit. this isn't true in the rationals, where you can define things like 1.4, 1.41, 1.412, ... that get really close, but don't converge to anything. • a complete ordered field is one in which every bounded subset has a least upper bound. intuitively, this means that you can always define a "maximal" element for any set. again, this isn't true in the rationals, where the set of all numbers whose square is less than sqrt(2) has no maximal element. what does this have to do with our definitions? the first definition of completeness corresponds to the video definition of exponentiation, and the second corresponds to the ln/exp definition. it's clear how the first definition relates; you can define a convergent cauchy sequence for any real number, so it's most natural to talk about real number exponents as limits of rational exponents. as to the second, the problem is how do we define ln and exp? a common way is to define ln(x) as the integral from 1 to x of dx/x, and then define exp as either the inverse of ln, a power series, or as the unique solution to f'(x)=f(x), f(0)=1. in any case, we're now talking about integration over some interval, i.e. a set of real numbers, and so now it's most natural to think of things in terms of the least upper bound property. this is all interesting in my opinion, but the primary motivation for this comment was to show that neither definition is better than the other. in fact, both definitions are quite useful, when applied to their own domains. the thing that you'd want to do as a mathematician is prove that they're equivalent, so that you can use either definition whenever you want.

And if the base is negative how can you do -2^(5/7)? What is the sign of your answer? Did you use complex to sole it?

Lets define the sequence r[n+1] = 0.5 * ( r[n] + (2/r[n])..Here r[n] converges to sqrt(2) if you start with good initial conditions. I think this sequence is more well defined than the one mentioned in the video.

Sir, how did the (2)^ root 2 became 2.66514? The last sequence in the video is (2)^ 1.4142 = 2.66512 Pls kindly help sir. I get stuck in these types of sums a lot.

Well you could evaluate the Taylor expansion of e^x at a certain approximate value: (sqrt 2)(ln 2)

Probably not a bad idea to point out that 2^x restricted to rational inputs x is an increasing function, without using calculus, so that one is assured the limit exists as an l.u.b.

is it valid to take the limit into the exponent ? Can we prove that 2^(limRn)=lim(2^Rn) as n goes to infinity?

The Taylor series for (2)^1/2 gives us 1 + 1/2 - 1/8 + 1/16 + 5/128 +... So this also can be presented as a pi series: 2 * 2^(1/2) * 2^(-1/8) * 2^(1/16) * 2^(5/128) *...

If a negative number to the power of an even number is positive and to the power of an odd number is negative, what is a negative number to the power of something such as 2.5?

Using Babylonian method, sqrt(2) can be iterated using the sequence (a_n + 2/a_n)/2 for some a_0>0. Then we can approach 2^sqrt(2) using the sequence (sqrt(2)^a_n)(2^(1/a_n)) for some a_n>0. Or we can use a series expansion of Σ(1/2 choose n, n from 0 to infinity), and convert 2^sqrt(2) to a partial product Π(2^(1/2 choose n), n from 0 to infinity). These are not "definition" though. Simply the methods to compute it.

could it be x^2^1/2 and then its x

Why can't we say 2^sqrt(2) = [e^ln(2)]^sqrt(2) = e^[sqrt(2)*ln(2)] ? Since e^x and ln(x) are well defined functions, that makes 2^x a well defined function?

Why isnt e^(ln(2)*sqrt(2)) used? Wouldnt that make the calculations 'easier', since e, ln and sqrt can be computed kind of easy?

Why not use the Pell sequence to approximate the square root of 2? (1/1, 3/2, 7/5, 17/12, 41/29, 99/70, 239/169...) Basically, starting with 1/1, for each Pell number n/d, the next one is (n+2d)/(n+d), and as it approaches infinity, it gives increasingly accurate approximations of sqrt2.

What is root 2 to the power root 2

We said that x^(1/n) = n root of x So 2 ^ sqrt(2) should be equal to 2 ^ 2 ^ (1/2), but 2*(1/2) = 1 so 2^sqrt(2) = 2^1 Where's the mistake?

Isnt sqrt2 just 2^(1/2)? Doesnt that rationalize the exponent?

And what's the meaning of imaginary exponentials?

thanks for the video....love from India

Would you calculate 2^(1/e) ?

Happy thank day BPRP and anyone else who reads this :)

nice video

A small doubt is 2^sqrt2 2 because a had a small proof for that somewhere that had something to do with how something to the power of something can be written differently which caused it to be 2 just can anyone prove me wrong?

Why not make a curve containing sqr(0), sqr(1), sqr(4), sqr(9), sqr(16), sqr(25)... to estimate the approximate value of sqr(2).

BUT: how do wo know the sequence 2^rn actually converges?

what about a negative number to an irrational exponent? It would be imaginary but say what is (-2)^sqrt(2) ?

My question is: what happens if we raise a negative irrational number to another irrational number? Shouldn't this diverge like the graph of y=x^x?

And of course, the difficulty in calculating, say, things like 2^1.41 = 2^(141/100) which requires a .. _hundredth root_ of a _141st power_ (!!!!) (*) ... is why that this definition, while theoretically valid and useful from that point of view, is not used in practice to actually _calculate_ such exponentiations to high precision. Instead we use that a^b = e^(b ln(a)), based on the exponential and logarithmic functions. _However_ , the key point of this definition is that it is in a sense the logical conclusion of the programme of extending exponentiation on the basis of exponent laws - in particular, the law (a^b)^c = a^(bc) - which suffices to define the exponential at rational exponents - while the exponential and logarithm function, though definitely clear as to their purpose and meaning in hindsight, are not at all so clear in "foresight" were they to be given first up (e.g. if I gave you the Taylor expansion or limit form of e^x = exp(x) off the bat, would you be able to guess, knowing nothing else, that this had anything at all to do with exponentiation? I'd think not.). Instead, it would be best to start this way from the intuitive point of view and then derive them and prove their expansions as a theorem. Note that also to make this definition valid, we have to prove that the necessary limit exists ... a nice little bit of elementary Real Analysis (the proof-based, theoretical version of calculus). Moreover it is interesting, and, I'd think, important, to note that this definition seems fairly directly tied to one of the basic ways to define the real numbers themselves in terms of suitable sequences of rational numbers. (*) To see how painful that is - this is an illustration. 2^141 = 2,787,593,149,816,327,892,691,964,784,081,045,188,247,552. That's a 43 digit number. No I did not work that by hand. Obviously. To do Newton's method you will need to raise _that_ to the 100th power just in getting the first approximating decimal. That's about 4300 digits in the result. Then you will need to do divisions by that. Long division by a 4300 digit number. By hand. **Yeckh.** You need to be an autistic savant of the right type to have any hope of carrying that through without errors and insane amounts of time - read megaseconds upon megaseconds. (Though actually if we were going to do it for seriousness, the trick would be to switch to "floating point", i.e. scientific notation, form, and work in that with a limited amount of precision, given that we are going to only be able to report the result to finite precision anyways, that is perhaps somewhat higher than the goal precision. This dramatically reduces the number of digits we need to work with at each step, and in fact would be how a computer might handle it, though it's worth nothing that the size of the power of 10 involved here (or really, power of 2 for computers, as they work in binary, and moreover note how nicely the choice of a "2" here gels with that fact) would blow the limit of the most common native computer hardware floating point formats and thus would require emulation in software - namely IEEE 754 double takes a binary exponent of only 1023 max, I believe, which is a power of 10 of log_10(2^1023) = 1023 log_10(2) ~ 308, versus 4300! On pen and paper, of course, we can make the exponents of 10 as large as we want. Still, we will have to do many self-multiplications and divisions involving about 6 digit numbers, by other 6 digit numbers, by hand if we're going to work this out by hand. And let's not then get started on 2^1.414 = 2^(1414/1000) = 2^(707/500) ... ☹)

Well that was disappointing. I was hoping for an analytical solution. I thought you would use the Taylor series expansion for square root of 2. Then you can express 2^sqrt(2) as the product of 2 raised to the power of each of the terms as n goes to infinity. Any reason why that shouldn't work? In fact, it should be generalizable to any square root exponent, n (n>0), where you use the expansion for f(x) = sqrt(x+1) where x = n -1 (that's why n has to be > 1)

where did you learn to write an r?

Please derangement of Mississippi ;) and Yes I also want to know how to estimate some integer to some power like 4/3, 5/9 ... Thanks! ;)

Sir, I am a big fan But I have a doubt here. 2^sqrt(2) can be written as 2^2^(1/2). Here 2^2 is 4 and replace it in our equation. It will be 4^(1/2) which is sqrt(4) which is 2. So I am thinking that 2^sqrt(2) can be 2. Thinking logically.

I thoroughly enjoy your content. You find fascinating and interesting bits of math and explain them beautifully. With this explanation, is there any other ways, besides numerically, to obtain this value without expanding Sqrt(2)?

How do you proof that all the exponential properties holds true for fractional powers as well

I know certain irrationals (maybe all) can be expressed as an infinite sum (e, pi, ln(2), etc.). Further discussion of this would be fun. See here for some good arguments regarding this: math.stackexchange.com/questions/1466622/is-it-possible-to-represent-every-irrational-number-as-a-limit-of-an-infinite For those irrationals that can be expressed as an infinite sum couldn't you use that fact to estimate a solution? Effectively x=2^(A+B+C+D+E+F+...) , becomes x=2^A*2^B*2^C*2^D*2^E*2^F... I'm assuming that B

why can't we do 2^sqrt(2) e^sqrt(2)*ln(2) Then use the series representation of e^x

(4/3)^-(4/3)^-(4/3) = ? Solve details please

I was wondering during whole video

I expected i but anyways great vids. New theorem

(pi^pi + i) when

Well that's really easy, we can write it as 2^[2^(1/2)] and the powers multiply, so 2^(2/2) = 2^1 = 2

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

What I found interesting is the thought of using the number in the exponent and write it down in some sort of summ equal to the number. For instance Sum n=0 to infinity (1/2^n) = 2 Now I try to write 2^2 as 2^(sum of 1/2^n) and get: 2^(1/2^0+1/2^1+1/2^2...) with the + being in the exponent i can do: 2^(1/2^0)*2^(1/2^1)*2^(1/2^2)... which is 2*sqrt2*4.sqrt2*8.sqrt2... and so on...4.sqrt means fourth sqaureroot... When I think of any number with decimal, I can write it down also as sum n=0 to k (a_n*10^(-1)) , then my number has k decimal places. with a_n being the number for the decimal 1.41... = 1*10^0+4*10^-1+1*10^-2... then I can say 2^sqrt(2) = 2^1+10.sqrt(2^4)+100.sqrt(2^1)+... and so on... when we speak of exponent, we usually imagine 2^2 = 2*2 or 2^3 =2*2*2 which means the formulation of the square root is: i am looking for the same number that multiplied by itself 2 times (other example 3) equals 4 (8)... a*b*c...=a*a*a*....=a^n I found it easier to use it for sth like 2^((pi^2)/6) = lim n-> infinity 2^(1/1^2+1/2^2+1/3^2....+1/n^2) product of n^2th sqrt2 How about 2^ln(2)= 2^(1/1-1/2+1/3-1/4+1/5-1/6+...) =(2*2^3*2^5...)/(sqrt(2)*4.sqrt(2)*6.sqrt(2)...)

And this is the general method to calculate powers with decimal numbers.

2^sqrt(2) = exp(log(2)*sqrt(2))

Or you could say : lim x->+∞ 2^(undervalue(V2 * 10^x)/10^x)

Well , it was seeming to approach 2.6666666..... So I think the answer is , 2^√2 = 8/3 Or maybe approximately , because there was no formal definition of the no. being rational or irrational

What 2^sqrt !7?

How they did it before the era of computer and calculator?

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

√2 = 2^(1/2) Then, 2^(2^(1/2)) Powers can be multiplied, as I read in class 6th. = 2^(2×(1/2)) = 2^(2/2) = 2^1 = 2. TADAN, LADIES AND GENTLEMAN. ANSWER IS 2.

Here's what I know: n=2^sqrt(2)=2×2^(.4142...)=2×2^(1/(1+sqrt(2))), so some x satisfies x^(1+sqrt(2))=2, and the answer n=2x. Then we find that x^(.4142...)=x^(1/(1+sqrt(2))), so there exists some y^(1+sqrt(2))=x, and so on. It's an infinite chain (the continued fraction of sqrt(2))

Gooooooooooooooooooood

But the real question is....if octopods had higher mathematics, would they use base 8?

Initially I thought you will use the f(x)=e^ln(f(x)) trick and pull out the exponent, but I see it would make it even more ugly.

#limit

You should probably mention that this is only dealing with principal roots.

A whole number >1 to the surd= transcendental number.

Nobody : BPRP doing complex math Me : 2^root2 = 2^2^1/2 = 2^2*1/2 = 2

Irrational numbers

for 2^sqrt(2) i got sqrt(2) 💀

It's kind of obvious that this definition has nothing to do with any real arithmetic, i.e. arithmetic a mortal man can do. Ok, you can calculate 2^1.4 as (2^14)^(1/10), and try to calculate 2^1.41 as (2^141)^(1/100). 2^141 is 2.79e42, 43 decimal digits (and I am not sure how to do root 100). 2^1414 has 426 decimal digits, and so forth. That's why you stopped writing fractions and used a calculator.

What if we use Taylor series. We know that sin(pi/4) is sqrt(2)/2 . So sqrt(2)= 2*sin(pi/4). Sin(x)= Sum from n= 0 to inf of ((-1)^n/(2n+1)!)*x^(2n+1). We can plug pi/4 in the formula,but we can do better and use Leibniz series pi/4= Sum from n= 0 to inf of (-1)^n/(2n+1). Finally 2^sqrt(2)= 2^(2* (Sum from m= 0 to inf of ((-1)^m/(2m+1)!)* ((Sum from n= 0 to inf of (-1)^n/(2n+1))^(2m+1))))

Yaaaaay

Well sqrt (2) = 2^1/2 therefore 2^(sqrt (2)) = 2^2^1/2. Since we a power to a power, we can multiply them. Therefore 2^(sqrt (2)) = 2^1 = 2 So we got 2^(sqrt (2)) = 2 We take log base 2 on both side sqrt (2) = 1 We square both sides and get 2=1. Nothing to see here, move along...

I like it. I think I'll use a calculator too.

I prefer letting 2^SqRt(2) = e^[Ln(2)*SqRt(2)] and evaluating the latter using the Taylor series. It can easily be rewritten in the form a + b*SqRt(2), where a and b are two infinite series with x = Ln(2). e^[Ln(2)*SqRt(2)] = 1 + Ln(2)*SqRt(2) + Ln(2)^2 + Ln(2)^3*SqRt(2)/3 + Ln(2)^4/6 + Ln(2)^5*SqRt(2)/30 + ••• = [1 + Ln(2)^2 + Ln(2)^4/6 + ••• ] + SqRt(2)*[Ln(2) + Ln(2)^3/3 + Ln(2)^5/30 + ••• ]. Both series can be calculated separately, especially given that Ln(2) is well known. I think this both more intuitive and more useful than trying to define 2^SqRt(2) as the limit of a sequence. The latter may be theoretically more satisfying, but much more difficult to use.

2^(Square root of 2) = irrational

Weo weo

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

🕊✋🌷

It is difficult to handle irrational powers. The you are explaining the method, which cannot be used without calculator!! Better, use infinite series to calculate irrational powers

2 i think i did something wrong 2 ^( 2 ^ 1/2) 2 ^(2 ^ 1/2) 2 ^ 1 2

Nooo. Where did your super cool beard go.😢😢

4:22 it's pretty clever to say something while writing something else :)

aaaaah no , approximations

Please bless me Blackpenredpen.