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))))
Yep ! Same as me... but these are transcental functions too, and the calculus is achieved in a similar way, using convergent series (if you forget 5' your calculator ;-)...
It makes sense! You could define exp and ln without reference to raising anything to a power... define exp as the function with exp(0)=1 which is equal to its own derivative, and ln as the functional inverse of that (or as the integral of 1/x)... then bootstrap from there: define e^y as exp(y), then x^y = e^(ln(x) * y). Now you've got a definition that is valid at least for positive real x and real y, and probably can be extended to the whole complex plane with a little care.
Fady Omari I prefer this, but yours works too; 2^x=i e^(xln(2))=i xln(2)=0.5pi(4n+1)i x=0.5pi(4n+1)i/ln(2) 2^sqrt(x)=i e^(Sqrt(x)ln(2))=i Sqrt(x)ln(2)=0.5pi(4n+1)i Sqrt(x)=i0.5pi(4n+1)/ln(2) x=(-0.25pi^2)(16n^2+8n+1)/ln(2)^2 The sad thing is this didn’t become a bprp vid :(
mahesh waran I’ll edit the “i” in there, but included 2n+1 (it’s actually 4n+1) because... well, Let me tell you why; ln(i)=i0.5pi(4n+1) because sin and cosine, but here’s the simplified explanation; e^(i0.5pi(4n+1))= (e^i0.5pi)(e^0.5i4npi) That second term equals e^i2npi; which is one. So we have 1(e^i0.5pi)=e^i0.5pi=I. I include 4n+1 to show 0.5i(pi) isn’t the only solution. I’ll edit in that i and turn 2n+1 into 4n+1.
The Gelfond-Schneider Theorem (GST) says that if a and b are algebraic numbers with a ≠ 0, a ≠ 1, and b is is not rational, then a^b is transcendental. So for example 2 is algebraic and √2 is irrational so GST says 2^√2 is transcendental. A proof of GST is at people.math.sc.edu/filaseta/gradcourses/Math785/Math785Notes8.pdf P.S. Another interesting bit about GST is that it applies to complex numbers as well as reals. So since e^π is a value of i^(-2i) it follows that e^π is transcendental. This last tidbit might actually make a good blackpenredpen video if he wants to give a complex power example of GST. :)
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.
Why not use a log-antilog method Let 2^√2 = x. => log 2^√2 = log x => √2•log2 = log x => 1.414214 × 0.30103 = log x => Log x = 0.4257207 => x = antilog 0.4257207 (i.e. 10^ 0.4257207) => x = 2.665144 Therefore, 2^√2 = 2.665144
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.
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)
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.
+K H another way of doing it, by using this video’s working out, is to rewrite (2)^sqrt(2) as (sqrt(2)^2)^sqrt(2) and then to swap the indices due to the index laws to produce: (sqrt(2)^sqrt(2))^2 I think the answer is pretty obvious from there.
@Michael Darrow For that, I thought about multiplying the exponent: a^b^c = a^bc, I guess there are some rules to this or limitations to its use that I am not aware of
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).
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.
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)
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) *...
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.
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.
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.
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.
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?
ib9rt I was going to say this too. e^[Ln(2)*SqRt(2)] = 1 + Ln(2)*SqRt(2) + Ln(2)^2 + Ln(2)^3*SqRt(2)/3 + •••, and each summand can be calculated. Ln(2) is a well know constant which can be calculated via the alternating harmonic series. Integer powers of SqRt(2) are fairly trivial to calculate, since if n = 2m + 1 for integers n and m, then SqRt(2)^n = 2^m*SqRt(2), and if n = 2m, then SqRt(2)^n = 2^m. Also, generally speaking, since (a + bi)^(c + di) is well-defines for all complex c and d, and all not simultaneously zero complex a and b, I think it is safe to use the expansion of this expression to calculate 2^SqRt(2). Simply let b = 0, a = 2, c = SqRt(2), and d = 0.
The idea is to look at how maths is taught. So imagine you are still at school and haven't learnt about e^x yet. How do you get there? First you learn about 2^n, 2^(-n), 2^(1/n), 2^(m/n). With this video, you learn what 2^x means for any real x, and so you have the exponential function 2^x. Similarly 3^x and indeed a^x for any real number. Once you have these exponential functions, you get e^x as the one whose gradient at 0 is 1. The point is that the way maths is usually taught to kids, you need to explain to them what 2^x means without reference to e^x. Later on, in a more rigorous analytical approach to maths, with power series etc. you can define e^x first. But pedagogically, 2^x comes first.
Michael Rothwell That is definitely not true. I do not know of a single school in the entire planet that teaches limits before teaching what e^x is, and when they teach you what 2^x is heuristically, then they tend to teach you what e^x is immediately after. And if they really are teaching limits at the time, then e^x = lim n -> infinity (1 + x/n)^n is still a superior definition that you can then use to learn what 2^x is. Also, I hate to break your bubble, but schools rarely discuss what 2^(n/m) is, and so students have a tendency to not understand what this is until they learn derivatives. This is why students freak out when the see radicals: they do not think of them as fractional exponents, because they never were taught to think of them that way. So when they differentiate radical expressions, they think they have to learn a new rule. My point is, even by pedagogical standards, learning what 2^x is for irrational x the way this video teaches it is still horrible, and there still are way better ways to teach the subject. I also don’t know of any teacher who actually teaches it this way, and to be fair, BPRP made somewhat of an acknowledgement to this.
This definition is given because you need a way to extend exponentiation to the reals. How would you define e^x for all x if you don't know what it means for the exponent to be real in the first place?
No kidding but I was wondering about this today and decided to ask my maths teacher about it, but forgot and now I get recommended this video. Does TH-cam listen to my thoughts?
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?
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) ... ☹)
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)
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)
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.
Why not use the well known sequence 1/1, 3/2, 7/5, 17/12, ... where each entry (except the first) can be calculated from its predecessor, i.e., ...p/q, (p+2q)/(p+q)... ?
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)
Because 2^sqrt(2) would be something different than a power of a power (which is kinda the procedure you took, but there was an error). 2^sqrt(2)=2^[(2)^(1/2)] The result you gave would have had to come from a different initial expression: [2^(2)]^(1/2) Consecutive exponentiation does not share the associative property that addition and multiplication have, so the order with which you group your elements will impact on the result. The highest (1/2) exponent belongs only to the 2 in the first exponent, and not to the whole number. The initial expression would be something similar to a tetration, but not quite yet a tetration. Kinda like saying 3•3+4 is not the same as 3•(3+4).
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
Yeah, I figured this out for myself before I found this, but I was hoping for some insight into a different conceptual approach. I came across n^e and wanted to backtrack as to why this came up in the sequence of equations. I agree with you: I don’t like this.
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))
@@blackpenredpenwhat if we define e^x as 1+x+x^2/2+x^3/3!+x^4/4!+...? This would get the right answer and avoids needing to do rational approximations for our irrational exponent.
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)...)
how do you prove that irrational exponents always converge, formally? even though it's "obvious" by intuition that it does converge because each number after the decimal point becomes less significant to the resulting number
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
My professor was mentioning how raising something to an exponent is actually doing something involving e/ln but he didn't really go in depth, anyone know what he was talking about? What *does* an exponent do? Like the real definition? *The lim n => ∞ seems to be connected somehow (I know 1^∞ is e)
Nosson Weissman well, we have a way to calculate e to any power to arbitrary precision (e^x = sum from n=1 to infinity of x^n/n!) and a way to calculate ln of something to arbitrary precision using a different series, and we can express any exponential problem say a^b as e^ln(a)*b so that is the best way to handle arbitrary exponents.
Nosson Weissman You definitely can, I do not know where you get the idea that you cannot. a^b = e^[b*Ln(a)] for all b and nonzero a. The Taylor expansion of e^x is valid for all x. Thus it is always valid for x = b*Ln(a). x^n is well defined for all x, and n! is, needless to say, well defined for all n > 0 and n = 0, which is what the series requires. Hence this is objectively the best way to evaluate 2^SqRt(2)
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?
Using the Taylor series for e^x, sin(x), and cos(x), you can prove that e^(i*x) = cos(x) + i*sin(x). We use this to define exponentiation for complex numbers. Now, to address your question, if we want to take some number b to the ith power, we rewrite b^i as e^(ln(b)*i), which we know from above is equal to cos(ln(b)) + i*sin(ln(b)), and we are done.
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.
If a_n converges, then for any continuous function f, f(a_n) also converges (and if the limit of a_n is L then the limit of f(a_n) is f(L)). This is very easy to prove just using the definition of continuity and convergence.
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...
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))))
For sqrt(2) you can also use the Alternating harmonic series. We know that Σ ( (-1)^n/(n+1))=ln(2) and that sqrt(2)=exp(1/2*ln(2)) So sqrt(2)= Σ ( Σ ( (-1)^n/(n+1))^k/k!)
I have to tell you that I would find it so interesting to watch you calculate something really hard without using a calculator. I remember you doing something extremely long with a gaussian matrix and I love it so much!
Note: 2^sqrt(2) is actually transcendental (thus irrational), using Gelfond-Schneider Theorem.
Happy Thanksgiving Everyone!
Thx :)
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))))
What if we need to find the derivative of x^√2? Is there a way of solving this? Thank you
@@neitoxotien2258 You just apply the normal rules: d/dx(x^sqrt(2))=sqrt(2)*x^(sqrt(2)-1)
What is the answer for (-1)^√2, it doesn't converge anything
My first instinct, from having watched so many of your videos, was to start with
exp(ln(2^sqrt(2))) and go from there.
WarpRulez I also thought this might be a valid way to go about it.
Yep ! Same as me... but these are transcental functions too, and the calculus is achieved in a similar way, using convergent series (if you forget 5' your calculator ;-)...
It makes sense! You could define exp and ln without reference to raising anything to a power... define exp as the function with exp(0)=1 which is equal to its own derivative, and ln as the functional inverse of that (or as the integral of 1/x)... then bootstrap from there: define e^y as exp(y), then x^y = e^(ln(x) * y). Now you've got a definition that is valid at least for positive real x and real y, and probably can be extended to the whole complex plane with a little care.
Eh, my instinct was to approximate with a fraction and use a nth root with a power, like he did. But I thought about that problem for myself long ago.
Yes 2^sqrt(2) = e^[ln(2^sqrt(2))] = e^[sqrt(2) * ln 2].
So take sqrt(2) * ln 2 and call it x.
e^x yields the result.
Much easier.
EXACTLY what I was looking for-for a while! Even asked my teacher and had no answer... Thank you for making this!
High school or university instructor? (I would feel embarrassed for him/her if it was university level.)
@@toddtrimble2555 it was high school
🐙
Tibees lol!
Hi tibees
@@blackpenredpen thank you so much.
I've been struggling so much time to find an explanation for this enigma
solve 2^(√x)=i and 2^x=i
Paulo Victor Fagundes Campos
2^x=i
X=ln(i)/ln(2)
X=0.5pi*i/ln(2)
X=pi*i/(2ln(2))
2^(sqrtx)=i
Sqrt(x)=ln(i)/ln(2)
Sqrt(x)=0.5pi*i/ln(2)
Sqrt(x)=pi*i/(2ln(2))
X=-(pi)^2/(4(ln2)^2)
Fady Omari I prefer this, but yours works too;
2^x=i
e^(xln(2))=i
xln(2)=0.5pi(4n+1)i
x=0.5pi(4n+1)i/ln(2)
2^sqrt(x)=i
e^(Sqrt(x)ln(2))=i
Sqrt(x)ln(2)=0.5pi(4n+1)i
Sqrt(x)=i0.5pi(4n+1)/ln(2)
x=(-0.25pi^2)(16n^2+8n+1)/ln(2)^2
The sad thing is this didn’t become a bprp vid :(
mahesh waran I’ll edit the “i” in there, but included 2n+1 (it’s actually 4n+1) because... well, Let me tell you why;
ln(i)=i0.5pi(4n+1) because sin and cosine, but here’s the simplified explanation;
e^(i0.5pi(4n+1))=
(e^i0.5pi)(e^0.5i4npi)
That second term equals e^i2npi; which is one. So we have
1(e^i0.5pi)=e^i0.5pi=I.
I include 4n+1 to show 0.5i(pi) isn’t the only solution. I’ll edit in that i and turn 2n+1 into 4n+1.
@@i_am_anxious02 now I got it!
Thanks
mahesh waran no problem
Very cool! I am not familiar with the Gelfond-Schneider Theorem so could you explain the proof of transcendence of 2^√2
Thanks.
The Gelfond-Schneider Theorem (GST) says that if a and b are algebraic numbers with a ≠ 0, a ≠ 1, and b is is not rational, then a^b is transcendental. So for example 2 is algebraic and √2 is irrational so GST says 2^√2 is transcendental.
A proof of GST is at people.math.sc.edu/filaseta/gradcourses/Math785/Math785Notes8.pdf
P.S. Another interesting bit about GST is that it applies to complex numbers as well as reals. So since e^π is a value of i^(-2i) it follows that e^π is transcendental. This last tidbit might actually make a good blackpenredpen video if he wants to give a complex power example of GST. :)
@@Bodyknock TYSM
@@Bodyknock first time hearing of this theorem and I'm shocked by how natural it feels. thank you
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.
You need this to define exp of irrational numbers.
Finally, for a long time, I've never seen you do transcendental numbers, and now you do! Thank you, blackpenredpen! :D
the only person who is not infected by tik tok
I hate tick tock
What’s tik tok?
@@mathlegendno12 it's probably best that you don't know
Haha seems like all math lovers hate tiktok
Jo E Actually, it seems like all smart people hate Tik Tok, probably because it’s a waste of time.
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.
2:09 that's a colon :
A semicolon is this ;
IT DOESN'T MATTER !!! DOES IT???
Why not use a log-antilog method
Let 2^√2 = x.
=> log 2^√2 = log x
=> √2•log2 = log x
=> 1.414214 × 0.30103 = log x
=> Log x = 0.4257207
=> x = antilog 0.4257207 (i.e. 10^ 0.4257207)
=> x = 2.665144
Therefore, 2^√2 = 2.665144
So effective. Thanks
It seems every time I get curious about a math question, bprp has a video for it. Great as always
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.
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)
Even on Thanksgiving we get a math lecture. Happy Thanksgiving.
Great video and your thumbnails are pieces of art.:)
Mayank Vats
Thank you!!!!
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.
More importantly, is sqrt(2)^sqrt(2) rational or irrational?
Using Gelfond-Schneider Theorem, sqrt(2)^sqrt(2) is transcendental thus irrational. : )
+K H another way of doing it, by using this video’s working out, is to rewrite (2)^sqrt(2) as (sqrt(2)^2)^sqrt(2) and then to swap the indices due to the index laws to produce: (sqrt(2)^sqrt(2))^2
I think the answer is pretty obvious from there.
Thanks everyone. I know the answer, I just really like the proof outlined by the last poster.
blackpenredpen I was more a statistician so I didn’t know about that part of algebra. Thanks for the reference
@@blackpenredpen th-cam.com/video/sMC9wyv3n0c/w-d-xo.html
what about that?
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.
@Michael Darrow For that, I thought about multiplying the exponent: a^b^c = a^bc, I guess there are some rules to this or limitations to its use that I am not aware of
That would be true if it was actually (2²)^(1/2) but in this case. It's actually a "power tower" 2^(2^1/2) so you can't use that rule in this case
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).
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.
How would the calculator know how to calculate it without a definition? I just define it to be lim n->sqrt2 of 2^n.
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)
Can you make a video about !(negative)?
Mtari ok
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) *...
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.
“Weo weo” - Blackpenredpen. Love ur videos ❤️
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.
You can do that, he's just using 1, 1.4, 1.41, 1.414, ...
The definition is cool in that it's the way we need to define it for our properties of exponents to be preserved.
GREAT! Now, can you do a video on the meaning of a+bi ^ c+di ???
pretty sure 2blue1brown had posted a video about it
I've been wondering about this for a very long time. thank you so much bprp.
Muhammad Agil Ghifari
: )
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.
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.
You’re an amazing mathematician. Did you have those values memorized or are you able to do those calculations in your head?
He did it by himself and memorized it ☺
In another video he calculated sqrt(2) and sin(10) without calculater
in this channel, power to complex number is more common than power to irrational number
3:28 好看we handle...
咱們 or 我們？ :D
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?
ib9rt I was going to say this too. e^[Ln(2)*SqRt(2)] = 1 + Ln(2)*SqRt(2) + Ln(2)^2 + Ln(2)^3*SqRt(2)/3 + •••, and each summand can be calculated. Ln(2) is a well know constant which can be calculated via the alternating harmonic series. Integer powers of SqRt(2) are fairly trivial to calculate, since if n = 2m + 1 for integers n and m, then SqRt(2)^n = 2^m*SqRt(2), and if n = 2m, then SqRt(2)^n = 2^m.
Also, generally speaking, since (a + bi)^(c + di) is well-defines for all complex c and d, and all not simultaneously zero complex a and b, I think it is safe to use the expansion of this expression to calculate 2^SqRt(2). Simply let b = 0, a = 2, c = SqRt(2), and d = 0.
The idea is to look at how maths is taught. So imagine you are still at school and haven't learnt about e^x yet. How do you get there? First you learn about 2^n, 2^(-n), 2^(1/n), 2^(m/n). With this video, you learn what 2^x means for any real x, and so you have the exponential function 2^x. Similarly 3^x and indeed a^x for any real number. Once you have these exponential functions, you get e^x as the one whose gradient at 0 is 1. The point is that the way maths is usually taught to kids, you need to explain to them what 2^x means without reference to e^x. Later on, in a more rigorous analytical approach to maths, with power series etc. you can define e^x first. But pedagogically, 2^x comes first.
Michael Rothwell That is definitely not true. I do not know of a single school in the entire planet that teaches limits before teaching what e^x is, and when they teach you what 2^x is heuristically, then they tend to teach you what e^x is immediately after. And if they really are teaching limits at the time, then e^x = lim n -> infinity (1 + x/n)^n is still a superior definition that you can then use to learn what 2^x is. Also, I hate to break your bubble, but schools rarely discuss what 2^(n/m) is, and so students have a tendency to not understand what this is until they learn derivatives. This is why students freak out when the see radicals: they do not think of them as fractional exponents, because they never were taught to think of them that way. So when they differentiate radical expressions, they think they have to learn a new rule.
My point is, even by pedagogical standards, learning what 2^x is for irrational x the way this video teaches it is still horrible, and there still are way better ways to teach the subject. I also don’t know of any teacher who actually teaches it this way, and to be fair, BPRP made somewhat of an acknowledgement to this.
This definition is given because you need a way to extend exponentiation to the reals. How would you define e^x for all x if you don't know what it means for the exponent to be real in the first place?
how is this any different than saying 2^(√2) = lim (n->√2) 2^n
No kidding but I was wondering about this today and decided to ask my maths teacher about it, but forgot and now I get recommended this video. Does TH-cam listen to my thoughts?
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?
compare it with
(-2)^(10/14),
and we dare nothing to say..
Happy thank day BPRP and anyone else who reads this :)
Thank you!
Well you could evaluate the Taylor expansion of e^x at a certain approximate value: (sqrt 2)(ln 2)
Very interesting, keep the good work up.
Thanks
Please post some for midway calc students
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?
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?
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) ... ☹)
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).
plz do newton's method for those I've never learned the technique
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)
Yes. He's just using the one he used as an example
How can we be sure that the sequence 2^(r_n) converges?
Pretty Much Physics 2^x is continuous.
@@blackpenredpen I see, thank you :)
How much does this definition by means of a limit differ from applying this formula: a^b=e^(b*ln(a)) ?
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)
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.
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?
Cmon math! Cant you come up with something better than this?
(4/3)^-(4/3)^-(4/3) = ?
Solve details please
For this one could you alternatively use Newton's method from calculus?
And this is the general method to calculate powers with decimal numbers.
Why not use the well known sequence 1/1, 3/2, 7/5, 17/12, ... where each entry (except the first) can be calculated from its predecessor, i.e., ...p/q, (p+2q)/(p+q)... ?
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)
Why 2^sqrt(2) = 2^[2^1/2] = 2^[2*1/2] = 2^[1] = 2 is wrong?
Because 2^(1/2) isn't 2*1/2
Because 2^sqrt(2) would be something different than a power of a power (which is kinda the procedure you took, but there was an error).
2^sqrt(2)=2^[(2)^(1/2)]
The result you gave would have had to come from a different initial expression:
[2^(2)]^(1/2)
Consecutive exponentiation does not share the associative property that addition and multiplication have, so the order with which you group your elements will impact on the result.
The highest (1/2) exponent belongs only to the 2 in the first exponent, and not to the whole number.
The initial expression would be something similar to a tetration, but not quite yet a tetration.
Kinda like saying 3•3+4 is not the same as 3•(3+4).
could it be x^2^1/2 and then its x
That's what I thought.
it's 2^(2^1/2)
No. You can multiply the exponents in this situation: (a^x)^y=a^xy, but that: a^x^y=a^xy, is a mistake. So x^2^1/2 is not x^(2×1/2)=x.
@@אלעדערוסי-צ4ד
it's the same thing, putting the parenthesis is just clarifying that it's not sqrt of 2^2 but it is 2^ of sqrt of 2.
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
Yeah, I figured this out for myself before I found this, but I was hoping for some insight into a different conceptual approach. I came across n^e and wanted to backtrack as to why this came up in the sequence of equations. I agree with you: I don’t like this.
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))
sqrt(2)=1+1/(2+1/(2+1/(2...)))
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?
Sky11631
Well, you still have irrational exponent. So you can't use that to define it.
@@blackpenredpen but if I compute e as series, I seem to only have natural numbers as exponents... or did I forget something?
Sky11631
Your exponent is still irrational. You cannot use irrational exponent to define irrational exponent.
@@blackpenredpenwhat if we define e^x as 1+x+x^2/2+x^3/3!+x^4/4!+...? This would get the right answer and avoids needing to do rational approximations for our irrational exponent.
why can't we do
2^sqrt(2)
e^sqrt(2)*ln(2)
Then use the series representation of e^x
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)...)
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
how do you prove that irrational exponents always converge, formally? even though it's "obvious" by intuition that it does converge because each number after the decimal point becomes less significant to the resulting number
And what's the meaning of imaginary exponentials?
That's a better question
How do you proof that all the exponential properties holds true for fractional powers as well
It's actually the opposite process. Fractional powers are defined the way that they are so they satisfy all the exponential laws
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 about a negative number to an irrational exponent? It would be imaginary but say what is (-2)^sqrt(2) ?
My professor was mentioning how raising something to an exponent is actually doing something involving e/ln but he didn't really go in depth, anyone know what he was talking about? What *does* an exponent do? Like the real definition?
*The lim n => ∞ seems to be connected somehow (I know 1^∞ is e)
Nosson Weissman well, we have a way to calculate e to any power to arbitrary precision (e^x = sum from n=1 to infinity of x^n/n!) and a way to calculate ln of something to arbitrary precision using a different series, and we can express any exponential problem say a^b as e^ln(a)*b so that is the best way to handle arbitrary exponents.
@@hedgechasing That's pretty cool. Tough you can't use x^n in the definition of what it is... I don't think
Nosson Weissman You definitely can, I do not know where you get the idea that you cannot. a^b = e^[b*Ln(a)] for all b and nonzero a. The Taylor expansion of e^x is valid for all x. Thus it is always valid for x = b*Ln(a). x^n is well defined for all x, and n! is, needless to say, well defined for all n > 0 and n = 0, which is what the series requires. Hence this is objectively the best way to evaluate 2^SqRt(2)
@@angelmendez-rivera351 Okay, I'll take your word for it... Gotta study for a Diff Eqs test
What does it mean to raise any number to e?
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?
What is root 2 to the power root 2
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?
Using the Taylor series for e^x, sin(x), and cos(x), you can prove that e^(i*x) = cos(x) + i*sin(x). We use this to define exponentiation for complex numbers.
Now, to address your question, if we want to take some number b to the ith power, we rewrite b^i as e^(ln(b)*i), which we know from above is equal to cos(ln(b)) + i*sin(ln(b)), and we are done.
thanks for the video....love from India
is it valid to take the limit into the exponent ? Can we prove that 2^(limRn)=lim(2^Rn) as n goes to infinity?
Yes, the limit of a function is the function of the limit, provided the function is continuous
How do we know that the sequence of function converges?
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.
BUT: how do wo know the sequence 2^rn actually converges?
If a_n converges, then for any continuous function f, f(a_n) also converges (and if the limit of a_n is L then the limit of f(a_n) is f(L)). This is very easy to prove just using the definition of continuity and convergence.
where did you learn to write an r?
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 still didn't get a sense of what it "means" to take an irrational exponent of an integer
2: 2^(2^1/2):: 8/ 4: 8/3, almost.
It's strangely close to e.
It is even closer to 8/3
Isnt sqrt2 just 2^(1/2)? Doesnt that rationalize the exponent?
I knew it was going to be the octopus!
Nitkola Nitkolajević : )
th-cam.com/channels/wZJkbfWLhLHF-FN6FZGrQQ.html
Or you could say : lim x->+∞ 2^(undervalue(V2 * 10^x)/10^x)
(pi^pi + i) when
no twist? :C
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))))
For sqrt(2) you can also use the Alternating harmonic series.
We know that Σ ( (-1)^n/(n+1))=ln(2) and that sqrt(2)=exp(1/2*ln(2))
So sqrt(2)= Σ ( Σ ( (-1)^n/(n+1))^k/k!)
How they did it before the era of computer and calculator?
I have to tell you that I would find it so interesting to watch you calculate something really hard without using a calculator. I remember you doing something extremely long with a gaussian matrix and I love it so much!
_fiNit{E_arth}
Ok!! I may try if I have time.
@@blackpenredpen you didn't do a live stream for a long time...
Nobody :
BPRP doing complex math
Me : 2^root2 = 2^2^1/2 = 2^2*1/2 = 2