Figuring out what 2^π means Resources on exponentiation: en.wikipedia.o... en.wikipedia.o... Resources about the exponent algorithms: stackoverflow.... netlib.org/fdl... Corrections:
On a broader version of this general idea, I remember my dad explaining that my calculator doesn't have trig tables in it. (Yes, I'm old enough to remember looking for trig values and logarithms in a table.) Once we covered power series in calculus class, it made sense, but there was a long gap before I made the connection between those and irrational exponents.
In the old days, before floating point hardware became cheap, calculators often used a method called Cordic for most math routines, like square roots, logarithms, and trig functions. After floating point hardware became available, calculators tended to use lookup tables, with the appropriate routines using those tables. I don't think they used power series like you probably learned in Calculus class, like the Taylor series, since those don't converge very quickly. They probably used something like a Chebyshev series, along with the lookup tables. According to Wikipedia, Cordic is still widely used for embedded processors, like fpga's.
@@charlesgantz5865 Good point. By the time I made the connection, I had indeed learned about other series such as Chebyshev, but the power series were the first ones that came to mind.
It's actually amazing the algorithmic optimizations from computer science and then the floating point and circuitry optimizations from computer engineering that make calculating a number like this so accessible and inexpensive for anyone to do.
Yes when calculations like this used to be so much work. Modern computers can do calculations that people spent their entire lives doing α few hundred years ago.
*groan* Can't BELIEVE nobody else has pointed out this obvious error to you yet, but the fourth "2" in your thumbnail is CLEARLY not accurate down to the full ratio! You need to enable transcendental mode on your editor so that it can render the length infinitely precisely. Beginner mistake!
thanks for posting videos like this. as a 14 year old sophomore, i shouldn't normally be interested in topics like these, but thanks to the ease of access it is made possible. very good explanation, i understood most of it, some of it i didnt, and that's on my part since I literally still am in high school math
@@somerandomcsgoplayerlol8977 how are you 14 and in 10th grade I'm 14 and just got into 9th grade tf Also I'm pretty sure you didn't understand limits and integrals in which case you should check out calculus. I took calc 1 and calc 2 and I'm halfway into calc 3 and I can tell you it's really fun. You're gonna be taught about limits as soon as you start calc 1, basically one of the first things you'll learn. Good luck.
In my real analysis class last semester I remember talking about definitions of real numbers to the power of rational numbers. I remember asking "what about 3^pi or pi^pi?" Prof replied "what about it?". I replied, "I have no idea what that means". Lead to this conversation. First time I've ever felt smart for saying I don't know anything about a topic
Excellent vid. Defining the exponential function as an infinite series seems to frequently be the way forward. It appears in a few theorems regarding the moments of the exponential family of distribution. 3Blue1Brown also has that great vid showing how it can make sense of matrices in the exponent. Btw, your channel is growing fast! I try to keep an eye out on all the math edu channels (I have one as well), and this one has the knowledge and animations to go places. Well done - im sure you’ll crush it
defining things in terms of power series whenever possible introduces a ton of nice behavior and simplifies computations, so it makes a lot of sense to use that technique if it's available
Excellent video indeed, I find pi always a victim of errors until 2 things are solved one being the elimination of 0 anywhere within it entire length as well as repeated numbers like 77 66 anywhere in its string. I dunno , maybe I’m wrong maybe I’m just crazy but pi does have several limitations one being a string of numbers after it is exhaustingly long only to find you reached a point where the string has repeated itself , that can indicate its ending amongst other factors like a circle so wide and possibly reaching beyond the known universe in diameter coming to a zenith where the curvature reaches flat , that point I theorize it collapses and reaches its eminent annihilation. Same as in the other direction where a circle reaches its Planck width and begins to vibrate in an attempt for it to avoid annihilation as well. So there you have my theory of the two ends of a circle . Maybe I’m wrong or maybe I’m just crazy . Who knows
I'd argue that it's probably not better to call the integration version the "definition" of exponentiation (at least for real numbers). Just because it's computationally faster, doesn't make it a better definition. This is why nobody defines multiplication with Harvey and Van Der Hoeven's O(n log n) multiplication algorithm even though it's technically faster. Edit: There's a lot of debate in the replies, and I think it all stems from different objectives, which is why we have different characterizations. My above comment naively repeated the definition I first learned for real numbers without further considerations as to it's advantages/disadvantages, which I now realize was presented first because it was the first definition that could be understood--as in a real analysis course, integration and power series are covered rather late. I'd argue that this definition is more natural for the real numbers, as real numbers themselves are sort of just limits of rational numbers (by one characterization of the real numbers, at least) and this is how many other operations are defined for the real numbers. The problem of well-definedness is a non-issue to me because the night after I wrote this comment I derived a relatively simple proof of well-definedness--one which would convince a fellow undergrad, but is really just a very standard proof, which might be delegated to the exercises section of a beginner analysis text. The definition presented in the video definitely has no questions of well-definedness, however exponent laws may cause trouble (and before you note that the exponent laws were proven by SackVideo in the replies, it's rather hard to prove that for f:R->R such that f(x) = e^xe^y/e^(x+y), its derivative f'(x) = (e^xe^ye^(x+y) - e^xe^ye^(x+y)/(e^(x+y))^2 without using any exponent laws, which would result in circular reasoning). The definition in the video has the advantage to generalizing to many fields, including the noteworthy complex numbers. This is probably why beyond real analysis it's used more often. For the question of which can prove more stuff... it's up to what you're proving. That's why we have different characterizations after all. Hope i didn't miss anything or mess anything up
The main reason we define x^y as e^(yln(x)) is because there are no concerns about well-definedness and its actually simpler to prove things with it. The fact that it's also better for computation is really just a benefit.
@SackVideo I think Edwin was talking about the definition of ln(x). As with many things in math, there are multiple equivalent ways to define ln(x) and it can be proven that they are all equivalent. The Wikipedia article for ln goes into a bit more detail: en.wikipedia.org/wiki/Natural_logarithm#Definitions
@@ASackVideo I think the most compelling reason for me is just how well it generalizes to other fields. However with a bit of reflection I certainly agree that the latter definition is better, with a few modifications. Also @NeatNit I was not referring to the definition of ln(x), however it definitely should just be defined as the inverse of e^x, as this is what generalizes best. In summary, I think my favourite definitions are: e^x = 1+x+x^2/2!+... e^ln(x)=x b^x=e^ln(b)*x. This however hides the inconvenience of the situation as you can no longer easily prove your exponent laws. This is a good thing however, as when generalizing to other fields it's important you don't just presuppose that the exponent laws hold. Also I just wanna note that I really liked this video cause it got me thinking about analysis again which i really liked learning about
Clicking around a bit more on Wikipedia, it turns out that equivalent definitions are called characterizations (en.wikipedia.org/wiki/Characterization_%28mathematics%29) and there is an article dedicated to the characterizations of exp(x): en.wikipedia.org/wiki/Characterizations_of_the_exponential_function Which is a bit of a tangent (no pun intended) but I thought it was worth sharing! Edit: and the article on exponentiation mentions two definitions for a^b when they are positive real numbers: en.wikipedia.org/wiki/Exponentiation#Real_exponents
@@edwinvlasics4047 Proving exponent laws is actually not too bad. x^1 = e^(1*ln(x)) = x. Let y be constant. Then for f(x) = e^xe^y/e^(x+y), f'(x) = (e^xe^ye^(x+y) - e^xe^ye^(x+y)/(e^(x+y))^2 = 0, so f(x) is constant. Then as f(0) = e^0e^y/(e^y) = 1, f(x) = 1 for all x, so e^xe^y = e^(x+y). x^a*x^b = e^(alnx)e^(blnx) = e^((a+b)lnx) = x^(a+b). x^(1+y) = x^1 * x^y = x*x^y. etc.
The turbo encabulator is an instrument that allows transmission that not only supplies inverse reactive current for use in unilateral phase detractors, but is also capable of automatically synchronizing cardinal grammeters. Basically, the only new principle involved is that instead of power being generated by the relative motions of conductors and fluxes, it is produced by the modial interaction of magnetoreluctance and capacitive directance. The original machine had a baseplate of prefabulated Amulite, surmounted by a malleable logarithmic casing in such a way that the two spurving bearings were in a direct line with the pentametric fan. The latter consisted simply of six hydrocoptic marzel vanes so-fitted to the ambifacient lunar wane shaft that side fumbling was effectively prevented. The main winding was of the normal lotus-o-delta type placed in panendermic semi-boloid slots in the stator, every seventh conductor being connected by a nonreversible trem'e pipe to the differential girdlespring on the up-end of the grammeters. The Turbo Encabulator has now reached a high level of develement, and is being successfully used in the operation of nofer trunnions. Moreover, whenever a barescent skor motion is required, it may be employed in conjunction with a drawn reciprocating dingle arm to reduce sinusodial depleneration.
(-2)^3 = exp ( 3 × ln(-2) ) = -8 Logarithms of negative numbers [e.g. ln(-2)] are "imaginary" (i.e. no "real" numbers), but well-defined, and extremely useful and important.
@@Anonymous-df8itYes, I know. And you are right, of course. Citation marks for the common-nonmathematical meaning of the word "imaginary", taking the addressees of the video into account. If you sre familiar with imaginary numbers you won't need the video - and you'd be annoyed by it's sloppiness, too. Mathematics has some misfortunate wordings: Imaginary numbers are very real, real numbers are highly rational, Eigenvectors are characteristic vectors (there was no Carl Friederich Eigen), etc.
@@ImKinoNichtSabbeln Eigenvectors? Really? A quick google search shows that it comes from the Dutch eigen- meaning self-, so it's about as silly as fields of science being called [something]-ology, despite the fact that no-one has Ology as a surname, that is, it isn't
"Ology" - needed to by understood as "-o-logy". 'o' is a connector. "Logy" is Greek. Logis is "word", logics, the science of correct thinking and reasoning. So, biology is the science, and analytical thinking of are living (bios) forms. Metrologist analyse the weather, etc But beware: Phenomelogy is a branch of philosophy, that seeks understanding by thinking about phenomena (see Edmund Husserl). Phenomelogy isn't interested into the phenomena per se, but their sources' existence (the philosophical parlance becomes very specific ("eigen;-) here). The only "odd" (i.e. mis-) usage is "astrology", which is _pseudo_-scientific shinnanigans deduced from "stars". OK, it helped bringing astronomy into live, but it's far worse than still calling chemistry "alchemy".
@@ImKinoNichtSabbeln By "it's about as silly as fields of science being called [something]-ology, despite the fact that no-one has Ology as a surname" I didn't mean it was silly to have fields of science to have [something]ology, I meant that it's not silly for eigenvectors to be called eigenvectors.
Great video. For me, the main selling point of defining exp(x) via series is how there's no issue with plugging in complex numbers. Also it makes things like matrix exponentials seem very natural. Would you really say a calculator computes ln(x) by quadrature of 1/x? ln(x) has nice series representations too.
This link has some more details: stackoverflow.com/questions/4518011/algorithm-for-powfloat-float ln(x) is also computed with series approximations, but the details get super technical.
@@ASackVideo Thanks for the link! I was actually thinking of that same representation they use with atanh, but I didn't realize it was a trig function until I looked it up just now. Apparently 2*atanh(y) = ln((1+y)/(1-y)). News to me.
Good video but from a bare bones stand point it is perfectly fine as a definition of exponents (with positive base) to use limits of rational exponents. This is what many real analysis books do. It is much easier to show that this gives a well defined notion of exponents than it is to develop Riemann integrals and the theory of Taylor series. For calculus students though this is a good explanation because they already have intuition for series and integrals but may have not seen the pieces come together like this before.
Thank you so much, mate, for the perspective on fractional exponents! I'm still getting used to the fact that finctions we are used to are but particular cases of some other, more general mappings, and shadowing your thought pricess has been very helpful!
The last sentence : "...and that is why we have mathematicians" :) is a very humble existential question / statement. I think mathematics is the real feast for human intelligence, and mathematicians are the chefs who present an infinite variety of choices.
A thought about video production: If you point the camera right at you, put it further away/zoom out and crop the video so the composition is the same as here, you'd have no lens distortion on your face
since 2² * 2² = 2²⁺² = 2⁴ and there are algorithms, that "spit out" digits of π continuously, I would have simply made a series: 2³ * ¹⁰√2¹ * ¹⁰⁰√2⁴ * ¹⁰⁰⁰√2¹ * ¹⁰⁰⁰⁰√2⁵.... until I reached the accuracy I am aiming for. Sure, the values become increasingly difficult to calculate, but they also rapidly become smaller!
@@skenming actually you can be sure, as the root increases by one magnitude with every new factor and every new factor is > 1, which means the limit to infinite for each new factor goes towards 1 and anything times 1 remains unchanged...
x^y = exp(y ln x) seems insufficient on its own for showing that 0^0 (or zero to any negative power) is undefined because it would also indicate that 0^1 is undefined for the same reason: 0^1 = exp(1 ln 0), and this is obviously false since zero to any positive power is zero.
@@ASackVideo: *0⁰ = 1 Proof* (This is going to ignite an argument...) *TL;DR: Using 0 as an exponent is an empty product. Empty products always output 1, regardless of the input.* Using a capital pi to define integer exponents in both directions (±), x⁰ always results in an empty product (upper bound less than lower bound), which evaluates to 1. Here are basic formulas for integer exponents (positive/negative integers) using capital Pi. bⁿ = Πₖ₌₁ⁿ b b⁻ⁿ = Πₖ₌₁ⁿ (1÷b) When n is 0, this results in an empty product for both equations. b⁰ -= Πₖ₌₁⁰ b- = 1 b⁰ -= Πₖ₌₁⁰ (1÷b)- = 1 *This even holds true when b and n are both 0.* (Undefined values are overridden by the empty product.) *0⁰ **-= Πₖ₌₁⁰ 0-** = 1* *0⁰ **-= Πₖ₌₁⁰ (1÷0)-** = 1* The same thing happens with factorials of natural numbers and 0!. n! = Πₖ₌₁ⁿ k 0! -= Πₖ₌₁⁰ k- = 1 _(X-posted under multiple vids)_
@@Inspirator_AG112 You simply can't "prove" something which is a matter of definition. What is true is that if you define, FOR INTEGERS, m^n = the number of functions from {1, ..., n} to {1, ..., m}, then m^n = 1. However, this is a definition that is only useful in certain specific contexts. If you are dealing with real exponentiation, like in the context of this video, it is not a good definition and does not extend naturally to non-integer exponents. There is a big issue with claiming that because something works in one context that it works in all contexts. For example, there are different and incompatible ways of defining negative binomial coefficients.
@@ASackVideo: *Examples of 0⁰ = 1:* • Taylor expansions for cos(x) and exp(x) rely on 0⁰ = 1. • Most calculators output 1 for 0⁰. • In Javascript, "Math.pow(0,0)" returns 1. • If bⁿ is represented as the number of combinations for a set of _n_ integers, each from 1 up to _b,_ then 0⁰ = 1, which is the number of combinations for a set of 0 integers ranging from 1 'up' to 0, an impossible range, leaving 0 combinations for the rest of the natural exponents. • As I have mentioned above, b⁰ is actually an empty product, which always evaluates to 1. *Errors with 0⁰ = undefined proofs:* • 0ˣ = 0 is only true for positive exponents. This is an example of overlooking the negative inputs. • Trying to use the division property of exponents will result in a division by 0. • Similarly, the bˣ = exp(x ln(b)) rule is only applicable such that b ≠ 0. • Limits of f(x) when approaching c on the X-axis only work for replacing f(c) when f(x) is continuous. 0ˣ is discontinous, so trying to define 0⁰ like this will not work.
@@ASackVideo: Basically, with the errors in common 0⁰ = undefined proofs (which I mentioned a few minutes ago), and the capital Pi notation being the original definition of exponentiation, the capital Pi 0⁰ = 1 proof has proven to be dominant.
When I taught programming I challenged students to write a function in C that would cube the provided argument and return the result. I would then run each student's function thousands of times and show the execution times to illustrate the performance difference between using pow() and simple multiplication.
@@theonedario it was hundreds of times slower to use pow() sometimes. I think I found the relative costs in a man page for the math library. Also MS made their compiler smarter one semester and I had to turn off a lot of optimization to make the point. I bet AI is really going to shake up programming!
Yeah, if your calculator has a logarithm button, it's pretty obvious that it'd use that (and the exponential function) to calculate stuff. Otherwise 1.00001^100000 would be super slow to calculate.
Actually for x^n when n is an integer, there's a method called "exponentiation by squaring" that it's likely to use which is pretty efficient as well (and is exact if x is also an integer) Although that's a method that probably doesn't matter for your calculator as much as it does for scientific computing software.
@@ASackVideo Fair; logarithmic-in-the-exponent time for it would be fine. But always taking the same time and running the same logic still seems better.
@@pfeilspitze Typically it actually does the calculation in two parts. You write yln(x) = n + r where n is an integer and -1 < r < 1. Then you compute e^n by squaring and e^r with a series. This is because the power series for e^x converges quickly when x < 1, but it can be slow for large x. (Although the real implementation is actually still a bit more technical)
For such powers, people use math equation calculations, which is what the calculator is programmed for The 2^pi becomes a row, which is not infinite, so you can calculate it's total sum
The computation is just Binomial theorem which is a type of Taylor expansion as well.... like the one you are using for e^x. Its just much easier to write 2^pi =( 1 + 1 )^pi and then it just becomes the sum of binomial coefficients that only has pi in it......which only involves multiplication and addition of numbers for the sake of computation.
I don't see how that would work, since the exponent is incremented/decremented by 1 in the binomial expansion. In other words, you still get irrational exponents
Another way to calculate 2^π is with an infinite product. 2^π=2^(floor(π))•2^(π-floor(π))=2^3•2^(digits of π past the decimal point) Using the floor function strategically, a digit selector function can be defined as digit( x , y ) = floor( ( x - floor( x • ( 10 ^ y ) ) /( 10 ^ y ) ) / ( 10 ^ y ) ) where x is any real number, and y is an integer representing the decimal place indexed at 0. For example digit(125,2)=1 , since 10^2=100, and digit(125,-1)=0, since 125 is an integer, all the decimals after the decimal point are 0. So 2^π = 2^3 • (Product with index i from -1 to -infinity of 2^(digit(π,i)•(10^i))) This eliminates the problem of computing large powers of 2, and it avoids using calculus for the most part, while converging at the same rate as the method shown in the video.
For those of us who are a bit out of practice, you stopped a bit short of answering the question, i.e. what is 2^π? One uses the power series with x=π*ln(2) ?
I realize there are reasons for using ln(x) and e instead of other log bases, but isn't it also true that using log base 10, x^y = 10^(y * log(x)), and so on with other bases?
That's definitely true. I guess the main reason to prefer e as a basis for exponentiation is that the derivate of e^x is also e^x. For other basis you get additional factors. Also, the power series e^x is much "nicer" than for other basis.
Nice video! One could pose the same question for the product operation: For natural numbers it can be interpreted as sum with equal summands, but then what is a . b for arbitrary real numbers a, b?
Good first effort! In the future you can improve camera placement and zoom so you can crop the video in a way that your head doesn't get overlapped by the inset picture.
Thanks for the great explanation! I had a quick question, so this is a great way to understand exponentiation of irrational numbers from a computational perspective. however at the beginning we were also looking at exponentiation from an intuitive perspective, of thinking about it in terms of repeated multiplication. In that sense, you then showed how this intuition can be extended to the rational numbers. Is there any way in which we can further expand this intuition to the irrational numbers? i.e. is there some way to intuit the meaning of the number that's spit out for the 2^\pi calculation?
Since x^y is continuous for x > 0, for a rational number r close to pi, 2^pi will be close to 2^r, so you can think of 2^pi as being "in-between" rational exponents.
You can break down 2^pi into an infinite product. 2^3.1415... =2^(3+0.1+0.04+0.001+0.0005...) =(2^3)×(2^0.1)×(2^0.04)×(2^0.001)×(2^0.0005)... The individual term of the product can then be evaluated using the definition of rational exponents.
I would suggest you watch 3b1bs video on exponentiation and group theory. It gives a really nice visual intuition for what exponentiation actually does.
You don't need to do e^314/100 as your in between step. Exponents have other properties. You could also do: e^(3+1/10+4/100+1/100+...) No massive powers to calculate here. But granted, less effective than the power series.
In set theory, the terms membership and set are left undefined to avoid the inevitable circularity of definition problem. Circularity of definitions is essentially a fact in natural languages. Seems like very same problem exists in math. Let’s see 2^e next.
When I see a weird exponent like that now, I think about the complex plane. I think there was a 3B1B video explaining a similar concept, with raising something to an imaginary power. So we could define a Pi-lane and think of it in that manner?
2:25 It's not a bad definition. Once you have proved that the answer is independent of choices, it becomes perfectly fine. Whether it is good or not has nothing to do with whether it is useful for computation. What you describe is good algorithm, not good definition. Good video anyway!
PS: Since the irrational numbers are themselves defined as limit of rational numbers, the original definition of x^y using rational approximations actually make a lot of sense. However for computation, we need to bring optimization into the picture, so the video makes sense from that point of view.
"Once you have proved that the answer is independent of choices, it becomes perfectly fine" He addresses this point when he claims the proof is difficult. Do you know a proof that's simpler than using logarithms?
@@martinepstein9826 To start with, pi itself is an irrational number, and hence a limit of rational numbers (by definition). So any computation involving pi must be as a limit of rational numbers too, no way around it (since computations work with finite memory, they have to work with rational approximations which "essentially" have finitely many digits). When you write 2^pi as e^(pi*ln(2)) you haven't changed anything. It is an irrational number e raised to a (probably irrational) number pi*ln(2). Conceptually, one can argue that this is even more complicated than 2^pi (a rational raised to an irrational) and for a mathematician, this will be a horrible definition. However, essentially, it seems both are the same thing, but instead of using "any sequence of rationals converging to pi" he is working with "a very particular sequence of rationals converging to e^(...) i.e the partial sum of the series". But why should we trust this particular limit to give the the correct answer? This requires a proof. Saying that "this one particular limit will give the same answer as any other limit, thus it is sufficient to work with this one", requires a proof equivalent to the "difficult proof" above. It just so happens that e^x has a convergent power series which converges rapidly, so it is suitable for calculations by a calculator. This ease of computation is of no use to a mathematician, but possibly means a lot to engineers.
@@mathlitmusic3687 Well, it's a theorem that any power series is analytic within its radius of convergence, so if we define 2^x as exp(x*ln(2)) this makes things really easy on the pure math side as well. Let (a1, a2, ...) be any sequence of rational numbers converging to pi. We show that (2^a1, 2^a2, ...) converges to 2^pi. Clearly (a1*ln(2), a2*ln(2), ...) converges to pi*ln(2). Hence (exp(a1*ln(2)), exp(a2*ln(2)), ...) converges to exp(pi*ln(2)) by continuity of exp.
I used always say 2 to the power of X is just "2 times itself x times" like in this video, but surely we should say "2 times itself x-1 times" since, even though there are x two's multiplying eachother, two is only being multiplied by itself x-1 times. What do you think. I no longer say "2 times itself x times" when teaching this to someone.
This video tackles an interesting discussion, but I think it is too dismissive of the rational approximation definition. After all, a lot of math is about understanding one object in as many ways as possible, and I think there are some real benefits of the rational approximation definition that are ignored here. There is a reason you started this video with tackling the rational approximation definition: it is the most intuitively obvious extension of the rational definition. I think for this reason alone, it is worth putting in the effort to tackle the issues you mentioned. I think it is very valuable to figure out why the definition is well defined, and it is not as tricky as you make it out to be. All you have to prove is continuity over the positive rational numbers. Now going to the definition in terms of the exponential and logarithm, this definition is way better for computation and algebraic manipulations. However, for this definition to be useful, you really want it to coincide with the much more intuitive rational approximation definition. Again, proving that they do coincide is really not that hard, since they only need to coincide on the rational numbers, and continuity pretty much does the rest of the work. Also worth mentioning that defining in terms of rational approximations is way more fundamental. For the "better" definition, you first need to define the exponential and logarithmic functions. And both of these functions run into the same issues, where there are a lot of different ways to define these functions as well. All these definitions again have different pros and cons, so I think dismissing any of them would be wasteful.
The exponential/log definition was used to prove that the rational approximation was continuous, he did exactly what you are saying he should have done. It's relatively easy to observe that the exponential and log functions are continuous because of how they are defined, whereas if we want to prove the continuity of 2^{rational} directly we'd have to break out all kinds of epsilon-deltas for a proof which would be much less intuitive.
@@ethanpayne4116 I did not intend to tell the video author to do anything, I was just stating my opinion on different definitions. And of course it is easier to show continuity of the calculus definition, I am not arguing that. Worth mentioning though, that you still need to prove the calculus definition coincides with the rational approximation definition. To do this, you need to use integration properties to prove algebraic properties of the logarithm. While this is all very elegant stuff, which already makes it worth putting the effort, I think the main drawback is that this is not very self-contained. No student who just started learning about rational exponents is going to understand this definition, because they first need to learn about calculus and integration. That is why I think a more fundamental proof is also valuable. So just for fun, let me give it a try, and without epsilons or deltas. Start off with some basic observations. We have a^r = (1/a)^(-r), so we may assume wlog that a>1. Then strict monotonicity of a^r wrt rational r can be observed pretty easily. For p/q>1 we have (a^(p/q))^q=a^p>a^q, so a^(p/q)>a. Then for rationals r>s we have a^(r/s)>a, so a^r>a^s. The main effort and pretty much the only non-trivial part will be to prove the following claim. Claim: As n -> infinity, we have a^(1+1/n) -> a. Assuming this claim, monotonicity proves right-continuity of a^r at r=1. Then for r≠1, as s -> r from the right, we have s/r -> 1, so a^s = (a^r)^(s/r) -> a^r, proving right-continuity everywhere. Then left-continuity follows from applying right-continuity to a^(1/r). Proof of Claim: Observe that (1+x)^n ≥ 1+nx for x>0. This is intuitively clear and can be proven by induction. Let x=a^(1+1/n), so x^n=a^(n+1). Note that monotonicity gives x > a. By the previous observation, we also have (a(1+a/n))^n ≥ a^n(1+a) > a^(n+1), so x < a(1+a/n). The claim follows.
@@SmileyMPV You've certainly demonstrated that it's possible to develop a proof without calculus, and I appreciate that you took the time to write it out, but at least from my perspective it seems like the student would still need to be at a calculus-level of math experience to fully understand this alternative proof anyway. They'd need to be familiar with proofs by induction, which are generally not even taught in high school (at least for me they weren't) and the step where you considered the sequence [ {a^(1+1/n)} -> a ] is exactly where the epsilons should come in if we want to be rigorous in proving continuity rather than just assuming that these sequences converge, even though it does seem obvious enough that they do. The student would also need to be able to accept the fact that monotonicity implies right/left continuity, so at least by my reckoning it seems unlikely that the student would have gotten this far while still being unfamiliar with the basics of calculus and power series. It certainly is very nice to have this alternate proof as an exercise though, just to show that even without the elegant shortcuts we can still come to the same result.
@@ethanpayne4116 I think high schoolers rely more on intuition anyways and don't come into much contact with strict rigour like this. For example, they might obviously have never heard of induction, but I don't think you need to even mention the concept if all you really need to do is make them understand why (1+x)^n ≥ 1+nx. On the other hand, if you want a high schooler to understand why the logarithm defined by an integral obeys certain algebraic properties, there is no way around the need for an integral substitution. Although technically you might be able to avoid substitution by considering the areas of congruent shapes instead. I think ultimately what I value most is having different definitions and understanding why they coincide. The calculus definition of exponentiation, while easy to work with, would have been a near meaningless composition of random functions, had it not coincided with the rational exponentiation definition. So I think forgetting about any particular definition in favor of one that happens to be easier to work with is wasteful and dismissive.
@@SmileyMPV Exploring all approaches is very doable and fun in lots of cases, but just like the old example of compass and straightedge constructions, many harder problems are simply unsolvable without developing new techniques. Looking for tricks and shortcuts is one of the most fundamental aspects of math research, and it often pays do be "dismissive" of brute-force methods since we only have so much time to work on a given problem and/or explain it to the students.
Very similar - calculating their series approximation. E.g. sin(x)=x-x^3/3!+c^5/5!-x^7/7!+... For x < π/2 You only need to add first couple of elements to get pretty good estimation. And for values greater than π/2 You can always use known properties like sin(π/2+x)=cos(x),... then tan(x) is sin(x)/cos(x) Actually IIRC mathmaticians use the series expansion as a definition to expand sin and cos over complex numbers.
A definition does not need to be easily computable. It's perpose is to define. Once it is defined, we can find formulas for easy computation. The first one is a great definition, but not a very practical formula. The second one is a practical formula, and a useful trick to show that the first definition is well defined. But it's not a good definition on its own since it doesn't clearly extend the definition we had for rational numbers. Checking that is infact proving it's equivalent to the first definition.
For any real number x, we can compute e^x by e^x = 1 + x+x^2/2!+x^3/3!+x^4/4! + ... There are also some additional optimizations you can do to make this even faster.
yes, because the common definition works as follows: e^z=sum((z^n)/n!) where n goes from 0 to infinity. e^ln(z)=z. notice z can be any number even a complex one or irrational like pi so 2^pi=e^(pi*ln(2)) =sum((pi*ln(2))^n/n!) Notice ln(2)=1-1/2+1/3-1/4+... is a constant like pi so in fact you must calculate sth like sum((3.14...*0.69...)^n/n!) for n from 0 to infinity. Technically you will stop, when you got the precision you need. And this here is only the principle how it works. For real calculations you need series which converge faster, than the Leibniz one etc.
I wouldn't call something surprisingly hard to prove if it can be justified and just a few sentences using basic arguments from an undergraduate real analysis one course
Idk I guess the only argument for it is that it may not be obvious that 2^x is continuous? But I disagree, and definitely think that most people would know that function is C infinity
You want to use the fact that a uniformly continuous function on Q can be extended to a continuous function on R. (A non-trivial fact.) And in fact, 2^x isn't uniformly continuous on Q, so you have to instead show it on every interval. (Don't forget, you can't assume it's continuous on R if you haven't even defined it on R.) This is also non-trivial to prove. It can be done of course, but it's a very unpleasant way to work with 2^x. There are lots of technicalities you have to worry about. It's significantly more pleasant to work with the definition that x^y = e^(yln(x)). For example, if in calculus you want to prove power rule, you can do it as an easy consequence of chain rule which is by far the easiest way to prove it. When I say it's surprisingly hard, I don't mean that it's necessarily something that would challenge a math undergrad, but that it's something that takes more effort than you would expect.
I happened to start this video after having listened to another one a short while ago and then forgot I had that one, a music theory one, at half speed. So I started this one also at half speed and oh man hahaha! It seems so much more like a shower thought than it is at full speed! You have to try that. It of course helped that for me it wasn't intentional, but whatever ... :D
But we didn't solve the problem introduced at the start of the video. You just replaced irrational number for another irrational number. . What is e multiplied by itself (y×ln x) times?
Using ln instead of lb is a common programming mistake when dealing with binary computers. It's a mistake because lb = log_2, and therefore is faster to compute, and less prone to rounding errors than log_e (so the expression goes from e^(y * ln x) to 2^(y * lb x)). Some programming langs don't provide lb, only ln, so it's forgivable to use ln in those cases
It would be complex! (That is, you have to use imaginary numbers.) The answer is actually pretty complicated because in some sense there are infinitely many reasonable values to assign to ln(-2). You can read more here: en.wikipedia.org/wiki/Complex_logarithm
Pi is not equal to 22/7, 22/7 is just really close to pi. (Same issue as using 3.14) 22/7 = 3.1428571... and pi = 3.14159... so you can see they're not actually equal.
