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Why does e to pi i equal -1? (2015) [video]

355 points by espeed 10y ago

datainplace 10y ago
Is it pure chance that this was posted only two or three days after I watched it along with a few other e to pi = -1 videos? Randomness aside. 3blue1brown makes some wonderful math videos that I find really explain the intuitiveness of some of the ideas. I was unfortunately cursed with a math teacher who for whatever reason required us to memorize until we passed the test. Imaginary numbers were taught as "something that will help you in college"
- vinchuco 10y ago
  Richard Feynman used to go up to people all the time and he'd say "You won't believe what happened to me today... you won't believe what happened to me" and people would say "What?" and he'd say "Absolutely nothing". I agree wholeheartedly with the quality assessment of 3blue1brown videos. All mathematics should be clear and intuitive, by definition :)
  - dvt 10y ago
    > All mathematics should be clear and intuitive, by definition :) I disagree with this completely, and so does all of higher mathematics. It's neither clear nor intuitive. In fact, as soon as you start learning about infinities (Calc I), intuition becomes hit and miss.
    - CN7R 10y ago
      Intuition always comes after the feeling of awe or confusion, in my opinion. Really depends on how you learn.
- CN7R 10y ago
  No, another link to 3Blue1Brown was posted earlier this week and hit the front page. My basic assumption is that when a topic is posted, people will explore that topic and, if it's interesting enough, post further content about that topic.
cousin_it 10y ago
1) e^x is a function whose derivative is equal to its value. 2) e^ix is a function whose derivative is equal to its value rotated by 90 degrees (ie^ix). 3) As x goes from 0 to pi, the trajectory of e^ix always has a velocity vector perpendicular to its current position. For example, when x = 0, the current position is 1 and the velocity vector is i. 4) So the trajectory a circle arc of length pi, which ends at -1.
- kmill 10y ago
  In a linear algebra course I helped teach last semester, I had the students go through the exercise of writing down the matrix for "multiplication by i," where the complex numbers are thought of as a two-dimensional vector space with {1,i} as a basis. Then I asked them to recognize the matrix (it's a 90-degree rotation of the plane), and then I asked them for an interpretation of that matrix squared (a 180-degree rotation). One thing to draw from this: for the real line, negation doesn't reflect the line, but instead it's a rotation through another dimension of numbers! When we went on to differential equations, I tried to convince them that e^(ix) is meaningful using your #2: it is a path around the unit circle since it's the solution to z'=iz, and we already agreed multiplication by i is a 90 degree rotation, and circles are what you get when velocity is orthogonal to vector position. It's a bit tougher to convince someone that x is arclength, though, especially when they aren't too comfortable with complex numbers yet.
  - deleted 10y ago
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  - godmodus 10y ago
    I love this answer. For some reason, I've been dreaming about negative numbers lately. I think they deserve their own number set notation.
    - dnautics 10y ago
      the standard IEEE floating point uses a 'sign and magnitude' system. Negative numbers are basically made by flipping a single bit. I've been playing around with a number system that expresses real numbers as two's complement (like how integers work) - xor all your bits and then add one. In this system, 0 is -0 and the number that is weird for signed integers: 0b100000...0000 is infinity. Now, there's an exponent and a fraction, too. I've been playing around with how to do these numbers with logic gates (and verilog!) and you can either two's complement the whole thing and work with absolute values, or you can keep the fraction part as a two's complement... So I just redid multiplication using two's complemented fractions! And addition/subtraction too, which for floating points in general is significantly harder than multiplication. The nice thing about two's complemented floating points is you don't need separate algorithms for addition and subtraction; you can just do everything with one algorithm.
hprotagonist 10y ago
This seemed like voodoo until i took signals and systems. Then it's just a normal observation about vector sums on the complex unit circle.
- gydfi 10y ago
  It seems like voodoo when you have high-school mathematics, because you've been taught about "to the power of" in terms of multiplying something by itself a certain number of times, and multiplying something by itself an imaginary number of times is nonsensical. Once you've done enough maths you think of it in a different way and e^ipi seems obvious. It's just a different conception of what the symbols mean, I suppose.
  - mstade 10y ago
    There are so many assumption that were hammered into me in high school (and below) that I find hold back my thinking, rather than help me. I kind of feel like the education I received was full of hacks to get me past whatever test was coming up, rather than truly teaching me. I've spent years playing what feels like catch up to actually truly learn the things I thought I knew – a lot of high school mathematics included.
    - fl0wenol 10y ago
      What was not obvious to me, and took a lot of higher math to get to, was that in the grouping of N=>I=>Q=>R=>C, at each stage you're pulling in a more complete mathematical representation (counting, rings, fields) until you're algebraically closed, and while you could go higher, you start losing properties you like again, like commutability over multiplication. Complex numbers are that nice saddle point, which is an interesting thing to ponder. I wish there was more focus on these "why" aspects in at least the optional advanced math or physics you could get in HS. It helps put some things in perspective.
neximo64 10y ago
For anyone that finds it beautiful isn't there a bit of humanisation and definition involved (for example the Sine function used in deriviation uses 'pi' instead of 90 degrees), not to mention Sine is a human created function. You could have e to pi (90 * -1) too.. or a different method to define angles instead of having 360 degrees (base 60)
- panic 10y ago
  Yeah, you can have the same thing with degrees if you use a different exponent base than e: e^((pi/180)ix) = (e^(pi/180))^ix = E^ix and therefore E^180i = -1 where E is about 1.0176065. The particular base e has some nice properties, though, like that its rate of change d/dx e^ix at a given point x is just i e^ix. The rage of change for E^ix, on the other hand, is d/dx E^ix = (pi/180) i E^ix, which is a little less "natural". This strange fact -- that using radians makes the expression e^ix = cos x + i sin x have a nice derivative -- is one of the reasons why mathematicians like to define these functions in terms of radians instead of degrees.
- deleted 10y ago
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- bbcbasic 10y ago
  Survivorship bias of beautiful math. Who want's to study systems that don't simply and compose well. The exception that proves the rule is the 4 color theorem - ugly big-arse proof generated by a machine. Who is going to look into that any further?
obastani 10y ago
The most elementary explanation I ever got for why e^(i pi) = -1 is from the definition e = lim_{n -> infinity} (1 + 1/n)^n First, we can generalize this to e^z = lim_{n -> infinity} (1 + 1/n)^(z n) = lim_{m -> infinity} (1 + z/m)^m using the substitution m = zn. Therefore, e^(i pi) = lim_{m -> infinity} (1 + i pi/m)^m Now, converting the complex number (1 + i pi/m) in terms of polar coordinates (r, theta) yields r = (1 + pi^2/m^2) ~ 1 theta = sin^{-1}(pi/m) ~ pi/m Since the product of two complex numbers is (r, theta) (r', theta') = (r r', theta + theta'), we have (1 + i pi/m)^m ~ (1^m, m * pi/m) = (1, pi) = -1.
phkahler 10y ago
It's a nice video but it goes so fast I don't think people who don't already understand this stuff will get it.
- sparky_z 10y ago
  Yeah, he's gotten much beeter about that since this video was released.
- vinchuco 10y ago
  "Young man, in mathematics you don't understand things. You just get used to them."
  - johnfn 10y ago
    I don't really like this, because it's a self limiting quote that leads to complacency over real understanding. Especially in the case of imaginary numbers and e where intuitive understanding only doesn't exist because mathematics has a history of being poorly taught.
    - E6300 10y ago
      Honestly, after spending months studying the subject, I don't think it's really possible to "get" complex numbers. I just view them as affine transformations written in an unusual notation. I don't think they make sense as anything but a recontextualization of R^2.
      - akvadrako 10y ago
        So what don't you get? Are you implying there is something more to complex numbers?
        E6300 10y ago
        What I don't get is why someone would bother with complex numbers and their silly notation when linear algebra would work perfectly fine. I know that in certain contexts they are useful. E.g. to avoid losing information when solving polynomials. But that's a minuscule fraction of their range of application. Nearly every practical use I've seen of complex numbers just use them as a vector representation.
hgdsraj 10y ago
Honestly the calculus explanation is the most clear to me (Taylor series expansion of e^x)
jjaredsimpson 10y ago
e^x = an infinite series if you substitute x = iz you can split the even and odd terms of the infinite series into cos and sin e^iz = cos z + i sin z evaluating at z = pi yields cos pi + i sin pi -1 + 0i = -1 the exponential function maps the imaginary axis to the unit circle. pi gets mapped to -1. This shows its true, but the "why" and real understanding requires calculus and the first week of complex analysis. Otherwise you are just parroting a set of facts.
- gizmo686 10y ago
  Math major here. This presentation does skip some nessasary legwork for complete rigor, but presents a valid non-standard construction of e^ipi. Specifically, he defines two sets of objects: adders and multipliers, and a function (written e^x for historical reasons) that maps adders into multipliers. He then generalizes this construction to work in 2 dimensions instead of 1 dimension. I would add to this construction that e^x is the particular converter that maps pi -> -1. However, I think this requirement is implicit in him stating that e^x is the most "natural" of the converters, and that pi -> -1 is the most natural of mappings. The only place where you might need calculus is to provide an explicit construction of e^x (and possibly to justify the notational choice of writing it like an exponential). EDIT: To make the point clearer. Under the construction presented by the video, e^x is not defined as an infinite some, but rather as a function satisfying certain properties. That this function is equal to a particular infinite sum is a statement that requires proof; and not a statement that is needed for many applications.
deleted 10y ago
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somewhat_slow 10y ago
At 2:37, right after finishing explaining adders and multipliers on the line it started going way too quick like where did e^x and some these infinite sums come from???
- gizmo686 10y ago
  He introduces e^x as a way of converting adders into multipliers. The infinite sum is not important. I believe he is showing it purely as a visual tool to indicate that e^x is just "some function". More specifically, based on the video, the important feature for a adder -> multiplier converter to have is that f(x+y) = f(x)f(y). However, many functions satisfy this requirement. He states (without justification [0]) that e^x is the most "natural" choice for such a function, and provides the infinite sum as a visual aid to show that it is just some function. [0] Rather, justification is provided by reference to another video.
1001101 10y ago
Gauss said that if the reason weren't immediately obvious to someone, they would never be a first rate mathematician. For the rest of us, there's YouTube.
lowpro 10y ago
I've been watching 3Blue1Brown a lot recently, and he's been linked to many times here too. His work is absolutely amazing, and he even wrote a python library to make the animations!
- espeed 10y ago
  Yes, the Python library is called Manim, and it's on GitHub: https://github.com/3b1b/manim https://github.com/3b1b/manim
skybrian 10y ago
If you prefer slides over a video, here is a slideshow I wrote: https://docs.google.com/presentation/d/1oMNjkDp-LieSGnZEwNpceG8KTIvbnus9olu3KqnM5bg/edit https://docs.google.com/presentation/d/1oMNjkDp-LieSGnZEwNpc... Or you might prefer Better Explained's article: https://betterexplained.com/articles/intuitive-understanding-of-eulers-formula/ https://betterexplained.com/articles/intuitive-understanding...
- mih 10y ago
  Upvoted for the Better Explained link. I always liked their explanation of e https://betterexplained.com/articles/an-intuitive-guide-to-exponential-functions-e/ https://betterexplained.com/articles/an-intuitive-guide-to-e...
gizmo686 10y ago
As always with this author, I followed a link to a math explanation video expecting to point out how it is only a superficial treatment that breaks down under rigorous analysis, only to be dissapointed in my inability to find any such flaws. Seriously, this guy is awesome. If you need to learn math, see if he has a video about it. I do have 1 gripe with this video though, and that is his handwaving around "natural". More specifically, as far as this particular case goes, there is nothing natural about the choice of e^x, or the significance of pi. As he identifies, we are interested in some function that maps adders into multipliers with the property f(x+y)=f(x)f(x). To uniquely identify such a function, we need to add an additional constraint. He chose to add f(pi) = -1. He justifies this by arguing that pi is the length you would travel along the unit circle to arrive at -1. This is true (and the underlying reason why pi and e end up being natural), however using this argument seems to break the abstraction for me. Under this construction, in the equation f(i * pi)=-1, "i * pi" is an object in the set of multipliers, and "-1" is an object in the set of adders. Specifically, "i * pi" is a function which takes a plane (or perhaps a point) and rotates it, while "-1" is a function which takes a plane (or point) and slides it. He then invokes an unstated mapping, g, to convert the multiplier [0] "i * pi" into the real number "pi". He than insists that g(x) gives the distance a point would travel along the unit circle when the multiplier x is applied to it. At this point, because arriving at the point "-1" [1] from the point "1" through rotation requires traveling pi distance, it makes sense that f(i * pi) = f(g^-1(pi)) = -1 I am still not convinced that (within this construction alone), this is a more natural choice that saying that f(i) = 1, but would agree that it is one of the two natural choices. To get to f(x) = e^x being the natural choice requires showing that it comes up in all sorts of unrelated parts of math, so it is probably more natural. [0] It might be better to speak of "rotational multipliers" here, as I am not sure how to natural define g(x) for multipliers that stretch the plane, instead of or in addition to simply rotating it. [1] This "1" is again distinct from the adder "1" and multiplier "1", but plays a central role in defining them, so I do not object to its usage.
dvt 10y ago
I've seen this a while ago, and while it's pretty instructive, it's actually also pretty confusing. The magic happens in a seemingly-innocuous throwaway sentence at around 4:20 (after being introduced to the 2D plane): > ... This can now include rotating along with some stretching and shrinking ... It's entirely non-obvious WHY we should be okay with rotating all of a sudden. The real answer is not super complicated, but it deals with a couple of amazing relationships between exponential functions and trigonometric identities[1]. So really, we don't have to accept "rotating" as some weird new action we can do when moving into the complex plane, we just have to accept that trigonometry is weirdly related to analysis due to some very cool properties. [1] https://www.phy.duke.edu/~rgb/Class/phy51/phy51/node15.html https://www.phy.duke.edu/~rgb/Class/phy51/phy51/node15.html
- gizmo686 10y ago
  >It's entirely non-obvious WHY we should be okay with rotating all of a sudden. Why shouldn't we? We have a set of object called multipliers that we want to generalize to the 2 dimensional plane. The presented generalization (eg, the multiplier identified by x maps the point at 1 to the point at x) seems natural; and it defiantly seems to be well defined. Additionally, it appears obvious that the presented generalization is equivalent to the original definition if considered only along the x axis. From the perspective of the video, the fact that e^x is written as an exponential is "a vestige of its relationship with repeated multiplication". In the construction presented in this video, we should think of e^x as the "natural" homomorphism from adders to multipliers. The fact that this function has anything to do with exponential, trigonometry, or analysis is neat, but not important in this context.
  - dvt 10y ago
    > Why shouldn't we? We have a set of object called multipliers that we want to generalize to the 2 dimensional plane. The presented generalization (eg, the multiplier identified by x maps the point at 1 to the point at x) seems natural Rotating all of a sudden does not feel natural to me at all. Why not start cutting up the 2D plane? Why not fold it? In analytic terms, we would cause discontinuities, whereas rotating is "smooth". But this is definitely nontrivial. Rotating is just so arbitrary. I think that skewing or flipping, for example is just as "natural" as rotating. > The fact that this function has anything to do with exponential, trigonometry, or analysis is neat, but not important in this context. It's actually at the heart of why rotation (and not some other geometric operation) is key to e^iπ.
jejones3141 10y ago
Look at the Taylor series for e^x, and see what you get for e^(i * theta). The terms with i raised to an even power are real, those with i raised to an odd power are imaginary. Group the odds together and the evens together, and compare with what the Taylor series for sin x and cos x give you for sin theta and cos theta. (I'm being a bad person here, ignoring convergence issues, but I want to say that all three are absolutely convergent.)
unholiness 10y ago
This explanation strikes me as a little too aggressive in throwing out the notation with the bathwater, only reaching its result by redefining the terms we already have intuition for into space-stretching operations that don't work like arithmetic does in my head. In that sense I think I have the same problem with this proof that I do with the standard one, where you add the Maclaurin series of cos(θ) and i * sin(θ) and match term-by-term with the series for e^(i * θ). The problem is, at the point you can actually show equality, the things on one side aren't obviously a rotation and the things on the other side aren't obviously an exponential. I'm not just hear to yell at clouds. I was given a proof that I truly love by a professor I adore, which I think really does give insight into what all these operators are doing. The best video I can find with it is here: https://www.youtube.com/watch?v=-dhHrg-KbJ0 https://www.youtube.com/watch?v=-dhHrg-KbJ0 (Skip to 7:30 if you're already comfortable with the limit definition of e^x) The basic summary is: 1) e^iθ is equal to (1 + iθ/n) ^ n for large n 2) That base, (1 + iθ/n), plotted as a complex number, has length approaching 1, angle approaching θ/n 3) The base squared, (1 + iθ/n)^2, by de moivre's theorem, forms another point as if the transformation from (0, 1) were repeated twice — that is, the length stays one, and another tiny angle is added for a total of 2θ/n 4) The full result is therefore n transformations, taking the path along the unit circle, traveling θ and arriving at cos(θ) + i * sin(θ)
- mistermumble 10y ago
  Thanks for the pointer to the video ("E to the Pi for Dummies" by Mathologer). It is definitely worth watching. Much more understandable, imho, than the one referenced at the top of this thread. Your text summary of the video is accurate and likely makes sense to some people but not so much for others (namely, me). Tour words are hard to grok until you see the visual depiction --- until you see the sequence of n transformations become a spiral arrangement of triangles that ends up approximating the (-1, 0i) point in the complex plane.
- jacobolus 10y ago
  Your (1) is just one of many possible (technically equivalent) definitions for the exponential function. And arguably not the most basic/natural/intuitive one. The primary motivation for the exponential function is to be the inverse of the logarithm function. And the motivation of logarithms is to convert multiplication problems to addition problems, so they could be solved with table lookups (later performed on a slide rule) instead of difficult arithmetic. That is, log ab = log a + log b. Which is to say, exp(c + d) = (exp c)(exp d). Just setting this constraint along with the derivative exp’ 0 = 1 is enough to characterize the exponential function. Or you can get to this function in many other ways, e.g. by solving the differential equation d/dx exp x = exp x; by the series 1 + x + x^2/2 + x^3/6 + ...; or by defining the logarithm as the definite integral of 1/x starting at 1, and then taking the exponential function to be its inverse. I like defining the (complex) exponential as a conformal mapping between the cylinder and the plane minus a point.
hawkice 10y ago
> (2015) I'd prefer if admins could change the link to a description of this mathematical fact that's more up to date, and fits the needs of me and my family in 2017.
- espeed 10y ago
  It originally said "Why does e to pi i equal -1? (visualized)".
timlod 10y ago
I love the Khan Academy video on this: https://www.youtube.com/watch?v=mgNtPOgFje0 https://www.youtube.com/watch?v=mgNtPOgFje0 Sal's enthusiasm at the end is contagious!
- mstade 10y ago
  I've learned so much from Khan Academy – not just whatever is explained by the video I'm watching, but different ways of educating people as well. In fact, I helped out with tutoring basic math a while back and I struggled at first because I didn't realize how much I wasn't teaching simply because it was second nature to me. After watching a bunch of Khan Academy videos on the same topics, I found different ways of explaining things, and was much more successful. Khan Academy has really challenged my own way of thinking about many things, math in particular. It's a great resource, and while I don't use Khan Academy much these days I still donate $100 every year.
ageofwant 10y ago
Why not just write it e^iτ=1 which conceptually and pedagogically makes just so much more sense.
mitchtbaum 10y ago
e^τi = 1. In words, this equation makes the following fundamental observation: The complex exponential of the circle constant is unity. Geometrically, multiplying by e^iθ corresponds to rotating a complex number by an angle θ in the complex plane, which suggests a second interpretation of Euler’s identity: A rotation by one turn is 1. The Tau Manifesto http://tauday.com/tau-manifesto#sec-euler_s_identity http://tauday.com/tau-manifesto#sec-euler_s_identity
- wodenokoto 10y ago
  Thanks for posting this. I recently saw another video about this equation than the one posted here and thought "maybe this is one of those cases where tau doesn't work out nicely", but didn't have the intuition to see if it was or not.
ttoinou 10y ago
BTW for thoses who like Complex Analysis here's how to visualize them as a 2D deformation : https://www.youtube.com/watch?v=CMMrEDIFPZY https://www.youtube.com/watch?v=CMMrEDIFPZY
0xbadf00d 10y ago
Very Nice Video - One of my favourite sub 5 minute explanations is Oliver Humpage's @ Ignite Bristol circa 2011: https://www.youtube.com/watch?v=vHaPyuEkK4A https://www.youtube.com/watch?v=vHaPyuEkK4A Hope you enjoy as much as I did.
faragon 10y ago
Pure beauty.
etatoby 10y ago
I think Mathologer's take is much more intuitive: https://youtu.be/-dhHrg-KbJ0 https://youtu.be/-dhHrg-KbJ0
cmollis 10y ago
I had never understood how imaginary numbers had any bearing on physics (probably because I'd never been taught). But lately I've thinking about how all mathematics must have some natural physical equivalent or relation. E.g. how does multiplication happen in nature.. what does it mean really to multiply something?...how does this happen in nature? For the most part, we take these things as facts and learn the mechanics, but in physics, it seems to me, you really need to understand how these algebraic interpretations translate into physical realities (also, perhaps sort of obvious for everyone reads HN). For someone who enjoys discovering these perhaps obvious things later in life, it's clear how mathematicians and physicists could clearly come to the same mathematical conclusions from completely opposite vectors.. I guess this probably happens all the time.
- tim333 10y ago
  It's an interesting and unresolved question whether imaginary numbers are fundamental to the nature of physics or just a handy tool to calculate stuff. They are certainly handy for calculating - they are an easy way to represent oscillations.
  - digler999 10y ago
    Maybe it's a mistake to allow imaginary to ever be separate from real numbers ? Another post here brought up how easily it is to work with imaginaries if you just treat them as the vector [re im]. Maybe (philosophically speaking), there are no purely real numbers. Could it be that all quantities in nature must contain a (sometimes zero) Im component ? That might be a more satisfying interpretation than just allowing them to creep in when they are absolutely required, such as polynomial equations.
- logfromblammo 10y ago
  You should read about Maxwell's original equations for electromagnetism in component form (1865), then in quaternion form (1873), and Heaviside's vector-based rewrite (1888), then look at how they can be rewritten in geometric algebra. If you invent enough crazy new mathematical constructs, eventually one of them will mirror a natural physical phenomenon. And then a pile of equations collapses into something incredibly simple.
nilkn 10y ago
In Lie theory, given a Lie group, there's a general notion of an "exponential function", which maps elements in the tangent space to the identity to "full" elements of the group. In this case, our group is the unit circle in the complex plane. This is not circular logic, by the way. If a, b are complex numbers on this circle, then |a| = |b| = 1 and so |ab| = |a||b| = 1. The identity of the group is the complex number 1 (with plane coordinates (1,0)). So the tangent space to the identity is the vertical line 1 + it, for t a real-valued parameter. In two-dimensional coordinates, that expression looks like (1,0) + t * (0,1). (To be completely precise, the tangent space is a vector space, not a line displaced from the origin. In particular, the tangent space must contain a zero vector.) If v is an element of this tangent space, then in Lie theory the exponential of v, exp(v), is defined to be g(1), where g is the unique geodesic on the circle (passing through the identity element 1) whose velocity at time 0 is v. Visually, this is sort of like taking the tangent vector v = ti, placing its base at 1, then wrapping it around the circle and marking where its endpoint ends up at. If the vector is v = pi*i, then it has length pi, so it will end up demarcating an arc length of pi on the circle. Since we're working with the unit circle, this takes us straight to (-1,0). I'm still leaving a lot out, of course -- most importantly why this notion of exponential has anything to do with the ordinary one.
- chombier 10y ago
  I was about to write a similar story, but then I realized you need a Riemannian metric (and exponential) for everything arclength-related. IIRC since the unit circle is a compact Lie group, there is a bi-invariant Riemannian metric whose exponential is the Lie group exponential and we land back on our feet, but the Lie group structure alone is not sufficient.
agumonkey 10y ago
I deeply thought the iterative nature was advanced .. I see how intuitive the scale / rotate mindset is; but I'm a bit sad that iteration is a mishap in their view.

355 points by espeed 10y ago

Is it pure chance that this was posted only two or three days after I watched it along with a few other e to pi = -1 videos? Randomness aside. 3blue1brown makes some wonderful math videos that I find really explain the intuitiveness of some of the ideas. I was unfortunately cursed with a math teacher who for whatever reason required us to memorize until we passed the test. Imaginary numbers were taught as "something that will help you in college"

vinchuco 10y ago
Richard Feynman used to go up to people all the time and he'd say "You won't believe what happened to me today... you won't believe what happened to me" and people would say "What?" and he'd say "Absolutely nothing". I agree wholeheartedly with the quality assessment of 3blue1brown videos. All mathematics should be clear and intuitive, by definition :)
- dvt 10y ago
  > All mathematics should be clear and intuitive, by definition :) I disagree with this completely, and so does all of higher mathematics. It's neither clear nor intuitive. In fact, as soon as you start learning about infinities (Calc I), intuition becomes hit and miss.
  - CN7R 10y ago
    Intuition always comes after the feeling of awe or confusion, in my opinion. Really depends on how you learn.
CN7R 10y ago
No, another link to 3Blue1Brown was posted earlier this week and hit the front page. My basic assumption is that when a topic is posted, people will explore that topic and, if it's interesting enough, post further content about that topic.

cousin_it 10y ago

1) e^x is a function whose derivative is equal to its value. 2) e^ix is a function whose derivative is equal to its value rotated by 90 degrees (ie^ix). 3) As x goes from 0 to pi, the trajectory of e^ix always has a velocity vector perpendicular to its current position. For example, when x = 0, the current position is 1 and the velocity vector is i. 4) So the trajectory a circle arc of length pi, which ends at -1.

kmill 10y ago
In a linear algebra course I helped teach last semester, I had the students go through the exercise of writing down the matrix for "multiplication by i," where the complex numbers are thought of as a two-dimensional vector space with {1,i} as a basis. Then I asked them to recognize the matrix (it's a 90-degree rotation of the plane), and then I asked them for an interpretation of that matrix squared (a 180-degree rotation). One thing to draw from this: for the real line, negation doesn't reflect the line, but instead it's a rotation through another dimension of numbers! When we went on to differential equations, I tried to convince them that e^(ix) is meaningful using your #2: it is a path around the unit circle since it's the solution to z'=iz, and we already agreed multiplication by i is a 90 degree rotation, and circles are what you get when velocity is orthogonal to vector position. It's a bit tougher to convince someone that x is arclength, though, especially when they aren't too comfortable with complex numbers yet.
- deleted 10y ago
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- godmodus 10y ago
  I love this answer. For some reason, I've been dreaming about negative numbers lately. I think they deserve their own number set notation.
  - dnautics 10y ago
    the standard IEEE floating point uses a 'sign and magnitude' system. Negative numbers are basically made by flipping a single bit. I've been playing around with a number system that expresses real numbers as two's complement (like how integers work) - xor all your bits and then add one. In this system, 0 is -0 and the number that is weird for signed integers: 0b100000...0000 is infinity. Now, there's an exponent and a fraction, too. I've been playing around with how to do these numbers with logic gates (and verilog!) and you can either two's complement the whole thing and work with absolute values, or you can keep the fraction part as a two's complement... So I just redid multiplication using two's complemented fractions! And addition/subtraction too, which for floating points in general is significantly harder than multiplication. The nice thing about two's complemented floating points is you don't need separate algorithms for addition and subtraction; you can just do everything with one algorithm.

hprotagonist 10y ago

This seemed like voodoo until i took signals and systems. Then it's just a normal observation about vector sums on the complex unit circle.

gydfi 10y ago
It seems like voodoo when you have high-school mathematics, because you've been taught about "to the power of" in terms of multiplying something by itself a certain number of times, and multiplying something by itself an imaginary number of times is nonsensical. Once you've done enough maths you think of it in a different way and e^ipi seems obvious. It's just a different conception of what the symbols mean, I suppose.
- mstade 10y ago
  There are so many assumption that were hammered into me in high school (and below) that I find hold back my thinking, rather than help me. I kind of feel like the education I received was full of hacks to get me past whatever test was coming up, rather than truly teaching me. I've spent years playing what feels like catch up to actually truly learn the things I thought I knew – a lot of high school mathematics included.
  - fl0wenol 10y ago
    What was not obvious to me, and took a lot of higher math to get to, was that in the grouping of N=>I=>Q=>R=>C, at each stage you're pulling in a more complete mathematical representation (counting, rings, fields) until you're algebraically closed, and while you could go higher, you start losing properties you like again, like commutability over multiplication. Complex numbers are that nice saddle point, which is an interesting thing to ponder. I wish there was more focus on these "why" aspects in at least the optional advanced math or physics you could get in HS. It helps put some things in perspective.

neximo64 10y ago

For anyone that finds it beautiful isn't there a bit of humanisation and definition involved (for example the Sine function used in deriviation uses 'pi' instead of 90 degrees), not to mention Sine is a human created function. You could have e to pi (90 * -1) too.. or a different method to define angles instead of having 360 degrees (base 60)

panic 10y ago
Yeah, you can have the same thing with degrees if you use a different exponent base than e: e^((pi/180)ix) = (e^(pi/180))^ix = E^ix and therefore E^180i = -1 where E is about 1.0176065. The particular base e has some nice properties, though, like that its rate of change d/dx e^ix at a given point x is just i e^ix. The rage of change for E^ix, on the other hand, is d/dx E^ix = (pi/180) i E^ix, which is a little less "natural". This strange fact -- that using radians makes the expression e^ix = cos x + i sin x have a nice derivative -- is one of the reasons why mathematicians like to define these functions in terms of radians instead of degrees.
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bbcbasic 10y ago
Survivorship bias of beautiful math. Who want's to study systems that don't simply and compose well. The exception that proves the rule is the 4 color theorem - ugly big-arse proof generated by a machine. Who is going to look into that any further?

obastani 10y ago

The most elementary explanation I ever got for why e^(i pi) = -1 is from the definition e = lim_{n -> infinity} (1 + 1/n)^n First, we can generalize this to e^z = lim_{n -> infinity} (1 + 1/n)^(z n) = lim_{m -> infinity} (1 + z/m)^m using the substitution m = zn. Therefore, e^(i pi) = lim_{m -> infinity} (1 + i pi/m)^m Now, converting the complex number (1 + i pi/m) in terms of polar coordinates (r, theta) yields r = (1 + pi^2/m^2) ~ 1 theta = sin^{-1}(pi/m) ~ pi/m Since the product of two complex numbers is (r, theta) (r', theta') = (r r', theta + theta'), we have (1 + i pi/m)^m ~ (1^m, m * pi/m) = (1, pi) = -1.

phkahler 10y ago

It's a nice video but it goes so fast I don't think people who don't already understand this stuff will get it.

sparky_z 10y ago
Yeah, he's gotten much beeter about that since this video was released.
vinchuco 10y ago
"Young man, in mathematics you don't understand things. You just get used to them."
- johnfn 10y ago
  I don't really like this, because it's a self limiting quote that leads to complacency over real understanding. Especially in the case of imaginary numbers and e where intuitive understanding only doesn't exist because mathematics has a history of being poorly taught.
  - E6300 10y ago
    Honestly, after spending months studying the subject, I don't think it's really possible to "get" complex numbers. I just view them as affine transformations written in an unusual notation. I don't think they make sense as anything but a recontextualization of R^2.
    - akvadrako 10y ago
      So what don't you get? Are you implying there is something more to complex numbers?
      - E6300 10y ago
        What I don't get is why someone would bother with complex numbers and their silly notation when linear algebra would work perfectly fine. I know that in certain contexts they are useful. E.g. to avoid losing information when solving polynomials. But that's a minuscule fraction of their range of application. Nearly every practical use I've seen of complex numbers just use them as a vector representation.

hgdsraj 10y ago

Honestly the calculus explanation is the most clear to me (Taylor series expansion of e^x)

jjaredsimpson 10y ago

e^x = an infinite series if you substitute x = iz you can split the even and odd terms of the infinite series into cos and sin e^iz = cos z + i sin z evaluating at z = pi yields cos pi + i sin pi -1 + 0i = -1 the exponential function maps the imaginary axis to the unit circle. pi gets mapped to -1. This shows its true, but the "why" and real understanding requires calculus and the first week of complex analysis. Otherwise you are just parroting a set of facts.

gizmo686 10y ago
Math major here. This presentation does skip some nessasary legwork for complete rigor, but presents a valid non-standard construction of e^ipi. Specifically, he defines two sets of objects: adders and multipliers, and a function (written e^x for historical reasons) that maps adders into multipliers. He then generalizes this construction to work in 2 dimensions instead of 1 dimension. I would add to this construction that e^x is the particular converter that maps pi -> -1. However, I think this requirement is implicit in him stating that e^x is the most "natural" of the converters, and that pi -> -1 is the most natural of mappings. The only place where you might need calculus is to provide an explicit construction of e^x (and possibly to justify the notational choice of writing it like an exponential). EDIT: To make the point clearer. Under the construction presented by the video, e^x is not defined as an infinite some, but rather as a function satisfying certain properties. That this function is equal to a particular infinite sum is a statement that requires proof; and not a statement that is needed for many applications.

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somewhat_slow 10y ago

At 2:37, right after finishing explaining adders and multipliers on the line it started going way too quick like where did e^x and some these infinite sums come from???

gizmo686 10y ago
He introduces e^x as a way of converting adders into multipliers. The infinite sum is not important. I believe he is showing it purely as a visual tool to indicate that e^x is just "some function". More specifically, based on the video, the important feature for a adder -> multiplier converter to have is that f(x+y) = f(x)f(y). However, many functions satisfy this requirement. He states (without justification [0]) that e^x is the most "natural" choice for such a function, and provides the infinite sum as a visual aid to show that it is just some function. [0] Rather, justification is provided by reference to another video.

1001101 10y ago

Gauss said that if the reason weren't immediately obvious to someone, they would never be a first rate mathematician. For the rest of us, there's YouTube.

lowpro 10y ago

I've been watching 3Blue1Brown a lot recently, and he's been linked to many times here too. His work is absolutely amazing, and he even wrote a python library to make the animations!

espeed 10y ago
Yes, the Python library is called Manim, and it's on GitHub: https://github.com/3b1b/manim https://github.com/3b1b/manim

skybrian 10y ago

If you prefer slides over a video, here is a slideshow I wrote: https://docs.google.com/presentation/d/1oMNjkDp-LieSGnZEwNpceG8KTIvbnus9olu3KqnM5bg/edit https://docs.google.com/presentation/d/1oMNjkDp-LieSGnZEwNpc... Or you might prefer Better Explained's article: https://betterexplained.com/articles/intuitive-understanding-of-eulers-formula/ https://betterexplained.com/articles/intuitive-understanding...

mih 10y ago
Upvoted for the Better Explained link. I always liked their explanation of e https://betterexplained.com/articles/an-intuitive-guide-to-exponential-functions-e/ https://betterexplained.com/articles/an-intuitive-guide-to-e...

gizmo686 10y ago

As always with this author, I followed a link to a math explanation video expecting to point out how it is only a superficial treatment that breaks down under rigorous analysis, only to be dissapointed in my inability to find any such flaws. Seriously, this guy is awesome. If you need to learn math, see if he has a video about it. I do have 1 gripe with this video though, and that is his handwaving around "natural". More specifically, as far as this particular case goes, there is nothing natural about the choice of e^x, or the significance of pi. As he identifies, we are interested in some function that maps adders into multipliers with the property f(x+y)=f(x)f(x). To uniquely identify such a function, we need to add an additional constraint. He chose to add f(pi) = -1. He justifies this by arguing that pi is the length you would travel along the unit circle to arrive at -1. This is true (and the underlying reason why pi and e end up being natural), however using this argument seems to break the abstraction for me. Under this construction, in the equation f(i * pi)=-1, "i * pi" is an object in the set of multipliers, and "-1" is an object in the set of adders. Specifically, "i * pi" is a function which takes a plane (or perhaps a point) and rotates it, while "-1" is a function which takes a plane (or point) and slides it. He then invokes an unstated mapping, g, to convert the multiplier [0] "i * pi" into the real number "pi". He than insists that g(x) gives the distance a point would travel along the unit circle when the multiplier x is applied to it. At this point, because arriving at the point "-1" [1] from the point "1" through rotation requires traveling pi distance, it makes sense that f(i * pi) = f(g^-1(pi)) = -1 I am still not convinced that (within this construction alone), this is a more natural choice that saying that f(i) = 1, but would agree that it is one of the two natural choices. To get to f(x) = e^x being the natural choice requires showing that it comes up in all sorts of unrelated parts of math, so it is probably more natural. [0] It might be better to speak of "rotational multipliers" here, as I am not sure how to natural define g(x) for multipliers that stretch the plane, instead of or in addition to simply rotating it. [1] This "1" is again distinct from the adder "1" and multiplier "1", but plays a central role in defining them, so I do not object to its usage.

dvt 10y ago

I've seen this a while ago, and while it's pretty instructive, it's actually also pretty confusing. The magic happens in a seemingly-innocuous throwaway sentence at around 4:20 (after being introduced to the 2D plane): > ... This can now include rotating along with some stretching and shrinking ... It's entirely non-obvious WHY we should be okay with rotating all of a sudden. The real answer is not super complicated, but it deals with a couple of amazing relationships between exponential functions and trigonometric identities[1]. So really, we don't have to accept "rotating" as some weird new action we can do when moving into the complex plane, we just have to accept that trigonometry is weirdly related to analysis due to some very cool properties. [1] https://www.phy.duke.edu/~rgb/Class/phy51/phy51/node15.html https://www.phy.duke.edu/~rgb/Class/phy51/phy51/node15.html

gizmo686 10y ago
>It's entirely non-obvious WHY we should be okay with rotating all of a sudden. Why shouldn't we? We have a set of object called multipliers that we want to generalize to the 2 dimensional plane. The presented generalization (eg, the multiplier identified by x maps the point at 1 to the point at x) seems natural; and it defiantly seems to be well defined. Additionally, it appears obvious that the presented generalization is equivalent to the original definition if considered only along the x axis. From the perspective of the video, the fact that e^x is written as an exponential is "a vestige of its relationship with repeated multiplication". In the construction presented in this video, we should think of e^x as the "natural" homomorphism from adders to multipliers. The fact that this function has anything to do with exponential, trigonometry, or analysis is neat, but not important in this context.
- dvt 10y ago
  > Why shouldn't we? We have a set of object called multipliers that we want to generalize to the 2 dimensional plane. The presented generalization (eg, the multiplier identified by x maps the point at 1 to the point at x) seems natural Rotating all of a sudden does not feel natural to me at all. Why not start cutting up the 2D plane? Why not fold it? In analytic terms, we would cause discontinuities, whereas rotating is "smooth". But this is definitely nontrivial. Rotating is just so arbitrary. I think that skewing or flipping, for example is just as "natural" as rotating. > The fact that this function has anything to do with exponential, trigonometry, or analysis is neat, but not important in this context. It's actually at the heart of why rotation (and not some other geometric operation) is key to e^iπ.

jejones3141 10y ago

Look at the Taylor series for e^x, and see what you get for e^(i * theta). The terms with i raised to an even power are real, those with i raised to an odd power are imaginary. Group the odds together and the evens together, and compare with what the Taylor series for sin x and cos x give you for sin theta and cos theta. (I'm being a bad person here, ignoring convergence issues, but I want to say that all three are absolutely convergent.)

unholiness 10y ago

This explanation strikes me as a little too aggressive in throwing out the notation with the bathwater, only reaching its result by redefining the terms we already have intuition for into space-stretching operations that don't work like arithmetic does in my head. In that sense I think I have the same problem with this proof that I do with the standard one, where you add the Maclaurin series of cos(θ) and i * sin(θ) and match term-by-term with the series for e^(i * θ). The problem is, at the point you can actually show equality, the things on one side aren't obviously a rotation and the things on the other side aren't obviously an exponential. I'm not just hear to yell at clouds. I was given a proof that I truly love by a professor I adore, which I think really does give insight into what all these operators are doing. The best video I can find with it is here: https://www.youtube.com/watch?v=-dhHrg-KbJ0 https://www.youtube.com/watch?v=-dhHrg-KbJ0 (Skip to 7:30 if you're already comfortable with the limit definition of e^x) The basic summary is: 1) e^iθ is equal to (1 + iθ/n) ^ n for large n 2) That base, (1 + iθ/n), plotted as a complex number, has length approaching 1, angle approaching θ/n 3) The base squared, (1 + iθ/n)^2, by de moivre's theorem, forms another point as if the transformation from (0, 1) were repeated twice — that is, the length stays one, and another tiny angle is added for a total of 2θ/n 4) The full result is therefore n transformations, taking the path along the unit circle, traveling θ and arriving at cos(θ) + i * sin(θ)

mistermumble 10y ago
Thanks for the pointer to the video ("E to the Pi for Dummies" by Mathologer). It is definitely worth watching. Much more understandable, imho, than the one referenced at the top of this thread. Your text summary of the video is accurate and likely makes sense to some people but not so much for others (namely, me). Tour words are hard to grok until you see the visual depiction --- until you see the sequence of n transformations become a spiral arrangement of triangles that ends up approximating the (-1, 0i) point in the complex plane.
jacobolus 10y ago
Your (1) is just one of many possible (technically equivalent) definitions for the exponential function. And arguably not the most basic/natural/intuitive one. The primary motivation for the exponential function is to be the inverse of the logarithm function. And the motivation of logarithms is to convert multiplication problems to addition problems, so they could be solved with table lookups (later performed on a slide rule) instead of difficult arithmetic. That is, log ab = log a + log b. Which is to say, exp(c + d) = (exp c)(exp d). Just setting this constraint along with the derivative exp’ 0 = 1 is enough to characterize the exponential function. Or you can get to this function in many other ways, e.g. by solving the differential equation d/dx exp x = exp x; by the series 1 + x + x^2/2 + x^3/6 + ...; or by defining the logarithm as the definite integral of 1/x starting at 1, and then taking the exponential function to be its inverse. I like defining the (complex) exponential as a conformal mapping between the cylinder and the plane minus a point.

hawkice 10y ago

> (2015) I'd prefer if admins could change the link to a description of this mathematical fact that's more up to date, and fits the needs of me and my family in 2017.

espeed 10y ago
It originally said "Why does e to pi i equal -1? (visualized)".

timlod 10y ago

I love the Khan Academy video on this: https://www.youtube.com/watch?v=mgNtPOgFje0 https://www.youtube.com/watch?v=mgNtPOgFje0 Sal's enthusiasm at the end is contagious!

mstade 10y ago
I've learned so much from Khan Academy – not just whatever is explained by the video I'm watching, but different ways of educating people as well. In fact, I helped out with tutoring basic math a while back and I struggled at first because I didn't realize how much I wasn't teaching simply because it was second nature to me. After watching a bunch of Khan Academy videos on the same topics, I found different ways of explaining things, and was much more successful. Khan Academy has really challenged my own way of thinking about many things, math in particular. It's a great resource, and while I don't use Khan Academy much these days I still donate $100 every year.

ageofwant 10y ago

Why not just write it e^iτ=1 which conceptually and pedagogically makes just so much more sense.

mitchtbaum 10y ago

e^τi = 1. In words, this equation makes the following fundamental observation: The complex exponential of the circle constant is unity. Geometrically, multiplying by e^iθ corresponds to rotating a complex number by an angle θ in the complex plane, which suggests a second interpretation of Euler’s identity: A rotation by one turn is 1. The Tau Manifesto http://tauday.com/tau-manifesto#sec-euler_s_identity http://tauday.com/tau-manifesto#sec-euler_s_identity

wodenokoto 10y ago
Thanks for posting this. I recently saw another video about this equation than the one posted here and thought "maybe this is one of those cases where tau doesn't work out nicely", but didn't have the intuition to see if it was or not.

ttoinou 10y ago

BTW for thoses who like Complex Analysis here's how to visualize them as a 2D deformation : https://www.youtube.com/watch?v=CMMrEDIFPZY https://www.youtube.com/watch?v=CMMrEDIFPZY

0xbadf00d 10y ago

Very Nice Video - One of my favourite sub 5 minute explanations is Oliver Humpage's @ Ignite Bristol circa 2011: https://www.youtube.com/watch?v=vHaPyuEkK4A https://www.youtube.com/watch?v=vHaPyuEkK4A Hope you enjoy as much as I did.

faragon 10y ago

Pure beauty.

etatoby 10y ago

I think Mathologer's take is much more intuitive: https://youtu.be/-dhHrg-KbJ0 https://youtu.be/-dhHrg-KbJ0

cmollis 10y ago

I had never understood how imaginary numbers had any bearing on physics (probably because I'd never been taught). But lately I've thinking about how all mathematics must have some natural physical equivalent or relation. E.g. how does multiplication happen in nature.. what does it mean really to multiply something?...how does this happen in nature? For the most part, we take these things as facts and learn the mechanics, but in physics, it seems to me, you really need to understand how these algebraic interpretations translate into physical realities (also, perhaps sort of obvious for everyone reads HN). For someone who enjoys discovering these perhaps obvious things later in life, it's clear how mathematicians and physicists could clearly come to the same mathematical conclusions from completely opposite vectors.. I guess this probably happens all the time.

tim333 10y ago
It's an interesting and unresolved question whether imaginary numbers are fundamental to the nature of physics or just a handy tool to calculate stuff. They are certainly handy for calculating - they are an easy way to represent oscillations.
- digler999 10y ago
  Maybe it's a mistake to allow imaginary to ever be separate from real numbers ? Another post here brought up how easily it is to work with imaginaries if you just treat them as the vector [re im]. Maybe (philosophically speaking), there are no purely real numbers. Could it be that all quantities in nature must contain a (sometimes zero) Im component ? That might be a more satisfying interpretation than just allowing them to creep in when they are absolutely required, such as polynomial equations.
logfromblammo 10y ago
You should read about Maxwell's original equations for electromagnetism in component form (1865), then in quaternion form (1873), and Heaviside's vector-based rewrite (1888), then look at how they can be rewritten in geometric algebra. If you invent enough crazy new mathematical constructs, eventually one of them will mirror a natural physical phenomenon. And then a pile of equations collapses into something incredibly simple.

nilkn 10y ago

In Lie theory, given a Lie group, there's a general notion of an "exponential function", which maps elements in the tangent space to the identity to "full" elements of the group. In this case, our group is the unit circle in the complex plane. This is not circular logic, by the way. If a, b are complex numbers on this circle, then |a| = |b| = 1 and so |ab| = |a||b| = 1. The identity of the group is the complex number 1 (with plane coordinates (1,0)). So the tangent space to the identity is the vertical line 1 + it, for t a real-valued parameter. In two-dimensional coordinates, that expression looks like (1,0) + t * (0,1). (To be completely precise, the tangent space is a vector space, not a line displaced from the origin. In particular, the tangent space must contain a zero vector.) If v is an element of this tangent space, then in Lie theory the exponential of v, exp(v), is defined to be g(1), where g is the unique geodesic on the circle (passing through the identity element 1) whose velocity at time 0 is v. Visually, this is sort of like taking the tangent vector v = ti, placing its base at 1, then wrapping it around the circle and marking where its endpoint ends up at. If the vector is v = pi*i, then it has length pi, so it will end up demarcating an arc length of pi on the circle. Since we're working with the unit circle, this takes us straight to (-1,0). I'm still leaving a lot out, of course -- most importantly why this notion of exponential has anything to do with the ordinary one.

chombier 10y ago
I was about to write a similar story, but then I realized you need a Riemannian metric (and exponential) for everything arclength-related. IIRC since the unit circle is a compact Lie group, there is a bi-invariant Riemannian metric whose exponential is the Lie group exponential and we land back on our feet, but the Lie group structure alone is not sufficient.

agumonkey 10y ago

I deeply thought the iterative nature was advanced .. I see how intuitive the scale / rotate mindset is; but I'm a bit sad that iteration is a mishap in their view.