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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
by 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 agoIn 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.
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- godmodus 10y agoI 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 agothe 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.
- espeed 10y agoHave you seen "Stanford Seminar: Beyond Floating Point: Next Generation Computer Arithmetic"? https://www.youtube.com/watch?v=aP0Y1uAA-2Y https://www.youtube.com/watch?v=aP0Y1uAA-2Y
- dnautics 10y agoWell, I'm the guy in the black shirt who did the demo. If you liked that lecture, I'm already starting on some verilog implementations. This is an example multiplication 8 bit * 8 bit -> 16 bit unpacked (20 bits). It differs from standard floating point in that the fractions are stored as two's complement. It takes a little bit of wrapping your head around, but the hidden bit for negative numbers is actually -2 ! Moment of zen. https://github.com/interplanetary-robot/mullinengine/blob/master/multiplier/posit_mult.v https://github.com/interplanetary-robot/mullinengine/blob/ma...
- AndrewOMartin 10y ago@espeed didn't ask if you contributed significantly to the lecture, they asked if you'd seen it. Please stick to the question!
- jacobolus 10y agoWhat do you mean by “number set notation”?
- k__ 10y agoSpeaking of number sets. I never understood why complex numbers are considered one. I mean, yes they "are" a set, but besides that they are completely different. All number sets I learned about did fill some gaps in one dimension, but complex numbers somehow added a new dimension. Like real numbers stood in an entirely different context to rational numbers than complex numbers stood to real numbers.
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- mr_pink 10y agoC (complex numbers) is a field, just like R (real numbers). It's an algebraically closed field, unlike R, so in some sense it's actually the most natural set to call "numbers". https://en.wikipedia.org/wiki/Algebraically_closed_field https://en.wikipedia.org/wiki/Algebraically_closed_field
- k__ 10y agoI'm not trying to argue for R being more a set than C. In my head a one dimensional thing like R is fundamentally different from a multidimensional thing like C.
- mr_pink 10y agoWell I was answering the question about why is C considered numbers - it's a field, hence elements of C behave exactly like numbers. Moreover, it's an 'algebraic closure' of R, hence a more natural choice for "all numbers", and it's a maximal one at that (i.e. quaternions and such are no longer fields and don't really behave like numbers, while C still does). Edit: FWIW, I consider the terminology of 'real' vs 'imaginary' completely stupid and misleading. This terminology didn't really make it to other languages, e.g. in Russian it's 'material' vs 'complex' numbers, but they don't use the term 'imaginary'.
- cousin_it 10y ago
- jacobolus 10y agoTo elaborate: The complex plane can be thought of as being made up of two-dimensional geometric transformations consisting of rotation and scaling (Tristan Needham calls such transformations “amplitwists”), with 1 as the identity transformation, and i as a quarter turn anticlockwise, and i^2 = –1 as a half turn. To compose two such transformations, you multiply the scales and add the angle measures of the rotations. Because such operations are linear, you can also break them into a part parallel to 1 and a part perpendicular to 1 (some multiple of i), and multiply two such transformations component-wise, using the distributive law (a + bi)(c + di) = (ac – bd) + (ad + bc)i. exp z is a complex function which maps (in an angle-preserving way, i.e. conformally) from an infinite two-ended cylinder to a whole plane minus one point (the “origin”). The exp function maps negative infinity on the cylinder to the origin on the plane, and it maps the zero point on the cylinder to a given “unit” point in the plane, and the “zero” circular slice through that point on the cylinder to the “unit circle” on the plane, containing the unit point and concentric with the origin. The coordinate system on the cylinder has 2πi measuring one loop around a circular slice, and 1 pointed along the cylinder axis. The coordinate system in the plane is the customary square grid. Addition of coordinates in the geometry of the cylinder (if you like, rotating and/or sliding the cylinder) corresponds to multiplication of complex numbers (composition of amplitwists) in the plane. That is, exp(w + z) = (exp w)(exp z). In particular, exp iy for some real number y maps points at distance y along the zero circle in the cylinder to points on the unit circle in the complex plane at a proportional distance around the circle; that is, to rotation operators which correspond to the given angle measure in radians. So πi is halfway around the zero slice in the cylinder, and exp maps it to the operator in the complex plane corresponding to a half-turn rotation, i.e. exp πi = –1. The log function is the inverse map, from the plane to the cylinder; it is a multi-valued function because we can make multiple “straight” helical connections between arbitrary points on the cylinder, which wrap around different numbers of times. Once we have this general concept for how we want the exp map to work, we can work out the details to find that the unique such function is the solution to a particular differential equation f'(z) = f(z), or alternately the Taylor series we are all familiar with, exp z = 1 + z + z^2/2 + z^3/6 + ...
- snogglethorpe 10y agoIs this just the contents of the video in text form? In any case, thanks! I don't need to watch the video now, as you've very clearly explained it in only four lines of text!
- gizmo686 10y agoI can't tell if you are being sarcastic, but no, parent's comment is not the contents of the video. The video provides a geometric construction, and makes no use of calculus.
- snogglethorpe 10y agoNot being sarcastic, I just didn't watch the video. I was wondering if it was basically the same thing with nice animations...
- recursive 10y agoIt's not really any clearer to me. I cannot generalize my understanding of exponents to anything that deals with imaginary numbers.
- chombier 10y agoFor me what made it clicked is realizing that complex numbers are 2D matrices: z = x + i y = [[x -y][y x]]. So really we should be writing z = x * [[1 0][0 1]] + y * [[0 -1][1 0]], but since it's tedious we just call 1 == [[1 0] [0 1]] the 2x2 identity matrix and i == [[0 -1][1 0]], and check that i^2 = -1. Then no more magical i number, the complex product can be derived from the matrix product, the exponential becomes the 2x2 matrix exponential, and so on.
- recursive 10y agoUsing a matrix as an exponent isn't any more comprehensible than an imaginary number to me. If it works for you, that's great, but it's not much help to me.
- jeffwass 10y agoWhen I was a physics TA, I used similar arguments with double derivatives show the connection between imaginary exponentials and sines/cosines. For f(x)=e^(kx), double derivative d^2f/dx^2 = k^2 f(x) Meanwhile for g(x)=sin(kx), d^2g/dx^2 = -k^2 g(x), and similarly for cosine. So if k is imaginary, from a differential equations point of view, the exponential behaves exactly like a sine or cosine. That shows the general idea, and further consideration of boundary conditions gives e^ix = cos(x) + isin(x).
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- amelius 10y agoI think you should explain step 2 in more detail.
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- niklasbuschmann 10y agoFirst time I understood this. Thank you for that.
- cousin_it 10y agoYou're welcome :-)
- JadeNB 10y agoMinor wording: e^x is a function whose derivative is equal to itself, or the value of whose derivative is equal to its own value. This might be considered the kind of excessive precision that obscures rather than clarifies (as would be, for example, writing x ↦ e^x in place of just e^x), but lots and lots of confusion can result (especially in a function-analytic setting) from failing to distinguish between a function and its values.
- ouid 10y agoThis isn't a proof, and while that may be obvious to you, it's not obvious to everyone who sees it. There exists no logical path to the statement from the previous definition of e^x because, up to this point, e^x is a function on real numbers. e^ix is nonsense until you define e^x in complex coordinates, at which point the proof needs to rely on properties of that definition. what you have posted misses the nature of the insight. If two analytic functions on the complex numbers agree on uncountably infinitely many points, then they agree everywhere they are defined. This means that there is a unique analytic function that maps (ix)[x \in R] to the circle and (x)[x \in R] to the natural exponential, and they are the same complex function. Without the knowledge that analytic continuations are unique, the statement is entirely (pun intended) meaningless.
- cousin_it 10y agoSure, step 1 relies on the existence of e^z which is not obvious. Step 2 relies on the chain rule over C, step 4 relies on having a unique solution to a differential equation, etc. My comment can be expanded to a proof, but it was more aimed to explain why Euler's identity is natural.
- ouid 10y agostep 1 isn't "not obvious"; it's nonsense. What, exactly, are you trying to prove exists? e^z? e^z is undefined at this point. If you're trying to prove that e^i\pi=-1, you need to start with a definition of e^z.