Math: Fleeting Thoughts
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Re: Math: Fleeting Thoughts
Yes, it's cool. People who think that pi's cooler than e don't understand this stuff.
Sure, it's "just" Taylor's series, but in introductory calculus we don't usually get taught it in terms of differential operators; this formula is probably the primary motivation for writing operators for higher derivatives in exponential form.
Deriving this for yourself is a great way to understand and remember it. So there's no need to feel embarrassed.
Sure, it's "just" Taylor's series, but in introductory calculus we don't usually get taught it in terms of differential operators; this formula is probably the primary motivation for writing operators for higher derivatives in exponential form.
Deriving this for yourself is a great way to understand and remember it. So there's no need to feel embarrassed.
Re: Math: Fleeting Thoughts
This is something that's been puzzling me for a few months, and I thought I'd toss it your way.
I've been researching the standard trig identities in preparation for my master's thesis, so I've been through more than my share of proofs of the Law of Sines and the Law of Cosines. So I was flipping through a 1960's collegiate textbook surveying what is now high school math up to precalc, and it asks the following two questions:
*20. Derive the Law of Cosines from the Law of Sines.
*21. Derive the Law of Sines from the Law of Cosines.
This fascinated me, because in all my research and Googlefu, I've never seen approaches to either theorem from this direction. But I've been confounded as to how the details of this proof would be worked out. Does anyone have any ideas?
I've been researching the standard trig identities in preparation for my master's thesis, so I've been through more than my share of proofs of the Law of Sines and the Law of Cosines. So I was flipping through a 1960's collegiate textbook surveying what is now high school math up to precalc, and it asks the following two questions:
*20. Derive the Law of Cosines from the Law of Sines.
*21. Derive the Law of Sines from the Law of Cosines.
This fascinated me, because in all my research and Googlefu, I've never seen approaches to either theorem from this direction. But I've been confounded as to how the details of this proof would be worked out. Does anyone have any ideas?
 jestingrabbit
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Re: Math: Fleeting Thoughts
I'd expect 20 to start with squaring the sine rule, and then a wall of algebra more tedious than anyone can imagine.
Or, perhaps you want to start with one triangle, and then put another triangle next to it, with the sum of two of the angles being pi/2. Then, cos of one angle is sine of the other.
What have you already tried?
Or, perhaps you want to start with one triangle, and then put another triangle next to it, with the sum of two of the angles being pi/2. Then, cos of one angle is sine of the other.
What have you already tried?
ameretrifle wrote:Magic space feudalism is therefore a viable idea.
Re: Math: Fleeting Thoughts
I started with the other one, taking the Law of Cosines in terms of two of the angles, adding them together, and then a wall of algebra more tedious than anyone could imagine to try to turn them into sines. It didn't lead to anything recognizable, but I don't know whether it was a failure of algebra at some step along the line.
Then I thought about translating between sine and cosine holistically rather than algebraically with something like this (assuming ABC is acute):
Now x²=b²+h²2bh sinα = b²+h²2abh(sinα/a) and y²=a²+h²2ah sinβ=a²+h²2abh(sinβ/b) since angle CAD is the complement of α and so on. This looked fruitful, but I couldn't think of how to combine x and y into a single equation in a manner that would be productive.
Then I decided to ask for advice.
Then I thought about translating between sine and cosine holistically rather than algebraically with something like this (assuming ABC is acute):
Now x²=b²+h²2bh sinα = b²+h²2abh(sinα/a) and y²=a²+h²2ah sinβ=a²+h²2abh(sinβ/b) since angle CAD is the complement of α and so on. This looked fruitful, but I couldn't think of how to combine x and y into a single equation in a manner that would be productive.
Then I decided to ask for advice.
 jestingrabbit
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Re: Math: Fleeting Thoughts
If you think about the proof of the sine rule that starts with the formula for the area using sine (ie A = a*b*sin(C)/2), can we really imagine a proof arriving quicker via the cos rule?
My "extra triangles" diagram had just one extra, and the two fitting inside a large right triangle. I doubt that's profitable, yours looks better.
My "extra triangles" diagram had just one extra, and the two fitting inside a large right triangle. I doubt that's profitable, yours looks better.
ameretrifle wrote:Magic space feudalism is therefore a viable idea.
Re: Math: Fleeting Thoughts
jestingrabbit wrote:If you think about the proof of the sine rule that starts with the formula for the area using sine (ie A = a*b*sin(C)/2), can we really imagine a proof arriving quicker via the cos rule?
My "extra triangles" diagram had just one extra, and the two fitting inside a large right triangle. I doubt that's profitable, yours looks better.
Yeah, it struck me in a "oh, duh" moment that I'm right at the doorstep. For the two equation, collect all the nontrig terms over to the left side and add in the Pythagorean theorem and you have 2h² = 2abh(sinα/a) and 2h² = 2abh(sinβ/b). And that would work equally well if ABC was obtuse, since we're not combining x and y. Pretty cool, that's a very different tactic than anything I'd seen up to this point.
ETA: Argh, no it isn't. It's just the traditional proof that sinα = h/b and sinβ = h/a, made more complex out of a desire to derive it from the LoC. I had a feeling that would be the case.
I'm a lot less sanguine about the converse, using an identity about two angles and turning into one about one angle. Maybe start with the law of sines in terms of angles A and B and then deriving the law of cosines relative to angle C, the supplement of the sum of the two angles? But I'd be thrilled if there was a proof that is as direct as the derivation of the Law of Sines we just thought up.
Re: Math: Fleeting Thoughts
The (Extended) Law of Sines tells us that a/sinA = b/sinB = c/sinC = 2R, where R is the circumradius, so we get that a = 2R sinA, b = 2RsinB, and c = 2RsinC. We want to establish the Law of Cosines, i.e. that c^2 = a^2 + b^2  2abcosC; using the substitutions and dividing through by 4R^2, we want to establish that sin^2 C = sin^2 A + sin^2 B  2 sinA sinB cosC. (The Extended Law of Sines isn't strictly necessary here: whatever constant a/sinA = b/sinB = c/sinC happens to be, we just end up dividing through by that constant squared.) Now, C = 180  A  B, so sinC = sin(A+B) = sinAcosB + cosA sinB and cosC = cos(A+B) = sinA sinB  cosA cosB, so we want that (sinA cosB + cosA sinB)^2 = sin^2 A + sin^2 B  2 sinA sinB (sinA sinB  cosA cosB). Expanding yields sin^2 A cos^2 B + cos^2 A sin^2 B + 2 sinA sinB cosA cosB = sin^2 A + sin^2 B + 2 sinA sinB cosA cosB  2 sin^2 A sin^2 B. Canceling and a slight rearrangement gives us sin^2 A cos^2 B + sin^2 B cos^2 A = sin^2 A  sin^2 A sin^2 B + sin^2 B  sin^2 A sin^2 B, which is true by the Pythagorean Identity.
The Law of Cosines tells us that c^2 = a^2 + b^2  2abcosC, or that cosC = (a^2 + b^2  c^2)/2ab. Square both sides to yield cos^2 C = (a^4 + b^4 + c^4 + 2a^2b^2  2a^2c^2  2b^2c^2)/4a^2b^2, and use the Pythagorean Identity to get sin^2 C = ( a^4  b^4  c^4 + 2a^2b^2 + 2a^2c^2 + 2b^2c^2)/4a^2b^2, so that sin^2 C/c^2 = ( a^4  b^4  c^4 + 2a^2b^2 + 2a^2c^2 + 2b^2c^2)/4a^2b^2c^2. This is symmetric in a, b, and c, so sin^2 C/c^2 = sin^2 A/a^2 = sin^2 B/b^2, which reduces to the Law of Sines. (This doesn't establish the Extended Law of Sines, though.)
The Law of Cosines tells us that c^2 = a^2 + b^2  2abcosC, or that cosC = (a^2 + b^2  c^2)/2ab. Square both sides to yield cos^2 C = (a^4 + b^4 + c^4 + 2a^2b^2  2a^2c^2  2b^2c^2)/4a^2b^2, and use the Pythagorean Identity to get sin^2 C = ( a^4  b^4  c^4 + 2a^2b^2 + 2a^2c^2 + 2b^2c^2)/4a^2b^2, so that sin^2 C/c^2 = ( a^4  b^4  c^4 + 2a^2b^2 + 2a^2c^2 + 2b^2c^2)/4a^2b^2c^2. This is symmetric in a, b, and c, so sin^2 C/c^2 = sin^2 A/a^2 = sin^2 B/b^2, which reduces to the Law of Sines. (This doesn't establish the Extended Law of Sines, though.)
(∫p^{2})(∫q^{2}) ≥ (∫pq)^{2}
Thanks, skeptical scientist, for knowing symbols and giving them to me.
Thanks, skeptical scientist, for knowing symbols and giving them to me.
 jestingrabbit
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Re: Math: Fleeting Thoughts
Not as hideous a wall of algebra as I'd thought.
ameretrifle wrote:Magic space feudalism is therefore a viable idea.
Re: Math: Fleeting Thoughts
My initial google of the matter revealed a presentation in which the spherical version of one was derived from the spherical version of the other, and the slide containing the derivation was pretty scary (and not terribly well formatted) which wasn't very inspiring.
I did gave a quick attempt at forcing the algebra, but as I did it in the confines of an unposted post rather than on paper, it was probably inevitably that I screwed it up somewhere.
I haven't squinted too closely at the above solutions, but at a glance it seems like basically two separate approaches to the derivation in each direction. Is there any reason why one couldn't just do the derivation one way, and then just do the opposite steps for the derivation in the other direction?
I did gave a quick attempt at forcing the algebra, but as I did it in the confines of an unposted post rather than on paper, it was probably inevitably that I screwed it up somewhere.
I haven't squinted too closely at the above solutions, but at a glance it seems like basically two separate approaches to the derivation in each direction. Is there any reason why one couldn't just do the derivation one way, and then just do the opposite steps for the derivation in the other direction?
Re: Math: Fleeting Thoughts
Dopefish wrote:Is there any reason why one couldn't just do the derivation one way, and then just do the opposite steps for the derivation in the other direction?
Well, my argument for "Sines implies Cosines" uses the Law of Sines but doesn't start from it, and my argument for "Cosines implies Sines" starts from (the three versions of) the Law of Cosines but drops a lot of information in the last step. Neither one reverses nicely at all.
(∫p^{2})(∫q^{2}) ≥ (∫pq)^{2}
Thanks, skeptical scientist, for knowing symbols and giving them to me.
Thanks, skeptical scientist, for knowing symbols and giving them to me.
Re: Math: Fleeting Thoughts
The proof that sin²C = sin²A + sin²B  2 sin A sin B cos C in a triangle "only" uses the fact that C = 180AB, the sum and difference identities for sine and cosine, and the Pythagorean identity. So, in theory, if you could show that all solutions to that equation have the same proportion, then you'd have proved the LoS from the LoC.
Not that it matters much in that direction, there are far more elegant proofs of LoS than this one could be.
Not that it matters much in that direction, there are far more elegant proofs of LoS than this one could be.
Re: Math: Fleeting Thoughts
Something of a firstworld problem, but my university offers far too many interesting modules for next year. I think I have a rough idea what I'm going to do, there's only really one or two things I'm undecided on. In fact, maybe some of you could be of help: Does this course (description quoted below) sound helpful, given that I'm strongly leaning towards doing some sort of computational maths or physics? The other options are various physics courses (either QFT, MHD, or Classical Field Theory)
(sadly, that's all the information I really have to go off about the course)
Numerical Solutions of Differential Equations wrote:The course will address modern algorithms for the solution of ordinary and partial differential equations, inclusive of finite difference and finite element methods, as well as broad mathematical principles underlying their analysis.
(sadly, that's all the information I really have to go off about the course)
(they pronouns please)
 doogly
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Re: Math: Fleeting Thoughts
Classical field theory isn't a prereq for QFT at your place? This seems odd but if not then I'd go for QFT. Numerical methods...
Well, that depends on how you like math / physics, and within math, math "methods." I'd always pick QFT or MHD and just do a numerical project for one or the other. Methods classes on their own sound very boring to me.
Numerical methods certainly are quite useful, but useful *for* something.
Well, that depends on how you like math / physics, and within math, math "methods." I'd always pick QFT or MHD and just do a numerical project for one or the other. Methods classes on their own sound very boring to me.
Numerical methods certainly are quite useful, but useful *for* something.
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Noc: A larval Doogly. They grow the tail and stinger upon reaching adulthood.
Keep waggling your butt brows Brothers.
Or; Is that your eye butthairs?
Re: Math: Fleeting Thoughts
Hmm, that is the impression I was getting from a couple of other people too, that numerical stuff is better to pick up as and when you need it. In that case I'd probably be most interested in either the classical or quantum field theory courses.
There's also one slight complication I haven't mentioned. Whichever one I pick I'm not actually taking for exams  it's a purelyforinterest addition to the actual modules I'm doing (which are mostly pure maths, but the parts which are useful for physics as well). Loose outlines of requirements:
Classical field theory: Supposed to be taken concurrently with GR and electrodynamics (I'll be doing GR and maybe electrodynamics myself) and after seeing basic lagrangian/hamiltonian particle mechanics, mostly covers more advanced things like the lagrangian and hamiltonian approaches to field theories, Noether's theorem and so on.
Quantum field theory: Only formal requirement is "an advanced course on quantum mechanics". Now, I've officially only done one QM course (out of three before QFT at my uni) but have been working through a book over the summer which covers everything up to basic QFT anyway. And it really doesn't look like you need that much. This is the one I'd be more interested in, but also the one which I'd probably find a bit harder.
There's also one slight complication I haven't mentioned. Whichever one I pick I'm not actually taking for exams  it's a purelyforinterest addition to the actual modules I'm doing (which are mostly pure maths, but the parts which are useful for physics as well). Loose outlines of requirements:
Classical field theory: Supposed to be taken concurrently with GR and electrodynamics (I'll be doing GR and maybe electrodynamics myself) and after seeing basic lagrangian/hamiltonian particle mechanics, mostly covers more advanced things like the lagrangian and hamiltonian approaches to field theories, Noether's theorem and so on.
Quantum field theory: Only formal requirement is "an advanced course on quantum mechanics". Now, I've officially only done one QM course (out of three before QFT at my uni) but have been working through a book over the summer which covers everything up to basic QFT anyway. And it really doesn't look like you need that much. This is the one I'd be more interested in, but also the one which I'd probably find a bit harder.
(they pronouns please)
Re: Math: Fleeting Thoughts
the Geometric Algebra approach to trig proofs may be new to some http://geocalc.clas.asu.edu/GA_Primer/G ... eting.html
http://geocalc.clas.asu.edu/GA_Primer/G ... index.html (has answer page)
http://geocalc.clas.asu.edu/GA_Primer/G ... index.html (has answer page)
Re: Math: Fleeting Thoughts
f5r5e5d wrote:the Geometric Algebra approach to trig proofs may be new to some http://geocalc.clas.asu.edu/GA_Primer/G ... eting.html
http://geocalc.clas.asu.edu/GA_Primer/G ... index.html (has answer page)
So geometric algebra basically defines a "nonsense" operation and derives geometric meaning from it? That is, the geometric product is the inner product plus the outer one: a scalar plus a bivector. That doesn't make any geometric sense at face value, does it? (I see how that wouldn't be a problem, but I'm just checking I got it right)
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Re: Math: Fleeting Thoughts
The meaning of "outer product" is clearly different. If ab is a scalar, then a∧b = ½(ab − ba) is also a scalar.
 doogly
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Re: Math: Fleeting Thoughts
But their ab is not a scalar.
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Keep waggling your butt brows Brothers.
Or; Is that your eye butthairs?
Re: Math: Fleeting Thoughts
Eebster the Great wrote:The meaning of "outer product" is clearly different. If ab is a scalar, then a∧b = ½(ab − ba) is also a scalar.
Their ab is not a scalar.
According to the definition of geometric product they give, the geometric product of two vectors (ai + bj)(ci + dj) (a,b,c,d real scalars; i,j the basis on the space) is ac + bd + acij + bdij: ac + bd is clearly a scalar and acij + bdij is equal to (ac + bd)ij, which is a bivector. The scalar is the inner product and the bivector is the outer product.
 Eebster the Great
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Re: Math: Fleeting Thoughts
Uh, so ab is a scalar only if a = b, or a^{2} ≠ aa ?
 doogly
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Re: Math: Fleeting Thoughts
The a^2 would be ambiguous; do you mean a dot a, a wedge a, or their geometric product aa?
It seems like a cute program, but there's trouble with generalizing it to arbitrary dimension and metric.
It seems like a cute program, but there's trouble with generalizing it to arbitrary dimension and metric.
LE4dGOLEM: What's a Doug?
Noc: A larval Doogly. They grow the tail and stinger upon reaching adulthood.
Keep waggling your butt brows Brothers.
Or; Is that your eye butthairs?
Noc: A larval Doogly. They grow the tail and stinger upon reaching adulthood.
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Or; Is that your eye butthairs?
 Eebster the Great
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Re: Math: Fleeting Thoughts
I mean the geometric product. According to that page, a^{2} is a scalar. But the geometric product ab for a ≠ b must not be a scalar, because if it were, a∧b = ½(ab − ba) would clearly be a scalar too.
Last edited by Eebster the Great on Sun Sep 21, 2014 11:35 pm UTC, edited 1 time in total.
 phlip
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Re: Math: Fleeting Thoughts
As long as we're working in two dimensions, their "outer product" is still a scalar. They're calling it a "bivector", but it's still just a number. They say it has "direction" in the article, but really it only has two directions: clockwise and anticlockwise, which map to positive and negative. Adding it to the dot product isn't necessarily something it makes sense to do, but it's not a type error like adding a scalar and a vector is... it's like, in 3D, adding a vector and a pseudovector... "a + a×b"... the types are compatible, the answer is just usually meaningless. Their "bivector" is basically the 2D equivalent of what the pseudovector is in 3D.
But the thing is, ultimately they're not actually doing anything with their geometric product... they're just defining it as these other two products added together, and showing that they can be decomposed to the two individual products... and the rest of those two pages is spent talking about the geometric applications of the two individual products. And then it just ends with:
It's like me defining f(x) = (sin(x) + x^{2})/2... given any number I can calculate sin(x) = f(x)  f(x) or I can calculate x^{2} = f(x) + f(x)... and both sin(x) and x^{2} have some geometric meaning... but that doesn't mean I can call f(x) "the geometric function" and assume I'm doing anything profound...
But the thing is, ultimately they're not actually doing anything with their geometric product... they're just defining it as these other two products added together, and showing that they can be decomposed to the two individual products... and the rest of those two pages is spent talking about the geometric applications of the two individual products. And then it just ends with:
Well... yeah. You'd think that'd be the first thing you do. Y'know, before you go around calling it a "geometric product" and all.The problem remains to assign geometric meaning to the quantity ab without expanding it into inner and outer products.
It's like me defining f(x) = (sin(x) + x^{2})/2... given any number I can calculate sin(x) = f(x)  f(x) or I can calculate x^{2} = f(x) + f(x)... and both sin(x) and x^{2} have some geometric meaning... but that doesn't mean I can call f(x) "the geometric function" and assume I'm doing anything profound...
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Re: Math: Fleeting Thoughts
Eebster the Great wrote:I mean the geometric product. According to that page, a^{2} is a scalar. But the geometric product ab for a ≠ b must not be a scalar, because if it were, a∧b = ½(ab − ba) would clearly be a scalar too.
Their product lives in the exterior algebra of the vector space (or whatever the word for the algebra of 0vectors, 1vectors, 2vectors, etc. is); so the product could, depending on circumstance, take a value as either a scalar, or a formal sum of scalars, vectors, bivectors (and more in higher dimension, but not on the plane). Since the wedge product a∧a is 0, in the case of aa, only scalars are present, so you can identify that with the scalar a^{2}. In particular, ab is a scalar exactly when a and b are parallel. (I have to say that that definition really threw me for the same reasons; it's just bad writing, really  a scalar is not the same as a formal sum involving 1).
phlip wrote:As long as we're working in two dimensions, their "outer product" is still a scalar. They're calling it a "bivector", but it's still just a number. They say it has "direction" in the article, but really it only has two directions: clockwise and anticlockwise, which map to positive and negative. Adding it to the dot product isn't necessarily something it makes sense to do, but it's not a type error like adding a scalar and a vector is... it's like, in 3D, adding a vector and a pseudovector... "a + a×b"... the types are compatible, the answer is just usually meaningless. Their "bivector" is basically the 2D equivalent of what the pseudovector is in 3D.
Just because the space of bivectors is only one dimensional doesn't mean it's just a number, unless you a particular basis in mind (which we don't). And they manage to show a sum of a scalar and bivector means something (a combination of a rotation & scaling)  which is reasonable enough.
The problem is that, they seem to implicitly assume that the algebra operators on any formal sum of a scalar, vector, and bivector. However, by taking the geometric product of an even number of vectors, you always get the sum of a bivector and scalar, and of an odd number of vectors yields a vector. So, it's reasonable to say that the system actually discusses two different domains  that of rotation/scaling transformation (bivector + scalar) and that of vectors. Obviously, transformations are composed when multiplied, and a transformation * a vector is the application of the transformation to the vector, but a vector * transformation is the conjugate of the transformation applied to the vector (i.e. the rotation the other way)  which is just how a particular subgroup of matrices work. (Also, the product on the space of transformations is actually the complex numbers with multiplication; they point the similarity out, but somehow don't seem to make the connection that they're isomorphic). The only particularly interesting thing is that ba is the transformation which takes a to ba (when premultiplied).
Mathematical hangover (n.): The feeling one gets in the morning when they realize that that short, elementary proof of the Riemann hypothesis that they came up with at midnight the night before is, in fact, nonsense.
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Re: Math: Fleeting Thoughts
Moole wrote:Just because the space of bivectors is only one dimensional doesn't mean it's just a number, unless you a particular basis in mind (which we don't).
I guess so... I was kinda assuming that there's a particular area that can be called "1", but I guess that isn't necessarily the case.
Moole wrote:And they manage to show a sum of a scalar and bivector means something (a combination of a rotation & scaling)  which is reasonable enough.
Can you link to where they show that? I'm not particularly in the mood for looking through a whole site full of this stuff to see whether there's some nugget in there that actually makes sense...
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Re: Math: Fleeting Thoughts
phlip wrote:Moole wrote:And they manage to show a sum of a scalar and bivector means something (a combination of a rotation & scaling)  which is reasonable enough.
Can you link to where they show that? I'm not particularly in the mood for looking through a whole site full of this stuff to see whether there's some nugget in there that actually makes sense...
The first bit of this page is relevant (especially when it talks about its idea of "rotors"). I suppose they don't outright state it, but from knowing that iv = vi is a rotation of v by a right angle, it's easy enough to see that that any value of the form (xi+y)v represents a rotation and scaling of v.
Mathematical hangover (n.): The feeling one gets in the morning when they realize that that short, elementary proof of the Riemann hypothesis that they came up with at midnight the night before is, in fact, nonsense.
Re: Math: Fleeting Thoughts
the Geometric Algebra site I pointed to is meant as an elementary introduction  there is more depth, history easily found by just quoting Geometric Algebra in search in that order
Some seem to object to "Geometric Algebra" being used as a name  but Grassman who invented the exterior product won the Leibniz prize for "an Algebra of Geometry", Clifford uses the same coinage in his writing on the applications of his Clifford Algebra, Hestenes is just honoring those historical precedents  which predate "Algebraic Geometry" as a named mathematics
for just the 2D plane the isomorphism with complex numbers may not be a big insight, the algebraic definition of the product of the Bivector, Complex Numbers/Rotors with Vectors is a step that does extend to higher dimension
the step up to 3D with 3 orthogonal Bivectors now isomorphic with Hamilton's Quaternions' Imagnary parts giving 3D Rotors, "unifying" Vector and Quaternion Algebras may be a bigger win
the "Geometric Product" is useful in solving equations that Gibbs "Vector Algebra" with its anticommutative, nonAssociative Cross Product has difficulty with (and Gibbs 3D specific Cross Product "trick" is explained with Duality in the 3D "Geometric Algebra")
a solution pattern if starting from a Gibbs based formula is to replace Cross Products with exterior product Duals, various "Geometric Algebra" identities, operations ~ "completing the Geometric Product", using its Associativity, inverses of Vectors, Rotors can give solutions to problems requiring "cracking open" the Cross Product
Clifford was early exploring mixed metrics, Hestenes "recovery" of the "Geometric Algebra" was for application in Special Relativity, Spacetime mixed metric
Some seem to object to "Geometric Algebra" being used as a name  but Grassman who invented the exterior product won the Leibniz prize for "an Algebra of Geometry", Clifford uses the same coinage in his writing on the applications of his Clifford Algebra, Hestenes is just honoring those historical precedents  which predate "Algebraic Geometry" as a named mathematics
for just the 2D plane the isomorphism with complex numbers may not be a big insight, the algebraic definition of the product of the Bivector, Complex Numbers/Rotors with Vectors is a step that does extend to higher dimension
the step up to 3D with 3 orthogonal Bivectors now isomorphic with Hamilton's Quaternions' Imagnary parts giving 3D Rotors, "unifying" Vector and Quaternion Algebras may be a bigger win
the "Geometric Product" is useful in solving equations that Gibbs "Vector Algebra" with its anticommutative, nonAssociative Cross Product has difficulty with (and Gibbs 3D specific Cross Product "trick" is explained with Duality in the 3D "Geometric Algebra")
a solution pattern if starting from a Gibbs based formula is to replace Cross Products with exterior product Duals, various "Geometric Algebra" identities, operations ~ "completing the Geometric Product", using its Associativity, inverses of Vectors, Rotors can give solutions to problems requiring "cracking open" the Cross Product
Clifford was early exploring mixed metrics, Hestenes "recovery" of the "Geometric Algebra" was for application in Special Relativity, Spacetime mixed metric
 doogly
 Dr. The Juggernaut of Touching Himself
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Re: Math: Fleeting Thoughts
Yeah, the GA program is just a sort of anti Goldilocks. It seems a bit too sophisticated for the high school level intros, and then once it's serious math time you are much better off with differential forms.
LE4dGOLEM: What's a Doug?
Noc: A larval Doogly. They grow the tail and stinger upon reaching adulthood.
Keep waggling your butt brows Brothers.
Or; Is that your eye butthairs?
Noc: A larval Doogly. They grow the tail and stinger upon reaching adulthood.
Keep waggling your butt brows Brothers.
Or; Is that your eye butthairs?
Re: Math: Fleeting Thoughts
Here are a couple of cute things I encountered earlier this year, but I forgot to put them here.
Belphegor's prime, aka the Devil's prime:
1000000000000066600000000000001. Note that there are 13 zeroes either side of the 666, and the total number of digits in Belphegor's prime is 31, which is 13 reversed. I'm not usually interested in basedependant properties of numbers, considering them somewhat numerological, but that one's just too cute.
...
Schizophrenic number
Belphegor's prime, aka the Devil's prime:
1000000000000066600000000000001. Note that there are 13 zeroes either side of the 666, and the total number of digits in Belphegor's prime is 31, which is 13 reversed. I'm not usually interested in basedependant properties of numbers, considering them somewhat numerological, but that one's just too cute.
...
Schizophrenic number
A schizophrenic number (also known as mock rational number) is an irrational number that displays certain characteristics of rational numbers.
The definition of schizophrenic numbers is given in The Universal Book of Mathematics as:[1]
An informal name for an irrational number that displays such persistent patterns in its decimal expansion, that it has the appearance of a rational number. A schizophrenic number can be obtained as follows. For any positive integer n let f(n) denote the integer given by the recurrence f(n) = 10 f(n − 1) + n with the initial value f(0) = 0. Thus, f(1) = 1, f(2) = 12, f(3) = 123, and so on. The square roots of f(n) for odd integers n give rise to a curious mixture appearing to be rational for periods, and then disintegrating into irrationality. This is illustrated by the first 500 digits of √f(49):
1111111111111111111111111.1111111111111111111111 0860
555555555555555555555555555555555555555555555 2730541
66666666666666666666666666666666666666666 0296260347
2222222222222222222222222222222222222 0426563940928819
4444444444444444444444444444444 38775551250401171874
9999999999999999999999999999 808249687711486305338541
66666666666666666666666 5987185738621440638655598958
33333333333333333333 0843460407627608206940277099609374
99999999999999 0642227587555983066639430321587456597
222222222 1863492016791180833081844 . . . .
The repeating strings become progressively shorter and the scrabbled strings become larger until eventually the repeating strings disappear. However, by increasing n we can forestall the disappearance of the repeating strings as long as we like. The repeating digits are always 1, 5, 6, 2, 4, 9, 6, 3, 9, 2, . . . .
[...]
 jestingrabbit
 Factoids are just Datas that haven't grown up yet
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Re: Math: Fleeting Thoughts
If you have a sphere, then you can do a stereographic projection in the usual way. Equivalently, you can place the sphere at the focus of a reflective paraboloid of revolution, project it, bounce it off the paraboloid, and catch the light at a plane at infinity... if you see what I mean.
ameretrifle wrote:Magic space feudalism is therefore a viable idea.
 jestingrabbit
 Factoids are just Datas that haven't grown up yet
 Posts: 5967
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 Location: Sydney
Re: Math: Fleeting Thoughts
A rewrite of Euclid's elements from 1847 where colour is used to indicate variable names.
https://openlibrary.org/books/OL2460864 ... _of_Euclid
https://openlibrary.org/books/OL2460864 ... _of_Euclid
ameretrifle wrote:Magic space feudalism is therefore a viable idea.
Re: Math: Fleeting Thoughts
cyanyoshi wrote:I kind of stumbled on an interesting property of analytic functions the other day, and I wondered if it had a proper name. Provided that the series converges,
f(x) + f'(x)*h/1! +f''(x)*h^{2}/2! + f'''(x)*h^{3}/3! + ... = f(x+h)
or alternatively, e^{hD}f = f(x+h)
where h is a constant and D is the differential operator, and D^{n}f is interpreted as the nth derivative of the function f evaluated at x.
As someone who has never taken a course on real or complex analysis (other than vanilla calculus), I just thought this was really cool to discover on my own!
Regarding this:
I distinctly remember seeing this a few days ago being called the <name>ian Identity, and I am somewhat sure it was Lagrange, but Lagrangian Identity seems to return something else. Ah well.

So in circuits, connecting three (n) resistors in series has the equivalent resistance of the sum of the resistances;
R_{series} = R_{1} + R_{2} + R_{3} = h_{1}(R_{1}, R_{2}, R_{3})/h_{0},
where h_{i} is the Complete Homogeneous Symmetric Polynomial of order i.
Similarly, connecting three (n) resistors in parallel has the resistance
R_{parallel} = h_{n}/h_{n1}.
Are there nice, "symmetrical" arrangements of resistors that have other hratios?
That is, is there a way of arranging three resistors such that the equivalent resistance is
h_{2}/h_{1} = (ab + bc + ca)/(a+b+c)?
 doogly
 Dr. The Juggernaut of Touching Himself
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 Location: Lexington, MA
 Contact:
Re: Math: Fleeting Thoughts
Maybe Hamilton? In quantum mechanics, the differential operator you use is the Hamiltonian, and that is how you get time evolution.
It is a big deal. Nice catch.
It is a big deal. Nice catch.
LE4dGOLEM: What's a Doug?
Noc: A larval Doogly. They grow the tail and stinger upon reaching adulthood.
Keep waggling your butt brows Brothers.
Or; Is that your eye butthairs?
Noc: A larval Doogly. They grow the tail and stinger upon reaching adulthood.
Keep waggling your butt brows Brothers.
Or; Is that your eye butthairs?
Re: Math: Fleeting Thoughts
Elmach wrote:cyanyoshi wrote:I kind of stumbled on an interesting property of analytic functions the other day, and I wondered if it had a proper name. Provided that the series converges,
f(x) + f'(x)*h/1! +f''(x)*h^{2}/2! + f'''(x)*h^{3}/3! + ... = f(x+h)
or alternatively, e^{hD}f = f(x+h)
where h is a constant and D is the differential operator, and D^{n}f is interpreted as the nth derivative of the function f evaluated at x.
As someone who has never taken a course on real or complex analysis (other than vanilla calculus), I just thought this was really cool to discover on my own!
Regarding this:
I distinctly remember seeing this a few days ago being called the <name>ian Identity, and I am somewhat sure it was Lagrange, but Lagrangian Identity seems to return something else. Ah well.
If I understand you correctly, you are refering to the Taylor series.
Be warned, though, there are functions for which the Taylor series does not equal the original function.
[edit] If only I had scrolled upwards before posting: it was identified/discussed in May already...
Re: Math: Fleeting Thoughts
Let me rephrase this:
I distinctly remember the e^{tD} = S_{t} (where S is the shift operator) being referenced as the <name>ian identity (or representation? or some other word... I am slowly forgetting.)
Ah well, I guess it doesn't really matter.
I distinctly remember the e^{tD} = S_{t} (where S is the shift operator) being referenced as the <name>ian identity (or representation? or some other word... I am slowly forgetting.)
Ah well, I guess it doesn't really matter.

 Posts: 154
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Re: Math: Fleeting Thoughts
I think you're talking about a Bernoulli flow. They're covered by Ornstein's Theorem.
 jestingrabbit
 Factoids are just Datas that haven't grown up yet
 Posts: 5967
 Joined: Tue Nov 28, 2006 9:50 pm UTC
 Location: Sydney
Re: Math: Fleeting Thoughts
Does anyone know of a game like this one
http://sciencevsmagic.net/geo/
where the progress is more modular ie one gets little buttons that do previous tasks (like bisect, or form a perpendicular etc) that ends with drawing a line tangent to two circles.
http://sciencevsmagic.net/geo/
where the progress is more modular ie one gets little buttons that do previous tasks (like bisect, or form a perpendicular etc) that ends with drawing a line tangent to two circles.
ameretrifle wrote:Magic space feudalism is therefore a viable idea.
Re: Math: Fleeting Thoughts
I think you're talking about Euclid: The Game.
Xenomortis wrote:O(n^{2}) takes on new meaning when trying to find pairs of socks in the morning.
 jestingrabbit
 Factoids are just Datas that haven't grown up yet
 Posts: 5967
 Joined: Tue Nov 28, 2006 9:50 pm UTC
 Location: Sydney
Re: Math: Fleeting Thoughts
You are correct. Thank you so much!
eta: its been spruced up a lot since I was last there, much better than it was.
eta: its been spruced up a lot since I was last there, much better than it was.
ameretrifle wrote:Magic space feudalism is therefore a viable idea.
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