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### Re: Math: Fleeting Thoughts

Posted: Wed Mar 02, 2016 11:35 pm UTC
Also, geometric algebra is gross.

I wouldn't try to sell it to mathematicians but Hestenes reboot of Clifford's, Grassman's ideas for physics, engineering math extends how far can be gotten in one system before Tensors are needed

Gibbs/Heaviside Vector Algebra is pretty lame in that context, even if a practical improvement over Hamilton's Quaternions which Maxwell rewrote his initial "Cartesian Triples" ~20 equations E&M in

Gibbs himself introduced Dyadics less than a decade after "Vector Algebra", wrote about "multiple algebras", late in life endorsed Grassman/Clifford ideas as the better approach

### Re: Math: Fleeting Thoughts

Posted: Thu Mar 03, 2016 11:39 pm UTC
After 6 years of college, I finally learned what a "group" is. Engineering courses ftw?

### Re: Math: Fleeting Thoughts

Posted: Thu Mar 03, 2016 11:54 pm UTC
And then, in your fourth class with specific title "Algebra," (5th or 6th if you count the HS one, which you shouldn't), you may learn what "an algebra" is.

### Re: Math: Fleeting Thoughts

Posted: Fri Mar 04, 2016 12:25 am UTC
If I've learnt anything, it's that algebra is the tactical mathematical breaking and setting of bones.

### Re: Math: Fleeting Thoughts

Posted: Mon Apr 18, 2016 5:31 pm UTC
Need help proving for non-int n:

Where;
ζ is taken as principle value (i.e. ζ(1):=γ)
H is generalised harmonic number
α is positive real
And at the moment n is positive int

E: the n-1 is order not exponent

### Re: Math: Fleeting Thoughts

Posted: Tue Apr 19, 2016 4:35 am UTC
What progress have you made so far? What problems are you having, where are you stuck?

### Re: Math: Fleeting Thoughts

Posted: Tue Apr 19, 2016 4:56 am UTC
doogly wrote:And then, in your fourth class with specific title "Algebra," (5th or 6th if you count the HS one, which you shouldn't), you may learn what "an algebra" is.

You called it. I was finally formally taught what an algebra is, in the context of Lie algebras.

### Re: Math: Fleeting Thoughts

Posted: Tue Apr 19, 2016 8:51 am UTC
http://abstrusegoose.com/394

I laughed so hard....

### Re: Math: Fleeting Thoughts

Posted: Wed May 11, 2016 1:49 am UTC
I was investigating rigid body motion in more than 3 spatial dimensions, and it yielded some very cool results. Just letting go and finally treating bivectors as bivectors is simply a wonderful experience. The moment of inertia tensor is ugly and stupid, and I'm glad that there's a prettier way to calculate these things.

Living in a three-dimensional world is sad because we have all these bad habits when it comes to thinking about rotations and stuff. Four-dimensional rotations are so cool. ### Re: Math: Fleeting Thoughts

Posted: Tue May 31, 2016 9:29 pm UTC
Is it valid to say that a set which is not open must be closed? Or is there some third possibility?

### Re: Math: Fleeting Thoughts

Posted: Tue May 31, 2016 9:54 pm UTC
There is some third possibility. For example, the set [0,1) is neither open nor closed in the real line with the standard topology.

### Re: Math: Fleeting Thoughts

Posted: Tue May 31, 2016 10:16 pm UTC
Oh, duh. Thank you.

### Re: Math: Fleeting Thoughts

Posted: Tue May 31, 2016 11:11 pm UTC
There's also the possibility of being both - the empty set and the set X are both clopen in X under pretty much any topology, I believe.

### Re: Math: Fleeting Thoughts

Posted: Wed Jun 01, 2016 2:29 am UTC
The empty set and the entire set must be open by the definition of a topology, and since they are complementary, that means they also must be closed. In every topology.

But many topologies have other closed open sets too. And every non-discrete, nontrivial topological space also has subsets which are neither closed nor open.

### Re: Math: Fleeting Thoughts

Posted: Thu Jun 02, 2016 10:51 am UTC
Eebster the Great wrote:And every non-discrete, nontrivial topological space also has subsets which are neither closed nor open.

I guess it depends on what you mean by non-trivial, but this doesn't seem right. Take the space on two points, A and B, whose open sets are {}, {A}, and {A, B}. More generally, take the open sets to be an ultrafilter on the underlying space, and then toss in the empty set.

### Re: Math: Fleeting Thoughts

Posted: Fri Jun 03, 2016 12:04 am UTC
Eebster the Great wrote:And every non-discrete, nontrivial topological space also has subsets which are neither closed nor open.

Spaces with no sets that are neither closed nor open (that's a mouthful) are called door spaces. Cauchy described one type, and the link gives a complete classification.

### Re: Math: Fleeting Thoughts

Posted: Fri Jun 03, 2016 2:00 am UTC
Yep, I was totally wrong on that point. The class of spaces containing sets that are neither closed nor open is large.

By the way, MostlyHarmless, that link is only accessible to U Washington students.

### Re: Math: Fleeting Thoughts

Posted: Fri Jun 03, 2016 3:47 am UTC
Even as a UW student on campus, it didn't show up for me. I just assumed that it was the school's network trying to redirect me or something. How many of us are there!?! ### Re: Math: Fleeting Thoughts

Posted: Fri Jun 03, 2016 7:12 am UTC
Meteoric wrote:Is it valid to say that a set which is not open must be closed? Or is there some third possibility?

Yes, there is a third possibility. (Also see mouse-over text.)

### Re: Math: Fleeting Thoughts

Posted: Fri Jun 03, 2016 8:21 am UTC
Eebster the Great wrote:By the way, MostlyHarmless, that link is only accessible to U Washington students.

Hmm, I'm not sure how that happened. I was off campus when I found it, but the link won't work for me anymore either. This one should work better.

### Re: Math: Fleeting Thoughts

Posted: Fri Jun 03, 2016 5:44 pm UTC
Gotta say, "minimally door" is a pretty funny description. I guess that's like when your dog has grown big enough so he can just barely fit through the dog door.

### Re: Math: Fleeting Thoughts

Posted: Tue Jun 07, 2016 8:33 pm UTC
Another dumb question: I'm struggling to parse this sentence. "A polynomial with integer coefficients has a root in Z_p if and only if it has an integer root modulo p^k for any k >= 1".

I get what it means to say two numbers are congruent mod n; it's a relationship between two numbers. I'm not sure what it means to "have a root mod n"; what number are we relating the root to?

P.S. The p-adics are delightfully weird. Square roots are integers! Except some numbers don't have square roots at all!

### Re: Math: Fleeting Thoughts

Posted: Tue Jun 07, 2016 9:17 pm UTC
Meteoric wrote:Another dumb question: I'm struggling to parse this sentence. "A polynomial with integer coefficients has a root in Z_p if and only if it has an integer root modulo p^k for any k >= 1".

I get what it means to say two numbers are congruent mod n; it's a relationship between two numbers. I'm not sure what it means to "have a root mod n"; what number are we relating the root to?

A polynomial f(x) has a root of a, mod n, when f(a) is congruent to zero mod n.

### Re: Math: Fleeting Thoughts

Posted: Wed Jun 08, 2016 3:11 am UTC
Oh, so it's an extended meaning of the term "root"! Thank you, that makes sense now.

### Re: Math: Fleeting Thoughts

Posted: Wed Jun 08, 2016 3:54 am UTC
While we're on meanings of the term "root", is there some link in meaning between square root and root of a polynomial? Or did we just end up using those terms by happenstance?

### Re: Math: Fleeting Thoughts

Posted: Wed Jun 08, 2016 3:57 am UTC
I don't know if this is the original etymology, but for example sqrt(6) is a root of the polynomial x^2 - 6.

### Re: Math: Fleeting Thoughts

Posted: Wed Jun 08, 2016 4:05 am UTC
Oh, of course, thanks! I was trying to work out whether there was something like that but got stuck in circles that kept coming back to "well obviously, that's because sqrt(x) is the inverse of x^2".
EDIT: Which was not helping me make the link.

### Re: Math: Fleeting Thoughts

Posted: Tue Jun 14, 2016 4:14 am UTC
Meteoric wrote:Oh, so it's an extended meaning of the term "root"! Thank you, that makes sense now.

I'd think of it more as shorthand for "a root of f when considering it as a polynomial over the integers mod n" than any sort of different use of the word root.

### Re: Math: Fleeting Thoughts

Posted: Sun Aug 07, 2016 1:25 am UTC
Given 3D eucledian space with n line segments of unit length that adjoin perpendicularly, can you make a loop? (like making a chain of 6-sided sodium atoms and rods with a molecular model set)

n<3 doesn't count; n=3 doesn't work because you're always √2 away from the starting point after two segments (and the plane of possible third segments doesn't intersect the starting point).
For n=4 this is simply a square. And for every larger even number n=2k>4 you can just pick two opposing sides of the square and insert line segments in between.
But does it also work for odd numbers n=2k+1>4? My gut feeling says no for at least n=5, but I can't prove it (at this time).

Also, does adding dimensions have any effect?

### Re: Math: Fleeting Thoughts

Posted: Sun Aug 07, 2016 5:40 am UTC
Flumble wrote:But does it also work for odd numbers n=2k+1>4? My gut feeling says no for at least n=5, but I can't prove it (at this time).

Also, does adding dimensions have any effect?

For the 3D case, the segments are unit vectors in the X, Y, and Z directions, and the oppositely-directed unit vectors -X, -Y, -Z. Because X, Y & Z are mutually perpendicular no combination of {X, -X, Y, -Y} sums to a scalar multiple of Z; similarly, you can't build a scalar multiple of Y just using scalar multiples of Xs & Zs, or a scalar multiple of X just using Ys & Zs. So if a path built from elements of {X, -X, Y, -Y, Z, -Z} returns to its starting point the number of X segments must be equal to the number of -X segments, and the same goes for Y & -Y and Z & -Z. So the total number of X & -X segments must be even, similarly for the Y & -Y and Z & -Z. And so the total number of all segments must be even.

### Re: Math: Fleeting Thoughts

Posted: Sun Aug 07, 2016 7:09 am UTC
PM 2Ring wrote:For the 3D case, the segments are unit vectors in the X, Y, and Z directions, and the oppositely-directed unit vectors -X, -Y, -Z.

I think the OP is asking for the more general case where the segments are not necessarily axis-aligned.

With 7 segments you'll get something more or less like this puzzle:

Clearly this particular arrangement is not a solution since the front vertex does not have a right angle between its segments.

I'm sure n=5 is not possible, and fairly sure that when you get to n=9 there are enough degrees of freedom to make a loop. I don't know about n=7.

Edit:
There are solutions for n=7. Here is a symmetric one, where the vertices are at the coordinates:
(1, 0, -2)
(1, 0, 0)
(1.4142, 1.9566, 0)
(0, 2.2560, 1.3822)
(-1.4142, 1.9566, 0)
(-1, 0, 0)
(-1, 0, -2)

BTW, I used segment length 2, to keep the numbers nicer.
The exact expression for those coordinates are:
1.4142 ~ sqrt(2)
1.9566 ~ sqrt(1 + 2sqrt(2))
2.2560 ~ (3 + sqrt(2)) / sqrt(1 + 2sqrt(2))
1.3822 ~ 2(2sqrt(2) - 1) / sqrt(7)

You can easily extend this to odd n>7 by adding onto the two parallel segments that form opposite sides of a square.

### Re: Math: Fleeting Thoughts

Posted: Sun Aug 07, 2016 12:45 pm UTC
jaap wrote:
PM 2Ring wrote:For the 3D case, the segments are unit vectors in the X, Y, and Z directions, and the oppositely-directed unit vectors -X, -Y, -Z.

I think the OP is asking for the more general case where the segments are not necessarily axis-aligned.

indeed (wait, am I "OP"? after 9 pages?)

Thanks for the n>=7 case, jaap. (I was doubtful whether n=7 even has enough degrees of freedom, or indeed whether any odd number were possible, as making it fit in practice/head doesn't entail that it has right angles in theory)

Now how to go about disproving the n=5 case?

### Re: Math: Fleeting Thoughts

Posted: Sun Aug 07, 2016 1:09 pm UTC
Flumble wrote:Now how to go about disproving the n=5 case?

Non-adjacent vertices must be a distance of sqrt(2) apart. So what you have is essentially this pentagon: right-pentagon.png (3.7 KiB) Viewed 11462 times

The three triangles in this pentagon all have known side lengths, so the only degrees of freedom left are the angle between the triangles, i.e. how far to fold along the two thin lines of the diagram. You can fold ABC up/down until angle C is 90 degrees, and similarly with triangle ADE to make D a right angle. You now have no freedom any more to make A a right angle, and it really would be very lucky if it were. If you fold both triangles upwards, or both downwards, then BCDE will lie in a plane, distance BE will be 1, and so A will be 60 degrees. I have not (yet) checked what angle A becomes when you fold the triangles in opposite directions, and it does look like it could be close to 90 degrees. (Edit: it's 110.9 degrees if I did it right)

### Re: Math: Fleeting Thoughts

Posted: Sun Aug 07, 2016 2:00 pm UTC
Flumble wrote:
jaap wrote:
PM 2Ring wrote:For the 3D case, the segments are unit vectors in the X, Y, and Z directions, and the oppositely-directed unit vectors -X, -Y, -Z.

I think the OP is asking for the more general case where the segments are not necessarily axis-aligned.

indeed (wait, am I "OP"? after 9 pages?)

Ah, right. That's a much more interesting problem. TBH, I was a bit surprised when I read the question, since I know you're no mathematics newbie. Now I feel a bit silly. Oh well, brain farts can happen to anyone. ### Re: Math: Fleeting Thoughts

Posted: Sun Aug 07, 2016 3:55 pm UTC
Thanks jaap, that proof is both easy to understand and rigorous (...enough to me; maybe a math professor will ask for an argument why rotating B over AC only has two angles for which BC⟂CD, but we know it's true). Now dare I ask: is it possible to make a 5-chain in 4D or higher? (I promise I won't ask about properties of these constructions in non-Eucledian space )

PM 2Ring wrote:Ah, right. No worries, I also know brain farts.

### Re: Math: Fleeting Thoughts

Posted: Sun Aug 07, 2016 7:01 pm UTC
Flumble wrote:Now dare I ask: is it possible to make a 5-chain in 4D or higher? (I promise I won't ask about properties of these constructions in non-Eucledian space :P )

Yes. Start with the same pentagon, with the two triangles folded in opposite directions, one in the positive z direction the other in the negative z direction. Now four of the angles are 90 degrees, only the angle at A is larger than that. You can fold the two triangles into the 4th dimension, to reduce the z-coordinates enough to make A a right angle.

What you get are the 5 vertices:

(-1, 0, 0 ,0)
(-1, 4/sqrt(7), 1, sqrt(5)/sqrt(7) )
(0, sqrt(7), 0 ,0)
(1, 4/sqrt(7), -1, sqrt(5)/sqrt(7) )
(1, 0, 0 ,0)

Again, I doubled the scale to simplify the numbers.
You can easily check that the distance between adjacent vertices is 2, and between non-adjacent vertices is 2sqrt(2).

### Re: Math: Fleeting Thoughts

Posted: Wed Aug 10, 2016 2:22 pm UTC
Well done, jaap!

Flumble wrote:No worries, I also know brain farts. Thanks for reminding me of the Goat Problem; I've just added some diagrams to that thread.

### Re: Math: Fleeting Thoughts

Posted: Thu Aug 11, 2016 4:05 pm UTC
Flumble wrote:(I promise I won't ask about properties of these constructions in non-Eucledian space )

Oh, in hyperbolic space it's a piece of cake. Remember that in non-Euclidean geometries the angle sum of a triangle isn't constant but instead is a linear function of the area. In hyperbolic space the sum is always less than pi; you can make the angle as small as you like by making the triangle large enough. Extending this to pentagons, we can therefore make a regular pentagon in hyperbolic space with each vertex angle being a right angle.

Here's a Poincaré disk representation of a hyperbolic tessellation of congruent regular pentagons, from Wikipedia. The angles in these pentagons are all right angles since 4 pentagons meet at a vertex.

Also see http://euler.slu.edu/escher/index.php/H ... y#Examples

### Re: Math: Fleeting Thoughts

Posted: Tue Aug 30, 2016 12:28 am UTC
Is there a name for the curve defined by the trajectory of a projectile experiencing gravity and quadratic drag? I was trying to come up with a concise way to describe the difference in the paths a fastball and curveball take. The former doesn't follow a parabolic trajectory of course, due to drag, but it still drops less than a curveball.

### Re: Math: Fleeting Thoughts

Posted: Tue Aug 30, 2016 9:31 am UTC
I don't know any, but I can tell you that wikipedia states that for low velocities (actually, low Reynolds numbers) you can approximate the trajectory with linear drag (so you can solve things analytically):
https://en.wikipedia.org/wiki/Trajectory_of_a_projectile#Trajectory_of_a_projectile_with_air_resistance wrote:The assumption that air resistance may be taken to be in direct proportion to the velocity of the particle is not correct for a typical projectile in air with a velocity above a few tens of meters/second, and so this equation should not be applied to that situation.

How fast do you throw?