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comment 1, 5, 9, 13, 17, 21,…
@JoshuaTaylor: That's the best.
2d
comment 1, 5, 9, 13, 17, 21,…
Why not flip it around? $\{n | n=4k-3 \; \forall k \in \mathbb{N}\}$. I think this is more straightforward. Or a little more simply, $\{n | n=4k-3, \; k \in \mathbb{N}\}$.
Aug
28
comment Show that from a group of seven people whose (integer) ages add up to 332 one can select three people with the total age at least 142.
@HagenvonEitzen: That's true, greater rigor would mean a little more complexity. However, I see this as one of those things that is intuitively true and simple, which also holds up under deeper analysis.
Aug
9
comment On the hyperboloid model, if the point $\mathbf{v}$ gets translated to the origin, then where does the point $\mathbf{x}$ go?
Finally solved it! I contest your claim that it is "an easy solution"! Took me several hours. But I did it! So, again, thank you very much for your input. Turned out to be just what I needed. :)
Aug
8
comment On the hyperboloid model, if the point $\mathbf{v}$ gets translated to the origin, then where does the point $\mathbf{x}$ go?
Had a key insight: $M\left[ \begin{array}{c}1\\0\\0\end{array} \right] = -\vec{v}$ and I've very nearly got the solution. Just need to find a bug or two.
Aug
8
comment On the hyperboloid model, if the point $\mathbf{v}$ gets translated to the origin, then where does the point $\mathbf{x}$ go?
After working on this for a couple hours, I challenge you to find a complete solution.
Aug
7
comment On the hyperboloid model, if the point $\mathbf{v}$ gets translated to the origin, then where does the point $\mathbf{x}$ go?
I thought about it more, and I realized that if I can get the last three equations to have the same three variables each, then that immediately leads to the full solution. Which is slightly less daunting than substitutions all the way down. Thanks for your answer! It's really helped me actually tackle my problem. :)
Aug
6
comment On the hyperboloid model, if the point $\mathbf{v}$ gets translated to the origin, then where does the point $\mathbf{x}$ go?
That was the first thing I did: get $a=\sqrt{1+d^2+g^2}$ and likewise for the next two (which are actually equal to $-1$, not $1$). But then the next equations get messy fast. $ab-de-gh=0 \Rightarrow (1+d^2+g^2)(-1+e^2+h^2) = d^2e^2 + 2degh + g^2h^2$. I can make this a quadratic in $d$ to eliminate the fourth variable, but I still have five more variables to go and I already have a really messy equation here. Doing the same for the fifth equation allows me to eliminate $g$, I think. Still need to eliminate $e,f,h,i$. It's horrible.
Aug
5
comment On the hyperboloid model, if the point $\mathbf{v}$ gets translated to the origin, then where does the point $\mathbf{x}$ go?
I let $M = \left[ \begin{array}{ccc}a&b&c \\ d&e&f \\ g&h&i \end{array} \right]$ and $\vec{v} = \left[ \begin{array}{c}x_1 \\ x_2 \\ x_3 \end{array} \right]$ and got a system of nine non-linear equations. $$\begin{align*}a^2-d^2-g^2 &= 1 \\ b^2-e^2-h^2 &= 1 \\ c^2-f^2-i^2 &= 1 \\ ab-de-gh &= 0 \\ ac-df-gi &= 0 \\ bc-ef-hi &= 0 \\ ax_1+bx_2+cx_3 &= 1 \\ dx_1+ex_2+fx_3 &= 0 \\ gx_1+hx_2+ix_3 &= 0 \end{align*}$$ I'm sure this is solvable, but I'd like to be able to solve it without the help of a software package. I started doing some substitutions, but it quickly gets very messy.
Aug
4
comment On the hyperboloid model, if the point $\mathbf{v}$ gets translated to the origin, then where does the point $\mathbf{x}$ go?
@LeeMosher: I'm having difficulty even figuring out what $SO(2,1)$ is, and I only vaguely understand Lorentz transformations at best. The paucity of examples is being a real problem too. I learn best when given an example, the general case, and then another example.
Jul
26
comment Why does the Pythagorean Theorem have its simple form only in Euclidean geometry?
I quite like this explanation. I wish I could give another +1 for teaching me that $\cos(ix) = \cosh(x)$; that's a very intriguing and fascinating link between the elliptical and hyperbolic geometries.
Apr
26
comment What is this pattern called?
I love that you posted this! I stumbled across this same pattern years ago but didn't think to ask about it online. (Then again, that might have been before I knew about Math.SE.) So, thanks for doing this! :)
Mar
16
comment Group Permutations Proof
@CoolNewFriends: You can't use the statement you're trying to prove as part of its own proof.
Jan
8
comment How can I prove whether a $9\times 9$ square can be filled with L-shaped pieces in a completely “regular” way?
@coffeemath: Mmm...it's an L-shape that's three times as large. I count 9 L-pieces, thus 27 squares, which can and does form an L-shape where each "square" is composed of 9 smaller squares.
Jan
8
comment How can I prove whether a $9\times 9$ square can be filled with L-shaped pieces in a completely “regular” way?
+1 for a simpler way to state my thought process and thus simplify the (false) proof.
Jan
8
comment How can I prove whether a $9\times 9$ square can be filled with L-shaped pieces in a completely “regular” way?
Aaand there's the counterexample! Right under my nose!
Nov
29
comment What would a Steiner tree look like for the vertices of a heptagon?
Darn! I was expecting something much more interesting. Maybe I'll make it an irregular heptagon. Nonetheless, thank you for answering my question! :)
Jul
8
comment Subgroups of Sufficiently Large Symmetric Groups / Cayley's Theorem explanation
Ah! That's just the sort of intuition I was looking for! Much appreciated!
Jul
8
comment Subgroups of Sufficiently Large Symmetric Groups / Cayley's Theorem explanation
@lhf: Basically, too much jargon and not enough intuition. I am new to the area of groups so I haven't gotten used to all the jargon. mesel's answer is pretty much exactly what I was looking for.
Jul
8
comment Subgroups of Sufficiently Large Symmetric Groups / Cayley's Theorem explanation
@David: Cayley's Theorem is exactly what I was suspecting.