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August 11

Tricky complex number / infinite series problem

Given $z={\frac {1}{2}}(\cos \theta +i\sin \theta )$ show that the imaginary part of the infinite series $1+z+z^{2}+z^{3}+...$ is ${\frac {2\sin \theta }{5-4\cos \theta }}$
Being an infinite geometric series of common ration ${\frac {e^{i\theta }}{2}}$ I expect I should evaluate $\Im ({\frac {1}{1-{\frac {e^{i\theta }}{2}}}})$ but no avail so far.--220.253.98.30 (talk) 08:52, 11 August 2010 (UTC)[reply]

Here's a start:

{\frac {1}{1-{\frac {e^{i\theta }}{2}}}}={\frac {2}{2-e^{i\theta }}}={\frac {2(2-e^{-i\theta })}{(2-e^{i\theta })(2-e^{-i\theta })}}={\frac {4-2e^{-i\theta }}{5-2(e^{i\theta }+e^{-i\theta })}}

I'll let you take it from there. Gandalf61 (talk) 09:01, 11 August 2010 (UTC)[reply]

It is simpler to keep z as

{\frac {1}{2}}(\cos \theta +i\sin \theta )

. Apply

{\frac {1}{x}}={\frac {\bar {x}}{|x|^{2}}}

to

{\frac {1}{1-z}}

.

Of course, the claim is true when θ is real, but not in general. -- Meni Rosenfeld (talk) 09:10, 11 August 2010 (UTC)[reply]

Duality (order theory)

Resolved

In the article Duality (order theory) it says that in the dual poset, meet becomes join and join becomes meet. While this seems intutively okay, what is a mathematical proof of this fact. Thanks-Shahab (talk) 16:50, 11 August 2010 (UTC)[reply]

Take the definition of a meet expressed in terms of ≤, replace all ≤ with ≥, and see what you get.—Emil J. 17:26, 11 August 2010 (UTC)[reply]

Ah! Its obvious after you told me.-Shahab (talk) 18:02, 11 August 2010 (UTC)[reply]

Puzzle Cube solutions

Resolved

How many possible combinations are there of the pieces of a Happy Cube to form a six-piece cube? I'm guessing that there are 122,880 (20*16*12*8*4), as there are 20 places to put the second piece, 16 places to put the third, etc. If this is correct, what is a better way to express this? I've never been very good with permutations/combinations. Other questions I'm thinking about include how the answer changes if you consider each side of a piece as distinct and how to tell how many solutions exist. Mannerisky (talk) 20:56, 11 August 2010 (UTC)[reply]

Assuming that the outside vs inside face of each piece is already known, that appears correct. If the sense of the faces is not initially known, the there would be 122,880 × 2⁶ = 2 × 40 × 32 × 24 × 16 × 8 = 7,864,320 possible combinantions. That assumes that the position and rotation of the first piece can be ignored, but the inward/outward sense of the first and all remaining pieces is relevant. -- Tom N (tcncv) talk/contrib 03:44, 12 August 2010 (UTC)[reply]

Thanks Tom! That's what I was looking for. Mannerisky (talk) 18:18, 12 August 2010 (UTC)[reply]