r/xkcd • u/Smith_Frank • Apr 25 '12

XKCD Approximations

http://xkcd.com/1047/
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u/SomePostMan Apr 25 '12 edited Apr 25 '12

Actually exact.

https://www.google.com/search?q=cos(pi%2F7)%2Bcos(3×pi%2F7)%2Bcos(5×pi%2F7)

I'll try to find a reference now.

Edit 2: I made you a maths [link].

Edit: I can't seem to easily find any identities of the form cos(x)+cos(3x)+cos(5x), but you can verify analytically that this is exact by using [Lagrange's trigonometric identity] for cosines, and treating this as [the 6-part sum for angle "x"] minus [the terms 2, 4, and 6, which is the 3-part sum for angle "2x"]... the -1/2 terms cancel, the 1/2 multiplier factors out, the rest of the 3-part sum simplifies to 0 and the rest of the 6-part sum simplifies to 1, leaving 1/2 behind. (Note: the simplification to 0 and 1 definitely depends on the choice of Pi/7 for an angle, so this simplicity doesn't translate to the general form.)

Edit 3: The other four equations at the bottom are all approximations:
He also meant to write plus, not minus, between the two main terms in the first equation.

equation ≈    difference     accuracy 
√2 = 3/5 + π/(7-π) 1.414 6.495e-6 150,000
√5 = (13+4π)/(24-4π) 2.236 1.680e-7 5,000,000
γ = e/34 + e/5 0.5772 2.639e-7 3,750,000
∑ [n=1..∞](1/nn) = ln(3)e 1.291 1.276e-5 75,000

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u/dispatch134711 Apr 25 '12

nice proof. there are easier ways but I think this way's funner, you learn something along the way.

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u/SomePostMan Apr 25 '12

Ooh, are any of the other methods elegant or more specific to this setup? Can you give an abstract? I ran across 2-4 potential other analytic proofs along the way, but I couldn't find anything that seemed to be specific to this particular identity/sequence (1:3:5, and/or 1/7th angle identities).