Editing Talk:1047: Approximations

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:: This explanation covers 42 adequately, and would probably be made slightly worse if such information were added. The very widely known cultural reference is to Adams's interpretation, not Dodgson's original obsession. Adding it would be akin to introducing the MPLM into the explanation for the hijacking of Renaissance artists' names by the TMNT. I definitely concede that it does not cover 42 exhaustively, but I think it can be considered complete and in working order without such an addition. If it really irks you, be bold and add it! --[[User:Quicksilver|Quicksilver]] ([[User talk:Quicksilver|talk]]) 00:37, 30 August 2013 (UTC)
 
:: This explanation covers 42 adequately, and would probably be made slightly worse if such information were added. The very widely known cultural reference is to Adams's interpretation, not Dodgson's original obsession. Adding it would be akin to introducing the MPLM into the explanation for the hijacking of Renaissance artists' names by the TMNT. I definitely concede that it does not cover 42 exhaustively, but I think it can be considered complete and in working order without such an addition. If it really irks you, be bold and add it! --[[User:Quicksilver|Quicksilver]] ([[User talk:Quicksilver|talk]]) 00:37, 30 August 2013 (UTC)
 
::: Additionally, the Lewis Carroll idea is only one of several theories about where Douglas Adams got the number from. [[Special:Contributions/162.158.158.87|162.158.158.87]] 20:47, 28 November 2019 (UTC)
 
  
 
"sqrt(2) is not even algebraic in the quotient field of Z[pi]" is not correct.  Q is part of the quotient field of Z[pi] and sqrt(2) is algebraic of it.  The needed facts are that pi is not algebraic, but the formula implies it is in Q(sqrt(2)).  --DrMath 06:47, 7 September 2013 (UTC)
 
"sqrt(2) is not even algebraic in the quotient field of Z[pi]" is not correct.  Q is part of the quotient field of Z[pi] and sqrt(2) is algebraic of it.  The needed facts are that pi is not algebraic, but the formula implies it is in Q(sqrt(2)).  --DrMath 06:47, 7 September 2013 (UTC)
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I'd add pi = (9^2 + (19^2)/22)^(1/4) [[Special:Contributions/198.41.230.73|198.41.230.73]] 02:41, 13 May 2015 (UTC)
 
I'd add pi = (9^2 + (19^2)/22)^(1/4) [[Special:Contributions/198.41.230.73|198.41.230.73]] 02:41, 13 May 2015 (UTC)
 
'''Yet another proof of cos(π/7) + cos(3π/7) + cos(5π/7) = 1/2''' — Use the multi-angle formula cos(7θ) = 64(cos θ)^7 − 112(cos θ)^5 + 56(cos θ)^3 − 7(cos θ),
 
and assume cos(7θ)=−1; then 7θ=π, 3π, 5π, 7π, etc.
 
Let x=cos θ, then x = cos(π/7), cos(3π/7), cos(5π/7), cos(7π/7), etc.<br />
 
Now one could actually solve 64x^7 − 112x^5 + 56x^3 − 7x + 1 = (x+1)(8x^3 − 4x^2 − 4x + 1)^2 = 0,
 
but it’s easier to argue that cos(π/7), cos(3π/7), cos(5π/7) are the 3 roots of the cubic equation 8x^3 − 4x^2 − 4x + 1,
 
and so (using the relationship of the roots and the coefficients) their sum is −(−4)/8 = 1/2.
 
[[User:Yosei|Yosei]] ([[User talk:Yosei|talk]]) 08:19, 17 February 2019 (UTC)
 

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