p********a
发帖数: 761 | | n***p
发帖数: 7668 | 2 sqrt(3).
【在 p********a 的大作中提到】
: 好像中学时做过，现在都忘了
: 谢了
| m**********e
发帖数: 1045 | | n***p
发帖数: 7668 | 4 其实是微积分里面的例题。
We need only consider S_n = sqrt( 2 + sqrt(2 + sqrt(2+ ...) ) )
with n level of sqrt and consider its limit as nto infty.
S_n is increasing as n increases, and Sn is bounded above (by 2). Then
S_n has a limit, say S.
Recursively S_n = sqrt(2 + S_{n-1}). Taking limit on both sides, we have
S = sqrt(2+S). Solving it we obtain S=2.
What LZ wants is just sqrt(1+S) = sqrt(3).
【在 m**********e 的大作中提到】
: 求详细解法
| z*********g
发帖数: 37 | 5 S=sqrt(2+....)
S*S=2+S
S=2
Sqrt(1+S)=sqrt(3) | G******i
发帖数: 163 | 6 \begin{array}{l}
2=\sqrt{2+2}\cr
2=\sqrt{2+\sqrt{2+2}}\cr
2=\sqrt{2+\sqrt{2+\sqrt{2+2}}}\cr
2=\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+2}}}}\cr
\vdots\cr
\end{array}
To see the equations, copy & paste the above to
http://www.codecogs.com/latex/eqneditor.php
【在 n***p 的大作中提到】
: 其实是微积分里面的例题。
: We need only consider S_n = sqrt( 2 + sqrt(2 + sqrt(2+ ...) ) )
: with n level of sqrt and consider its limit as nto infty.
: S_n is increasing as n increases, and Sn is bounded above (by 2). Then
: S_n has a limit, say S.
: Recursively S_n = sqrt(2 + S_{n-1}). Taking limit on both sides, we have
: S = sqrt(2+S). Solving it we obtain S=2.
: What LZ wants is just sqrt(1+S) = sqrt(3).
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