Ning Zhu edited Appendix.tex  over 8 years ago

Commit id: 7ae70c62c21241dfa4b34f738535d30dbd1898fb

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$$\Delta_{E_{n}}=E_{n}-E_{n-1}$$  $$\Delta_{E_{n}}=n(E_{a}+\delta_{n})-(n-1)(E_{a}+\delta_{n-1})$$  $$\Delta_{E_{n}}=n(E_{a}+n\frac{\lambda}{L} E_{a})-(n-1)(E_{a}+(n-1)\frac{\lambda}{L} E_{a})$$  $$\Delta_{E_{n}}=E_{a}\frac{\lambda}{L} (2n-1)+E_{a}$$ Therefore, when n=\frac{1}{2}:  $$\Delta_{E_{n}}=E_{a}$$