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 and use induction.


{$$R = \left( 1+{1 \over n}\right) ^n = \sum_{k=0}^n {n \choose k} \left( {1 \over n} \right) ^k$$}

Every term of the binomial expansion is positive: the coefficient must be positive and since n is positive, 1/n is positive for all powers.

For {$n=1, R_n \lt R_{n+1}, \text{ to wit:}$}

{\begin{align} 2 &\lt {9 \over 4} \cr 1 + 1 &\lt 1 + 1 + {1 \over 4} \cr 1({1 \over 1})^0 + 1({1 \over 1})^1 &\lt 1({1 \over 2})^0 + 2({1 \over 2})^1 + 1({1 \over 2})^2 \cr {1 \choose 0}({1 \over 1})^0 + {1 \choose 1}({1 \over 1})^1&\lt {2 \choose 0}({1 \over 2})^0 + {2 \choose 1}({1 \over 2})^1 + {2 \choose 2}({1 \over 2})^2 \cr \sum_{k=0}^1 {1 \choose k} \left( {1 \over 1} \right) ^k &\lt \sum_{k=0}^2 {2 \choose k} \left( {1 \over 2} \right) ^k \cr \text{for } n=1 & \cr \sum_{k=0}^n {n \choose k} \left({1 \over n} \right) ^k &\lt \sum_{k=0}^{n+1} {{n+1} \choose k} \left( {1 \over {n+1}} \right) ^k \cr \left( 1+{1 \over n}\right) &\lt \left( 1+{1 \over {n+1}}\right) ^{n+1} \cr R_n &\lt R_{n+1}\end{align}}

Well, that was interesting, but pointless as it appears induction is not required at all. Recalling that we are doing the integer values for n > 0:

{\large \begin{align} (1-e^{i\theta} )^{-{1 \over 2}}+(1-e^{-i\theta} )^{-{1 \over 2}}&=(1+\csc(\theta/2) ^{1 \over 2} \cr {1 \over {\sqrt{1-e^{i\theta}}}}+ {1 \over {\sqrt{1-e^{-i\theta}}}} &= \cr { { \sqrt{1-e^{-i\theta}} +\sqrt{1-e^{i\theta}} } \over {\sqrt{1-e^{i\theta}}\sqrt{1-e^{-i\theta}}}} &= \cr { { \sqrt{1-e^{-i\theta}} +\sqrt{1-e^{i\theta}} } \over { \sqrt{ (1-e^{i\theta})(1-e^{-i\theta} ) } }} &= \cr { { \sqrt{1-e^{-i\theta}} +\sqrt{1-e^{i\theta}} } \over { \sqrt{ (1-\cos\theta-i\sin\theta)(1-\cos\theta + i\sin\theta ) } }} &= \cr { { \sqrt{1-e^{-i\theta}} +\sqrt{1-e^{i\theta}} } \over { \sqrt{ (2-2\cos\theta ) } }} &= \cr \left[ { { \sqrt{1-e^{-i\theta}} +\sqrt{1-e^{i\theta}} } \over { \sqrt{ (2-2\cos\theta ) } }} \right] ^2 &= 1+\csc{\theta \over 2} \cr { { (\sqrt{1-e^{-i\theta}} +\sqrt{1-e^{i\theta}})^2 } \over { 2-2\cos\theta }} &= \cr { { 1-e^{-i\theta} +2\sqrt{1-e^{-i\theta}}\sqrt{1-e^{i\theta}} + 1-e^{i\theta} } \over { 2-2\cos\theta }} &= \cr { { 2-2\cos\theta +2\sqrt{1-e^{-i\theta}}\sqrt{1-e^{i\theta}} } \over { 2-2\cos\theta }} &= \cr {{2-2\cos\theta } \over { 2-2\cos\theta }}+{ { 2\sqrt{1-e^{-i\theta}}\sqrt{1-e^{i\theta}} } \over { 2-2\cos\theta }} &= \cr 1 +{ { 2\sqrt{1-e^{-i\theta}}\sqrt{1-e^{i\theta}} } \over { 2-2\cos\theta }} &= 1+\csc{\theta \over 2}\cr { { 2\sqrt{1-e^{-i\theta}}\sqrt{1-e^{i\theta}} } \over { 2-2\cos\theta }} &= \csc{\theta \over 2}\end{align}}

{\large \begin{align} { { 2\sqrt{1-e^{-i\theta}}\sqrt{1-e^{i\theta}} } \over { 2-2\cos\theta }} &= \csc{\theta \over 2} \cr { { \sqrt{ (1-e^{-i\theta})(1-e^{i\theta}) } } \over { 1-\cos\theta }} &=\end{align}}

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This is a student's notebook. I am not responsible if you copy it for homework, and it turns out to be wrong.

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March 17, 2018

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