Μιγαδικό άθροισμα
Δημοσιεύτηκε: Δευ Αύγ 01, 2011 6:06 pm
Έστω 
και δίνεται ότι : 
και 
όπου 
με 
.
Να δείξετε ότι :
και δίνεται ότι : και όπου με .Να δείξετε ότι :
![\displaystyle{\begin{array}{l}
A = \left[ {\begin{array}{*{20}{c}}
a & b \\
d & c \\
\end{array}} \right] \\
\det (A - {\varphi ^k}{I_2}) = \det \left[ {\begin{array}{*{20}{c}}
{a - {\varphi ^k}} & b \\
d & {c - {\varphi ^k}} \\
\end{array}} \right] = ... = \det A - trA \cdot {\varphi ^k} + {\varphi ^{2k}} = {\varphi ^2} - {\varphi ^{k + 1}} + {\varphi ^{2k}} \\
\sum\limits_{k = 1}^n {{\varphi ^k}} \cdot\det \left( {A - {\varphi ^k}{I_2}} \right) = \sum\limits_{k = 1}^n {{\varphi ^k}} \cdot({\varphi ^2} - {\varphi ^{k + 1}} + {\varphi ^{2k}}) = {\varphi ^2}\sum\limits_{k = 1}^n {{\varphi ^k}} - \varphi \sum\limits_{k = 1}^n {{\varphi ^{2k}}} + \sum\limits_{k = 1}^n {{\varphi ^{3k}}} \\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\varphi ^2}\varphi \frac{{{\varphi ^n} - 1}}{{\varphi - 1}} - \varphi {\varphi ^2}\frac{{{\varphi ^2}^n - 1}}{{{\varphi ^2} - 1}} + {\varphi ^3}\frac{{{\varphi ^3}^n - 1}}{{{\varphi ^3} - 1}} = \,\,\mathop = \limits^{[{\varphi ^n} = {{\left( {\cos \left( {\frac{{2\pi }}{n}} \right) + i\sin \left( {\frac{{2\pi }}{n}\,\,\,} \right)\,} \right)}^n} = 1]\,\,\,\,\,\,} 0 \\
\end{array}}](/forum/ext/geomar/texintegr/latexrender/pictures/5172fa4594a4d61de05fc3de82f8fffc.png "\displaystyle{\begin{array}{l}
A = \left[ {\begin{array}{*{20}{c}}
a & b \\
d & c \\
\end{array}} \right] \\
\det (A - {\varphi ^k}{I_2}) = \det \left[ {\begin{array}{*{20}{c}}
{a - {\varphi ^k}} & b \\
d & {c - {\varphi ^k}} \\
\end{array}} \right] = ... = \det A - trA \cdot {\varphi ^k} + {\varphi ^{2k}} = {\varphi ^2} - {\varphi ^{k + 1}} + {\varphi ^{2k}} \\
\sum\limits_{k = 1}^n {{\varphi ^k}} \cdot\det \left( {A - {\varphi ^k}{I_2}} \right) = \sum\limits_{k = 1}^n {{\varphi ^k}} \cdot({\varphi ^2} - {\varphi ^{k + 1}} + {\varphi ^{2k}}) = {\varphi ^2}\sum\limits_{k = 1}^n {{\varphi ^k}} - \varphi \sum\limits_{k = 1}^n {{\varphi ^{2k}}} + \sum\limits_{k = 1}^n {{\varphi ^{3k}}} \\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\varphi ^2}\varphi \frac{{{\varphi ^n} - 1}}{{\varphi - 1}} - \varphi {\varphi ^2}\frac{{{\varphi ^2}^n - 1}}{{{\varphi ^2} - 1}} + {\varphi ^3}\frac{{{\varphi ^3}^n - 1}}{{{\varphi ^3} - 1}} = \,\,\mathop = \limits^{[{\varphi ^n} = {{\left( {\cos \left( {\frac{{2\pi }}{n}} \right) + i\sin \left( {\frac{{2\pi }}{n}\,\,\,} \right)\,} \right)}^n} = 1]\,\,\,\,\,\,} 0 \\
\end{array}}")
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Re: Μιγαδικό άθροισμα
Δημοσιεύτηκε: Δευ Αύγ 01, 2011 6:42 pm