Eufisky - The lost book

## 幸子解答的李炯生习题

$\left\{ {\begin{array}{*{20}{c}}{v\left( x \right) = {v_1}\left( x \right) + {{\left( {x + 1} \right)}^n}\alpha \left( x \right)}\\{w\left( x \right) = {w_1}\left( x \right) - {{\left( {x - 1} \right)}^n}\alpha \left( x \right)}\end{array}} \right.,$

$$\left\{ {\begin{array}{*{20}{c}}{v\left( x \right) = {v_0}\left( x \right) + {{\left( {x + 1} \right)}^n}\alpha \left( x \right)}\\{w\left( x \right) = {w_0}\left( x \right) - {{\left( {x - 1} \right)}^n}\alpha \left( x \right)}\end{array}} \right.$$

\begin{align*}{v^{\left( i \right)}}\left( { - 1} \right) &= \sum\limits_{k = 0}^i {{{\left[ {1 - w\left( x \right){{\left( {x + 1} \right)}^n}} \right]}^{\left( k \right)}}{{\left[ {{{\left( {x - 1} \right)}^{ - n}}} \right]}^{\left( {i - k} \right)}}} {|_{x = - 1}} = {\left( { - 1} \right)^n}{2^{ - n - i}}n \cdots \left( {n + i - 1} \right)\\{w^{\left( i \right)}}\left( 1 \right) &= \sum\limits_{k = 0}^i {{{\left[ {1 - v\left( x \right){{\left( {x - 1} \right)}^n}} \right]}^{\left( k \right)}}{{\left[ {{{\left( {x + 1} \right)}^{ - n}}} \right]}^{\left( {i - k} \right)}}} {|_{x = 1}} = {\left( { - 1} \right)^i}{2^{ - n - i}}n \cdots \left( {n + i - 1} \right).\end{align*}

$f\left( x \right) = 2\left[ {{{\left( {x - 1} \right)}^n}\alpha \left( x \right) - {w_0}\left( x \right)} \right]{\left( {x + 1} \right)^n} + 1= 2\left[ {{v_0}\left( x \right) + {{\left( {x + 1} \right)}^n}\alpha \left( x \right)} \right]{\left( {x - 1} \right)^n} - 1$均满足题设.

$T = UA{U^H} = \frac{1}{2}\left( {\begin{array}{*{20}{c}}{B + G + \left( {C - D} \right)i}&{B - G - \left( {C + D} \right)i}\\{B - G + \left( {C + D} \right)i}&{B + G - \left( {C - D} \right)i}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}P&Q\\{\bar Q}&{\bar P}\end{array}} \right).$

$\left( {\begin{array}{*{20}{c}}P&Q\\{\bar Q}&{\bar P}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{I_n}}&O\\O&{ - {I_n}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{P^H}}&{{Q^T}}\\{{Q^H}}&{{P^T}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{{I_n}}&O\\O&{ - {I_n}}\end{array}} \right).$