Problem:
Solution:
Part a is obvious. The given formula is the parametric line and by definition it is in the convex set $ C $.
Part a is the basis of the hypothesis. Consider it is true for $ n = k $, we have
$ \begin{eqnarray*} & & \sum\limits_{i=1}^{k+1} t_i \left(\begin{array}{c} x_i \\ y_i \end{array}\right) \\ &=& \sum\limits_{i=1}^{k} t_i \left(\begin{array}{c} x_i \\ y_i \end{array}\right) + t_{k+1}\left(\begin{array}{c} x_{k+1} \\ y_{k+1} \end{array}\right) \\ &=& (\sum\limits_{i=1}^{k} t_i)\sum\limits_{i=1}^{k} \frac{t_i}{\sum\limits_{i=1}^{k} t_i} \left(\begin{array}{c} x_i \\ y_i \end{array}\right) + t_{k+1}\left(\begin{array}{c} x_{k+1} \\ y_{k+1} \end{array}\right) \end{eqnarray*} $
The inner sum is actually a point in $ C $ because it is a convex combination.The outer sum is also a point in $ C $ because that is also a convex combination!
Solution:
Part a is obvious. The given formula is the parametric line and by definition it is in the convex set $ C $.
Part a is the basis of the hypothesis. Consider it is true for $ n = k $, we have
$ \begin{eqnarray*} & & \sum\limits_{i=1}^{k+1} t_i \left(\begin{array}{c} x_i \\ y_i \end{array}\right) \\ &=& \sum\limits_{i=1}^{k} t_i \left(\begin{array}{c} x_i \\ y_i \end{array}\right) + t_{k+1}\left(\begin{array}{c} x_{k+1} \\ y_{k+1} \end{array}\right) \\ &=& (\sum\limits_{i=1}^{k} t_i)\sum\limits_{i=1}^{k} \frac{t_i}{\sum\limits_{i=1}^{k} t_i} \left(\begin{array}{c} x_i \\ y_i \end{array}\right) + t_{k+1}\left(\begin{array}{c} x_{k+1} \\ y_{k+1} \end{array}\right) \end{eqnarray*} $
The inner sum is actually a point in $ C $ because it is a convex combination.The outer sum is also a point in $ C $ because that is also a convex combination!
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