Problem:
Solution:
The case for $ k = 1 $ is trivial. The case for $ k = 2 $ is given as definition. Using induction, we can assume $ \theta_1 x_1 + \cdots + \theta_{k-1} x_{k-1} \in C $ work for any set of $ \theta $ that sums to 1.
Now consider
$ \begin{eqnarray*} & & \theta_1 x_1 + \cdots + \theta_k x_k \\ &=& \theta_1 x_1 + \cdots + \theta_{k-1} x_{k-1} + \theta_k x_k \\ &=& \frac{1 - \theta_k}{1 - \theta_k}(\theta_1 x_1 + \cdots + \theta_{k-1} x_{k-1}) + \theta_k x_k \\ &=& (1 - \theta_k)(\frac{\theta_1}{1 - \theta_k} x_1 + \cdots + \frac{\theta_{k-1}}{1 - \theta_k} x_{k-1}) + \theta_k x_k \end{eqnarray*} $
Now we know the sum inside the bracket is in $ C $ by the induction hypothesis, and therefore the point we wanted to prove is also in $ C $ because of the definition.
Solution:
The case for $ k = 1 $ is trivial. The case for $ k = 2 $ is given as definition. Using induction, we can assume $ \theta_1 x_1 + \cdots + \theta_{k-1} x_{k-1} \in C $ work for any set of $ \theta $ that sums to 1.
Now consider
$ \begin{eqnarray*} & & \theta_1 x_1 + \cdots + \theta_k x_k \\ &=& \theta_1 x_1 + \cdots + \theta_{k-1} x_{k-1} + \theta_k x_k \\ &=& \frac{1 - \theta_k}{1 - \theta_k}(\theta_1 x_1 + \cdots + \theta_{k-1} x_{k-1}) + \theta_k x_k \\ &=& (1 - \theta_k)(\frac{\theta_1}{1 - \theta_k} x_1 + \cdots + \frac{\theta_{k-1}}{1 - \theta_k} x_{k-1}) + \theta_k x_k \end{eqnarray*} $
Now we know the sum inside the bracket is in $ C $ by the induction hypothesis, and therefore the point we wanted to prove is also in $ C $ because of the definition.
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