Problems 1-9
1 To show that the $\operatorname{mean}(\{kx\})$ equals $k\operatorname{mean}(\{x\})$ comes from the definition of mean,
\begin{equation} \label{eq:mean} \operatorname{mean}({x}) = \frac{1}{N} \sum_{i = 1}^{N} x_i, \end{equation}
because
$$ \begin{array}{r c l} \operatorname{mean}(\{kx\}) & = & \dfrac{1}{N} \displaystyle\sum_{i=1}^N kx_i \\ & = & \dfrac{1}{N} (kx_1 + kx_2 + \cdots + kx_n) \\[0.5em] & = & \dfrac{1}{N} k(x_1 + x_2 + \cdots + x_n) \\[0.5em] & = & k \dfrac{1}{N} (x_1 + x_2 + \cdots + x_n) \\[0.5em] & = & k \displaystyle\sum_{i=1}^N x_i \\[0.5em] & = & k\operatorname{mean}(\{x\}). \end{array} $$so
\begin{equation} \operatorname{mean}({kx}) = k \operatorname{mean}({x}). \end{equation}
2 Using the definition \eqref{eq:mean} to show that
\begin{equation} \label{eq:meanconst} \operatorname{mean}({ x + c }) = \operatorname{mean}({x}) + c \end{equation}
goes as follows:
$$ \begin{array}{r c l} \operatorname{mean}(\{x + c\}) & = & \dfrac{1}{N}\displaystyle\sum_{i=1}^N (x + c) \\[0.5em] & = & \dfrac{1}{N}\left((x_1 + c) + (x_2 + c) + \cdots + (x_n + c)\right) \\[0.5em] & = & \dfrac{1}{N}(x_1 + c + x_2 + c + \cdots + x_n + c) \\[0.5em] & = & \dfrac{1}{N}(x_1 + x_2 + \cdots + x_n + c + c + \cdots + c) \\[0.5em] & = & \dfrac{1}{N}(x_1 + x_2 + \cdots + x_n + Nc) \\[0.5em] & = & \dfrac{1}{N}(x_1 + x_2 + \cdots + x_n) + \dfrac{1}{N}Nc \\[0.5em] & = & \dfrac{1}{N}(x_1 + x_2 + \cdots + x_n) + c \\[0.5em] & = & \dfrac{1}{N} \displaystyle\sum_{i=1}^N x_i + c \\[0.5em] & = & \operatorname{mean}(\{x\}) + c. \end{array} $$3
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