Can it be shown that
$U_{2} = \sum_{i=1}^{n} [i*g(Y_{i})]$
is a function of
$U_{1}=\sum_{i=1}^{n} g(Y_{i})$ ?
My intuition tells me that this is not true because of the changing (for lack of a better word) coffecient, $i$, in $U_{2}$. But this is not a sound mathematical explanation.
Any helpful hints about how I can prove or disprove this are greatly appreciated. For example, is there a property of Series or Transformations I can look at to show that this is true or not true?
I have another value
$U_{3} = \sum_{i=1}^{n}[(n-i)g(Y_{i})] = n*\sum_{i=1}^{n}g(Y_{i}) - \sum_{i=1}^{n}[i*g(Y_{i})] = n*U_{1}-U_{2}$
Again, my intuition tells me that $U_{3}$ is a function of $U_{1}$ & $U_{2}$ because $U_{1}$ & $U_{2}$ appear in $U_{3}$. However, what property should I appeal to?
EDIT: For clarification, $g(Y_{i})$ is any constant function, g, of only the $i^{th}$ variable I am considering where $i \in \{1,...,n\}$. Does the specific function, g, matter? For example, is there a an important difference between $g(Y_{i})=Y_{i}$ and $g(Y_{i})=log_{e}(Y_{i})/(c-Y_{i})$ (for some constant c)? Besides the simple algebra of the functions and resulting sereis.
Let $U_1(a)=\sum a_i$ and $U_2(a)=\sum ia_i$.
Now consider $a=(a_i)=\begin{cases}1\quad i=1\\0\quad i\neq1\end{cases}$ and $b=(b_i)=\begin{cases}1\quad i=2\\0\quad i\neq2\end{cases}$
$U_1(a)=U_1(b)$ but $U_2(a)\neq U_2(b)$. Thus $U_2$ is not a function of the individual numbers produced by $U_1$.