Problem source

\begin{questionparts}
\item Let $r$ be a real number with $\vert r \vert<1$ and let 
\[ S = \sum_{n=0}^\infty r^n\,. \]
 You may assume  without proof that $S = \displaystyle \frac{1}{1-r}\, $.
Let $p= 1 + r +r^2$. Sketch the graph of the function $1+r+r^2$ and deduce that $\frac{3}{4} \le  p < 3\,$. Show that, if $1 <  p < 3$, then the value of $ p$  determines $r$, and hence $S$, uniquely. 
Show also that, if $\frac{3}{4} <  p < 1$, then there are two possible values of  $S$ and these values  satisfy the equation $(3-p)S^2-3S+1=0$.
\item Let $r$ be a real number with $\vert r \vert<1$ and let 
\[ T =\sum_{n=1}^\infty nr^{n-1}\,. \]
 
You may assume without proof that $T = \displaystyle \dfrac{1}{(1-r)^2}\,.$
Let $ q= 1+2r+3r^2$. Find the set of values of $ q$ that  determine $T$ uniquely. Find the set of values of $q$ for which $T$ has two possible values. Find also a quadratic equation, with coefficients depending on $ q$, that is satisfied by these two values.
\end{questionparts}

Solution source

\begin{questionparts}

\item $\,$ 

\begin{center}
    \begin{tikzpicture}
    \def\functionf(#1){1+(#1)+(#1)^2};
    \def\xl{-5};
    \def\xu{5};
    \def\yl{-3};
    \def\yu{7};
    
    % Calculate scaling factors to make the plot square
    \pgfmathsetmacro{\xrange}{\xu-\xl}
    \pgfmathsetmacro{\yrange}{\yu-\yl}
    \pgfmathsetmacro{\xscale}{10/\xrange}
    \pgfmathsetmacro{\yscale}{10/\yrange}
    
    % Define the styles for the axes and grid
    \tikzset{
        axis/.style={very thick, ->},
        grid/.style={thin, gray!30},
        x=\xscale cm,
        y=\yscale cm
    }
    
    % Define the bounding region with clip
    \begin{scope}
        % You can modify these values to change your plotting region
        \clip (\xl,\yl) rectangle (\xu,\yu);
        

        \filldraw (-.5, 0.75) circle (2pt) node[below] {$(-\tfrac12, \tfrac34)$};
        % \filldraw (0, 1) circle (2pt) node[right] {$1$};
        \filldraw (1, 3) circle (2pt) node[right] {$(1,3)$};
        \filldraw (-1, 1) circle (2pt) node[left] {$(-1,1)$};
        
        \draw[thick, blue, smooth, domain=\xl:\xu, samples=100] 
            plot (\x, {\functionf(\x)});

    \end{scope}
    
    % Set up axes
    \draw[axis] (\xl,0) -- (\xu,0) node[right] {$r$};
    \draw[axis] (0,\yl) -- (0,\yu) node[above] {$p$};
    
    \end{tikzpicture}
\end{center}

Notice that $1+r+r^2$ ranges from $\frac34$ to $3$ over $(-1,1)$ therefore $\frac34 \leq p < 3$ attaining its minimum but not its maximum.

If $p > 1$ we know we must be on the right branch and hence we can determine $r$ and $S$ uniquely.

If $\frac34 < p < 1$ then we must have two possible values for $r$, satisfying $r_i^2 + r+1 = p$ and so our two possible values for $S_i = \frac{1}{1-r_i}$ or $r_i = 1-\frac{1}{S_i}$ and so

\begin{align*}
&& p &= \left ( 1 - \frac{1}{S} \right)^2 + 1 - \frac1S + 1 \\
\Rightarrow && pS^2 &= (S-1)^2 + S^2-S + S^2 \\
\Rightarrow && 0 &= (3-p)S^2 -3S + 1
\end{align*}

\item $\,$



\begin{center}
    \begin{tikzpicture}
    \def\functionf(#1){1+2*(#1)+3*(#1)^2};
    \def\xl{-2};
    \def\xu{2};
    \def\yl{-3};
    \def\yu{7};
    
    % Calculate scaling factors to make the plot square
    \pgfmathsetmacro{\xrange}{\xu-\xl}
    \pgfmathsetmacro{\yrange}{\yu-\yl}
    \pgfmathsetmacro{\xscale}{10/\xrange}
    \pgfmathsetmacro{\yscale}{10/\yrange}
    
    % Define the styles for the axes and grid
    \tikzset{
        axis/.style={very thick, ->},
        grid/.style={thin, gray!30},
        x=\xscale cm,
        y=\yscale cm
    }
    
    % Define the bounding region with clip
    \begin{scope}
        % You can modify these values to change your plotting region
        \clip (\xl,\yl) rectangle (\xu,\yu);
        

        \filldraw ({-1/3}, {2/3}) circle (2pt) node[below] {$(-\tfrac13, \tfrac23)$};
        % \filldraw (0, 1) circle (2pt) node[right] {$1$};
        \filldraw (1, 6) circle (2pt) node[right] {$(1,6)$};
        \filldraw (-1, 2) circle (2pt) node[left] {$(-1,2)$};
        
        \draw[thick, blue, smooth, domain=\xl:\xu, samples=100] 
            plot (\x, {\functionf(\x)});

    \end{scope}
    
    % Set up axes
    \draw[axis] (\xl,0) -- (\xu,0) node[right] {$r$};
    \draw[axis] (0,\yl) -- (0,\yu) node[above] {$q$};
    
    \end{tikzpicture}
\end{center}

So $\frac23 \leq q < 6$ and $r$ and hence $T$ is uniquely determined if $2 \leq q < 6$. if $\frac23 <q < 2$ then there are two possible values, and they must satisfy:

$r= 1-1/\sqrt{T}$ so 
\begin{align*}
&& q &= 1 +2 \left( 1-\frac{1}{\sqrt{T}} \right)+3\left (  1-\frac{1}{\sqrt{T}} \right)^2 \\
&&&= 6 - \frac{2}{\sqrt{T}} - \frac{6}{\sqrt{T}} + \frac{3}{T} \\
&&&= 6 - \frac{8}{\sqrt{T}} + \frac3T \\
\Rightarrow && (q-6)T-3 &= -8\sqrt{T} \\
\Rightarrow && (q-6)^2T^2-6(q-6)T+9 &= 64T \\
\Rightarrow && 0 &= (q-6)^2T^2 -(6q+28)T+9
\end{align*}

\end{questionparts}