- The formula
(*)
\[ \begin{array}{l} h (1) = 1, \\ h (2) = 2, \\ h (n) = [ h (n + 1) ] ^ {2} - [ h (n - 1) ] ^ {2} \quad \text { for } n \geq 2 \end{array} \]
is not one to which the principle of recursive definition applies. Show that nevertheless there does exist a function \(h: \mathbb{Z}_+ \to \mathbb{R}\) satisfying this formula. [Hint: Reformulate (*) so that the principle will apply and require \(h\) to be positive.]
Show that the formula (*) of part (a) does not determine \(h\) uniquely. [Hint: If \(h\) is a positive function satisfying (*), let \(f(i) = h(i)\) for \(i \neq 3\), and let \(f(3) = -h(3)\) .]
Show that there is no function \(h: \mathbb{Z}_{+} \to \mathbb{R}\) satisfying the formula
\[ \begin{array}{l} h (1) = 1, \\ h (2) = 2, \\ h (n) = [ h (n + 1) ] ^ {2} + [ h (n - 1) ] ^ {2} \quad \text { for } n \geq 2. \end{array} \]
Solution
Rewrite the recurrence as \[ [h(n+1)]^2=h(n)+[h(n-1)]^2. \] Require all values to be positive, and define recursively \[ h(1)=1,\qquad h(2)=2, \] \[ h(n+1)=\sqrt{h(n)+[h(n-1)]^2}\qquad(n\ge2), \] where the positive square root is chosen. The radicand is positive, so the recursive definition is valid. Squaring the defining equation gives \[ h(n)=[h(n+1)]^2-[h(n-1)]^2 \] for every \(n\ge2\). Hence a solution exists.
The printed hint suggesting that one merely negate \(h(3)\) and leave all later values unchanged is not correct: the equation for \(n=3\) would then fail. Instead define a second solution \(f\) by \[ f(1)=1,\qquad f(2)=2,\qquad f(3)=-\sqrt3, \] and then, for \(n\ge3\), recursively set \[ f(n+1)=\sqrt{f(n)+[f(n-1)]^2}. \] For \(n=3\), the radicand is \(4-\sqrt3>0\); thereafter the newly defined terms are positive, so every later radicand is positive as well. At \(n=2\), \[ [f(3)]^2-[f(1)]^2=3-1=2=f(2). \] For every \(n\ge3\), squaring the recursive definition gives exactly \[ f(n)=[f(n+1)]^2-[f(n-1)]^2. \] Thus \(f\) is a solution distinct from the positive solution of part (a), so the original formula does not determine \(h\) uniquely.
Suppose instead that \[ h(n)=[h(n+1)]^2+[h(n-1)]^2. \] At \(n=2\), \[ 2=[h(3)]^2+1, \] so \([h(3)]^2=1\), hence \(h(3)=\pm1\). But at \(n=3\), \[ h(3)=[h(4)]^2+[h(2)]^2=[h(4)]^2+4\ge4, \] contradicting \(h(3)=\pm1\). Therefore no such function exists.