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Practice Paper 4 Question 7

The function \(f\) is defined recursively for all integers \(n>1,p>0\) as follows: \[ \begin{align} f(1,p) &= C + p, \\ f(n,1) &= C, \\ f(n,p) &= f(n-1, f(n,p-1)) \end{align} \] where \(C>0\) is a real constant. Give a non-recursive expression for \(f(4,p).\) Proof is not required. Hint: You may want to try \(f(2,p)\) first.

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  • Hint 1
    Try writing out and spotting patterns in \(f(2,p).\)
  • Hint 2
    Try writing this in a non-recursive form.
  • Hint 3
    ... by noticing it is only recursive in the second argument, \(p\) (the \(2\) is fixed).
  • Hint 4
    How might you do the same for \(f(3,p)?\)
  • Hint 5
    Remember to use results you've already shown.


Using the hint provided in the question, we try to find an expression for \(f(2,p).\) From the definition we get that \(f(2,p) = f(1, f(2, p-1)) = C + f(2, p-1).\) Since the first argument is fixed at \(2,\) we can unravel it to get \(f(2,p)=C+\ldots+C+f(2,1)=C \cdot p.\)

We can apply the same strategy to find \(f(3,p).\) We start with \(f(3,p) = f(2, f(3, p-1)),\) at which point we use our previous result to get \(f(3,p)=C \cdot f(3,p-1).\) Similarly, this unravels to \(f(3,p)=C^p.\)

Finally, we get to \(f(4,p) = f(3, f(4, p-1)) = C^{f(4,p-1)},\) which expands to \(f(4, p) = C^{C^{\cdot^{\cdot^{\cdot^C}}}},\) which is a tower of height \(p.\)

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