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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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## Hints

• 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.

## Solution

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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