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An apple can only be sliced through its center. When 0 slices have been made, the apple is in 1 piece. When 1 slice is made, the apple is in 2 pieces. When 2 slices are made, the apple is in 4 pieces. Find a formula to describe number of pieces the apple will be in when sliced $n$ times.

I have tried to do this problem. After 1 slice, the formula for the total number of pieces is $2n$. However, the formula does not work when n = 0. When $n = 0$, there is only 1 piece, so the formula does not work.

Is it impossible to find a formula for this question? I'm trying to find the number of exposed surfaces of an apple when cut n times through the center, but I cannot even seem to find a formula for this basic problem.

  • it depends on what you call a slice for what the answer is each time. but that's more a technicality. also there's a pattern I can see right now that may suggest the pattern for the whole thing. –  Sep 01 '17 at 00:59
  • To be clear: by slice you mean only a single stroke of the knife, not going through a previous cut? So, for $n=3$ you have 6 pieces, right? – Randall Sep 01 '17 at 01:00
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    Yes, there is always 2n pieces above 0 slices. But the problem is that at 0, there is not 0 slices but 1. I am trying to find the formula for the number of exposed surface on an apple, so I decided to try a simpler version first, which I can't seem to solve. –  Sep 01 '17 at 01:23
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    Why is that a problem? What is unsatisfying about $f(n)=\begin{cases}1&\text{if}~n=0\2n&\text{otherwise}\end{cases}$? – JMoravitz Sep 01 '17 at 01:26
  • @JMoravitz's answer is honestly the neatest way of writing it. As you said, your formula does hold for any value of n above 0. So if you didn't feel like writing it as a piecewise function, you could always just say that the number of pieces is $2n$, for $n > 0$ – Etched Sep 01 '17 at 01:41
  • Thank you, I guess that is the only formula that would work. I wanted to make sure that there isn't a single line formula that would be able to describe all the results. –  Sep 01 '17 at 02:07

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If you want a single line, I think people generally do this by an indicator function or using the Kronecker delta: $$ \delta_{ij} = \begin{cases} 1 & \text{ if } i=j \\ 0 & \text{ otherwise } \end{cases} $$ Then, if you accept $0^0=1$, you can write: $$ f(n) = (2n)^{1-\delta_{0n}} $$ Or, more simply: $$ f(n) = \delta_{0n} + 2n(1-\delta_{0n}) $$

user3658307
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