Natural numbers apparently don’t necessarily involve the number zero

```
Definition of Natural Numbers
Base case: n=0
n=0 is a natural number
Successive case: n_{i+1} = n_i + 1
if n_i is a natural number, then n_i + 1 is a natural number.
```

There was the rabbit Fibonacci sequence analogy

```
F(0) = 1
F(1) = 1
F(2) = F(1) + F(0)
F(3) = F(2) + F(1)
F(4) = F(3) + F(2)
F(n) = F(n-1) + F(n-2), for n > 1
```

```
Algorithm Fib(n)
Input: n //rank of fib. number
Output: actual fib. number
if n < 2 then
return 1;
end if
return F(n-1) + Fib(n-2)
```

`Fib(5) = (Fib(4) + Fib(3)) = ((Fib(3) + Fib(2)) + (Fib(2) + Fib(1)))`

`

Do the analysis to speed it up.

```
T(n)
T(0) = 1
T(1) = 1
T(n-1) + T(n-2)
So T(n) = T(n-1) + T(n-2) + 1
T(n) = 2F(n) - 1
```

Note: `T(n-2) < T(n-1), for n >= 3`

`T(n) < T(n-1) + T(n-1) +1`

`T(n) < 2T(n-1) + 1`

Note: `T(n-1) < 2T(n-2) + 1`

`T(n) < 2^2 T(n-2) + 2 + 1`

Note: `T(n-2) < 2T(n-3) + 1`

`T(n) < 2^3T(n-3) + 2^2 + 2 + 1`

…

`T(n) < 2^k(n-k) + (2^{k-1} + ... + 2^2 + 2 + 1)`

Side note: how to solve
`S(k) = (2^{k-1} + … + 2^2 + 2 + 1)`

`2S(k) = (2^k + ... + 2^3 + 2^2 + 2)`

`2S(k) - S(k) = 2^k + 2^{k-1} - 2^{k-1} + … + 2^2 - 2^2 + 2 - 2 + 1`

So `S(k) = 2^k + 1`

Therefore

`T(n) < 2^kT(n-k) + (2^k + 1)`

`T(n) < 2^{n-1}T(1) + (2^{n-1} - 1)`

`T(n) < 2^{n-1} + 2^{n-1} - 1`

`T(n) < 2^n - 1`

- Start with
`F(0)`

and`F(1)`

- Always keep the last two value,
`F(i-1)`

and`F(i-2)`

- Compute until
`F(i=n)`

In this case, `T(n) = n`

This is called dynamic programming, when we save data that is frequently used.

How can we do better?

`F(n) = \frac{(golden ratio)^n - (golden ratio conjugate)^n}{\sqrt{5}}`

`T(n) = 2*(the cost of the power function)`

The actual `T(n)`

for the recursive function is `O((golden ratio)^n)`

`a^n = a*a*a*a*a* … *a`

Which would be O(n) and worse than the dynamic programming version

`a^n = a^{n/2} * a^{n/2}`

if n is even
`a^n = a^{n/2} * a^{n/2} * a`

if n is odd

- Solve the power function analysis

`