# SICP - Solution: Exercise 1.45

##### October 29, 2018

## Exercise 1.45 #

We saw in 1.3.3 that attempting to compute square roots by naively finding a fixed point of ${y\mapsto x/y}$ does not converge, and that this can be fixed by average damping. The same method works for finding cube roots as fixed points of the average-damped ${y\mapsto x/y^2}$. Unfortunately, the process does not work for fourth roots—a single average damp is not enough to make a fixed-point search for ${y\mapsto x/y^3}$ converge. On the other hand, if we average damp twice (i.e., use the average damp of the average damp of ${y\mapsto x/y^3}$) the fixed-point search does converge. Do some experiments to determine how many average damps are required to compute $n^{\text{th}}$ roots as a fixed-point search based upon repeated average damping of $y\mapsto x/y^{n-1}$. Use this to implement a simple procedure for computing $n^{\text{th}}$ roots using

`fixed-point`

,`average-damp`

, and the`repeated`

procedure of Exercise 1.43. Assume that any arithmetic operations you need are available as primitives.

## Solution #

### Experimentation #

First, let’s implement a function `nth-root-damped`

that allow us to quickly test the parameters `nth`

and `damping`

:

```
(define (compose f g)
(lambda (x)
(f (g x))))
(define (repeated f n)
(if (= n 1)
f
(compose f (repeated f (- n 1)))))
(define (power x n)
(if (= n 1)
x
(* x (power x (- n 1)))))
(define (nth-root-damped x nth damp)
(fixed-point
((repeated average-damp damp)
(lambda (y)
(/ x (power y (- nth 1)))))
1.0))
```

By varying the parameters, we can see for $x=2$ the minimal number of damping needed for the function to converge:

nth | minimal damp |
---|---|

2 | 1 |

3 | 1 |

4 | 2 |

5 | 2 |

6 | 2 |

7 | 2 |

8 | 3 |

9 | 3 |

10 | 3 |

11 | 3 |

12 | 3 |

13 | 3 |

14 | 3 |

15 | 3 |

16 | 4 |

17 | 4 |

… | 4 |

30 | 4 |

31 | 4 |

32 | 5 |

… | 5 |

63 | 5 |

64 | 6 |

The pattern where the number if minimal damp do converge looks to increase every time the nth is reaching a power of 2. By searching the
documentation, it seems that we can compute minimal damp using a base 2 logarithms and the `floor`

function to provide an integer (needed for computing the power):

```
(floor (log 63 2))
> 5
(floor (log 64 2))
> 6
```

### Solution #

We can insert that in our solution that becomes:

```
(define (nth-root x nth)
(fixed-point
((repeated average-damp (floor (log nth 2)))
(lambda (y)
(/ x (power y (- nth 1)))))
1.0))
```

We can try the solution with large `nth`

to check if it holds:

```
(display (nth-root 2 258))
> 1.0026902132630033
```

### Open question #

- Where does this pattern of convergence for log base 2 comes from?