Exercise 1.40

This is the $40^{th}$ question in SICP.

The Question

Exercise 1.40: Define a procedure cubic that can be used together with the newtons-method procedure in expressions of the form:

(newtons-method (cubic a b c) 1)

to approximate zeros of the cubic $x^{3} + ax^{2} + bx + c$.

The Answer

In order to solve this question, we need to understand what exactly is Newton’s method.

Newton’s Method

In Newton’s method, what essentially happens is a fixed point search to find $g(x) = 0$.

Inside, a derivative function $Dg$ of the form:

$$Dg(x) = \frac{g(x + dx) - g(x)}{dx}$$

where $g$ is the function the user inputs, and $d$x is a small number is used to create a function of the form:

$$f(x) = x - \frac{g(x)}{Dg(x)}$$

A fixed point function is then done on this function $f(x)$ to find $g(x) = 0$

Writing cubic

Now that we have got what Newton’s method is in place, we can now start thinking about about cubic.

Since, all we need to do is approximate zeros of the polynomial $x^{3} + ax^{2} + bx + c$, we do not need to worry about anything other than writing a function acceptable to Newton’s Method as of right now.

cubic will be of the form (cubic a b c) to approximate $x^{3} + ax^{2} + bx + c$. And because all we need to do is write a function of Newton’s Method, we just need return a lambda with a, b, and c hard coded. This means cubic is the following:

(define (cubic a b c)
  (lambda (x)(+ (cube x) (* a (square x)) (* b x) c)))

Testing cubic

For simplicity’s sake, let’s just let a, b, c be 1. Thus, we get the following simplification: $x^{3} + x^{2} + x + 1$. So a few example inputs for x:

1. For $x = 1$, we will get 4.
2. For $x = 2$ we will get 15.
3. For $x = 3$ we will get 40.

Let us test the above scenarios:

> ((cubic 1 1 1) 1)
4
> ((cubic 1 1 1) 2)
15
> ((cubic 1 1 1) 3)
40

Writing a function to find the zero

Now that we have got cubic, all we need to do a procedure to find the zero. This is as simple as:

(define (zero a b c)
  (newtons-method (cubic a b c) 1))

Testing

Let us take some random numbers for a, b, c. Perhaps 1, 2, and 3. Let’s see what is the result we get:

1 *** 1
2 *** 5.714266429257542e-06
3 *** -1.4999839284734562
4 *** -1.3043428073335799
5 *** -1.276209090923146
6 *** -1.27568238137649
-1.2756822036498454

And now trying that with the function return by cubic:

> ((cubic 1 2 3) -1.275)
0.002953125000000334

And that’s it! You can find my source file here