Exercise 1.35 asks us to show that the golden ratio ϕ is a fixed point of the transformation x → 1 + 1 / x. We're then asked to use this fact to compute ϕ by means of the

`fixed-point`

procedure defined earlier in the chapter.First we can show that ϕ is one of the roots of x → 1 + 1 / x. We learned the value of ϕ in section 1.2.2 is (1 + √5) / 2, or approximately 1.618.

x = 1 + 1 / x

If we multiply both sides by x we get:

x

x

^{2}= x + 1x

^{2}- x - 1 = 0Now if we use the quadratic equation (or cheat and use WolframAlpha like I did) we find that the roots of the equation are:

x = 1/2(1 - √5)

x = 1/2(1 + √5)

x = 1/2(1 + √5)

Computing ϕ by means of the

`fixed-point`

procedure is fairly straightforward since the procedure is already given.```
(define tolerance 0.00001)
(define (fixed-point f first-guess)
(define (close-enough? v1 v2)
(< (abs (- v1 v2)) tolerance))
(define (try guess)
(let ((next (f guess)))
(if (close-enough? guess next)
next
(try next))))
(try first-guess))
```

The

`fixed-point`

procedure takes a function and an initial guess. Since we're trying to find a point where
x = 1 + 1 / x

that's the function we need to pass. The initial guess can really be just about anything, but we already know that we're trying to prove the fixed point is around 1.6, so let's try an initial value close to that.

```
> (fixed-point (lambda (x) (+ 1 (/ 1 x))) 2.0)
1.6180327868852458
```

Related:

For links to all of the SICP lecture notes and exercises that I've done so far, see The SICP Challenge.

## 2 comments:

"Since we're trying to find a point where

f(x) = 1 + 1 / x"

Aren't you trying to find a point where f(x) = x, i.e. x = 1 + 1 / x? The function is equal to that in

everypoint since that's its definition.Jan,

Yes, you're right. That line should have read:

"Since we're trying to find a point where

x = 1 + 1 / x"

I'll update the post. Thanks for the correction.

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