# SICP - Solution: Exercise 1.35

## Oct 28, 2018 03:03 · 125 words · 1 minute read

**Exercise 1.35:** Show that the golden ratio $\varphi$ (1.2.2) is a fixed point of the transformation ${x\mapsto1+1/x}$, and use this fact to compute $\varphi$ by means of the `fixed-point`

procedure.

**Solution**

Per definition, a fixed point for ${x\mapsto1+1/x}$ is:

$$x=1+\frac1x$$

Which can be rewritten as:

$$x^2=x+1$$ $$x^2-x-1=0$$

This is a second order polynomial whose solution is:

$$x=\frac{1+\sqrt5}2=\varphi$$

In order to compute $\varphi$ by means of the `fixed-point`

, you can just insert the function with a `lambda`

in `fixed-point`

:

```
(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))
(display (fixed-point (lambda (x) (+ 1 (/ 1 x))) 1.0))
```

Which evaluates to:

```
1.6180327868852458
```