sicp -- Marginalia

e1-40-41-42 toc
Exercises 1.40 - 1.42	(ns e1-40-41-42 (:use [util.util :only (repfun compose square)]) (:use [clojure.contrib.test-is :only (deftest is run-tests)]))
1.40 We must define a function which will use newtons-method to find the approximate 0 (zero) of the cubic equation of the form $$x^3 + ax^2 + bx + c$$ I.e. a function which approximates to a `sqrt(x)` is defined as $$y \rightarrow y^2 - x, y0 = 1$$ Newtons method idea is to start with a reasonable guess of the result and iterate further by using (following math snippet doesn't work as marginalia processes underscores using markdown): $$X{n+1} = X{n} - f(X{n}) / f'(X{n})$$ This means we have to express given differentiable function in a way that's iteratable by the (newtons-method) function. In our case: $$g(x) = x^3 + ax^2 + bx + c = y$$ $$Dg(x) = 3x^2 + x^2 + b$$ Actually, this exercise doesn't need any of this stuff. Disregard it and carry on.	(defn cubic [a b c] (fn [x] (+ (* x x x) (* x x a) (* x b) c)))
$$2^3 + 2^3 + 2^2 + 2 = 22$$	(deftest cubic-test (is (= ((cubic 2 2 2) 2) 22)))
1.41
Essentially a `(repfun f 2)`	(defn doubl [f] (fn [x] (f (f x))))
	(deftest doubl-test-1 (is (= ((doubl inc) 1) 3)))
This one is tricky. It's obvious, that `((double inc) x)` will perform an `(inc (inc x))`. Let's see what is going to happen with `(((doubl doubl) inc) x)`. Lets expand doubles one by one: `(((doubl doubl) inc) x)` `(fn [x] (doubl (doubl inc)) x)` `(fn [x] (doubl (inc (inc x))))` `(fn [x] (inc (inc (inc (inc x)))))` So we can say that nesting `doubl`s once will result in the argument function being applied 4 times.	(deftest doubl-test-2 (is (= (((doubl doubl) inc) 1) 5)))
And what about `(((doubl (doubl doubl)) inc) x)` ? Oh, it will clearly apply the inc function 8 times, will it? Actually, no. For the sake of clarity we'll call `(doubl doubl)` a `(quad)`. Lets see what happens here: `(((doubl quad) inc) x)` -> `((quad (quad inc)) x)` That's not an 8x application, but a 16x! We apply `(doubl doubl)` to the `inc` once and then apply `(doubl doubl)` to the result of the previous application.	(deftest doubl-test-2 (is (= (((doubl (doubl doubl)) inc) 5) 21)))
1.42 `compose` is defined in util/util.clj	(deftest compose-test (is (= ((compose square inc) 6) 49)))