本节主要讲述无穷流。
3.53，显然
(define s (cons
-
stream
1
(add
-
stream s s)))
定义是2的n次方组成的无穷数列，2,4,8,16,32...
3.54，定义阶乘组成的无穷序列：
(define (mul
-
streams s1 s2) (stream
-
map
*
s1 s2)) (define factorials (cons
-
stream
1
(mul
-
streams factorials (stream
-
cdr integers))))
3.55解答，比较有趣，也是不难的题目，列出来找出规律就成了，就是将(stream-car s)加到生成的序列中的每个元素上，通过stream-map，最后的结果就是每个元素都是前n个元素累积的结果，我的解答：
(define (partial
-
sums s) (cons
-
stream (stream
-
car s) (stream
-
map (
lambda
(x) (
+
x (stream
-
car s))) (partial
-
sums (stream
-
cdr s)))))
3.56，有了merge就好办了，根据条件合并起3种情况来就好：
(define S (cons
-
stream
1
(merge (scale
-
stream s
2
) (merge (scale
-
stream s
3
) (scale
-
stream s
5
)))))
3.57,略过
3.58，观察到，num每次都与radix相乘并且radix保持不变，那么radix可以认为是一个基数，den也保持不变作为除数，那么这个序列就是以radix为基数对den求整数商的序列，不明白num为什么每次要变换成余数？这个序列有啥特别的用途呢？未解。
(expand 1 7 10)
=> 1 4 2 8 5 7 1 4 2 8
(expand 3 8 10)
=> 3 7 5 0 0 0 0 0 0 03.59解答：
a)只要将序列通过前面定义的mul-streams与整数的倒数序列相乘：
(define (integrate
-
series s) (mul
-
streams (stream
-
map (
lambda
(x) (
/
1
x)) integers) s))
b)照着定义来了，cons的级数注意使用scale-stream乘以-1：
(define sine
-
series (cons
-
stream 0 (integrate
-
series cosine
-
series))) (define cosine
-
series (cons
-
stream
1
(scale
-
stream (integrate
-
series sine
-
series)
-
1
)))
3.64解答：
(define (stream
-
limit s tolerance) (define (stream
-
limit
-
iter stream current) (cond ((
or
(stream
-
null? stream) (null? (stream
-
car stream)))
#
f)
(
else
(let ((next (stream
-
car stream))) (
if
(
<
(abs (
-
next current)) tolerance) next (stream
-
limit
-
iter (stream
-
cdr stream) next)))))) (stream
-
limit
-
iter (stream
-
cdr s) (stream
-
car s)))
习题3.65：
(define (ln
-
summands n) (cons
-
stream (
/
1.0
n) (stream
-
map
-
(ln
-
summands (
+
n
1
))))) (define ln
-
stream (partial
-
sums (ln
-
summands
1
))) (define ln
-
stream2 (euler
-
transform ln
-
stream)) (define ln
-
stream3 (accelerated
-
sequence euler
-
transform ln
-
stream))
经过欧拉变换加速过的级数收敛的很快，测测就知道
文章转自庄周梦蝶 ，原文发布时间2008-05-13
相关资源：sicp第二章练习题的解答
