Edit this page Backlinks This page is read only. You can view the source, but not change it. Ask your administrator if you think this is wrong. ====== Lazy evaluation ====== 1. Consider the following recurrence scheme described informally below. Use it to build the sequence $math[1, 2, \ldots, n!, ...] <code> 4 4*5 4*5*6 ... 3 3 3 3 ---------------------- 3 3*4 3*4*5 3*4*5*6 ... </code> 2. Define the sequence $math[1, \frac{1}{2}, \ldots, \frac{1}{k!}, \ldots ] 3. Write a function which takes a sequence $math[(a_n)_{n\geq 0}] and computes the sequence $math[(s_n)_{n\geq 0}] where $math[s_k = \sum_{k\geq 0} a_k]. Use a strategy similar to that from exercise 1. 4. Write the stream of approximations of $math[e] [[https://en.wikipedia.org/wiki/E_(mathematical_constant) | details ]]. 5. Write a function which takes a value $math[d], a sequence of approximations $math[(a_n)_{n\geq 0}] and returns that value $math[a_k] from the sequence which satisfies the condition $math[\mid a_k - a_{k+1}\mid \leq d] 6. Write a function which takes an $math[f], a value $math[a_0] and computes the sequence $math[a_0, f(a_0), f(f(a_0)), \ldots] 7. The sequence $math[(a_n)_{n\geq 0}] defined as $math[a_{k+1} = \frac{a_k + \frac{n}{a_k}}{2}]