I suppose it depends on your definition of straightforward but you can use the iterate function from Data.List to quickly define sequences like this. a = iterate (\x -> (1/5) * (x**2)) 10 On Mon, Mar 4, 2013 at 9:19 PM, Christopher Howard < christopher.howard@frigidcode.com> wrote:

Hi. My Haskell is (sadly) getting a bit rusty. I was wondering what would be the most straightforward and easily followed "procedure" for translating a recursively defined sequence into a Haskell function. For example, this one from a homework assignment.

quote: -------- a_1 = 10 a_(k+1) = (1/5) * (a_k)**2 --------

(The underscore is meant to represent subscripting what follows it.)

-- frigidcode.com

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On 13-03-05 12:19 AM, Christopher Howard wrote:

Hi. My Haskell is (sadly) getting a bit rusty. I was wondering what would be the most straightforward and easily followed "procedure" for translating a recursively defined sequence into a Haskell function. For example, this one from a homework assignment.

quote: -------- a_1 = 10 a_(k+1) = (1/5) * (a_k)**2 --------

(The underscore is meant to represent subscripting what follows it.)

1. decode subscripts back to arguments a 1 = 10 a (k+1) = (1/5) * (a k)**2 2. normalize LHS arguments sometimes, some arguments on the LHS (k+1 here) are not accepted by Haskell 2010; therefore, we need an equivalent definition with another argument form. a 1 = 10 a k = (1/5) * (a (k-1))**2 3. translate to Haskell (that's right, the above two steps are pure math, not Haskell) a 1 = 10 a k = (1/5) * (a (k-1))**2 The result may or may not be an efficient algorithm (which depends on how you use it). But it gives the correct answer. An efficient algorithm requires further study. Here is an example where step 3 makes change. b_0 = 0 b_(k+1) = sqrt k * b_k 1. decode subscripts back to arguments b 0 = 0 b (k+1) = sqrt k * b k 2. normalize LHS arguments b 0 = 0 b k = sqrt (k-1) * b (k-1) 3. translate to Haskell b 0 = 0 b k = sqrt (fromIntegral (k-1)) * b (k-1)