Summation of a List Using Foldl

Asked Apr 28 '18 at 05:17

Active Apr 28 '18 at 06:27

Viewed 708 times

I'm new to Haskell and I'm trying to implement the sine approximation using MacLaurin's series using foldl on Haskell. I've already generated the list, and now I'm trying to get the summation using fold.

This is my code. I know I'm doing something wrong with the sin function, but I'm not sure what to do. I also have to pass x. If sin receives x as input, list will use the same value of x.

factorial n = product [1..n]
list x = [-1**n * ((x**(2*n+1))/factorial(2*n+1)) | n <- [0..29] ]
sin x = foldl (+) 0 list x

SOLVED: Correct solution:

factorial n = product [1..n]
list x = [(-1)**n * ((x**(2*n+1))/factorial(2*n+1)) | n <- [0..29] ]
sine x = foldl (+) 0 (list x)

edited Apr 28 '18 at 06:27

asked Apr 28 '18 at 05:17

zuzu

I didn't try, but maybe `foldl (+) 0 (list x)` ? Otherwise `factorial(2*n+1)` is an integer and the division is not valid. – Stéphane Laurent Apr 28 '18 at 06:21
Wow. It worked! Haha, I can't believe it's that simple. Thank you! :) – zuzu Apr 28 '18 at 06:26
Why not use `sum` ? – Elmex80s Apr 28 '18 at 06:59
This is for class! We're required to use foldl :) – zuzu Apr 28 '18 at 07:00
If you can afford using [`foldl1`](http://hackage.haskell.org/package/base-4.11.1.0/docs/Prelude.html#v:foldl1) then you don't even need the zero value. – Regis Kuckaertz Apr 28 '18 at 09:46

Summation of a List Using Foldl

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