New to Haskell? This menu will give you a first impression. Don't read all the explanations, or you'll be starved before the meal.

Apéritifs

Foretaste of an excellent meal.

tsort :: Ord a => [a] -> [a]
tsort []     = []
tsort (x:xs) = tsort (filter (<x) xs) ++ [x] ++ tsort (filter (>=x) xs))
Treesort in three lines (!). Sorts not only integers but anything that can be compared.
fibs = 1:1:zipWith (+) fibs (tail fibs)
The infinite list of fibonacci numbers. Just don't try to print all of it.
linecount = interact $show . length . lines wordcount = interact$ show . length . words
Count the number of lines or words from standard input.

Entrées

square x = x*x
is the function $f(x)=x\cdot x$ which maps a number to its square. While we commonly write parenthesis around function arguments in mathematics and most programming languages, a simple space is enough in Haskell. We're going to apply functions to arguments all around, so why clutter the notation with unnecessary ballast?
square :: Int -> Int
square x = x*x
Squaring again, this time with a type signature which says that squaring maps integers to integers. In mathematics, we'd write $f:\mathbb{Z}\to\mathbb{Z},\ f(x)=x\cdot x$. Every expression in Haskell has a type and the compiler will automatically infer (= figure out) one for you if you're too lazy to write down a type signature yourself. Of course, parenthesis are allowed for grouping, like in square (4+2) which is 36 compared to square 4 + 2 which is 16+2=18.
square :: Num a => a -> a
square x = x*x
Squaring yet again, this time with a more general type signature. After all, we can square anything (a) that looks like a number (Num a). By the way, this general type is the one that the compiler will infer for square if you omit an explicit signature.
average x y = (x+y)/2
The average of two numbers. Multiple arguments are separated by spaces.
average :: Double -> Double -> Double
average x y = (x+y)/2
Average again, this time with a type signature. Looks a bit strange, but that's the spicey currying. In fact, average is a function that takes only one argument (Double) but returns a function with one argument (Double -> Double).
power a n = if n == 0 then 1 else a * power a (n-1) $a^n$, defined with recursion. Assumes that the exponent n is not negative, that is n >= 0.
Recursion is the basic building block for iteration in Haskell, there are no for or while-loops. Well, there are functions like map or foldr that provide something similar. There is no need for special built-in control structures, you can define them yourself as ordinary functions (later).
power a 0 = 1
power a n = a * power a (n-1)
Exponentiation again, this time with pattern matching. The first equation that matches will be chosen.
length []     = 0
length (x:xs) = 1 + length xs
Calculate the length of a list. What's a list? Well, a list may either be empty ([]) or be an element (x) prepended (:) to another list (xs). Read "xs" as the plural of "x", that is as "ex-es". It's a list of other such elements x, after all.
length :: [a] -> Int
length []     = 0
length (x:xs) = 1 + length xs
Length of a list again, this time with type signature. [a] is the type of lists with elements of type a. length can be used for any such element type.
First element of a list. Undefined for empty lists.
sum []     = 0
sum (x:xs) = x + sum xs
Sum all elements of a list.
average xs = sum xs / (fromIntegral (length xs))
Arithmetic mean. fromIntegral converts the integer result of length into a decimal number for the division /.
(++) :: [a] -> [a] -> [a]
(++) []     ys = ys
(++) (x:xs) ys = x:(xs ++ ys)
Concatenate two lists. Custom infix operators can be defined freely.

Soupes

The best soup is made by combining the available ingredients.

(.) :: (b -> c) -> (a -> b) -> (a -> c)
(.) f g x = f (g x)

fourthPower = square . square
The dot f . g is good old function composition $f \circ g$. First apply g, then apply f. Simple example: squaring something twice.
To find the least element of a list, first sort and then take the first element. You think that takes too much time ( $O(n\cdot\log n)$ instead of $O(n)$)? Well, thanks to lazy evaluation, it doesn't! In Haskell, expressions are evaluated only as much as needed. Therefore, the sorting won't proceed further than producing the first element of the sorted list. Ok, the sorting function has to play along and produce that one quickly, but many like treesort (in the average case) or mergesort do so.