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Functions in mathematics and Haskell.
Suppose \(f : Int \to Int \) sends an integer to its square, \(f(x) := x^2 \) and that \(g : Int \to Int \) sends an integer to its successor, \(g(x) := x+ 1\).
(a) Write \(f \) and \(g \) in Haskell, including their type signature and their implementation.
(b) Let \(h := f◦g \). What is \(h(2) \)?
(c) Let \(i := f ; g \). What is \(i(2) \)?
Comments
My answer: https://paste.sr.ht/~leif/42cc64b988ce44a69dbfc16ab88a0c37237665ba
My answer: https://paste.sr.ht/%7Eleif/42cc64b988ce44a69dbfc16ab88a0c37237665bawhat is meant by $f;g$?
what is meant by $f;g$?@HanifBinAriffin: \(f;g\) means apply \(f\), then apply \(g\). It's the opposite order of operations than \(f \circ g\), which means apply \(g\) then \(f\).
In \(f;g\) the data flows "forwards" from left to right (forward only because English reads from left to right), whereas in \(f \circ g\) it flows "backwards" from right to left.
@HanifBinAriffin: \\(f;g\\) means apply \\(f\\), then apply \\(g\\). It's the opposite order of operations than \\(f \circ g\\), which means apply \\(g\\) then \\(f\\). In \\(f;g\\) the data flows "forwards" from left to right (forward only because English reads from left to right), whereas in \\(f \circ g\\) it flows "backwards" from right to left.