C9 Lectures: Dr. Erik Meijer - Functional Programming Fundamentals Chapter 6 of 13

Again, nice lecture.  I'm lloking forward to the rest.

Here is my take on Append:

public static IEnumerable<T> Append<T>(this IEnumerable<T> a, IEnumerable<T> b)
{
    foreach (var t in a)
    {
        yield return t;
    }
    foreach (var t in b)
    {
        yield return t;
    }
}

More homework:

1) Prove that fac n = recfac n

Definition of recfact n:

(1)        recfac 0 = 1

(2)        recfac (n+1) = (n+1) * recfact n

Definition of fac n:

(3)        fac n = product [1...n]

                        where

(4)                                product [] = 1

(5)                                product [1..n] = 1 * 2 * … (n-1) * n

We want to prove that:

                    fact n = recfac n

Using induction:

Case n = 0

            fac 0 = recfac 0

            fac 0 = 1 (Eq. 1)

            recfac 0 = product [1 .. 0] = product [] = 1 (Eq. 4)

            1 = 1

Case (m + 1):

(7)        fac (m+1) = recfac (m+1)

fac (m+1) = product [  1..(m+1)] = 1 * 2 *….* m * (m+1)

1. Using Eq.5:       1 * 2 *….* m  = product [1..m]

            fac (m+1) = 1 * 2 *….* m * (m+1)  =(product [1..m]) * (m+1)

(8)        fac (m+1) = (product [1..m]) * (m+1)

Using Eq 2 :  recfac (m+1) = (m+1) * recfac m

Replacing Eq 8 into Eq 7:

            fac (m+1) = (product [1..m]) * (m+1) = recfac (m+1)

Using Eq 2 on the right side:

            (product [1..m]) * (m+1) = (m+1) * recfac m

Dividing by (m+1) on both sides:

            product [1..m] = recfac m

Using Eq 3 on the left side:

            fac m = recfac m

Q.E.D.!