Answers to review problems for the final
1. Let f(x) = 1/(1 - (x + x^2 + x^3)). Show that the coefficient of x^n is positive for each n, when we expand
f(x) as a power series in x. What's a recurrence relation for the coefficient of x^n?
1A. The geometric series says f(x) = sum_n (x+x^2+x^3)^n. Each of those summands is a polynomial with
positive coefficients, so when they're added together the result has positive coefficients.
The homework problem this is closest to is the stamp problem. In those terms, we're trying to get n cents
postage, using a bunch of 1,2,3 cent stamps in any order (and order matters). Thought of that way, the ways
to get such postage break three cases, based on the last stamp. That gives us the recurrence
F_n = F_(n-1) + F_(n-2) + F_(n-3).
(You can also get this from the equation f(x) = 1 + (x+x^2+x^3)f(x) and comparing coefficients.)
2. Let X = {1,2,3}, Y = {1,2,3,4,5,6}, and Doub : X -> Y be the doubling function. (1->2, 2->4, 3->6)
Can you find two functions f,g from Y -> X,
not equal
, such that f o Doub = g o Doub? (These are functions
X->X.)
Can you find two functions f,g from Y -> X,
not equal
, such that Doub o f = Doub o g? (These are functions
Y->Y.)
If you can, give an explicit example of an f and a g.
2A. Well, yes and no. That is to say, yes, and no. Here's a hint: f o Doub = g o Doub is really three equations,
f(Doub(n)) = g(Doub(n)), n=1,2,3. That's not enough to determine a function from Y->X, which is six
choices.
But we can (and must) be more specific. For the first one, let f:Y->X take everybody to 1, whereas g:Y->X
takes even numbers (the image of Doub) to 1 -- this is what the equations tell us must be true -- and does
something else with the odd numbers. For example, maybe g(1)=1, g(3)=3, g(5)=1.
(For the "no" part, if two numbers are equal after they're doubled, then they were already equal. So