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If X has n elements, then the number of its subsets is the sum of the
number of 0-subsets (there is only 1, the empty set), the number of 1subsets, the number of 2-subsets, etc. But n is the number of r -subsets,
as we have seen in the text, and so the total number of subsets is the sum
of all the binomial coefﬁcients, which is 2n , by Exercise 1.31.
1.40 A weekly lottery asks you to select 5 different numbers between 1 and 45.
At the week’s end, 5 such numbers are drawn at random, and you win the
jackpot if all your numbers match the drawn numbers. What is your chance
Solution. The answer is ”45 choose 5”, which is 455 = 1, 221, 759. The
odds against your winning are more than a million to one.
1.41 Assume that “term-by-term” differentiation holds for power series: if f (x )
= c0 + c1 x + c2 x 2 +· · ·+ cn x n +· · · , then the power series for the derivative
f (x ) is
f (x ) = c1 + 2c2 x + 3c3 x 2 + · · · + ncn x n −1 + · · · .
(i) Prove that f (0) = c0 .
Solution. f (0) = c0 , for all the other terms are 0. (If one
wants to be fussy–this is the wrong course for analytic fussiness–
then the partial sums of the series form the constant sequence
c0 , c0 , c0 , . . ..) (ii) Prove, for all n ≥ 0, that
f (n ) (x ) = n !cn + (n + 1)!cn +1 x + x 2 gn (x ),
where gn (x ) is some power series .
Solution. This is a straightforward proof by induction on n ≥ 0.
The base step is obvious; for the inductive step, just observe that
f n +1 (x ) = ( f (n ) (x )) . As f (n ) (x ) is a power series, by assumption, its derivative is computed term by term.
(iii) Prove that cn = f (n ) (x )(0)/ n ! for all n ≥ 0. (Of course, this is
Solution. If n = 0, then our conventions that f (0) (x ) = f (x ) and
0! = 1 give the result. For the inductive step, use parts (i) and (ii).
1.42 (Leibniz) Prove that if f and g are C ∞ -functions, then
( f g) (n ) n (x ) =
k =0 n (k )
f (x ) · g (n −k ) (x ).
k Solution. The proof is by induction on n ≥ 1. ...
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