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9). Then bk = ak . Proof From (b) we know that z is also a zero of q with multiplicity m. Therefore bk is uniquely determined. For x ∈ C\{z1 , . . 9) that n mj j=1 k=1 ajk p p(x) = = (x) = (x − zj )k q(x) q n mj j=1 k=1 ajk . 8. Suppose now that r := p/q for p, q ∈ R[X] with deg(p) < deg(q) is a real rational function. 10(c), the summand a a (a + a)X − (za + za) Re(a)X − Re(za) + = =2 2 , X −z X −z (X − z)(X − z) X − 2 Re(z)X + |z|2 and we have D := (Re z)2 − |z|2 < 0. 14) for x√∈ I, where the coeﬃcients A and B follow uniquely from Re a, Re z, Re(za) and −D.

D) (the Wallis product) This says4 π = 2 ∞ k=1 4k 2 = lim 4k 2 − 1 n→∞ n k=1 4k 2 4k 2 − 1 2k 2k 2 2 4 4 6 6 · · ··· . 3) on the interval [0, π/2]. We set π/2 sinn x dx An := for n ∈ N . 3), it follows that A0 = π/2 , A1 = 1 , An = n−1 An−2 n for n ≥ 2 . A simple induction argument gives A2n = (2n − 1)(2n − 3) · · · · · 3 · 1 π · , 2n(2n − 2) · · · · · 4 · 2 2 A2n+1 = 2n(2n − 2) · · · · · 4 · 2 . (2n + 1)(2n − 1) · · · · · 5 · 3 From this follows the relations 2 A2n+1 2n · 2n(2n − 2)(2n − 2) · · · · · 4 · 4 · 2 · 2 · = A2n (2n + 1)(2n − 1) (2n − 1)(2n − 3) · · · · · [5 · 3][3 · 1] π = 2 π n k=1 (2k)2 (2k)2 − 1 and lim n→∞ A2n+2 2n + 1 = lim =1.

C) Suppose f : I → E, and Z := (α0 , . . , αn ) is a partition of I with between points ξj ∈ [αj−1 , αj ]. 6 in Volume III). 3 The Cauchy–Riemann Integral 23 the Riemann sum. If f is Riemann integrable, then n β f (ξj )(αj − αj−1 ) , f dx = lim Z →0 α j=1 expresses its integral symbolically. Exercises 1 Deﬁne [·] to be the ﬂoor function. 2 and also 1 0 β (iv) sign x dx . 7. 2 Compute 3 Suppose F is a Banach space and A ∈ L(E, F ). Then show for f ∈ S(I, E) that Af := x → A(f (x)) ∈ S(I, F ) and A β α f= β α Af .