Math 1010 HW12 solutions p.294, #4. Use the ratio test. p.294, #5. We have 1 1 = = 1 − (t − 1) + (t − 1)2 + · · · + (−1)k (t − 1)k + . . . t 1 + (t − 1) which has radius of convergence 1. Integrating term by term gives (t − 1)2 (−1)k+1 + ··· + (t − 1)k + . . . . 2 k The constant term is 0 since log 1 = 0. This also has radius of convergence 1 (a power series and its derivative have the same radius of convergence). log t = (t − 1) − p.294, #6. A function is analytic if it has a power series expansion (of positive radius of convergence) at every point c in its (open) domain. One can get a power series for F near any point by integrating the corresponding series for f term by term. The only thing to check is that the constant terms can be chosen consistently. However whenever two power series have overlapping domains of convergence, on their common domain they agree up to an additive constant; thus one can assign the constants in a consistent manner. P∞ 2+2k (b.) Differentiate 1/(1 − x) term by term to p.294, #7. (a.) k=0 x P∞ get k=1 kxk−1 (c.) See problem 4. p.294, #10. Use Theorem 7.4.1 relating the radius of convergence R to the growth of the coefficients: For any R0 < R and large enough n we have (n) |an |1/n < 1/R0 , so |f n!(x0 )| = |an | < 1/R0n . Choose M so that |ak | < M/R0k for k ≤ n. We conclude that |f (n) (x0 )| ≤ M n!rn for all n, where r = 1/R0 . p.307, #5. Suppose f is zero outside [a, b]. It suffices to show that (f ∗g)0 = f ∗ g 0 and f ∗ g 0 = f 0 ∗ g. Then we can differentiate k + m times, putting k derivatives on f and m on g. We have Z Z b Z b d d (f ∗g)0 (x) = f (t)g(x−t) dt = f (t)g(x−t) dt = f (t)g 0 (x−t) dt. dx R dx a a Here we can differentiate under the integral sign since the difference quotients g(x+h−t)−g(x−t) converge uniformly to g 0 (x − t) as h → 0, since g 0 (x) h is continuous. To show that f ∗ g 0 = f 0 ∗ g, use integration by parts. R1 Rx p.307, #15. If 0 ≤ x ≤ 1 then 0 f (t)f (x − t)dt = 0 1 · 1dt = x and if R1 R1 1 < x ≤ 2 then 0 f (t)f (x − t)dt = x−1 1dx = 2 − x. If x 6∈ [0, 2] then f ∗f (x) = 0. I would call this a “tent function” rather than a “hat function”. 1