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By Reinhard Diestel

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The series k=1 ak , where a3k+1 = 1, a3k+2 = 0 and a3k+3 = −1. Compute the Ces` aro means σn and show that the series has the Ces` aro sum 23 . 6 and the previous exercise can be generalized as follows. , that there is a positive integer p such that sn+p = sn for all n. Then the series is summable (C, 1) to the sum σ = (s1 + s2 + · · · + sp )/p. Prove this! 12 Show that if ak has a ﬁnite (C, 1) value, then lim n→∞ sn = 0. n What can be said about lim ak /k ? 13 Prove that if ak ≥ 0 and vergent in the usual sense.

19 Invert the following Laplace transforms: (c) ln s+3 s+2 (d) ln s2 + 1 s(s + 3) (e) s+1 s4/3 1 + e−s √ s s−1 (f) . 20 Solve the initial value problem y + y = 2et , t > 0, y(0) = y (0) = 2. 22 Solve y (t) − 2y (t) + y(t) = t et sin t, y(0) = 0, y (0) = 0. y (3) (t) − y (t) + 4y (t) − 4y(t) = −3et + 4e2t , y(0) = 0, y (0) = 5, y (0) = 3. 23 Solve the system x (t) − y(t) = e−t , x(0) = 3, x (0) = −2, y(0) = 0. 24 Solve the system x (t) + 2y (t) + x(t) = 0, x(0) = x (0) = y(0) = 0. 25 Solve the problem y (t) − 3y (t) + 2y(t) = 1, t > 2 ; 0, t < 2 y(0) = 1, y (0) = 0.

Then f (u) sin λu du = 0. lim λ→∞ I Proof. We do it in four steps. First, assume that the interval is compact, I = [a, b], and that f is constant and equal to 1 on the entire interval. Then b b sin λu du = − f (u) sin λu du = a a cos λu λ u=b = u=a 1 (cos λa − cos λb), λ which gives b f (u) sin λu du ≤ a 2 −→ 0 λ as λ → ∞. 26 2. Preparations The assertion is thus true for this f . Now assume that f is piecewise constant, which means that I (still assumed to be compact) is subdivided into a ﬁnite number of subintervals Ik = (ak−1 , ak ), k = 1, 2, .