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MCS 351 ENGINEERING MATHEMATICS SOLUTION OF ...

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1-3 Uniform Convergence :<br />

1. Let<br />

∞∑<br />

(z − 2i) 2n and |z − 2i| ≤ 0.999. We know that<br />

n=0<br />

1<br />

1 − z = 1 + z + z2 + z 3 + . . .<br />

1<br />

∞<br />

1 − (z − 2i) 2 = ∑<br />

(z − 2i) 2n = 1 + (z − 2i) 2 + (z − 2i) 4 + (z − 2i) 6 + . . .<br />

n=0<br />

Because |z − 2i| ≤ 0.999, this serie converges uniformly.<br />

2. Let<br />

∞∑<br />

n=0<br />

z 2n+1<br />

(2n + 1)! and |z| ≤ 1010 .999. Since<br />

sinh z =<br />

∞∑<br />

n=0<br />

z 2n+1<br />

(2n + 1)! = z + z3<br />

3! + z5<br />

5! + . . .<br />

this Maclaurin series of sinh z converges uniformly on every bounded set.<br />

3. Let<br />

∞∑<br />

n=0<br />

π n<br />

n 4 z2n and |z| ≤ 0.56. From ratio test, we have<br />

lim L = lim<br />

n→∞ n→∞<br />

∣<br />

π n+1<br />

(n+1) 4<br />

π n<br />

n 4<br />

∣ ∣ = lim<br />

n→∞<br />

and |z| ≤ 0.56, this serie converges uniformly on everywhere.<br />

πn 4<br />

(n + 1) 4 = π > 1,<br />

9-11 Power series :<br />

9. Let<br />

∞∑ (z + 1 − 2i) n<br />

4 n . By ratio test, we get<br />

n=0<br />

lim L = lim<br />

n→∞ n→∞<br />

∣<br />

1<br />

4 ( n+1)<br />

1<br />

4 n ∣ ∣ = lim<br />

n→∞<br />

then we can say that this serie converges uniformly on everywhere.<br />

10. Let<br />

∞∑<br />

n=0<br />

(z − i) 2n<br />

. Since<br />

(2n)!<br />

cosh z =<br />

∞∑<br />

n=0<br />

1<br />

4 = 1 4 < 1,<br />

z 2n<br />

(2n)! = 1 + z2<br />

2! + z4<br />

4! + . . .<br />

this Taylor series of cosh z with center i converges uniformly on every bounded set.<br />

49

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