By Leif Mejlbro
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Additional info for Complex Functions Examples c-4 - Power series
3! 4! 2! 4! = 1+z+ z2 + 0 · z3 + · · · . 6 1 π is analytic for |z| < , so cos z 2 1 = a0 + a 1 z + a 2 z 2 + a 3 z 3 + · · · . cos z Since cos(−z) = cos z is an even function, it follows immediately that a1 = 0 and a3 = 0, and the series expansion is reduced to 1 = a0 + a 2 z 2 + o z 3 . cos z π Therefore, if |z| < , then 2 1 · cos z = a0 + a2 z 2 + a4 z 4 + · · · cos z a0 = a0 + − + a2 z 2 + o z 3 , 2 1− 1 = z4 z2 + − ··· 2! 4! and it follows from the identity theorem that a0 = 1 and a2 = 1 a0 = , 2 2 thus 1 1 = 1 + z2 + 0 · z3 + · · · , cos z 2 |z| < π .
A Taylor series is of the form +∞ f (z) = f (n) (z0 ) n (z − z0 ) . n! z −n , n ∈ N. 5 –1 Figure 6: The series is convergent in the larger disc, and it converges towards Log z in the smaller disc, actually in that part of the larger disc which lies in the second quadrant. 2−n · (1 + i)n = −(n − 1)! n exp i n π , 4 and we get the Taylor series from z0 = −1 + i, +∞ Log z (n − 1)! n! n=1 = Log(−1 + i) − = 1 1 3π ln 2 + i − n 2 4 n=1 +∞ 1 √ 2 1 √ 2 n exp i n n exp i n π 4 π 4 (z + 1 − i)n (z + 1 − i)n .
2! 4! 6! n=1 z ∈ C, hence (if |1 − cos z| < 1) 1 cos z z2 · z2 z2 z4 z6 z2 z4 − + − ··· + −2· · + ··· 2! 4! 6! 2! 2! 4! 2! + ··· when |1 − cos z| < 1. Note that because the power series expansion is unique in the larger domain π |z| < , the same expansion holds here. 15 Find the ﬁrst three terms of the power series expansion from z 0 = 0 for the solution w of the transcendental equation w e−w = z. Hint: Find w(0), w (0), w (0), . . , and then apply Taylor’s formula. When we diﬀerentiate the equation (with respect to z) z = w e−w , then 1 = e−w − w e−w dw dw = e−w (1 − w) .
Complex Functions Examples c-4 - Power series by Leif Mejlbro