By Bernard W. Roos
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8. series exp(x) = ∞ ∑ xn n=0 n! 44) is uniformly convergent and hence diﬀerentiable term by term. Thus, d exp(x) ∑ x(n−1) = n ???????? n! n=0 ∞ = ∞ ∑ x(n−1) (n − 1)! 49) so exp(−x) is a strictly decreasing, positive function of x for x > 0. t. 4. ′ g (x) = g(x) then g(x) = c exp(x) where c is a constant. Proof: Let f (x) = g(x) exp(x) then f ′ (x) = 0; hence f (x) = c where c is a constant; hence, c exp(x) = g(x). t. g ′ (x) = g(x) and g(0) = 1, we know that g(x) = c exp(x). Substituting the values at 0 shows that c = 1; hence, g(x) = exp(x).
KGaA. Published 2016 by Wiley-VCH Verlag GmbH & Co. KGaA. 6) We can also deﬁne calculus for complex numbers. 7) where x, y are diﬀerentiable functions of t then we have the following deﬁnition. 3. 9) Here, we are talking about the derivate of the complex function of a real variable, which looks a lot like the derivative of a vector. The important case of a derivative of a complex function of a complex variable is discussed later in Chapter 8. 14) Motivated by this analysis, we can deﬁne the exponent of a complex number z in terms of a power series.
2n+1) ???? n (????t) + ????t + … + (−) +… ???? (2n + 1)! 34) Thus, we see that the general solution of Eq. 35) where A and B are determined by the boundary conditions. 36) where N and ???? are constants determined by the boundary conditions. 38) are all equally good, and we can use the one which is the most convenient. Notice that in each case we have two constants 37 38 2 Complex Numbers (????, ????), (A, ????), (A, ????), which is all we need for a general solution. 37). 43) where C, D are constants to be determined by the boundary conditions.
Analytic functions and distributions in physics and engineering by Bernard W. Roos