exp(ix) = cos(x) + i sin(x),
where i = (-1)^(1/2) , and which is widely used in different items of mathematics is usually deduced from the Maclaurin expansions of the functions involved.
But the theory of Taylor (Maclaurin) expansions is a part of more general theory developed in the course of the functions of complex variable. As the Moivre-Laplace formula has numerous applications outside this theory, it seems reasonable to deduce it without references to Maclaurin series.
To prove the Moivre-Laplace formula: exp(ix) = cos(x) + i sin(x) without use of the Maclaurin expansions.© BrainMass Inc. brainmass.com October 24, 2018, 5:22 pm ad1c9bdddf
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e ix = cos(x) + i sin(x),
where i = , and which is widely used in different items of mathematics is usually deduced from the Maclaurin expansions of the functions involved:
cos(x) = ; sin(x) = .
As -1 = i2 , we have:
cos(x) + i sin(x) = + and the sum of these two series can obviously be written as a single series
and the latter series is just the Maclaurin expansion of e ix .
But the theory of Taylor (Maclaurin) expansions is a part of more general ...
The Moivre-Laplace Formula is prven without the use of Maclaurin expansions.
Evaluating an Integral with 2nd Order Pole : Moivre-Laplace Fomulation
Problem: Evaluate the integral from 0 to INF of: (x^a)/(x^2 +4)^2 dx, -1 < a < 3 We are to use f(z)= (z^a)/(z^2 +4)^2, with z^a = e^(a Log z), Log z= ln|z| + i Arg z, and -pi/2 < Arg z < 3pi/2. I have found the residue at 2i to be: [2^a(1-a)/16]*[cos ((pi*a)/2) + i sin ((pi*a)/2). Please let me know if this is correct and how to solve this problem. Many thanks for your help.
Use the branch cut of the negative Im axis, and use the following curve:
C= -R to R, -p to p, p to R, and Cr from 0 to pi.
Can anyone do this problem using these parameters?View Full Posting Details