By Derek F. Lawden
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Additional info for A Course in Applied Mathematics, Vol.1, 2
Consider the series with the parameter x in it. f (x) = x − x3 x5 x7 x9 + − + − ··· 3 5 7 9 (24) If I differentiate this with respect to x I get f (x) = 1 − x2 + x4 − x6 + x8 − · · · That looks a bit like the geometric series except that it has only even powers and the signs alternate. Is that too great an obstacle? If 1/(1 − x) has only the plus signs, then change x to * finite mean and variance 2—Infinite Series 35 −x, and 1/(1 + x) alternates in sign. Instead of x as a variable, use x2 , then you get exactly what you’re looking for.
That’s a parabola. 2—Infinite Series 38 The dots are the points where the intensity goes to zero, nλ/a. Between these directions it reaches a maximum. How big is it there ? These maxima are about halfway between the points where (ka sin θ)/2 = nπ. This is ka sin θ = (n + 1/2)π, 2 n = ±1, ±2, . . At these angles the value of I is, from Eq. (27), I= ka 2 2 1 (2n + 1)π/2 2 The intensity at θ = 0 is by Eq. 016, . . 9 Checking Results When you solve any problem, or at least think that you’ve solved it, you’re not done.
Can you rearrange the terms of an infinite series? Sometimes yes and sometimes no. If a series is convergent but not absolutely convergent, then each of the two series, the positive terms and the negative terms, is separately divergent. In this case you can rearrange the terms of the series to converge to anything you want! Take √ the series above that converges to ln 2. I want to rearrange the terms so that it converges to 2. Easy. Just start adding the positive terms until √ you’ve passed 2.
A Course in Applied Mathematics, Vol.1, 2 by Derek F. Lawden