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ORIGIN 1 := Solution: a. Because sin(4 π )=0, this reduces to 1 16 43.8 + + 1 20 i ii 1 + () [] = 0.00284 = b. For r 0.1 0.3 , 0.9 .. := r 0.1 0.3 0.5 0.7 0.9 = r is a vector. r 3 = does not work 0 k 2r k = 2.2222222222222222222 2.8571428571428571429 4.0 6.6666666666666666667 20.0

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A variation on this approach (that I prefer): i1 2 , 5 .. := r i 0.2 i 0.1 := r 0.1 0.3 0.5 0.7 0.9 = r is a column matrix. r 3 0.5 = 0 k 2r i () k = 2.2222222222222222222 2.8571428571428571429 4.0 6.6666666666666666667 20.0 Second approach: we define the function fx 0 k 2x k = := Then we can define the results (with more typing) as: f 0.1 ( ) 2.2222222222222222222 Notice that f has to be evaluate using the symbolic arrow. f 0.3 ( ) 2.8571428571428571429 f 0.5 ( ) 4.0 f 0.7 ( ) 6.6666666666666666667 f 0.9 ( ) 20.0 More efficient from MathCad's viewpoint is to first evaluate 0 k k = 1x if 2 x1 x 1 < if Then define g x 2 := ,for |x|<1, and fird the values using g. . g 0.7 ( ) 6.66667 =
Third approach: We write a short program that returns the vector of solutions corresponding to different values of r. BUT you cannot evaluate the infinite sum symbollically inside the program.

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