Page 857 - Advanced_Engineering_Mathematics o'neil
P. 857

Answers to Selected Problems  837


                                                             a
                                                         u v =  (−a cosh(u)sin(v)i + sinh(u)cos(v)j),
                                                            h 1
                                                                       u z = k
                                                                   1  
  ∂      ∂        ∂ F 3
                                                       ∇· F(u,v, z) =   (F 1 h 1 ) +  (F 2 h 1 ) +
                                                                   h 2 1  ∂u   ∂v         ∂z

                                                                      1 ∂ F 3  ∂ F 2
                                                        ∇× F(u,v, z) =     −      u u
                                                                      h 1 ∂v  ∂z

                                                                      ∂ F 1  1 ∂ F 3
                                                                   +     −        u v
                                                                      ∂z   h 1 ∂u
                                                                     1     ∂     ∂
                                                                   +       (F 2 h 1 −  (F 1 h 1 ) u z
                                                                     h  2 1  ∂u  ∂v
                                                                                2
                                                                          2
                                                                                      2
                                                                      1    ∂ f  ∂ f     ∂ f
                                                            2
                                                           ∇ f (u,v, z) =    +     +
                                                                      h 2 1  ∂u  2  ∂v 2  ∂z  2
                            4.
                                                                       √
                                                                             2
                                                                          2
                                                                h u = h v =  u + v ,h z = 1
                                                                     1 ∂ f   1 ∂ f   ∂ f
                                                          ∇ f (u,v, z) =  u u +  u v +  u z
                                                                     h u ∂u  h v ∂v  ∂z
                                                                     1     ∂      ∂
                                                         ∇· F(u,v, z) =    (h u F 1 ) +  (h v F 2 )
                                                                     h 2 u  ∂u    ∂v
                                                                     1 ∂    2
                                                                   +      h F 3
                                                                           u
                                                                      2
                                                                     h ∂z
                                                                      u
                                                                     1     ∂    ∂
                                                        ∇× F(u,v, z) =    (F 3 ) −  (h u F 2 ) u 1
                                                                    h u  ∂u     ∂z
                                                                     1     ∂      ∂
                                                                  +       (h u F 1 ) −  (F 3 ) u 2
                                                                    h  2 v  ∂z   ∂u
                                                                     1     ∂      ∂
                                                                  +       (h v F 2 ) −  (h u F 1 ) u 3
                                                                    h  2  ∂u     ∂v
                                                                     v
                                                                     2
                                                                          2
                                                                1     ∂ f  ∂ f  ∂     ∂ f
                                                            2                      2
                                                           ∇ f =       +    +     h  u
                                                                h 2 u  ∂u 2  ∂v  2  ∂z  ∂z
                            CHAPTER THIRTEEN FOURIER SERIES
                            Section 13.1 Why Fourier Series?
                            3. If p(x) has degree k, then differentiating p(x), k + 1 times yields the zero function, while    N n=1  b n sin(nx) can be
                               differentiated any number of times, and none of these derivatives is identically zero on [0,π].
                            Section 13.2 The Fourier Series of a Function
                             1. 4; the series (consisting of one term) converges to 4 on [−3,3].
                             3.
                                                                            ∞
                                                          1        2          (−1) n
                                                           sinh(π) +  sinh(π)      cos(nπx),
                                                                               2
                                                          π        π          n + 1
                                                                           n=1
                                converging to cosh(πx) for −1 ≤ x ≤ 1.


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                                   October 14, 2010  17:50  THM/NEIL    Page-837        27410_25_Ans_p801-866
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