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December 28, 2020 10:30 ampts 30 pts
Fig. ns of Taylor's series expansion of f(z) = (z-3)(z-4) a0 The 1st abce (z-3)(z-4) region of convergence. 1(z-3)(z-4) f2) f(2) +F(2) (z-2) + 2 (z - 2 This process is very lengthy. So, breaking f(z) in terms of partial fraction Z A B (z-3)(z-4) z 3 z-4 z A(z -4) + B(z 3) At z = 3, 3 = A(1) A=-3 z A(z -4) + B(z 3) z= 4, 4 = 0 + B B-4 -3 4 (z-3)(z-4) z-3 z-4 z =(-3) (z 4) + 4(z 3) fz2 f(z) = f(2) + f(z) (Z 2) z 2) .CO 2! 3(-2)(2-2 (-3)-(2-2)+-321 2 fz) = (2-3)(2-4) (z-3) 6 2 4 -1(-2) (2-3) (Z 2)2(2-3(2- 2) 5-2)-3(2-2 f(z) = 1 + 4(z - 2) + 3(z -2) I Tador's series expansion of f(z) at z = 2 sir please explain it step by step.especially f`(z) and f''(z)
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  • Indrajeet maurya thankyou
    see the attachment
    • cropped4627093337137598617.jpg
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