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6 2b give the taylor series expression of gx fx answer

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Unformatted text preview: m −1 m xm . 6 (2b) Give the Taylor series expression of +gx fx Answer for (2b) ∞ : ζm −1 m m=2 . xm −1 x + −1 x + ζ = ζ + −1 ζ x + −1 ζ x −1 + ··· . ζ x : Work for (2b) +gx n = lim − n− ∞ → fx k=2 = lim − n− ∞ → n k=2 1 x x − ln 1 − k 2 1 x− k n ln 1− k=2 1− x k x 3 · ··· · 1− x n . In this last quantity, substitute x with −x. Then multiply −1 with it. Then the outcome is −f x + g x . Namely: 7 f x +g x =− +g −f . Hence the answer for (2b) is simply − ∞ m=2 −1 m−1 m ζm −1 m , which is simplified as ∞ m=2 (3) ζm −1 m Define the Euler’s Gamma function Γ x xm . as follows: +∞ Γx tx−1 e−t dt = x>0 t=0 Clearly, this function Γ x . is a generalization of the notion of factorials, in the following sense: For a non-negative integer m, Γ m+1 = m! . Also, know the following: Γ 2m + 1 2 = 2m − 1 ! ! √ π 2m where for m = 0 you should read m : non-negative integer , 2m − 1 !! as 1. 8 (3a) Γ0 = ‘divergent’ Γ , Γ1 = 1 Γ , Γ2 = 1 Γ , Γ3 = 2 Γ , Γ4 = Γ , Γ5 = Γ , 1 2 = 3 2 = 5 2 = 7 2 = 9 2 = 11 2 √ = π , 1√ π 2 , 3√ π 4 , , , . Compare the above with the results of “Extra Credit Homework – VI”. Γ (3b) 3 2 = 1 2 ·Γ 1 2 , Γ2 = 1 · Γ 1), Γ 5 2 = 3 2 ·Γ 3 2 , Γ3 = 2 · Γ 2), Γ 7 2 = 5 2 ·Γ 5 2 , Γ4 = · Γ 3), Γ 9 2 = ·Γ 7 2 , Γ5 = · Γ 4), Γ 11 2 = ·Γ 9 2 , Γ6 = · Γ 5), Γ 13 2 = ·Γ 11 2 9 . (4) (Extra 10pts) Let Γ x be as in (3). Prove that, for an arbitrary real number x with x > 0, the following identity holds: Γ x+1 = xΓ x. Show proof on a separate sheet. Attach that sheet with this copy. Hint for (...
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