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 nonnegative 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 : nonnegative 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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This note was uploaded on 01/12/2014 for the course MATH 116 taught by Professor Scholle,minho during the Fall '08 term at Kansas.
 Fall '08
 SCHOLLE,MINHO
 Calculus

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