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: prove int^x _0 xf'(x)dx = xf(x)-int^x _0 f(x)dx using taylor expansion at x = 0
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x ' x
: PROVE ∫ xf (x)dx = xf(x)-∫ f(x)dx using Taylor expansion at x=0
: 0 0
(proof)
Taylor expansion for f(x) at x=0
(n)
f (0) n
f(x)=Σ------- x
n!
(n) (n)
' f (0) n x ' f (0) n+1
=> xf(x) = Σ-------- x => ∫xf(x)dx = Σ------------ x
(n-1)! 0 (n-1)!(n+1)
(n)
f (0) n+1
xf(x) = Σ-------- x
n!
(n)
x f (0) n+1
∫f(x)dx = Σ-------- x
0 (n+1)!
Substituting these into the equation
x ' x
∫xf(x)dx = xf(x) - ∫f(x)dx
0 0
yields
(n) (n) (n)
f (0) n+1 f (0) n+1 f (0) n+1
Σ------------ x = Σ--------- x - Σ-------- x
(n-1)!(n+1) n! (n+1)!
1 1 1
=> ------------- = --- - ------- -------------(*)
(n-1)!(n+1) n! (n+1)!
n
Left hand side of Eq.(*) = -------
(n+1)!
(n+1)-1 n
Right hand side of Eq.(*) = ---------- = ------- Holds !
(n+1)! (n+1)!
END !
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※ 編輯: phs 來自: 140.112.101.4 (11/04 17:27)