Use taylor expansion on [sin(h)-h*cos(h)] / h
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Use taylor expansion on [sin(h)-h*cos(h)] / h

[From: ] [author: ] [Date: 11-09-08] [Hit: ]
- 1/(2n)!],= Σ(n = 0 to ∞) (-1)^n h^(2n) [1/(2n+1)! - (2n+1)/(2n+1)!= Σ(n = 0 to ∞) (-1)^n h^(2n) * (-2n)/(2n+1)!= Σ(n = 0 to ∞) 2(-1)^(n+1) h^(2n)/(2n+1)!......
(sin h - h cos h)/h
= (1/h) sin h - cos h
= (1/h) * Σ(n = 0 to ∞) (-1)^n h^(2n+1)/(2n+1)! - Σ(n = 0 to ∞) (-1)^n h^(2n)/(2n)!
= Σ(n = 0 to ∞) (-1)^n h^(2n)/(2n+1)! - Σ(n = 0 to ∞) (-1)^n h^(2n)/(2n)!
= Σ(n = 0 to ∞) (-1)^n h^(2n) [1/(2n+1)! - 1/(2n)!], combining the series
= Σ(n = 0 to ∞) (-1)^n h^(2n) [1/(2n+1)! - (2n+1)/(2n+1)!]
= Σ(n = 0 to ∞) (-1)^n h^(2n) * (-2n)/(2n+1)!
= Σ(n = 0 to ∞) 2(-1)^(n+1) h^(2n)/(2n+1)!.

I hope this helps!
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