power series

2013-10-19 5:30 am

回答 (2)

2013-10-21 2:37 am
✔ 最佳答案
a > 0
f(x)=x^(-6) + 1
f'(x)=(-1) 6!/5! x^(-7) , f'(a)=(-1) 6!/5! a^(-7)
f''(a) = (-1)^2 7!/5! a^(-8)
(d/dx)^n f(x) |_(x=a) = (-1)^n (n+5)!/5! a^(-n-6)
Taylor series
f(x)= f(a) + f'(a) (x-a) + (1/2!)f"(a) (x-a)^2+ ...+(1/n!)f^(n)(a) (x-a)^n + ... ---(*)
(Ratio test)
| [1/(n+1)! f^(n+1)(a) (x-a)^(n+1)]/[1/n! f^(n)(a) (x-a)^n] |
= | -(n+6)/(n+1) (x-a)/a | -> |x-a| / a

If |x-a| / a < 1, then (*) converges, so the radius of conv. is a

In case of a=1:
f(x)= 2 - C(6,5) (x-1)+ C(7,5) (x-1)^2 - C(8,5)(x-1)^3 + ...
Radius of conv. = 1 [interval of conv. = (0, 2) ]

2013-10-20 10:33 pm
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2013-10-20 14:34:10 補充:
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