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更新1:
To myisland8132: why it is 1/7?
Maclaurin Series 2
2014-05-15 12:22 am
Find the coefficient of the x^7 term in the Maclaurin series (Taylor series centered at 0) of the function f(x) = ∫ (from 0 to x) 1/(1+t^3) dt, where −1 < x < 1. [Answer: 1/7]
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回答 (2)
2014-05-15 7:43 am
✔ 最佳答案
1/(1 + x) = 1 - x + x^2 - x^3 + ...∫ (from 0 to x) 1/(1 + t^3) dt
= ∫ (from 0 to x) (1 - t^3 + t^6 - t^9 + ...) dt
= 1 - t^4/4 + t^7/7 - t^10/10 + ...
The coefficient of the x^7 term in the Maclaurin series is 1/7
2014-05-16 11:51 pm
myisland8132 係正解呀！！
To Mf Wong, 因為 你見到 個 integral 係 approximated by
1 - x⁴/4 + x⁷/7 - x¹⁰/10 + ...
= 1 - (1/4)x⁴ + (1/7)x⁷ - (1/10)x¹⁰ + ...
所以, x⁷ 的 係數 是 1/7 。
2014-05-16 16:02:19 補充：
myisland8132, I have acknowledged you here:
https://hk.knowledge.yahoo.com/question/question?qid=7014051200135
To Mf Wong, 因為 你見到 個 integral 係 approximated by
1 - x⁴/4 + x⁷/7 - x¹⁰/10 + ...
= 1 - (1/4)x⁴ + (1/7)x⁷ - (1/10)x¹⁰ + ...
所以, x⁷ 的 係數 是 1/7 。
2014-05-16 16:02:19 補充：
myisland8132, I have acknowledged you here:
https://hk.knowledge.yahoo.com/question/question?qid=7014051200135
收錄日期: 2021-04-24 23:35:50
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20140514000051KK00084