Find the binomial expansion of (1 - 4x)^(-1/2) and the interval on which it converges.?
回答 (1)
2015-01-26 6:23 pm
As previously stated we have:
(1 + x)^n ≈ 1 + nx + n(n - 1)x²/2! + n(n - 1)(n - 2)x³/3! + ....
so, (1 - 4x)⁻¹ʹ² ≈ 1 + (-1/2)(-4x) + (-1/2)(-3/2)(-4x)²/2! + (-1/2)(-3/2)(-5/2)(-4x)³/3! + ...
i.e. (1 - 4x)⁻¹ʹ² ≈ 1 + 2x + 6x² + 20x³ + ...
The expansion is valid, or converges, when -1 ≤ 4x ≤ 1
i.e. when -1/4 ≤ x ≤ 1/4
:)>
(1 + x)^n ≈ 1 + nx + n(n - 1)x²/2! + n(n - 1)(n - 2)x³/3! + ....
so, (1 - 4x)⁻¹ʹ² ≈ 1 + (-1/2)(-4x) + (-1/2)(-3/2)(-4x)²/2! + (-1/2)(-3/2)(-5/2)(-4x)³/3! + ...
i.e. (1 - 4x)⁻¹ʹ² ≈ 1 + 2x + 6x² + 20x³ + ...
The expansion is valid, or converges, when -1 ≤ 4x ≤ 1
i.e. when -1/4 ≤ x ≤ 1/4
:)>
收錄日期: 2021-04-21 01:17:53
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20150126095450AAMOHWb