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Question
Write the first 4 terms of the logarithmic series
`log((1 + 3x)/(1 -3x))` Find the intervals on which the expansions are valid.
Solution
`log((1 + 3x)/(1 -3x))` = log(1 + 3x) – log(1 – 3x)
= `[3x - (3x)^2/2 + (3x)^3/3 - (3x)^4/4 ...] - [- 3x - (3x)^2/2 - (3x)^3/3 - (3x)^4/4 ...]`
= `3x - (3x)^2/2 + (3x)^3/3 - (3x)^4/4 .... + 3x + (3x)^2/2 + (3x)^2/3 + ....`
= `2(3x + (3x)^3/3 + (3x)^5/5 + (3x)^7/7 ...)`
Hence |3x| < 1
⇒ |x < `1/3`
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