Advertisements
Advertisements
Question
Write the first 4 terms of the logarithmic series
`log((1 - 2x)/(1 + 2x))` Find the intervals on which the expansions are valid.
Solution
`log((1 - 2x)/(1 + 2x))` = log(1 – 2x) – log(1 + 2x)
= `[ - 2x - (2x)^2/2 - (2x)^3/3 - (2x)^4/4 ....] - [2x- (2x)^2/2 + (2x)^3/3 - (2x)^4/4 ...]`
= `- 2x - (2x)^2/2 - (2x)^3/3 - (2x)^4/4 .... - 2x +(2x)^2/2 - (2x)^3/3+ (2x)^4/4 ...`
= `-2(2x + (2x)^3/3 + (2x)^5/5 + (2x)^7/7...)`
Hence |2x| < 1
⇒ |x| < `1/2`
APPEARS IN
RELATED QUESTIONS
Find `root(3)(10001)` approximately (two decimal places
Prove that `root(3)(x^3 + 6) - root(3)(x^3 + 3)` is approximately equal to `1/x^2` when x is sufficiently large
Write the first 6 terms of the exponential series
e5x
Write the first 6 terms of the exponential series
`"e"^(-2x)`
Write the first 6 terms of the exponential series
`"e"^(1/2x)`
Write the first 4 terms of the logarithmic series
log(1 + 4x) Find the intervals on which the expansions are valid.
Write the first 4 terms of the logarithmic series
`log((1 + 3x)/(1 -3x))` Find the intervals on which the expansions are valid.
If y = `x + x^2/2 + x^3/3 + x^4/4 ...`, then show that x = `y - y^2/(2!) + y^3/(3!) - y^4/(4) + ...`
If p − q is small compared to either p or q, then show `root("n")("p"/"q")` ∼ `(("n" + 1)"p" + ("n" - 1)"q")/(("n"- 1)"p" +("n" + 1)"q")`. Hence find `root(8)(15/16)`
Choose the correct alternative:
The coefficient of x6 in (2 + 2x)10 is
Choose the correct alternative:
The coefficient of x8y12 in the expansion of (2x + 3y)20 is
Choose the correct alternative:
If (1 + x2)2 (1 + x)n = a0 + a1x + a2x2 + …. + xn + 4 and if a0, a1, a2 are in AP, then n is
Choose the correct alternative:
The value of the series `1/2 + 7/4 + 13/8 + 19/16 + ...` is
Choose the correct alternative:
The sum of an infinite GP is 18. If the first term is 6, the common ratio is
Choose the correct alternative:
The coefficient of x5 in the series e-2x is
Choose the correct alternative:
The value of `1 - 1/2(2/3) + 1/3(2/3)^2 1/4(2/3)^3 + ...` is
