In the expansion of log_e frac{1}{1 - x - x^2 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given expression: $f(x) = \log_e \left[ \frac{1}{1 - x - x^2 + x^3} \right]$Factorizing the denominator: $1 - x - x^2 + x^3 = (1 - x) - x^2(1 - x)

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(B) Given expression: $f(x) = \log_e \left[ \frac{1}{1 - x - x^2 + x^3} \right]$Factorizing the denominator: $1 - x - x^2 + x^3 = (1 - x) - x^2(1 - x) = (1 - x)(1 - x^2) = (1 - x)(1 - x)(1 + x) = (1 - x)^2(1 + x)$So,$f(x) = \log_e \left[ (1 - x)^{-2}(1 + x)^{-1} \right]$Using logarithmic properties: $f(x) = -2 \log_e(1 - x) - \log_e(1 + x)$Using the series expansions $\log_e(1 - x) = -x - \frac{x^2}{2} - \frac{x^3}{3} - \dots$ and $\log_e(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \dots$$f(x) = -2(-x - \frac{x^2}{2} - \dots) - (x - \frac{x^2}{2} + \dots)$$f(x) = (2x + x^2 + \dots) - (x - \frac{x^2}{2} + \dots)$$f(x) = (2 - 1)x + (1 + 0.5)x^2 + \dots = x + 1.5x^2 + \dots$The coefficient of $x$ is $1$.

In the expansion of $\log_e \frac{1}{1 - x - x^2 + x^3}$,the coefficient of $x$ is

Similar Questions

$1 + \frac{2}{3} - \frac{2}{4} + \frac{2}{5} - \dots \infty = $

$(0.5) - \frac{(0.5)^2}{2} + \frac{(0.5)^3}{3} - \frac{(0.5)^4}{4} + \dots$

The expression $\log_{e} 2 + \log_{e} \left( 1 + \frac{1}{2} \right) + \log_{e} \left( 1 + \frac{1}{3} \right) + \dots + \log_{e} \left( 1 + \frac{1}{n - 1} \right)$ is equal to

The sum of the series $\frac{1}{2 \times 3} + \frac{1}{4 \times 5} + \frac{1}{6 \times 7} + \dots = $

If $y = 2x^2 - 1$,then $\left[ \frac{1}{y} + \frac{1}{3y^3} + \frac{1}{5y^5} + \dots \right]$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module