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

For $|x| < 1$,the coefficient of $x^3$ in the expansion of $\log(1+x+x^2)$ in ascending powers of $x$ is:

$\frac{1}{1 \cdot 2 \cdot 3} + \frac{1}{3 \cdot 4 \cdot 5} + \frac{1}{5 \cdot 6 \cdot 7} + \dots \infty = $

$\frac{1}{2 \cdot 3} + \frac{1}{4 \cdot 5} + \frac{1}{6 \cdot 7} + \frac{1}{8 \cdot 9} + \dots$ is equal to

If $y = x - \frac{x^2}{2} + \frac{x^3}{3} - \dots \infty$,then $x = $

Let $\alpha$ and $\beta$ be the roots of $5x^2 - 3x - 1 = 0$. Then the expression $\left[ (\alpha + \beta)x - \left( \frac{\alpha^2 + \beta^2}{2} \right)x^2 + \left( \frac{\alpha^3 + \beta^3}{3} \right)x^3 - \dots \right]$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module