If y = 2x^2 - 1,then left[ frac{1}{y} + frac{ | vedclass

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

Question

(B) We know the logarithmic series expansion: $\frac{1}{2} \ln \left( \frac{1+z}{1-z} \right) = z + \frac{z^3}{3} + \frac{z^5}{5} + \dots$ for $|z| <

Vedclass · Accepted Answer

$\frac{1}{2} \left[ \frac{1}{x^2} + \frac{1}{2x^4} + \frac{1}{3x^6} + \dots \right]$

Vedclass · Answer

(B) We know the logarithmic series expansion: $\frac{1}{2} \ln \left( \frac{1+z}{1-z} \right) = z + \frac{z^3}{3} + \frac{z^5}{5} + \dots$ for $|z| Let $z = \frac{1}{y}$. Then the given expression is $\frac{1}{2} \ln \left( \frac{1 + 1/y}{1 - 1/y} \right) = \frac{1}{2} \ln \left( \frac{y+1}{y-1} \right)$.Substitute $y = 2x^2 - 1$:$\frac{1}{2} \ln \left( \frac{2x^2 - 1 + 1}{2x^2 - 1 - 1} \right) = \frac{1}{2} \ln \left( \frac{2x^2}{2x^2 - 2} \right) = \frac{1}{2} \ln \left( \frac{x^2}{x^2 - 1} \right) = -\frac{1}{2} \ln \left( \frac{x^2 - 1}{x^2} \right) = -\frac{1}{2} \ln \left( 1 - \frac{1}{x^2} \right)$.Using the expansion $\ln(1-u) = -(u + \frac{u^2}{2} + \frac{u^3}{3} + \dots)$ where $u = \frac{1}{x^2}$:$-\frac{1}{2} [ -(\frac{1}{x^2} + \frac{1}{2x^4} + \frac{1}{3x^6} + \dots) ] = \frac{1}{2} [ \frac{1}{x^2} + \frac{1}{2x^4} + \frac{1}{3x^6} + \dots ]$.

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

Similar Questions

If $x = \operatorname{sech}^{-1} \frac{1}{2} + \tanh^{-1} \frac{1}{2}$,then $\cosh x =$

If $|x| < 1$ and $y = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \ldots$,then $x$ is equal to :

The value of the infinite series $\log _4 2 - \log _8 2 + \log _{16} 2 - \dots \infty$ is:

$\frac{4}{1 \times 3} - \frac{6}{2 \times 4} + \frac{12}{5 \times 7} - \frac{14}{6 \times 8} + \dots \infty = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module