frac{m - n}{m + n} + frac{1}{3}left( frac{m - | vedclass

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

Question

(D) We use the logarithmic series expansion: $\frac{1}{2}\log_e\left( \frac{1+x}{1-x} \right) = x + \frac{x^3}{3} + \frac{x^5}{5} + \dots$ where $x =

Vedclass · Accepted Answer

$\frac{1}{2}\log_e\left( \frac{m}{n} \right)$

Vedclass · Answer

(D) We use the logarithmic series expansion: $\frac{1}{2}\log_e\left( \frac{1+x}{1-x} \right) = x + \frac{x^3}{3} + \frac{x^5}{5} + \dots$ where $x = \frac{m-n}{m+n}$.Substituting $x$ into the formula:$= \frac{1}{2}\log_e\left( \frac{1 + \frac{m-n}{m+n}}{1 - \frac{m-n}{m+n}} \right)$$= \frac{1}{2}\log_e\left( \frac{\frac{m+n+m-n}{m+n}}{\frac{m+n-(m-n)}{m+n}} \right)$$= \frac{1}{2}\log_e\left( \frac{2m}{2n} \right)$$= \frac{1}{2}\log_e\left( \frac{m}{n} \right)$.

$\frac{m - n}{m + n} + \frac{1}{3}\left( \frac{m - n}{m + n} \right)^3 + \frac{1}{5}\left( \frac{m - n}{m + n} \right)^5 + \dots \infty = $

Similar Questions

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

If $x, y, z$ are three consecutive positive integers,then $\frac{1}{2}\log_e x + \frac{1}{2}\log_e z + \frac{1}{2xz + 1} + \frac{1}{3}\left( \frac{1}{2xz + 1} \right)^3 + \dots = $

Evaluate: $\log _e(x + 1) - \log _e(x - 1) = $

$\frac{2}{1} \cdot \frac{1}{3} + \frac{3}{2} \cdot \frac{1}{9} + \frac{4}{3} \cdot \frac{1}{27} + \frac{5}{4} \cdot \frac{1}{81} + \dots \infty = $

If $|x| < 1$,then the coefficient of $x^5$ in the expansion of $(1 - x) \ln(1 - x)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module