frac{(a - 1) - frac{(a - 1)^2}{2} + frac{(a - | vedclass

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

Question

(A) We know that the logarithmic series expansion is given by $\ln(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \dots \infty$ for $-1 < x \le 1$. Subs

Vedclass · Accepted Answer

$\log_b a$

Vedclass · Answer

(A) We know that the logarithmic series expansion is given by $\ln(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \dots \infty$ for $-1 Substituting $x = a - 1$ in the numerator,we get $(a - 1) - \frac{(a - 1)^2}{2} + \frac{(a - 1)^3}{3} - \dots \infty = \ln(1 + a - 1) = \ln(a)$. Similarly,substituting $x = b - 1$ in the denominator,we get $(b - 1) - \frac{(b - 1)^2}{2} + \frac{(b - 1)^3}{3} - \dots \infty = \ln(1 + b - 1) = \ln(b)$. Thus,the expression becomes $\frac{\ln(a)}{\ln(b)}$. Using the change of base formula $\frac{\log_k a}{\log_k b} = \log_b a$,we get $\frac{\ln(a)}{\ln(b)} = \log_b a$.

$\frac{(a - 1) - \frac{(a - 1)^2}{2} + \frac{(a - 1)^3}{3} - \dots \infty}{(b - 1) - \frac{(b - 1)^2}{2} + \frac{(b - 1)^3}{3} - \dots \infty} = $

Similar Questions

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

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

The expansion $\log_e(1 + x) = \sum\limits_{i = 1}^\infty \left[ \frac{(-1)^{i + 1}x^i}{i} \right]$ is defined for:

$1+\frac{1}{3 \cdot 2^2}+\frac{1}{5 \cdot 2^4}+\frac{1}{7 \cdot 2^6}+\ldots$ is equal to

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module