If S = sumlimits_{n = 0}^infty frac{(log x)^{ | vedclass

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

Question

(C) We know that the Taylor series expansion for $\cosh(y)$ is given by $\sum\limits_{n=0}^\infty \frac{y^{2n}}{(2n)!} = \frac{e^y + e^{-y}}{2}$.Given

Vedclass · Accepted Answer

$\frac{1}{2}(x + x^{-1})$

Vedclass · Answer

(C) We know that the Taylor series expansion for $\cosh(y)$ is given by $\sum\limits_{n=0}^\infty \frac{y^{2n}}{(2n)!} = \frac{e^y + e^{-y}}{2}$.Given $S = \sum\limits_{n=0}^\infty \frac{(\log x)^{2n}}{(2n)!}$,we substitute $y = \log x$.Thus,$S = \frac{e^{\log x} + e^{-\log x}}{2}$.Since $e^{\log x} = x$ and $e^{-\log x} = e^{\log(x^{-1})} = x^{-1}$,we get $S = \frac{x + x^{-1}}{2}$.

If $S = \sum\limits_{n = 0}^\infty \frac{(\log x)^{2n}}{(2n)!}$,then $S$ =

Similar Questions

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

If $0 < x < 1$,then $\frac{3}{2} x^{2} + \frac{5}{3} x^{3} + \frac{7}{4} x^{4} + \ldots$ is equal to:

The coefficient of $n^{-r}$ in the expansion of $\log_{10}\left(\frac{n}{n-1}\right)$ is

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

$\log _4 2 - \log _8 2 + \log _{16} 2 - \ldots$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module