1 + frac{(log_e n)^2}{2!} + frac{(log_e n)^4} | vedclass

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

Question

(C) We know the expansion of the hyperbolic cosine function: $\cosh(x) = 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \dots = \frac{e^x + e^{-x}}{2}$.Substit

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) We know the expansion of the hyperbolic cosine function: $\cosh(x) = 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \dots = \frac{e^x + e^{-x}}{2}$.Substituting $x = \log_e n$,we get:$1 + \frac{(\log_e n)^2}{2!} + \frac{(\log_e n)^4}{4!} + \dots = \frac{e^{\log_e n} + e^{-\log_e n}}{2}$.Since $e^{\log_e n} = n$ and $e^{-\log_e n} = e^{\log_e(n^{-1})} = n^{-1}$,the expression becomes:$\frac{n + n^{-1}}{2}$.

$1 + \frac{(\log_e n)^2}{2!} + \frac{(\log_e n)^4}{4!} + \dots = $

Similar Questions

$\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 = $

Evaluate: $\log_e \sqrt{\frac{1+x}{1-x}}$

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 value of $\log_e \left( 1 + ax^2 + a^2 + \frac{a}{x^2} \right)$ is

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module