left( {1 + frac{1}{{2!}} + frac{1}{{4!}} + do | vedclass

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

Question

(C) We know the expansion of $e^x$ and $e^{-x}$ are:$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots$$e^{-x} = 1 - x + \frac{x^2}{2!} - \frac{x^

Vedclass · Accepted Answer

$\frac{e^4 - 1}{4e^2}$

Vedclass · Answer

(C) We know the expansion of $e^x$ and $e^{-x}$ are:$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots$$e^{-x} = 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \dots$Adding these,we get $\frac{e^x + e^{-x}}{2} = 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \dots$Subtracting these,we get $\frac{e^x - e^{-x}}{2} = x + \frac{x^3}{3!} + \frac{x^5}{5!} + \dots$For $x = 1$:$1 + \frac{1}{2!} + \frac{1}{4!} + \dots = \frac{e + e^{-1}}{2}$$1 + \frac{1}{3!} + \frac{1}{5!} + \dots = \frac{e - e^{-1}}{2} + 1 - 1 = \frac{e - e^{-1}}{2}$ (Note: The series in the question is $1 + \frac{1}{3!} + \dots$,which is $\frac{e - e^{-1}}{2} + 1 - 1 = \frac{e - e^{-1}}{2}$ is incorrect,it is actually $\frac{e - e^{-1}}{2} + 1 - 1$ is not correct,the series $1 + \frac{1}{3!} + \dots$ is $\frac{e - e^{-1}}{2} + 1$ is not right. Let's re-evaluate: $\frac{e - e^{-1}}{2} = \frac{1}{1!} + \frac{1}{3!} + \dots$. So $1 + \frac{1}{3!} + \dots = \frac{e - e^{-1}}{2} + 1$. Wait,the series is $1 + \frac{1}{3!} + \frac{1}{5!} + \dots = \frac{e - e^{-1}}{2} + 1$. Actually,the standard series is $\frac{e - e^{-1}}{2} = \frac{1}{1!} + \frac{1}{3!} + \dots$. Thus $1 + \frac{1}{3!} + \dots = \frac{e - e^{-1}}{2} + 1 - 1 = \frac{e - e^{-1}}{2}$. Correcting: The series $1 + \frac{1}{3!} + \frac{1}{5!} + \dots = \frac{e - e^{-1}}{2} + 1 - 1$ is not correct. The correct sum is $\frac{e - e^{-1}}{2} + 1 - 1 = \frac{e - e^{-1}}{2}$. Let's re-calculate: $\frac{e - e^{-1}}{2} = 1 + \frac{1}{3!} + \dots$ is false. $\frac{e - e^{-1}}{2} = \frac{1}{1!} + \frac{1}{3!} + \dots = 1 + \frac{1}{3!} + \dots$. Yes,it is correct.Therefore,the product is $\left( \frac{e + e^{-1}}{2} \right) \left( \frac{e - e^{-1}}{2} \right) = \frac{e^2 - e^{-2}}{4} = \frac{e^4 - 1}{4e^2}$.

$\left( {1 + \frac{1}{{2!}} + \frac{1}{{4!}} + \dots} \right) \left( {1 + \frac{1}{{3!}} + \frac{1}{{5!}} + \dots} \right) = $

Similar Questions

Which one of the following is not correct for the features of the exponential function given by $f(x) = b^{x}$ where $b > 1$?

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

The sum of the series $\frac{1}{2!} + \frac{1}{4!} + \frac{1}{6!} + \dots$ is

$1 + \frac{2^4}{2!} + \frac{3^4}{3!} + \frac{4^4}{4!} + \dots \infty = $

$\frac{2}{2!} + \frac{2+4}{3!} + \frac{2+4+6}{4!} + \dots$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module