Question

1. Given $e^{\omega} = \sum_{n=0}^{\infty} \frac{\omega^n}{n!}$ prove \\ (a) $cosh \omega \equiv \frac{1}{2}(e^{\omega} + e^{-\omega}) = \sum_{n=0,2,4,...}^{\infty} \frac{\omega^n}{n!} $ \\ (b) $sinh \omega \equiv \frac{1}{2}(e^{\omega} - e^{-\omega}) = \sum_{n=1,3,5,...}^{\infty} \frac{\omega^n}{n!}$

          1. Given $e^{\omega} = \sum_{n=0}^{\infty} \frac{\omega^n}{n!}$ prove \\
(a) $cosh \omega \equiv \frac{1}{2}(e^{\omega} + e^{-\omega}) = \sum_{n=0,2,4,...}^{\infty} \frac{\omega^n}{n!} $ \\
(b) $sinh \omega \equiv \frac{1}{2}(e^{\omega} - e^{-\omega}) = \sum_{n=1,3,5,...}^{\infty} \frac{\omega^n}{n!}$

1. Given e^ω = ∑n=0^∞(ω^n)/(n!) prove

(a) cosh ω≡(1)/(2)(e^ω + e^-ω) = ∑n=0,2,4,...^∞(ω^n)/(n!)

(b) sinh ω≡(1)/(2)(e^ω - e^-ω) = ∑n=1,3,5,...^∞(ω^n)/(n!)

Added by Benjamin V.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Sri K

10/09/2023

Step 1

Step 1: Start with the definition of cosh(ω) and sinh(ω): cosh(ω) = (1/2)(e^(ω) + e^(-ω)) sinh(ω) = (1/2)(e^(ω) - e^(-ω)) Show more…

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Given e^(omega )=sum_(n=0)^(infty ) (omega ^(n))/(n!) prove (a) coshomega =(1)/(2)(e^(omega )+e^(-omega ))=sum_(n=0,2,4...)^(infty ) (omega ^(n))/(n!) (b) sinhomega =(1)/(2)(e^(omega )-e^(-omega ))=sum_(n=1,3,5...)^(infty ) (omega ^(n))/(n!) l. Given e" = Sn wn (b) wn i=1.3.5,... n!