HOROSCOPE EXPLORER 3.81 (8 LANGUAGES INCLUDING HINDI) [2021]

HOROSCOPE EXPLORER 3.81 (8 LANGUAGES INCLUDING HINDI) [2021]





             

HOROSCOPE EXPLORER 3.81 (8 LANGUAGES INCLUDING HINDI)


» Download Horoscope Explorer: » Horoscope Explorer: Prove $\sum_{k=0}^{\infty} \frac{\mathrm{sign}(k)}{k!} x^k = \sinh (x)$ How can I prove that : $$ \sum_{k=0}^{\infty} \frac{\mathrm{sign}(k)}{k!} x^k = \sinh (x) $$ where $\mathrm{sign}(k)$ is the sign of the integer $k$ and $x \in \mathbb{R}$. More generally, how can I prove that : $$ \sum_{k=0}^{\infty} \frac{\mathrm{sign}(k)}{k!} x^k = \cosh (x) $$ where $\mathrm{sign}(k)$ is the sign of the integer $k$ and $x \in \mathbb{R}$. Can this be proved with the help of the Cauchy product, or Taylor series? A: Introduce: $$ L(x):=\frac{\cosh(x)}{x}\cdot\sum_{k\ge 0}\frac{\mathrm{sign}(k)}{k!}x^k. $$ With the Cauchy product, $L(x)=\frac1{x\sinh x}$. On the other hand, the Cauchy product applied to $$ \sinh(x)=\sum_{k\ge 0}\frac{x^{2k+1}}{(2k+1)!} $$ gives $$ \sinh(x)=\sum_{k\ge 0}\frac{x^{2k+1}}{

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