Strona 1 z 1

Dobiński, parzyste i nie

: 20 lip 2024, o 05:00
autor: Eariu52
\(\displaystyle{ \frac{1}{e} \sum_{n=0}^{ \infty } \frac{\left( 2n+1\right) ^{k+1} }{\left( 2n+1\right)! } = \frac{1}{2e} \sum_{n=0}^{ \infty } \frac{n ^{k+1} +\left( -1\right) ^{n} \left( n+1\right) ^{k} }{n!} }\)

\(\displaystyle{ \frac{1}{e} \sum_{n=0}^{ \infty } \frac{\left( 2n\right) ^{k+1} }{\left( 2n\right)! } = \frac{1}{2e} \sum_{n=0}^{ \infty } \frac{n ^{k+1} -\left( -1\right) ^{n} \left( n+1\right) ^{k} }{n!} }\)

Re: Dobiński, parzyste i nie

: 20 lip 2024, o 08:56
autor: a4karo
Jaki jest sens pisania `e` po obu stronach tych równań?