-
+1
Ovi korisnici su zahvalili autoru
Daniel za post:
ubavic
Reputacija: 4.55%
od Daniel » Utorak, 16. Jul 2013, 16:15
Funkcija [inlmath]e^{-x}[/inlmath] razvijena u Tejlorov red bila bi
[dispmath]f(x)=e^{-x}=1-\frac{x}{1!}+\frac{x^2}{2!}-\frac{x^3}{3!}+\cdots=\sum_{k=0}^\infty(-1)^k\frac{x^k}{k!}[/dispmath] Odatle je
[dispmath]\sum_{k=0}^\infty(-1)^k\frac{(\ln4)^k}{k!}=f(\ln4)=e^{-\ln4}=e^{\ln4^{-1}}=4^{-1}=\frac{1}{4}[/dispmath]
I do not fear death. I had been dead for billions and billions of years before I was born, and had not suffered the slightest inconvenience from it. – Mark Twain