Szereg Taylora
: 13 wrz 2010, o 10:05
Mam znalezc rozwiniecie szeregu taylora i mam maly problem
\(\displaystyle{ f(x) = \sqrt{x}
f ^{0} (x) = \sqrt{x}
f ^{1} (x) = \frac{1}{2} x ^{-\frac{1}{2} }
f ^{2} (x) = -\frac{1}{2}\frac{1}{2} x ^{-\frac{3}{2} }
f ^{3} (x) = \frac{1}{2}\frac{1}{2}\frac{3}{2} x ^{-\frac{5}{2} }
f ^{4} (x) = -\frac{1}{2}\frac{1}{2}\frac{3}{2}\frac{5}{2} x ^{-\frac{7}{2} }
f ^{n} (x) = (-1) ^{n+1} x ^{ \frac{1}{2} - n }
ale jak zapisac ten iloczyn ulamkow 1/2 , 1/2 1/2, 1/2 1/2 3/2...}\)
\(\displaystyle{ f(x) = \sqrt{x}
f ^{0} (x) = \sqrt{x}
f ^{1} (x) = \frac{1}{2} x ^{-\frac{1}{2} }
f ^{2} (x) = -\frac{1}{2}\frac{1}{2} x ^{-\frac{3}{2} }
f ^{3} (x) = \frac{1}{2}\frac{1}{2}\frac{3}{2} x ^{-\frac{5}{2} }
f ^{4} (x) = -\frac{1}{2}\frac{1}{2}\frac{3}{2}\frac{5}{2} x ^{-\frac{7}{2} }
f ^{n} (x) = (-1) ^{n+1} x ^{ \frac{1}{2} - n }
ale jak zapisac ten iloczyn ulamkow 1/2 , 1/2 1/2, 1/2 1/2 3/2...}\)