piotr: f(x)=((x+2)
2−11)
1/3
g(t) = (t−11)
s, gdzie s=1/3
g(0) = (−11)
s
g'(t) = s (−11 + t)
−1 + s,⇒ g'(0) = s (−11)
−1 + s
g''(t) = (−1 + s) s (−11 + t)
−2 + s, ⇒g''(0) = (−1 + s) s (−11)
−2 + s
g'''(t) = (−2 + s) (−1 + s) s (−11 + t)
−3 + s, ⇒g'''(0) = (−2 + s) (−1 + s) s (−11)
−3 + s
...
g
(n)(t) = s(s−1)(s−2)...(s−n+1)(−11+t)
s−n
g
(n)(0) = s(s−1)(s−2)...(s−n+1)(−11)
s−n
| s (−11)−1 + s | |
((x+2)2−11)1/3 = (−11)s + |
| ((x+2)2)1 + |
| 1! | |
| s(−1+s)(−11)−2+s | | s(−1+s)(−2+s)(−11)−3+s | |
|
| ((x+2)2)2 + |
| ((x+2)2)3 +...+ |
| 2! | | 3! | |
| s(−1+s)(−2+s)...(s−n+1)(−11)−n+s | |
|
| ((x+2)2)n |
| 3! | |