- #1
emc92
- 33
- 0
Use the error estimate for Taylor polynomials to find an n such that
| e - (1 + (1/1!) + (1/2!) + (1/3!) + ... + (1/n!) | < 0.000005
all i have right now is the individual components...
f(x) = ex
Tn (x) = 1/ (n-1)!
k/(n-1)! |x-a|n+1 = 0.000005
a = 0
x = 1
I don't know where to go from here
| e - (1 + (1/1!) + (1/2!) + (1/3!) + ... + (1/n!) | < 0.000005
all i have right now is the individual components...
f(x) = ex
Tn (x) = 1/ (n-1)!
k/(n-1)! |x-a|n+1 = 0.000005
a = 0
x = 1
I don't know where to go from here