MATH1051 Lecture Notes - Lecture 25: Taylor Series, Ratio Test
Lecture #25 – Taylor Series
Taylor Series
●The Taylor Series for near is (x)f a (x)∑
∞
n=0 n!
f(a)
(n)
−an
● when the series converges(x) (x)f= ∑
∞
n=0 n!
f(a)
(n)
−an
Example #1
●when (x) f= 1
1−xa= 0
●(x) (x)f= ∑
∞
n=0 n!
f(0)
(n)
− 0 n
●(x) f= 1
1−x= 1
●(x) (1 ) (− ) (1 )f′= − 1 − x−2 1 = − x2= 1
●(x) (1 ) (− ) (1 )f′′ = − 2 − x−3 1 = 2 − x−3 = 2
●(x) (1 ) (− ) (1 )f′′′ = − 3 *2 − x−4 1 = 3 *2 − x−4 = 6
●Once we find the pattern, we can determine (x)f(n)
○(x) (n)...3 (1 )f(n)=n− 1 *2 − x−(n+1)
○In other words, (a) n!f(n)=
●x on(− , )∑
∞
n=0 n!
n!n= ∑
∞
n=0
xn1 1
Example #2
●near (x) ef = xa= 0
●(x) ef = x= 1
●(x) ef′= x= 1
●(x) ef′′ = x= 1
○The nth derivative of this function is ex
○Plugging in zero, no matter what is, yields n1
Document Summary
The taylor series for f (x) near a is n=0 f (n) (a) n! (x. A n f (x) n=0 f (n) (a) n! (x. Example #1 (x) f f (x f (x) (x) f when (n) (0) n! a = 0 f (x) = 3 * 2 x 4. 1 = 2 x 3 = 2. Once we find the pattern, we can determine. )3 f (n) f (x) f (x) n! f (n) (x) (1 (1 (n (1 (1 (1 (1. 1 = 3 * 2 x 4 = 6 f (n) (x) n=0 n! n = n! n=0 x xn on( , ) Example #2 f (x) (x) f f (x) (x) f . The nth derivative of this function is ex. Plugging in zero, no matter what n is, yields. P ower series that represents e on ( , x f (n) n=1. | x n+1 x| n+1 = 0 < 1.
