MTH351 20130401 week 1 Monday April 1, 2013 page 1
transcendental functions
sec 1.1 Taylor polynomials
background: Suppose f(x) has continuous derivative of degree n+1.
then f(x)≅f(a)+
f ' (a)(x-a) f"(a)(x-a)2
f (n) (a)(x-a)n
+
+…
1!
2!
n!
f(a) is center
Taylor approximation is local approximation so the farther from a we need more terms to get a good
answer.
example: f(x) = ex
f(-k)(x)=ex
a=0
e0 x e0 x 2
e0 x n
e ≅e +
+
+…+
1!
2!
n!
x
0
ex ≅1+x+
x2 x3
xn
+ +…+
2! 3!
n!
n
Taylor polynomial approximation of f(x)is f(x)= ∑
i=0
f (n) (a)(x-a)i
i!
first set of homework is ch 1.1 #10,11 due Friday
How much error due to polynomial approximation?
ch 1.2 error in using Taylor polynomial
How much error is involved in approximating a function by the Taylor polynomial?
If doing a Taylor series on polynomials the derivatives eventually becomes zero.
(1-11) 1.2.1 Taylor’s theorem: Assume that f(x) has n+1 continuous derivatives on closed interval [α,β].
Let aϵ[α,β] a is center ϵ means element of
Let Rn(x) = f(x) – n(x) denote the remainder in approximating f(x) by Pn(x).
Taylor series. Then
R n (x)=f(x)-Pn (x)=
P is polynomial, same as
(x-a)n+1 (n+1)
(cx ) where cx between x and a(n+1)! and xϵ[α,β]
f
(n+1)!
f(n+1)(cx) is n+1th derivative
example: suppose f(x) = ex and a=0.
Pn (x)=1+x+
x2 x3
xn
+ +…+
2 3!
n!
Then by Taylor’s theorem:
ex -Pn (x)=
(x-0)n+1 (n+1)
x n+1 c
(cx )=
f
ex
(n+1)!
(n+1)!
cx between 0 and x
1 1
1
Thus, if x=1, then e≅Pn (1)=1+1+ + +…+
2 3!
n!
e-Pn (1)=
Since 0≤c≤1, we have e0≤ec≤e1
Thus
ec x
(n+1)!
0≤c1 ≤1
c1 is just c in book
since ex↑(increasing)
1
ec
e
3
≤
≤
<
(n+1)! (n+1)! (n+1)! (n+1)!
since e<3
Now if we want to approximate Rn(1) with an accuracy of 10-9, we get the error bound of
3
≤10-9 ⟹n≥12
(n+1)!
Thus, P12(1) is an approximation of e with error ≤ 10-9
in calculator: sum(seq(1/N!,N,0,12))
red through book for examples
section 1.2 homework is #5,9
section 1.3 polynomial evaluation, ch2 tomorrow
graph on page 52(noise in function)