Do Now: Find both the local linear and the local quadratic approximations of f(x) = e x at x = 0...
25
Do Now: Find both the local linear and the local quadratic approximations of f(x) = e x at x = 0 Aim: How do we make polynomial approximations for a given function ?
-
Upload
naomi-hamilton -
Category
Documents
-
view
219 -
download
0
Transcript of Do Now: Find both the local linear and the local quadratic approximations of f(x) = e x at x = 0...
Do Now:
Find both the local linear and the local quadratic approximations of f(x) = ex at x = 0
Aim:How do we make polynomial approximations for a given function ?
we want to find a second degree polynomial of the form:
20 1 2 P x a a x a x
xf x e at 0x that approximates the behavior of
If we make , and the first, and the second, derivatives the same, then we would have a pretty good approximation.
0 0P f
20 1 2 P x a a x a x xf x e
xf x e
00 1 f e
20 1 2 P x a a x a x
00P a 0 1a
xf x e
00 1 f e
1 22 P x a a x
10P a 1 1a
xf x e
00 1 f e
22 P x a
20 2P a 2
1
2a
x
2
L(x) 1 x
1Q(x) 1
f(x) e
x x2
Suppose we wanted to find a fourth degree polynomial of the form:
2 3 40 1 2 3 4P x a a x a x a x a x
ln 1f x x at 0x that approximates the behavior of
If we make , and the first, second, third and fourth derivatives the same, then we would have a pretty good approximation.
0 0P f
2 3 40 1 2 3 4P x a a x a x a x a x ln 1f x x
ln 1f x x
0 ln 1 0f
2 3 40 1 2 3 4P x a a x a x a x a x
00P a 0 0a
1
1f x
x
10 1
1f
2 31 2 3 42 3 4P x a a x a x a x
10P a 1 1a
2
1
1f x
x
10 1
1f
22 3 42 6 12P x a a x a x
20 2P a 2
1
2a
2 3 40 1 2 3 4P x a a x a x a x a x ln 1f x x
3
12
1f x
x
0 2f
3 46 24P x a a x
30 6P a 3
2
6a
4
4
16
1f x
x
4 0 6f
4424P x a
440 24P a 4
6
24a
2
1
1f x
x
10 1
1f
22 3 42 6 12P x a a x a x
20 2P a 2
1
2a
2 3 40 1 2 3 4P x a a x a x a x a x ln 1f x x
2 3 41 2 60 1
2 6 24P x x x x x
2 3 4
02 3 4
x x xP x x ln 1f x x
P x
f x
If we plot both functions, we see that near zero the functions match very well!
This pattern occurs no matter what the original function was!
Our polynomial: 2 3 41 2 60 1
2 6 24x x x x
has the form: 42 3 40 0 0
0 02 6 24
f f ff f x x x x
or: 42 3 40 0 0 0 0
0! 1! 2! 3! 4!
f f f f fx x x x
Maclaurin Series:
(generated by f at )0x
2 30 00 0
2! 3!
f fP x f f x x x
If we want to center the series (and it’s graph) at some point other than zero, we get the Taylor Series:
Taylor Series:
(generated by f at )x a
2 3
2! 3!
f a f aP x f a f a x a x a x a
Brook Taylor1685 - 1731
Taylor Series
Brook Taylor was an accomplished musician and painter. He did research in a variety of areas, but is most famous for his development of ideas regarding infinite series.
Greg Kelly, Hanford High School, Richland, Washington
Write the Taylor Series for f(x) = cos x centered at x = 0.
What are the Taylor Series for some common functions ?
2 3 4 5 61 0 1 0 1
1 0 2! 3! 4! 5! 6!
x x x x xP x x
cosy x
cosf x x 0 1f
sinf x x 0 0f
cosf x x 0 1f
sinf x x 0 0f
4 cosf x x 4 0 1f
2 4 6 8 10
1 2! 4! 6! 8! 10!
x x x x xP x
2
0
( 1)
(2 )!
n n
n
x
n
cosy x 2 4 6 8 10
1 2! 4! 6! 8! 10!
x x x x xP x
The more terms we add, the better our approximation.
For
use the Ratio Test to determine the interval of convergence.
2 4 6 2
0
( 1)1 ...
2! 4! 6! (2 )!
n n
n
x x x x
n
22 4 6
0
1cos 1
2! 4! 6! 2 !
n n
n
xx x xx
n
2 2
2
(2 )!lim
(2 2)!
n
nn
x n
n x
2 2
2
(2 )!lim
(2 2)!
n
nn
n x
n x
21lim
(2 2)(2 1)
nx
n n0
If the limit of the ratio between consecutive terms is less than one, then the series will converge.
11
1nn
n n
aa
a a
The interval of convergence is ,
When referring to Taylor polynomials, we can talk about number of terms, order or degree.
2 4
cos 12! 4!
x xx
This is a polynomial with 3 positive terms.
It is a 4th order Taylor polynomial, because it was found using the 4th derivative.
It is also a 4th degree polynomial, because x is raised to the 4th power.
The 3rd order polynomial for is , but it is degree 2.
cos x2
12!
x
The x3 term drops out when using the third derivative.
This is also the 2nd order polynomial.
A recent AP exam required the student to know the difference between order and degree.
2 30 00 0
2! 3!
f fP x f f x x x
cos x
cos x
sin x
cos x
sin x
cos x
1
0
1
0
1
0nf nf x
2 3 4cos2! 3!
1 0
4!
11 0x x x x x
2 4 6
cos 1 2! 4! 6!
x x xx
Both sides are even functions.
Cos (0) = 1 for both sides.
2 30 00 0
2! 3!
f fP x f f x x x
sin x
sin x
cos x
sin x
cos x
sin x
0
1
0
1
0
0nf nf x
2 3 4sin2
0 1 00 1
! 3! 4!x x x x x
3 5 7
sin 3! 5! 7!
x x xx x
Both sides are odd functions.
Sin (0) = 0 for both sides.
What is the interval of convergence for the sine series?
3 5 7
...3! 5! 7!
x x xx
2 1
0
( 1)
(2 1)!
n n
n
x
n
Find the Taylor Series for ( ) lnf x x centered at
a=1And determine the interval of convergence
2 3 41 1 1 2 1 6 10
0! 1! 2! 3! 4!
x x x xP x
lny x
lnf x x 1 0f
1f x
x 1 1f
2
1f x
x 1 1f
3
2f x
x 1 2f
4
4
6f x
x 4 1 6f
2 3 41 1 1
ln 12 3 4
x x xx x
1
1
11
nn
n
x
n
2 3 41 1 1ln 1 1 1 1
2 3 4x x x x x 1
1
11 1n n
n
xn
2 1
1
1 1lim
1 1 1
n n
n nn
x nL
n x
1 1lim
1 1
n
nn
x x n
n x
1lim
1n
x n
n
1x
If the limit of the ratio between consecutive terms is less than one, then the series will converge.
11
1nn
n n
aa
a a
1 1x
1 1 1x
0 2x
The interval of convergence is (0,2].
The radius of convergence is 1.
If the limit of the ratio between consecutive terms is less than one, then the series will converge.
2 4 6 2
0
( 1)cos( ) 1 ...
2! 4! 6! 2 !
n n
n
x x x xx
n
2 3
0
1 ...2! 3! !
nx
n
x x xe x
n
3 5 7 2 1
0
( 1)sin( ) ...
3! 5! 7! (2 1)!
n n
n
x x x xx x
n
2 3
0
11 ...
1n
n
x x x xx
2 3 4 1
1
( 1) ( 1) ( 1) ( 1) ( 1)ln( ) ( 1) ...
2 3 4
n n
n
x x x xx x
n
Power series for elementary functions
Interval of Convergence
( , )
( , )
( , )
( 1,1)
(0,2]
