25
Taylor Series Chapter 1 Lecture 2
Taylor SeriesChapter 1 Lecture 2
Taylor Series
Taylor series states that any smooth function can be approximated as a polynomial. The Taylor series then provides a mean to predict a function value at one point in terms of the function value and its derivatives at another point.
Taylor Series
The Taylor series expansion of the function f(x) at xi+1 is given by
1ii xx
)(
)( 1
i
i
xf
xf
)(xf
x
nni
nii
ii
ii
Rhn
xfh
xfh
xfhxfxf
hxfxf
!
)(...
!3
)('"
!2
)(")(')(
)()()(
32
1
h
Taylor Series
where and is a remainder term to account for all terms from n+1 to infinity:
where is a value of x that lies between
Usually is expressed as
where means the truncation error is of the order of
ii xxh 1 nR
1)1(
1
)(
)!1(
)(
!
)(
n
nk
nk
ik
n hn
fh
k
xfR
. and 1ii xx
nR
)( 1 nn hOR
)( 1nhO .1nh
Taylor Series
The zero-order Taylor series approx. is
The first-order Taylor series approx. is
The second-order Taylor series approx. is
and so on.
In general, the nth-order Taylor series approximates the function with an nth-order polynomial.
)()( 1 ii xfxf
hxfxfxf iii )(')()( 1
21 !2
)(")(')()( h
xfhxfxfxf i
iii
Taylor Series
Substitute and thus we get
This series is called Taylor series expansion of f(x) around the point xi.
iii xxhhxx 1
nn
ii
n
ii
iii Rxxn
xfxx
xfxxxfxfxf )(
!
)(...)(
!2
)("))((')()(
)(2
h
xi
)(
)(
ixf
xf
xx
Maclaurin Series
For the special case where the Taylor series becomes
This series is called Maclaurin series expansion of f(x).
0ix
...!
)0(...
!2
)0(")0(')0()(
)(2 n
n
xn
fx
fxffxf
Example 1
Use zero through fourth-order Taylor series expansions to approximate the function
at on the basis of the values at .
2.125.05.015.01.0)( 234 xxxxxf
11 ix 0ix
Solution
The step size
We have
1011 ii xxh
4.2)(
9.04.2)(
0.19.02.1)("
25.00.145.04.0)('
2.125.05.015.01.0)(
)4(
)3(
2
23
234
xf
xxf
xxxf
xxxxf
xxxxxf
4.2)0(
9.0)0(
0.1)0("
25.0)0('
2.1)0(
)4(
)3(
f
f
f
f
f
Solution
The true value of f(x) at is
The zero-order Taylor series approx. is
The true error is
The first-order approx. is
The true error is
11 ix
2.02.125.05.015.01.0)1( f
2.1)0()1( ff
0.1|2.12.0| tE
95.0)0(')0()1( hfff
75.0|95.02.0| tE
Solution
The second-order approx. is
The true error is
The third-order approx. is
The true error is
45.0!2
)0(")0(')0()1( 2 h
fhfff
25.0|45.02.0| tE
3.0!3
)0('"
!2
)0(")0(')0()1( 32 h
fh
fhfff
1.0|3.02.0| tE
Solution
The fourth-order approx. is
The true error is
Thus, the 4th-order Taylor series approx. of f(1) is equal to the true value. We observed that the truncation error is decreased by the addition of terms.
2.0!4
)0(
!3
)0('"
!2
)0(")0(')0()1( 4
)4(32 hf
hf
hf
hfff
0|2.02.0| tE
Example 2
Use Taylor series expansions with n=0 to 5 to approx.
on the basis of the value at
31 at cos)( ixxxf
.4ix
Solution
The step size
The true value at is
The zero-order Taylor series approx. is
The true percent relative error is
12431 ii xxh
31ix 5.0)cos()( 33 f
707106781.0)cos()()(
)()(
443
1
ff
xfxf ii
%4.41%1005.0
707106781.05.0
t
Solution
The first-order approx. is
The true percent relative error is
521986659.0))(sin()cos())((')()(
)(')()(
124412443
1
fff
hxfxfxf iii
%40.4%1005.0
521986659.05.0
t
Solution
The second-order approx. is
The true percent relative error is
497754491.0)(2
)cos())(sin()cos()(
!2
)(")(')()(
212
412443
21
f
hxf
hxfxfxf iiii
%449.0%1005.0
497754491.05.0
t
SolutionThe third-order approx. is
The true percent relative error is
499869147.0
)(!3
)sin()(
!2
)cos())(sin()cos()(
!3
)('"
!2
)(")(')()(
312
4212
412443
321
f
hxf
hxf
hxfxfxf iiiii
%0262.0%1005.0
499869147.05.0
t
Solution
The fifth-order approx. is
The true percent relative error is
Thus, the inclusion of additional terms results in an improved estimate.
500000304.0)(!5
)sin()(
!4
)cos(
)(!3
)sin()(
!2
)cos())(sin()cos()(
512
4412
4
312
4212
412443
f
%000060838.0%1005.0
500000304.05.0
t
Example 3
A chemical solution is tested in a temperature chamber. Microwave signals are transmitted towards the solution and the reaction is reported with the following function
(a) Find the sixth-order Maclaurin series for the reaction that happened between t=0 and 0.5 seconds
(b) Calculate the true value of k from the function definition.
5.0
5.00)(
3
tk
tettf
t
Example 3
(c) Starting with the simplest version of the Maclaurin series obtained in part (a), add terms one at a time to estimate k by considering the intersection point t=0.5 between the two steps of the function. After each new term is added, compute the true and approximate percent relative errors. Add terms until the absolute value of the approximate error estimate falls below a pre specified error criterion %.10s
Solution
(a) Here
The Maclaurin series expansion of f(t) is
t
t
t
t
t
t
t
ettttf
ettttf
ettttf
ettttf
ettttf
etttf
ettf
)1890120()(
)156060()(
)123624()(
)9186()(
)66()("
)3()('
)(
32)6(
32)5(
32)4(
32)3(
32
32
3
120)0(
60)0(
24)0(
6)0(
0)0("
0)0('
0)0(
)6(
)5(
)4(
)3(
f
f
f
f
f
f
ft
6615
2143)( tttttf
Solution
(b) The true value of k from the function definition is
(c) The pre specified error estimate is
First estimate:
The true percent relative error is
206090158.0)5.0()5.0( 5.03 efk
%.10s
125.0)5.0()5.0(
)(3
3
fk
ttf
%35.39%100206090158.0
125.0206090158.0
t
Solution
Second estimate:
The true percent relative error is
The approx. percent relative error is
1875.0)5.0()5.0()5.0(
)(43
43
fk
tttf
%0204.9%100206090158.0
1875.0206090158.0
t
%10%33.33%1001875.0
125.01875.0
sa
Solution
Third estimate:
The true percent relative error is
The approx. percent relative error is
203125.0)5.0()5.0()5.0()5.0(
)(5
2143
52143
fk
ttttf
%4388.1%100206090158.0
203125.0206090158.0
t
%10%6923.7%100203125.0
1875.0203125.0
sa
Solution
Thus, after three terms are included, the approximation error falls below and the approx. value of k is 0.203125.
%10s