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Taylor Series
&
Error
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Series and Iterative methods
Any series xncan be turned into an
iterative method by considering the
sequence of partial sumssn. nnn
n
n
k
kn
xss
xxxxxs
1
210
0
Functions and Taylor Expansions
From Intermediate Calculus we know any function that has
n+1derivatives at a
point ahas a nthTaylor Polynomial expansion centered at aand an
expression for
the error.
xRaxnaf
ax
af
ax
af
afxf
erroraxk
afxf
n
nn
n
k
kk
!!2!1
!
2
'''
0
The valuef(k)(a) is the kthderivative
evaluated at a. The functionRn(x)
represents the error where cis a value
betweenxand a.
1
1
!1
n
n
n axn
cfxR
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To find the Taylor series for the function f(x) = arctan(x).
75312
1
arctan
111
1
111
11
1
753
0
12
642
0
2
2
32
0
32
0
xxxxk
xx
xxxxx
xxxxx
xxxxx
k
kk
k
kk
k
k
k
kStart with known Taylor Series
Substitutexforx
Substitutex2forx
Integrate both sides.
!6!4!2!2cosh
!6!4!21
!2cosh
!6!4!21!2cosh
654
3
0
3
3
32
0
642
0
2
xxxx
k
xxx
xxx
k
xx
xxx
k
xx
k
k
k
k
k
k
To find the Taylor series for the function xxxg cosh3
Start with known Taylor Series
Substitute
Multiply byx3
xxfor
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A second version of Taylor Polynomials
For any functionf(x) consider the function
f(x+h) where we think ofxas a constant and h
is the variable. Consider the Taylor expansion
forf(x+h) (with variable h) centered at 0. The
kthderivatives are found to the right.
xfhxfdh
hxfd kh
k
h
k
k
0
0
1
1
2
'''
0 !1!!2!1!
k
kk
kn
k
kk
h
k
cxfh
k
xfh
xfh
xfxfh
k
xfhxf
This is a more computationally useful way of applying Taylor
Series to
approximate functions. We think of the value of has small (close
to zero) and
will represent the distance between computable derivatives andx.
Consider the
following example.
Approximate sin(30.01) with an error of less than
0.00000001.
We first need to convert to radians to get the value for h.
180
01.
6180 sin01.30sin01.30sin01.30sin
180
01. h
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We need to figure out the value of kthat
gives the desired accuracy. The exact
error is given by the expression to the
right.
1
66
180
01.
!1
cosorsinError
k
k
cc
00000001.
4500
1
!1
1
180
01.
!1
cosorsinError
1
66
kk
kk
cc
We know the values for sine and cosine are always less than 1 in
absolute
value and is less than 4. Try different values of kthat make
this inequality true.
When we try k=2 we find the right side is 0.000000000001829.
This means
we need go out to the k=2 term.
2
180
01.
6180
01.
66180
01.
6
sincossinsin01.30sin
47161480.50015113180
01.
2
1
180
01.
2
3
2
1 2
2330817...0.5001511401.30sin
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Alternating Series and Error
From intermediate calc remember that a series (-1)kakconverges
if the
sequence akare positive terms that decrease to zero, this was
called the
alternating series test. The series (-1)kakis called an
Alternating Series. The
nice thing about alternating series is that keeping tract of the
error is very easy asgiven by the theorem below.
Alternating Series Error Theorem
In any convergent alternating series (-1)kakthe error is always
less than the
first term omitted. In other words:
|Error| < ak+1
Example: Compute 1/eaccurate to .0001
!4
1
!3
1
!2
1
11
1 1
ee The series for 1/eis alternating so the theorem applies.
.0001160.00002480!8
1 We need to go to the k=7 term.
290.36785714!7
1
!6
1
!5
1
!4
1
!3
1
!2
111
1
e10.36787944
1
e
