Convergence of Taylor Series
Objective: To find where a Taylor Series converges to the original
function; approximate trig, exponential and logarithmic functions
The Convergence Problem
• Recall that the nth Taylor polynomial for a function f about x = xo has the property that its value and the values of its first n derivatives match those of f at xo. As n increases, more and more derivatives match up, so it is reasonable to hope that for values of x near xo the values of the Taylor polynomials might converge to the value of f(x); that is
kn
k
k
nxx
k
xfxf )(
!
)(lim)( 0
0
0
The Convergence Problem
• However, the nth Taylor polynomial for f is the nth partial sum of the Taylor series for f, so the formula below is equivalent to stating that the Taylor series for f converges at x, and its sum is f(x).
kn
k
k
nxx
k
xfxf )(
!
)(lim)( 0
0
0
The Convergence Problem
• This leads to the following problem:
The Convergence Problem
• One way to show that this is true is to show that
• However, the difference appearing on the left side of this equation is the nth remainder for the Taylor series. Thus, we have the following result.
0)(!
)()(lim
00
0
n
k
kk
nxx
k
xfxf
The Convergence Problem
• Theorem:
Estimating the nth Remainder
• It is relatively rare that one can prove directly that as . Usually, this is proved
indirectly by finding appropriate bounds on and applying the Squeezing Theorem. The Remainder Estimation Theorem provides a useful bound for this purpose. Recall that this theorem asserts that if M is an upper bound for on an interval I containing xo , then
n0)( xRn|)(| xRn
|)(| 1 xf n
10 ||
)!1(|)(|
n
n xxn
MxR
Example 1
• Show that the Maclaurin series for cosx converges to cosx for all x; that is
),(...;!6!42
1)!2(
)1(cos642
0
2
xxx
k
xx
k
kk
Example 1
• Show that the Maclaurin series for cosx converges to cosx for all x; that is
• From Theorem 10.9.2, we must show that for all x as . For this purpose let f(x) = cosx, so that for all x, we have or
n0)( xRn
),(...;!6!42
1)!2(
)1(cos642
0
2
xxx
k
xx
k
kk
xxf n cos)(1 xxf n sin)(1
Example 1
• Show that the Maclaurin series for cosx converges to cosx for all x; that is
• From Theorem 10.9.2, we must show that for all x as . For this purpose let f(x) = cosx, so that for all x, we have or
• In all cases, we have so we will say that M = 1 and xo = 0 to conclude that
n0)( xRn
),(...;!6!42
1)!2(
)1(cos642
0
2
xxx
k
xx
k
kk
xxf n cos)(1 xxf n sin)(1
1|)(| 1 xf n
1||)!1(
1|)(|0
n
n xn
xR
Example 1
• Show that the Maclaurin series for cosx converges to cosx for all x; that is
• However, it follows that
• So becomes1||)!1(
1|)(|0
n
n xn
xR
),(...;!6!42
1)!2(
)1(cos642
0
2
xxx
k
xx
k
kk
0||)!1(
1lim 1
n
nx
n
0|)(|0 xRn
Example 1
• Show that the Maclaurin series for cosx converges to cosx for all x; that is
• The following graph illustrates this point.
),(...;!6!42
1)!2(
)1(cos642
0
2
xxx
k
xx
k
kk
Example 2
• Use the Maclaurin series for sinx to approximate sin 3o to five decimal-point accuracy.
Example 2
• Use the Maclaurin series for sinx to approximate sin 3o to five decimal-point accuracy.
• In the Maclaurin series
• The angle is assumed to be in radians. Since 3o = /60 it follows that
...!7!5!3)!12(
)1(sin753
0
12
xxxx
k
xx
k
kk
...!7
)60/(
!5
)60/(
!3
)60/(
6060sin3sin
7530
Example 2
• Use the Maclaurin series for sinx to approximate sin 3o to five decimal-point accuracy.
• In the Maclaurin series
• We must now determine how many terms in the series are required to achieve five decimal-place accuracy. We have two choices.
...!7
)60/(
!5
)60/(
!3
)60/(
6060sin3sin
7530
Example 2
• Use the Maclaurin series for sinx to approximate sin 3o to five decimal-point accuracy.
• The Remainder Estimation Theorem.• For five decimal-place accuracy, we need
• Using M = 1, x = /60, and xo = 0, we have
000005.60
nR
000005.)!1(
)60/( 1
n
n
Example 2
• Use the Maclaurin series for sinx to approximate sin 3o to five decimal-point accuracy.
• The Remainder Estimation Theorem.• This happens at n = 3, so we have
05234.!3
)60/(
603sin
30
Example 2
• Use the Maclaurin series for sinx to approximate sin 3o to five decimal-point accuracy.
• The Alternating Series Test.• Let sn denote the sum of the terms up to and including
the nth power of /60. Since the exponents in the series are odd integers, the integer n must be odd, and the exponent of the first term not included in the sum sn must be n + 2.
)!2(
)60/(|3sin|
20
ns
n
n
Example 2
• Use the Maclaurin series for sinx to approximate sin 3o to five decimal-point accuracy.
• The Alternating Series Test.• This means that for five decimal-place accuracy we
must look for the first positive odd integer n such that
• Again, this happens at n = 3.
000005.)!2(
)60/( 2
n
n
Example 3
• Show that the Maclaurin series for ex converges to ex for all x; that is
),(...;!
...!2
1!
2
0
k
xxx
k
xe
k
k
kx
Example 3
• Show that the Maclaurin series for ex converges to ex for all x; that is
• Let f(x) = ex, so that• We want to show that as for all x.
It will be useful to consider the cases x < 0 and x > 0 separately. If x < 0, then the interval we will look at is [x, 0] and if x > 0, the interval is [0, x].
),(...;!
...!2
1!
2
0
k
xxx
k
xe
k
k
kx
xn exf )(1
0)( xRn n
Example 3
• Show that the Maclaurin series for ex converges to ex for all x; that is
• Let f(x) = ex, so that• Since f n+1 (x) = ex is an increasing function, it follows
that if c is in the interval [x, 0], then• If c is in the interval [0, x] then
),(...;!
...!2
1!
2
0
k
xxx
k
xe
k
k
kx
xn exf )(1
1|)0(||)(| 011 efcf nn
xnn exfcf |)(||)(| 11
Example 3
• Show that the Maclaurin series for ex converges to ex for all x; that is
• We apply the Theorem with M = 1 or M = ex yielding
• In both cases, the limit is 0.
),(...;!
...!2
1!
2
0
k
xxx
k
xe
k
k
kx
0;)!1(
|||)(|0
1
xn
xxR
n
n 0;)!1(
|||)(|0
1
xn
xexR
nx
n
Approximating Logarithms
• The Maclaurin series
is the starting point for the approximation of natural logs. Unfortunately, the usefulness of this series is limited because of its slow convergence and the restriction -1 < x < 1. However, if we replace x with –x in this series, we obtain
)11(...;432
)1ln(432
xxxx
xx
)11(...;432
)1ln(432
xxxx
xx
Approximating Logarithms
• The Maclaurin series
taking the top equation minus the bottom gives
)11(...;432
)1ln(432
xxxx
xx
)11(...;432
)1ln(432
xxxx
xx
)11(...;432
)1ln(432
xxxx
xx
11;...753
21
1ln
753
xxxx
xx
x
Approximating Logarithms
• This new series can be used to compute the natural log of any positive number y by letting
or equivalently
and noting that -1 < x < 1.
x
xy
1
1
1
1
y
yx
Approximating Logarithms
• For example, to compute ln2 we let y = 2 in which yields x = 1/3. Substituting this value in
• gives
...
7
)(
5
)(
3
)(
3
122ln
7315
313
31
1
1
y
yx
11;...753
21
1ln
753
xxxx
xx
x
Binomial Series
• If m is a real number, then the Maclaurin series for (1 + x)m is called the binomial series; it is given by
...!
)1)(1(...
!3
)2)(1(
!2
)1(1 32
kx
k
kmmmx
mmmx
mmmx
Binomial Series
• If m is a real number, then the Maclaurin series for (1 + x)m is called the binomial series; it is given by
• In the case where m is a nonnegative integer, the function f(x) = (1 + x)m is a polynomial of degree m, so
...!
)1)(1(...
!3
)2)(1(
!2
)1(1 32
kx
k
kmmmx
mmmx
mmmx
0...)0()0()0( 321 mmm fff
Binomial Series
• If m is a real number, then the Maclaurin series for (1 + x)m is called the binomial series; it is given by
• In the case where m is a nonnegative integer, the function f(x) = (1 + x)m is a polynomial of degree m, so
• The binomial series reduces to the familiar binomial expansion
...!
)1)(1(...
!3
)2)(1(
!2
)1(1 32
kx
k
kmmmx
mmmx
mmmx
0...)0()0()0( 321 mmm fff
mxxmmm
xmm
mx
...!3
)2)(1(
!2
)1(1 32
Binomial Series
• It can be proved that if m is not a nonnegative integer, then the binomial series converges to (1 + x)m if |x| < 1. Thus, for such values of x
or in sigma notation
...!
)1)(1(...
!3
)2)(1(
!2
)1(1)1( 32
km x
k
kmmmx
mmmx
mmmxx
1||;!
)1)...(1(1)1(
1
xxk
kmmmx k
k
m
Example 4
• Find the binomial series for (a) (b)2)1(
1
x x11
Example 4
• Find the binomial series for (a) (b)
(a) Since the general term of the binomial series is complicated, you may find it helpful to write out some of the beginning terms of the series to see developing patterns.
2)1(
1
x x11
Example 4
• Find the binomial series for (a) (b)
(a) Substitution m = -2 in the formula yields
2)1(
1
x x11
...!3
)4)(3)(2(
!2
)3)(2()2(1)1(
)1(
1 3222
xxxxx
...!3
!4
!2
!321 32 xxx ...4321 32 xxx
Example 4
• Find the binomial series for (a) (b)
(a) Substitution m = -2 in the formula yields
2)1(
1
x x11
...!3
)4)(3)(2(
!2
)3)(2()2(1)1(
)1(
1 3222
xxxxx
...!3
!4
!2
!321 32 xxx ...4321 32 xxx
k
k
k xk )1()1(0
Example 4
• Find the binomial series for (a) (b)
(b) Substitution m = -1/2 in the formula yields
2)1(
1
x x11
...!3
)2/5)(2/3)(2/1(
!2
)2/3)(2/1()2/1(1)1(
1
1 322/1
xxxxx
Example 4
• Find the binomial series for (a) (b)
(b) Substitution m = -1/2 in the formula yields
2)1(
1
x x11
...!3
)2/5)(2/3)(2/1(
!2
)2/3)(2/1()2/1(1)1(
1
1 322/1
xxxxx
...!32
531
!22
31
2
11 3
32
2
xxx
kk
k
k xk
k
!2
)12(531)1(1
0
Homework
• Pages 702-703
• 1, 3, 5, 9, 11, 17
• Look at page 701. There is a list of several important Maclaurin series. Be familiar with them.