Jiwen He 1.1 Geometric Series and Variations
[Pages:6]Lecture 2711.7 Power Series 11.8 Differentiation and
Integration of Power Series
Jiwen He
1 Power Series
1.1 Geometric Series and Variations
Geometric Series
Geometric Series:
k=0
xk
xk = 1 + x + x2 + x3 + ? ? ?
1 , if |x| < 1,
1-x
k=0
diverges, if |x| 1.
Power Series
Define a function f on the interval (-1, 1)
f (x) = xk = 1 + x + x2 + x3 + ? ? ? =
1
1-x
k=0
for |x| < 1
As the Limit
f can be viewed as the limit of a sequence of polynomials:
f
(x)
=
lim
n
pn(x),
where pn(x) = 1 + x + x2 + x3 + ? ? ? + xn.
Variations on the Geometric Series (I) Closed forms for many power series can be found by relating the series to the geometric series Examples 1.
f (x) = (-1)kxk = 1 - x + x2 - x3 + ? ? ?
k=0
= (-x)k =
1
1
=
,
1 - (-x) 1 + x
k=0
for |x| < 1.
f (x) = 2kxk+2 = x2 + 2x3 + 4x4 + 8x5 + ? ? ?
k=0
= x2 (2x)k =
x2
1 - 2x
k=0
for |2x| < 1.
1
Variations on the Geometric Series (II) Closed forms for many power series can be found by relating the series to the geometric series Examples 2.
f (x) = (-1)kx2k = 1 - x2 + x4 - x6 + ? ? ?
k=0
=
(-x2)k
=
1 1 - (-x2)
=
1 1 + x2 ,
k=0
for |x| < 1.
f (x) =
x2k+1 3k
= x + 1 x3 + 1 x5 + 1 x7 + ? ? ? 3 9 27
k=0
x2 k
x
3x
=x
3 = 1 - (x2/3) = 3 - x2
k=0
for |x2/3| < 1.
1.2 Radius of Convergence
Radius of Convergence There are exactly three possibilities for a power series:
ak xk .
Radius of Convergence: Ratio Test (I) The radius of convergence of a power series can usually be found by applying the ratio test. In some cases the root test is easier.
2
Example 3.
f (x) = k2xk = x + 4x2 + 9x3 + ? ? ?
k=1
Ratio Test :
ak+1 ak
=
(k + 1)2xk+1 k2xk
(k + 1)2 = k2 |x| |x| as k
Thus the series converges absolutely when |x| < 1 and diverges when |x| > 1.
Radius of Convergence: Ratio Test (II)
The radius of convergence of a power series can usually be found by applying the ratio test. In some cases the root test is easier. Example 4.
f (x) = (-1)k xk = 1 - x + 1 x2 - 1 x3 + ? ? ? = e-x
k!
26
k=1
Ratio Test :
ak+1 ak
=
xk+1/(k + 1)! xk /k!
k! xk+1
1
= (k + 1)!
xk
=
|x| 0 < 1
k+1
for all x
Thus the series converges absolutely for all x.
Radius of Convergence: Ratio Test (III) The radius of convergence of a power series can usually be found by applying the ratio test. In some cases the root test is easier. Example 5.
f (x) =
k+1
k2
xk = 2x + (3/2)4x2 + (4/3)9x3 + ? ? ?
k
k=1
1
Ratio Test :(|ak|) k =
1
k+1
k2
|x|k
k
=
k+1 k |x|
k
k
1k = 1 + |x| e|x| < 1 if |x| < 1/e
k
Thus the series converges absolutely when |x| < 1/e and diverges when |x| > 1/e.
3
Interval of Convergence
For a series with radius of convergence r, the interval of convergence can be [-r, r], (-r, r], [-r, r), or (-r, r).
Example 6. In general, the behavior of a power series at -r and at r is not predictable. For example, the series
xk ,
(-1)k xk, k
1 xk, k
1 k2
xk
all have radius of convergence 1, but the first series converges only on (-1, 1), the second converges on (-1, 1], but the third converges on [-1, 1), the fourth on [-1, 1].
IEnxtaemrpvlael 7o.f Convergence
f (x) = (-1)k-1 xk k
k=1
Ratio Test :
ak+1 ak
=
xk+1/(k + 1) xk /k
k
=
|x| |x|
k+1
Thus the series converges absolutely when |x| < 1 and diverges when |x| > 1.
So the radius of convergence is 1
x = -1 : (-1)k-1 (-1)k = -1 diverges
k
k
k=1
k=1
x = 1 : (-1)k-1 (1)k = (-1)k-1 converges conditionally
k
k
k=1
k=1
The interval of convergence is (-1, 1].
2 Differentiation and Integration
2.1 Differentiation and Integration
Differentiation and Integration Theorem Let f (x) = akxk be a power series with a nonzero radius of convergence r.
Then
f (x) = ak k xk-1 for |x| < r
f (x) dx =
ak xk+1 + C for |x| < r k+1
4
Geometric series:
1
= xk
1-x
k=0
for |x| < 1
Differentiation:
1 (1 - x)2 =
k xk-1
(k + 1) xk
for |x| < 1
k=0
k=0
Integration: - ln(1 - x) =
1 xk+1 = 1 xk for |x| < 1
k+1
k
k=0
k=1
2.2 Examples
Power Series Expansion of ln(1 + x)
d Note: ln(1 + x) =
1
= (-1)kxk for |x| < 1
dx
1+x
k=0
Integration: ln(1 + x) = (-1)k xk+1(+C = 0) k+1
k=0
= (-1)k xk = x - 1 x2 + 1 x3 - 1 x4 + ? ? ?
k
234
k=1
The interval of convergence is (-1, 1]. At x = 1,
(-1)k
111
ln 2 =
= 1- + - +???
k
234
k=1
Power Series Expansion of tan-1 x
Note:
d dx
tan-1 x
=
1 1 + x2
=
(-1)k x2k
for |x| < 1
k=0
Integration: tan-1 x = (-1)k x2k+1(+C = 0) 2k + 1
k=0
= x - 1 x3 + 1 x5 - 1 x7 + ? ? ? 357
The interval of convergence is (-1, 1]. At x = 1,
tan-1 1 =
(-1)k
111
= 1- + - +??? =
2k + 1
357
4
k=1
Outline
5
Contents
1 Power Series
1
1.1 Geometric Series . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.2 Radius of Convergence . . . . . . . . . . . . . . . . . . . . . . . . 2
2 Diff and Integ
4
2.1 Diff and Integ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
2.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
6
................
................
In order to avoid copyright disputes, this page is only a partial summary.
To fulfill the demand for quickly locating and searching documents.
It is intelligent file search solution for home and business.
Related download
- ln 1 x edu
- second order linear differential equations
- lektion 6 logaritmefunktioner
- 3 6 derivatives of logarithmic functions
- second fundamental theorem of calculus
- ma 16100 final exam practice problems
- lecture 2 the natural logarithm
- techniques of integration whitman college
- table of basic integrals basic forms
- properties of common functions properties of ln x
Related searches
- 1 or 2 374 374 1 0 0 0 1 168 1 1 default username and password
- 1 or 3 374 374 1 0 0 0 1 168 1 1 default username and password
- 1 or 2 711 711 1 0 0 0 1 168 1 1 default username and password
- 1 or 3 711 711 1 0 0 0 1 168 1 1 default username and password
- 1 or 2 693 693 1 0 0 0 1 168 1 1 default username and password
- 1 or 3 693 693 1 0 0 0 1 168 1 1 default username and password
- 1 or 2 593 593 1 0 0 0 1 or 2dvchrbu 168 1 1 default username and password
- 1 or 3 593 593 1 0 0 0 1 or 2dvchrbu 168 1 1 default username and password
- 1 or 2 910 910 1 0 0 0 1 168 1 1 default username and password
- 1 or 3 910 910 1 0 0 0 1 168 1 1 default username and password
- 1 or 2 364 364 1 0 0 0 1 168 1 1 admin username and password
- 1 or 3 364 364 1 0 0 0 1 168 1 1 admin username and password