d. r. wilton ece dept. ece 6382 power series representations 8/24/10

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D. R. WiltonECE Dept.

ECE 6382 ECE 6382

Power Series Representations

8/24/10

Geometric SeriesGeometric Series

Consider

2

0

2 1

1

1

11 1 1

2

0

1

1 1

1

1

0 1

1li 1

1

m S

NN n

Nn

NN

NN N N

N

N

i N NN N N

nN

N n

S z z z z

zS z z z ,

S zS z S z

zS

z

z r e r r z

z z zz

,

• Consider the sum

Noting that

we have that and hence

• Since iff

1

1

11 1

1

z

z

z

,

Geometric Series (G.S.)

• The above series converges inside, but diverges outside the unit circle. But there exists

another series representing that is valid outside the unit circle :

2 3 2 31

1 1 1 1 1 1 1 1 11 1 1

1 z

zz z z r zz z z z z

G.S.iff i.e.,

• The above series may or may not converge at points on the unit circle

• Note the interior infinite series is an expansion in (po z

z

sitive) powers of ; the exterior series

is an expansion in reciprocal powers of

x

y

1

1

1z

1z

Geometric Series, cont’dGeometric Series, cont’d

Consider 2 30 0 0 0

00

2 3

0 0 0 0 0 00

0

0

01 1 1

1 11

1 1 11 1

1

z zz z z z

zz z z z z z zz

z

z z z zz

z z z z z zzz

zz

z

• Note that if , i.e.

Similarly, if , i.e.

x

y

0z z

0z z

0z Radius of convergence

0z

Geometric Series, cont’dGeometric Series, cont’d

Consider

20 0

00 0

0

1 1 1 11

1

z .

z

z z z zz zz z z z z z z z z z z z

z zz z

:

• The above series were expanded about the origin, But we can also expand about another

point, say

30

00

2 3

0 0 0 0 0 00

0

0

1

1 1 1 11

1

1

z z

z z

z zz z z z

z z

z z z z z z

z z z z z z z z z z z z z zz zz z

z z

z zz

z z

if , i.e.

Similarly,

if , i.e. 0z z z

x

y

z z

0z z Radius of convergence

0z

z

z

Factor out the largest term!

Uniform ConvergenceUniform Convergence

Consider2 3

3 2 1

3

3

11

1

10 0 10 0 10 0

10 0

11 00 0 001 0 000001 0 000000001

1 10

z z zz

z i , i , i

z i

. . . .

.

:

• Consider the infinite geometric series,

Let's evaluate the series for some specific values, say

2

2

1 001001001001001

10 0

11 00 0 01 0 0001 0 000001

1 101 0101010101

.

z i

. . . .

.

:

Clearly, every additional term adds 3 more significant figures to the final result.

Here, however, each additional term a1

1

10 0

11 00 0 1 0 01 0 001

1 101 11111

z i

. . . .

.

:

dds only 2 more significant figures to the result.

And here each additional term adds only 1 more significant figure to the result.

In general, f z .| |or a given accuracy, the number of terms increases with

Uniform Convergence, cont’dUniform Convergence, cont’d

Consider2 31

11

0

S z z zz

z z

!

• For the infinite geometric series,

only the first term is needed to produce an exact result for But as increases

the number of terms needed to provide a fixed nu

12

111

rel

1 0

11

1

1

NN

N

NNN N

N N

z i .

zS z z z

z

z S SS S e S z z

z S

N

mber of significant figures increases,

approaching infinity as

• Since , the partial sum error is

; hence the relative error is

rel

rel

log1 ceil(n)

log

1

nz

z .

z R

( Note denotes )

Note the number of terms needed depends

on and The relationship is

plotted in the figure.

• On the other hand if we lim

it th

both

relrel

log1

logN ,

R

z

en

which depends on but

on (see next slide) not

Number of geometric series terms N vs. |z|

0

100

200

300

400

500

0 0.2 0.4 0.6 0.8 1

|z|

N

2 sig. digits

4 sig. digits

6 sig. digits

8 sig. digits

10 sig. digits

Uniform Convergence, cont’dUniform Convergence, cont’d

Consider 1

N

z

• As the figure shows, it is impossible to find a fixed value of which yields a specified accuracy

over the entire region , i.e., the series is in this region

• Note the

.

G

non -uniformly convergent

0

0

0

9

1

5

nn

z

z R .

f z

.

g z

,

R

.S. is uniformly

convergent, say, for ,

as shown, or for

region

• A series is in a region if corresponding to an

there exists a numbe

any

uniformly convergent

0

N

nn

N z N N

f z g z z .

,

R

r , dependent on but such that

implies for all in

independent of

050

100150200250300350400450500

0 0.2 0.4 0.6 0.8 1

N

|z|

Number of geometric series terms N vs. |z|

2 sig. digits

4 sig. digits

6 sig. digits

8 sig. digits

10 sig. digits

0.95

N1

N8

N6

N4

N2x

y

1

1 1z

0 95z .

Key Point: Term-by-term integration of a series is allowed over any region where it is uniformly convergent.

Taylor Series Expansion of an Analytic FunctionTaylor Series Expansion of an Analytic Function

0 0

00

0

0

0 00

1

2

1

2

1

21

1

2

1

2

C

C

C

n

nC

f zf z dz

i z z

f zdz

i z z z z

f zdz

i z zz z

z z

f z z zdz

i z z z z

zi

uniform convergence

• Write the Cauchy integral formula in the form

0 10 0

( )0 ( )

0 0 10 0

0 00

( )0

10

!

! 2

1

2 !

n

nn C

nn n

nn C

nn

n

n

n nC

f zz dz

z z

f z f znz z f z dz

n i z z

f z a z z f z z

f zf za dz

i nz z

( )

derivative formulas

recall

Taylor series expansion of about

where

(both forms are used!)

x

y

0z z

0z

zz

0z z

z z

C

sz

R

0 0z z z z

Taylor Series Expansion of an Analytic Function, Cont’dTaylor Series Expansion of an Analytic Function, Cont’d

0 0 0

0

0

0s

s

s

z z z z z z

z z

z z z z

;

• Note the construction is valid for any

where is the singularity nearest hence the region of convergence is

x

y

0z z

0z

z

0z z

z z

C

sz

R

0 0z z z z

z

The Laurent Series ExpansionThe Laurent Series Expansion

ConsiderThis generalizes the concept of a Taylor series, to include cases where the function is analytic in an annulus.

z0 a

b

0n

nn

f z a z z

0

0 1 0

1nn n n

n n

f z a z z bz z

or

n nb awhere

Converges for

0 0bz z b z z

Converges for

0 0

0

a

a

z z a z zz z

(we often have )

z

Key point: The point z0 about which the expansion is made is arbitrary, but determines the region of convergence of the Laurent or Taylor series.

za

zb

The Laurent Series Expansion, cont’dThe Laurent Series Expansion, cont’d

ConsiderExamples:

z0a

b 0

cos0 0

zf z z , a , b

z

0 1 01

zf z z , a , b

z

01

0 0f z z , a , bz

0 0 1 21 2

zf z z , a , b

z z

This is particularly useful for functions that have poles.

z

0 0 0a bz z a z z b z z

Converges in region

But the expansion point z0 does not have to be at a singularity, nor must the singularity be a simple pole:

022 3 4

2 1

zf z z , a , b

z z

y

x

0 2z

z

2 1 1 2

branch cut

pole

zbza

The Laurent Series Expansion, cont’dThe Laurent Series Expansion, cont’d

Consider

z0a

b

Theorem: The Laurent series expansion in the annulus region is unique.

(So it doesn’t matter how we get it; once we obtain it by valid steps, it must be correct.)

0

cos0 0

zf z z , a , b

z

0

2 4 611

2! 4! 6!

zz

z z zf z

z

analytic valid for for

3 51

02! 4! 6!

z z zf z , z

z Hence

Example:

The Laurent Series Expansion, cont’dThe Laurent Series Expansion, cont’d

ConsiderWe next develop a general method for constructing the coefficients of the Laurent series.

0n

nn

f z a z z

1

0

1

2n nC

f za dz

i z z

z0a

b

C

Note: If f (z) is analytic at z0, the integrand is analytic for negative values of n.

Hence, all coefficients for negative n become zero (by Cauchy’s theorem).

Final result:

(This is the same formula as for the Taylor series, but with negative n allowed.)

Consider

The Laurent Series Expansion, cont’dThe Laurent Series Expansion, cont’d

Pond, island, & bridge

Consider

11

1

2C c

f zf z dz

i z z

By Cauchy's Integral Formula,

2c

21

2

0 0

01

0 0 00 00

0

1

2 210 0

1 1

2 2

1 1 1

1

1 1

n

n

C

CC

n

f z f zdz dz

i z z i z z

z z z z

z z

z z z z z z z z z zz zz z

C

C ,z z z z

z

C C

z

, ,

,

where on

and on (note the convergence regions of overlap!)

0 01 1

0 0 0 10 0 00

0

1

1

1

n n ,n nn n

n nn n

z z z z

z z z z z z z z z zz zz z

x

y simply - connected regionR

1C2C

1c2cz

0z

1sz

2sz

z

z

• Contributions from the paths c1 and c2 cancel!

The Laurent Series Expansion, cont’dThe Laurent Series Expansion, cont’d

Pond, island, & bridge

The Laurent Series Expansion, cont’dThe Laurent Series Expansion, cont’d

Consider

1 1

1

2cC

f zf z dz

i z z

• Hence,

2c

1

2

2

0 10 0

0 11 0

0

01 20

1

1 2

1

2

1

2

1

2

nn

n

nn

n

n

n

n

C

n

n

C

C

C

C

f zz z dz

i z z

f zz z dz

i z z

f z a z z

f za dz C z

i z z

C C

C

.

, ,

uniformconvergence

where and encircles

Note we can deform to a s

2 1 1 2

10

0 0 0

1 2

n

s s s s

f zC z

z zz z z z z z z z

C C

,

,

,

ingle contour since is - independent

and analytic at least for where are the nearest

singularities to respectively.

x

y multiply - connected regionR

1C2C

z

0z

1sz

2szz

C

Examples of Taylor and Laurent Series Examples of Taylor and Laurent Series ExpansionsExpansions

Consider

0

1 2 20

1

1

0

1 1 1 1 11

2 2 21

1

2

nn

n

mn n n n

mC C C

f zz z

a z z

f za dz dz z dz , z

i i iz z z z

Obtain all expansions of about the origin :

The series will have the form (since )

where

( )

Example 1:

20

2 2

12 120 00 0

2 3

0 12 1

1

0 11 1 1 1

1 12 2

11 0 1

i in m

mC

i

n n mi n m i n mn mm m

, m n, m n

dz z re , dz ire di z

, nirea d d

, ni rr e e

f z z z z zz

; let

,

Examples of Taylor and Laurent Series Examples of Taylor and Laurent Series Expansions,cont’dExpansions,cont’d

Consider

1 2 31

30

30

0

1 1 1 1 1

2 2 21 1

1 1 11

2

1 1

2

1

2

n n n nC C C z

n mmC

i in m

mC

i

nm

f za dz dz dz

i i iz z z z

dz , zi z z

dz z re , dz ire di z

irea

i

On the other hand,

( )

; let

Example 1, cont'd

2 2

23 2300 0

2 3 4

0 22 2

0 21 1 1

1 22

1 1 11

n mi n m i n mn mm

, m n, m n

, nd d

, nrr e e

f z zz z z

,

In practice . To illustrate, we

the contour integral approach is rarely used

f z reconsider expanding as a partial fraction and using the geometric series.

Examples of Taylor and Laurent Series Expansions,cont’dExamples of Taylor and Laurent Series Expansions,cont’d

Consider

0 0

1

1

1

1 1

lim limz z

f zz z

A Bf z

z z z z

zA z f z

Expand about the origin (we use partial fractions and G.S.) :

;

Example 1, cont'd

z

1 1

11

1lim 1 limz z

z

zB z f z

1z z

2 3

1

1 1 1 1 1

1 1 1

11 0 1

1 1 1 1

1 1 1

1

f zz z z z z z

f z z z z zz

f zz z z z z

f zz

,

11

z

2 2 3 4

1 1 1 1 11z

z z z z z

,

Examples of Taylor and Laurent Series Expansions,cont’dExamples of Taylor and Laurent Series Expansions,cont’d

Consider

0

1

2 3

1

0 1 1

1 1 2

1 2

0 1 1

f zz z

z ,

z

z

z

z

f z

Expand in a Taylor / Laurent series

about valid in the annular regions

(1) ,

(2) ,

(3) .

For :

Using partial fraction expansion and G.S.,

Example 2

22

2

2 3

1 1 1

2 3 3 2

1 1 1 1

1 2 1 1 1 12 1 1 2

1 111 1 1 1

2 2 2

1 3 7 151 1 1 0 1 1

2 4 8 16

z z z z

z z zz

z zz z

f z z z z , z

(Taylor series)

y

1 2 3x

1 1 2z

z

1 2z

1 1z

Examples of Taylor and Laurent Series Expansions,cont’dExamples of Taylor and Laurent Series Expansions,cont’d

Consider

2

2 2

1 1 2

1 1 1 1

1 2 1 1 2 1 1 2 1 1 1 1

1 11 1 1 11 1

2 2 1 12 1

1 2

1 1 1 1

1 2 1 1 1 1 2 1

z

f zz z z z z

z zf z

z z z

z

f zz z z z z

For :

(Laurent series)

For :

Example 2,cont'd

1 1 1 1

11

1

z

z

2

2

2 2 11

1 11z zz

2

2 3 4

1 1

1 1

1 3 7

1 1 1

z z

f zz z z

(Laurent series)

Examples of Taylor and Laurent Series Expansions,cont’dExamples of Taylor and Laurent Series Expansions,cont’d

Consider

2

0

1 cos0

01 2 0

1 cos 1

z

z, z

f z zz, z

z

Find the series expansion about :

( is a "removable" singularity)

Example 3

1

2 4 6 2 4 6

2 4

2 4

2! 4! 6! 2! 4! 6!

1

2! 4! 6!

sin1

3! 5!

z z z z z z

z zf z z

zf z z

z,

z z

,

Similarly, we have

Examples of Taylor and Laurent Series Expansions,cont’dExamples of Taylor and Laurent Series Expansions,cont’d

Consider

3 5

sin

sin sin

sin cos cos sin sin

3! 5!

sin 0

z z

f z z z

z z z

z zf z z , z

f

Find the series for about :

Alternatively, use the derivative formula for Taylor series :

Example 4

3 5

cos 1

sin 0

cos 1

sin 0

cos 1

3! 5!

iv

v

f

f

f

f

f

z zf z z , z

Examples of Taylor and Laurent Series Expansions,cont’dExamples of Taylor and Laurent Series Expansions,cont’d

Consider

2

2

2 3 4

0

2 3 4

23 52

sin ln 1 0

11 1

1

1ln 1 1

1 2 3 4

ln 1 12 3 4

sin3! 5!

z

z z z

z z , zz

z z zdz z z , z

z

z z zz z , z

z zz z

Find the first few terms of the series for about :

Since then

Also

Example 5

42 6

4 2 3 42 2 6

43 5

2

3 45

2sin ln 1

3 45 2 3 4

0 12

zz z

z z z zz z z z z

zz z , z

Hence

(why?)

Summary of Methods for Generating Taylor and Laurent Summary of Methods for Generating Taylor and Laurent Series Expansions Series Expansions

Consider

0 0 0

0 0n n

n nn n

n n

z z , f z f z z z

f z a z z g z b z z

f z g z a b z

To expand about first write in the form , rearrange

and expand using known series or methods.

Note that if

t

,

he

n

0

00

0 !

n

n

nn

n nn

z

f zf z a z z a

n

Taylor ( Laurent) series, , can be generated using

Use partial fraction expansion and geometric series to generat

e serie

in their common region of convergence.

not

s for rational functions

(ratios of polynomials, degree of numerator less than degree of denominator).

Laurent / Taylor series can be integrated or differentiated term - by - term within their radius

o

0 00 0

0 00 0

n mn m

n m

n mn m

n m

f z a z z g z b z z

f z g z a z z b z z

f convergence

Two Taylor series can be multiplied term - by - term :

,

within their common region of convergence

00 0

nn

n n p n pn p

c z z c a b

wher e = .

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