2019학년도 2학기

                   반도체 공학과 공학수학2


   주교재: Erwin Kreyszig, Engineering Mathematics, 10th Edition

   부교재: 이상구 외 4인, 최신공학수학 Ⅱ,  1st Edition  

   실습실: http://www.hanbit.co.kr/EM/sage/

          http://matrix.skku.ac.kr/2018-album/R-Sage-Stat-Lab-1.html  

          http://matrix.skku.ac.kr/2018-album/R-Sage-Stat-Lab-2.html 

   강의시간:  공학수학2, (화16:30-17:45)   (목15:00-16:15)

   담당교수:  김응기 박사



Week 12


12주차

*15.1: Sequences. Series. Convergence Tests


*15.2: Power Series


*15.3: Functions given by Power Series

  15.1 복소 수열과 복소 무한급수


15.4: Taylor and Maclaurin Series

  15.2 테일러 급수와 로랑급수

  http://www.hanbit.co.kr/EM/sage/2_chap14.html




Chapter 15. Power Series, Taylor Series


*15.1. Sequences, Series, Convergence Tests


Sequences


A sequence (or infinite sequence) is obtained by assigning to each positive integer a number called a term of the sequence, and is written

       or      briefly      or   .

We may also write or or start with some other integer if convenient.


A real sequence is one whose terms are real.



Convergence


A convergent sequence is one that has a limit , written

       or  

(1)         for all .

   


A divergent sequence is one that does not converge.


Example 1  Convergent and Divergent Sequences

The sequence is convergent.

    .

The sequence is divergent.

, is divergent.


Example 2  Sequences of the Real and Imaginary Parts

The sequence with

    

    

    

 is convergent.

The sequence with is

    , , , .

It converges with the limit . Observe that has the limit and has the limit . That is typical. It illustrates the following theorem by which the convergence of a complex sequence can be referred back to that of the two real sequences of the real parts and the imaginary parts. 


Sage Coding

http://math3.skku.ac.kr/  http://sage.skku.edu/  http://mathlab.knou.ac.kr:8080/


Theorem 1  Sequences of the Real and Imaginary Parts

A sequence of complex numbers and

          , .

Proof

Convergence implies convergence and because of

 then lies within the circle of radius about , so that(Fig. )

    ,   .

Conversely, if and as , then for a given we can choose

so large that, for every ,

    ,   .

These two inequalities imply that lies in a square with center and side .

Hence, must lie within a circle of radius with center .(Fig. )

    


Series


Given a sequence we may from the sequence of the sums

    ,   ,   ,  

and in general

(2)        ().

 is called the -th partial sum of the infinite series or series

(3)    .

The are called the terms of series.


A convergent series is one whose sequence of partial sums converges, say

    .   Then we write

and call the sum or value of the series. A series that is not convergent is called a divergent series.


If we omit the terms of from , there remains

(4)   

This is called the remainder of the series after the term . Clearly, if converges and has the sum , then

       thus   .

Now by the definition of convergence hence .


Theorem 2  Real and Imaginary Parts

A series with converges and has the sum if and only if converges and has the sum and converges and has the sum .

    , , .



Test for Convergence and Divergence of Series


Theorem 3  Divergence

If a series converges, then . Hence, if this does not hold, the series diverges.

Proof

If   converges, , then ,

    .


Theorem 4  Cauchy’s Convergence Principle Series

A series is convergent   

(5)       for every    and .



Absolute Convergence


 is convergent

 is convergent     is absolutely convergent

 is diverges     is conditionally convergent.


Example 3  A Conditionally Convergent Series

 is converges, By the alternating series test.

 is divergent series().

 is conditionally convergent.


Theorem 5  Comparison Set

If a series is given and we can find a convergent series with non-negative real terms such that , then the given series converges, even absolutely.

Proof

By Cauchy’s principle, since converges, for any given we can find an

 such that

       for every    and .

From this and we conclude that for those and ,

    .

Hence, again by Cauchy’s principle, is absolutely convergent.


Theorem 6  Comparison Set

The geometric series

(6*)   

converges with the sum if and diverges if .

Proof

If , then and Theorem implies divergence.

Now let . The partial sum is .

From this, .

On subtraction, most terms on the right cancel in parts, and we are left with

    .

Now since , and we may solve for , finding

(6)    .

Since , the last term approaches zero as .

Hence if , the series is convergent and has sum .


Ratio Test


Theorem 7  Ratio Test

If a series with has the property that for every greater than some ,

(7)       

(where is fixed), the series converses absolutely. If for every ,

(8)        ,

the series diverges.

Proof

If holds, then for , so that divergence of the series follows

from Theorem

If holds, then for , in particular,

    ,   ,   etc.,

and in general, .

Since , we obtain from this and Theorem

    .

Absolute convergence of now follows from Theorem .


Theorem 8  Ratio test

If a series with is such that , then

(a) If , the series converges absolutely.

(b) If , the series diverges.

(c) If , the series is inconclusive.

Proof

(a) We write and let .

Then by the definition of limit, the must eventually get closed to , say,

     for all greater than some .

Convergence of now follows from Theorem .

(b) Similarly, for we have for all (sufficiently large),

which implies divergence of by Theorem .

(c) The harmonic series has ,

hence , and diverges. The series

     has ,

hence , but it converges. Convergence and is monotone increasing

(since the terms of the series are all positive);


Both properties together are sufficient for the convergence of the real sequence

    .

(In calculus this is proved by the so-called integral test, whose idea we have used).

    


Example 4  Ratio test

Is the following series convergent or divergent?

    .

Solution

By Theorem8, the series is convergent, since

    .


Sage Coding

http://math3.skku.ac.kr/  http://sage.skku.edu/  http://mathlab.knou.ac.kr:8080/


Example 5  Theorem 7 More General than Theorem 8

Let and .

Is the following series convergent or divergent?

    .

    

Solution

The ratios of the absolute values of successive terms are .

Hence convergence follows from Theorem .

Since the sequence of these ratio has no limit, Theorem is not applicable.

    



Root Test


Theorem 9  Root Test

If a series is such that for every greater than some ,

(9)      

(where is fixed), this series converges absolutely.

If for infinitely many ,

(10)    ,

the series diverges.

Proof

If holds, then for all . Hence the series

converges by comparison with the geometric series, so that the series

 converges absolutely.

If holds, then for infinitely many . Divergence of now follows

from Theorem .


Theorem 10  Root Test

If a series is such that , then :

(a) If , the series converges absolutely.

(b) If , the series diverges.

(c) If , the series is inconclusive.

Proof

The proof parallels that of Theorem .

(a) Let . Then by the definition of limit we have

 for all greater than some (sufficiently large).

Hence for all . Absolute convergence of

the series now follows by the comparison with the geometric series.

(b) If , then we also have for all sufficiently large .

Hence for those . Theorem now implies that diverges.

(c) Both the divergent harmonic series and the convergent series

     give .

This can be seen from and

    ,    .




*15.2. Power Series


A power series in powers of is a series of the form

(1)   

where is a complex variable are complex (or real) constants, called the coefficients of the series, and is a complex (or real) constant, called the center of the series. This generalizes real power series of calculus. If , we obtain as a particular case a power series in powers of :

(2)    .

where is a complex variable and the are constants called the coefficients of the series.



Convergence Behavior of Power Series


Example 1  Convergence in a Disk, Geometric Series

The geometric series

    

    .


Example 2  Convergence for Every z

The power series(which will be the Maclaurin series of in Sec.)

    

is absolutely convergent for every . In fact, by the ratio test, for any fixed ,

    .


Example 3  Convergence Only at the Center.

The following power series converges only at , but diverges for every , as we shall show.

    

In fact, from the ratio test we have

    .


Theorem 1  Convergence of a Power Series

(a) Every power series converges at the center .

(b) If converges at a point , it converges absolutely for every closer to than , that is, . See Fig. .

(c) If diverges at a it diverges for every farther away from than .

See Fig. .

    

Proof

(a) For the series reduces to the single term .

(b) Convergence at gives by Theorem 3 in Sec.

       as   .

This implies boundedness in absolute value,

       for every .

Multiplying and dividing by we obtain from this

    .

Summation over gives

(3)    .

Now our assumption implies that .

Thus the series on the right side of is a converging geometric series.

Absolute convergence of as stated in now follows by the comparison test.

(c) If this were false, we would have convergence at a farther away from than .

This would imply convergence at , by , a contradiction to our assumption of

divergence at .



Radius of Convergence of a Power Series


For a given power series there are only three possibilities

(i) The series converges only when .

(ii) The series converges for all .

(iii) There is a positive number such that the series converges if and diverges if .

 is called the radius of convergence of the power series

    

                 interval of convergence

 is an endpoint of the interval : .

The interval of convergence of a power series is the interval that consists of all values of for which the series converges

 is converge at both endpoints

    

 is converge at once endpoints

    ,  

 is diverge at both endpoints

    


Convergence for every (the nicest case, Ex) or for (the useless case, Ex) needs no further discussion, and we put these cases aside for a moment. We consider the smallest circle with center that includes all the points at which a given power series converges. let denote its radius. The circle

        (Fig )

is called the circle of convergence and its radius the radius of convergence of Theorem then implies convergence everywhere within that circle, that is, for all for which

(4)    

(the open disk with center and radius ). Also, since is as small as possible, that series diverges for all for which

(5)     .

No general statements can be made about the convergence of a power series on the circle of convergence itself. The series may converges at some or all or none of these points. Details will not be essential to us. Hence a simple complex may just gives us the idea.

    


Example 4  Behavior on the Circle of Convergence

On the circle of convergence(radius in all three series),

 converges everywhere since converges,

 converges at (by Leibniz’s test) but diverges at ,

 diverges everywhere.


Notations and . To incorporate these two exclude cases in the present notation, we write if  the series converges for all (as in Ex ), if converges only at center (as in Ex )

These convenient notations, but nothing else.



Real Power Series


In this case in which powers, coefficients and center real, formula gives the convergence interval of length on the real line.


Theorem 2  Radius of Convergence R

Suppose that the sequence , , converges with limit . If , then ; that is, the power seres converges for all . If (hence ), then

(6)        (Cauchy-Hadamard formula).

If , then (convergence only at the center ).

Proof

For the ratio of the terms in the ratio test is

    .   The limit is

Let thus . We have convergence if ,

thus , and divergence if .

By and this shown that is the convergence radius and proves .

If , then for every , which gives convergence for all by the ratio test.

If , then for any and all sufficiently large .

This implies for all by the ratio test.


Example 5  Radius of Convergence

By the radius of convergence of the power series is

    

The series converges in the open disk of radius and center .


Sage Coding

http://math3.skku.ac.kr/  http://sage.skku.edu/  http://mathlab.knou.ac.kr:8080/


Example 6  Extension of Theorem 2

Find the radius of convergence of power series

    .

Solution

The sequence of the ratios , , ,

does not converges, so that Theorem is of no help.

It can be shown that

(6*)    ,    .

This still does not help here, since does not converge because

 for odd , whereas for even we have

       as   .

so that has the two limit points and .

It can further be shown that

(6**)    ,    the greatest limit point of the sequence .

Here , so that . The series converges for .




*15.3. Function Given by Power Series


A power series in powers of is a series of the form

    .

We takes and write

(1)    .

This is no restriction because a series in power of with any can always be reduced to the from if we set .



Terminology and Notation.


If any given power series has a non-zero radius of converges (thus ), its sum is a function of , say . Then we write

(2)       .

 is represented by the power series.

 is developed in the power series.

The geometric series in the interior of the unit circle .


Theorem 1  Continuity of the Sum of a Power Series

If a function can be represented by a power series with radius of convergence , then is continuous at .

Proof

From with we have . Hence by the definition of continuity

we must show that .

That is, we must show that for a given there is a

such that implies .

Now converges absolutely for with any such that ,

by Theorem in Sec. . Hence the series

    

converges. Let be its sum.(S = 0 is trivial.)

Then for ,

    

and when , where is less than and less than .

Hence .


Theorem 2  Theorem for Power Series. Uniqueness

Let the power series and both be convergent for , where is positive.

      .

A function can be represented by a power series with any center , this representation is unique.

Proof

We proceed by induction. By assumption,

       .

The sum of these power series are continuous at , by Theorem .

Hence if we consider and let on both sides, we see that :

the assertion is true for .

Now assume that for .

Then on both sides we may omit the terms that are equal and divide the result by

; this gives

    .

Similarly as before by letting we conclude from this that .



Operations on Power Series


Termwise addition and subtraction of two power series with radii of convergence and yields a power series with radius of convergence at least equal to the smaller of and .

Proof

Add (or subtract) the partial sum and term by term and use

    .

Termwise multiplication of two power series

    

and 

    

mean the multiplication of each term of the first series by each term of the second series and the collection of like powers of .

This gives a power series, which is called Cauchy product of the two series and is given by

  .

We mention without proof that this power series converges absolutely for each within the circle of convergence of each of the two given series and has the sum

    .

Termwise differentiation and integration of power series is permissible, as we show next. We called derived series of the power series the power series from by termwise differentiation, that is,

(3)    .


Theorem 3  Termwise Differentiation of a Power Series

The derived series of power series has the same radius of convergence as the original series.

Proof

This follows from in Sec. because

    

or, if the limit does not exist, from in Sec.

by noting that as .


If the power series has radius of convergence , that the function defined by

    

is differentiable on the interval and

    

         

         

The radius of convergence the power series is .



Example 1  Application of Theorem 3

Find the radius of convergence of the following series by applying Theorem

    .

Solution 1

Differentiate the geometric series twice term by term and multiply the result .

This yields the given series. Hence by Theorem .

Solution 2

    .

The radius of convergence the power series is .


Theorem 4  Termwise Integration of a Power Series

The power series

    

obtained by integrating the series term by term has the same radius of convergence as the origin series.


If the power series has radius of convergence , that the function defined by

    

is continuos on the interval and

    

             

             

The radius of convergence the power series is .



Power Series Represent Analytic Functions


Theorem 5  Analytic Function. Their Derivatives

A power series with a nonzero radius of convergence represents an analytic function at every point interior to its circle of convergence. The derivatives of this function are obtained by differentiating the original series term by term. All the series thus obtained have the same radius of convergence as the original series. Hence, by the first statement, each of them represents an analytic function.

Proof

We consider any power series with positive radius of convergence .

Let be its sum and the sum of its derived series ; thus

(4)       and   .

We show that is analytic and has the derivative in the interior of

the circle of convergence.

We do this by proving that for any fixed with and

the difference quotient approaches .

By termwise addition we first have from

(5)    .

Note that the summation starts with since the constant term drops out in taking

the difference , and so does the linear term when we subtract

from the difference quotient.

(b) We claim that the series in can be written

(6)    .

The somewhat technical proof of this is given in App. .

(c) We consider . The brackets contain terms, and largest coefficient is .

Since , we see that for and , ,

the absolute value of this series cannot exceed

(7)     .

This series with instead of is the second derivative series of at and

converges absolutely by Theorem of this section and Theorem 1 of  Sec. .

Hence our present series converges.

Let the sum of (without the factor ) be .

Since is the right side of , our present result is

    .

Letting and noting that is arbitrary, we conclude that is analytic

at any point interior to the circle of convergence and its derivative is represents

by the derived series.

From this the statements about the higher derivatives follow by induction.




15.4. Taylor and Maclaurin Series


The Taylor series of a function , the complex analog of the reeal Taylor series is

(1)       where

or by

(2)    .

In we investigate counterclockwise a simple closed path that contains in its interior and is such that is analytic in a domain and every point inside .


The remainder of the Taylor series after the term is

(3)    .

Writing out the corresponding partial sum of , we thus have

(4)    .

This is called Taylor’s formula with remainder.


Theorem 1  Taylor’s Theorem

Let be analytic in a domain , and let be any point in .

Then there exist precisely one Taylor series with center that represent .

This representation is valid in the largest open disk with center in which in analytic.

The remainders of can be represented in the from .

The coefficients satisfy the inequality

(5)   

where is the maximum of on a circle in whose interior is also in .

The key tool is Cauchy’s integral formula in Sec. ; writing and instead of and (so that is the variable of integration), we have

(6)    .

 lies inside , for which we take a circle of radius r with and interior in (Fig. ).

We develop in in powers of .

By a standard algebraic manupulation  we first have

(7)    .

For later use we note that since is on while is inside , we have

(7*)      (Fig.).

    


To we now apply the sum formula for a finite geometric sum

(8*)       .

which we use in the form (take the last term to the other side and interchange sides)

(8)     .

Applying this with to the right side of , we get

    .

We insert this into . Powers of do not depend on the variable of integration ,

so that we may take them out from under the integral sign. This yields

  

with given by . The integrals are those in related to the derivatives, so that we have proved the Taylor formula .

Since analytic functions have derivatives of all orders, we can take in as large as we please. If we let approach infinitely, we obtain . Clearly, will converge and represent if and only if

(9)     .

We prove as follows. Since lie on , whereas lie inside (Fig. ),

we have. Since is analytic inside and on , it is bounded, and so is the function , say,

    

for all on . Also, has the radius and the length .

HENCE BY THE inequality(Sec.) we obtain from

(10)

 .

Now because lie inside . Thus ,

so that the right side approaches as . This proves the convergence of the Taylor series. Uniqueness follows from Theorem in the last section. Finally, follows from and the Cauchy inequality in Sec. .


Accuracy of Approximation

We can achieve any preassigned accuracy in approximation by a partial sum of by choosing large enough. This is the partial aspect of formula .


Singularity, Radius of Convergence

One the circle of convergence of there is at least one singular point of , that is, a point at which is not analytic (but such that every disk with center contains points at which is analytic).

We also say that is singular at or  has a singularity at .

Hence the radius of convergence of is usually equal to the distance from to the nearest singular point of .



Power Series as Taylor Series


Theorem 2   Relation to the Last Section

A power series with a non-zero radius of convergence is the Taylor series of its sum.

Proof

Given the power series

    .

Then . We obtain

    ,   thus  

    ,   thus  

and in general . With these coefficients the given series becomes the

Taylor series of with center .



Important Special Taylor Series


Example 1  Geometric Series

Let . Then we have , .

Hence the Maclaurin expansion of is the geometric series

(11)      

 is singular at ; this point lies on the circle of convergence.


Example 2  Exponential Function

The exponential function is analytic for all , and .

Hence from with we obtain the Maclaurin series

(12)     .

This series is also obtained if we replace in the familiar Maclaurin series of by .

Solution

    

    

The Maclaurin series is

    .


Furthermore, by setting in and separating the series into the real and imaginary parts we obtain

    .

(13)              

This is the Euler formula.


Sage Coding

http://math3.skku.ac.kr/  http://sage.skku.edu/  http://mathlab.knou.ac.kr:8080/


Example 3  Trigonometric and Hyperbolic Functions

When these are the familiar Maclaurin series of the real functions and . Similarly, by substituting into , Sec. we obtain

(15)    

By substituting into of Sec. we obtain

(14)    

Solution

                         

                      

                      

                       

                       

          

The Maclaurin series is

    

                          

                         

                       

                      

                        

          

The Maclaurin series is

    


When these are the familiar Maclaurin series of the real functions and . Similarly, by substituting into , Sec. we obtain

(15)    

    

    

     :

     :


Example 4  Logarithm

From it follows that

(16)        .

Replacing by and multiplying both sides by , we get

(17)        .

By adding both series we obtain

(18)        .


Find the Maclaurin series of the function

     (where, ).

Solution

    

    

             

             

             

    

    

    


Sage Coding

http://math3.skku.ac.kr/  http://sage.skku.edu/  http://mathlab.knou.ac.kr:8080/


Find the Maclaurin series of the function

(where, ).

Solution

    

    

              

              

              

    

    

    

     :

                       


Practical Methods

Example 5  Substitution

Find the Maclaurin series of .

Solution

By substituting for in we obtain

(19)     .

     .


Example 6  Integration

Find the Maclaurin series of .

Solution

    

    

          

          

To find we put and obtain . Therefore

      .

this series represents the principal value of defined

as that value for which .

The radius of convergence of the series for is .


Sage Coding

http://math3.skku.ac.kr/  http://sage.skku.edu/  http://mathlab.knou.ac.kr:8080/


Example 7  Development

Develope in powers of , where .

Solution

By Theorem 1, where . Use of with replaced by ;

    

        .

This series converges for , that is, .



Binomial series

    

           


Example  Expand as a power series.

Solution

We use the binomial series with . The binomial coefficient is

    

         

and so, when

    


Example 8  Binomial Series, Reduction by Partial Fractions

Find the Taylor series of the following function with center

    .

Solution

We develope in partial fractions and the first fraction in a binomial series

    

with and the second fraction in a geometric series,

and then add the two series term by term.

This gives

    

        

        

        

        

        

        .

The first series converges for .

The second for .

 is singular at and is singular at .

Point have distance from the center .

Point have distance from the center .

Hence the whole series converges for .


Sage Coding

http://math3.skku.ac.kr/  http://sage.skku.edu/  http://mathlab.knou.ac.kr:8080/




http://matrix.skku.ac.kr/sglee/ 


[한빛 아카데미] Engineering Mathematics with Sage:

[저자] 이상 구, 김영 록, 박준 현, 김응 기, 이재 화


Contents

 A. 공학수학 1 – 선형대수, 상미분방정식+ Lab

Chapter 01 벡터와 선형대수 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-1.html


Chapter 02 미분방정식의 이해 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-2.html

Chapter 03 1계 상미분방정식 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-3.html    

Chapter 04 2계 상미분방정식 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-4.html

Chapter 05 고계 상미분방정식 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-5.html

Chapter 06 연립미분방정식, 비선형미분방정식 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-6.html

Chapter 07 상미분방정식의 급수해법, 특수함수 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-7.html  

Chapter 08 라플라스 변환 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-8.html

 

B. 공학수학 2 - 벡터미적분, 복소해석 + Lab

Chapter 09 벡터미분, 기울기, 발산, 회전 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-9.html

Chapter 10 벡터적분, 적분정리 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-10.html

Chapter 11 푸리에 급수, 적분 및 변환 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-11.html

Chapter 12 편미분방정식 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-12.html

Chapter 13 복소수와 복소함수, 복소미분 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-13.html

Chapter 14 복소적분 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-14.html

Chapter 15 급수, 유수 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-15.html

Chapter 16 등각사상 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-16.html



Made by Prof. Sang-Gu LEE  sglee at skku.edu

http://matrix.skku.ac.kr/sglee/   with Dr. Jae Hwa LEE


Copyright @ 2019 SKKU Matrix Lab. All rights reserved.
Made by Manager: Prof. Sang-Gu Lee and Dr. Jae Hwa Lee
*This research was supported by Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education (2017R1D1A1B03035865).