6. Power Series

8/10/2019 6. Power Series

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Power Series


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Power Series

Power series representsanalytic functions and are the

most important series incomplex analysis:

Their sums are analyticfunctions andEvery analytic functions can berepresented

by power series.


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A power series is a series in the form,

Where x is a complex variable, , a0,anare

complex (or real) constants, called thecoefcients of the series, and x0is a

complex (or real)constant, called the centerof the series.ifx0= 0, we obtain as a particular case a

power series in powers of z:


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for some reason we could alwys write

the power series as,

ifx0= 0, we obtain as a particular case a

power series in powers of z:


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convergence o the power series

Since power series are functions ofx andthat not every series exist,

It then maes sense to as if a powerseries will exist for allx.

!his "uestion is answered by looin# atthe convergenceof the power series.


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A power series converges forx = c,if the series,

conver#es, or

is a $nite number.


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%ote that a power series will alwaysconver#e if

x = x0.

In this case the power series will become


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&iven a power series, there will exist a

number' so that the power series will conver#eforx - xo| <

and diver#e for|x - xo| > .

!his number is called the radius oconvergence.

Radius o convergence.


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the interval of all xs, incluing t!een points if nee be, for w!ic! t!epower seriesconver#es.

interval of convergence of the series.


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*eterminin# the radius of conver#encefor most power series is usually "uitesimple if we use the ratio test.

Ratio Test

&iven a power series compute,

then,


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Example 1

*etermine the radius of conver#ence andinterval of conver#ence for the followin#power series.


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Solution


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notice that since x is not epenent ont!e li"it an so it can be factored out ofthe limit.


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+y the ratio test.

!he radius of conver#ence for thispower series is # = $ .


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the interval of conver#ence


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!o determine conver#ence at these pointsis to simply plu# them into the ori#inalpower series and see if the seriesconver#es or diver#es

x = -% &

!his series is diver#ent by the *iver#ence!est since .


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x ='&

- x


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Example 2

*etermine the radius of conver#ence forthe followin# power series.

Solution

So, in this case we have,


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0emember that to compute an('all we do is

replace all the n/s in anwith n('. 1sin# theratio test then #ives,


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and the series will diver#e if,

In other words, the radius of theconver#ence for this series is,

%ow we now that the series will conver#e if,


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Addition and Subtraction.

a. the two series start at the same placeandb. both have the same exponent of thex-

x0.

In other words all we do is add or subtractthe coe2cients and we #et the newseries.


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where c is some constant

3ultiplication


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%ote that in order to do this both thecoe2cient in front of the series and the

term inside the series must be in theformx-x0.

All we need to do is to multiply theterm in front into the series and addexponents.


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*i4erentiation of a power series

!he derivative of a power series will be,

All that is needed to do is di4erentiate theterm inside the series and we/re done


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the second derivative is

In this case since the n=0 an n=' termsare both 5ero we can start at any of threepossible startin# points as determined bythe problem that we/re worin#.


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ndex shifts

Index shifts themselves really aren/t

concerned with the exponent on the xter", t!e) instea are concerned withwhere the series starts as the followin#example shows


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Example 2

Write the followin# as a series that startsat n=0 instead of n=*.

Solution

An index shift is a fairly simplemanipulation to perform. 6irst we willnotice that if we de$ne i=n-* then when

n=* we will have i=0. +o what we/ll do isrewrite the series in terms of instead ofn. We can do this by notin# that n=i(*.So, everywhere we see an n in the actualseries term we will replace it with an i(*.


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*oin# this #ives,

!he upper limit won/t chan#e in thisprocess since in$nity minus three is stillin$nity.


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!he $nal step is to reali5e that the letterwe use for the index doesn/t matter and sowe can 7ust switch bac to ns.


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notice that we dropped the startin#point in the series by 8 and everywhereelse we saw an n in the series we

increased it by 8.

In other words, all the ns in the seriesmove in the opposite direction that we

moved the startin# point.


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Example 3

Write the followin# as a series that startsat n= instea of n=*.

Solution!o start the series at n= all we nee to ois notice t!at t!is "eans we will increaset!e startin# point by 9 and so all the other

ns will nee to ecrease b) . /oing t!isfor t!e series in the previous examplewould #ive,


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Example 4

Write each of the followin# as a sin#leseries in terms of

Solution


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6irst, notice that there are two series hereand the instructions clearly as for only asin#le series. So, we will need to subtractthe two series at some point in time.

!he vast ma7ority of our wor will be to #etthe two series prepared for thesubtraction.

!his means that the two series can/t haveany coe2cients in front of them (other

than one of course), they will need tostart at the same value of n and they willneed the same exponent on thex-x0.


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We/ll almost always want to tae care of

any coe2cients $rst. So, we have one infront of the $rst series so let/s multiplythat into the $rst series. *oin# this #ives,


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%ow, the instructions specify that the new

series must be in terms of (x - x' )n, sothat/s the next thin# that we/ve #ot to taecare of.We will do this by an index shift on each of

the series. !he exponent on the $rst seriesneeds to #o up by two so we/ll shift the$rst series down by 9.;n the second series will need to shift up

by < to #et the exponent to move down by


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6inally, in order to subtract the two serieswe/ll need to #et them to start at thesame value of n.*ependin# on the series in the problem

we can do this in a variety of ways.In this case let/s notice that since there isan n-' in the second series we can in factstart the second series at n=' without

chan#in# its value.Also note that in doin# so we will #et bothof the series to start at n=' and so wecan do the subtraction. ;ur $nal answeris then,


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In this part the main issue is the fact that

we can/t 7ust multiply the coe2cient intothe series this time since the coe2cientdoesn/t have the same form as the terminside the series. !herefore, the $rst thin#that we/ll need to do is correct thecoe2cient so that we can brin# it into theseries.We do this as follows,


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We can now move the coe2cient into the

series, but in the process of we mana#edto pic up a second series. !his willhappen so #et used to it. 3ovin# thecoe2cients of both series in #ives,


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We now need to #et the exponent in both

series to be an n. !is will "ean s!iftingt!e 1rst series up by > and the secondseries up by 8. *oin# this #ives,


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In this case we can/t 7ust start the $rstseries at n=* because there is not an n-*sittin# in that series to mae the n=*term 5ero. So, we won/t be able to do this

part as we did in the $rst part of thisexample.What we/ll need to do in this part is stripout the n=* from the second series sothey will both start at n=$. We will then

be able to add the two series to#ether.Strippin# out the n=* term from thesecond series #ives,


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We can now add the two series to#ether.


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!his is what we/re looin# for. We won/tworry about the extra term sittin# in front

of the series. When we $nally #et aroundto $ndin# series solutions to di4erentiale"uations we will see how to deal withthat term there.


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!here is one $nal fact that we need taecare of before movin# on. +efore #ivin# thisfact for power series let/s notice that theonly way for

to be 5ero for allx is to !ave a=b=c=0.


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We/ve #ot a similar fact for power series

!his fact will be ey to our wor withdi4erential e"uations so don/t for#et it.

!act

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6. Power Series