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in this video we're going to talk about

power series

and let me give you an example of a

power series

here's one that starts from zero goes to

infinity

and a power series is basically an

infinite series with the variable x in

it

and x is raised to the n power

so this is a power series

that is centered

at zero

so c is equal to zero for this one

and here's another power series

but this one

is not centered

at zero in this case

the center

is c

now the first thing you need to be able

to do is determine where it's centered

at so let me give you some examples

so for each of these examples

determine where

each power series is centered at

feel free to pause the video and try

now let's consider the first one on the

upper left where is the power series

sent to that

so just by looking at it you could see

that it's centered at c equals two

now what about the second one

on the bottom

left another way you could find the

answer is by setting the inside in this

case x plus four

equal to zero and solve for x

so at negative four this whole thing

becomes zero

so it's centered at c equals negative

four

now for this one if you don't see a c

value if it's just x to the n

then it's centered at zero

you can write this as x

minus zero to the n and that's the same

as x to the n

now what about the last example

if you don't see what it is immediately

set the inside part 3x minus 2

equal to 0 and solve for x

and so you should get x is 2 over 3

and so it's centered

at that value

now the next thing that we need to be

able to do

is we need to determine

the radius of convergence

and the interval of convergence of a

power series

and so you need to use the ratio test to

do that

so let's say if you use a ratio test you

take the limit as n goes to infinity

of u sub n plus 1

divided by u sub n

now let's say if you get zero

what does this mean these three

scenarios you need to

be familiar with this is the first one

so if the ratio test gives you zero

then it means that the series

it converges

for all

x values it's always convergent

so the radius of convergence

that's r that's equal to infinity

and the interval of convergence

it's from negative infinity to infinity

so what this means is that at any x

value

the power series will converge

now the second situation

is

after doing the ratio tests

let's say you don't get zero

let's say this time

you get infinity instead of zero

so what does that mean

now if you recall

in order for a series to converge

the ratio test have to give you an

answer that's less than one

so if it's greater than one or if it's

infinity that means that it diverges

for all x values except

one particular x value

it will only converge

when x is equal to c

at this point

you're going to get a limit

with

some expression of n times zero so the

whole limit is equal to zero

and because according to the ratio test

is less than one

it's going to converge for that point

but for everywhere else

for all other x values that is not equal

to c

you're going to get infinity

and so the series will diverge for all

other x values

but for this

particular x value where x is equal to c

it's going to converge

and because it only converges at one

finite number

the radius of convergence is zero

and the interval of convergence

is only one number it's not a range of

numbers

so the interval of convergence

is just

the x equals c value or c itself

so let me separate

the first situation from the second

let's move on to the third situation

so for the third scenario if we use

the ratio test

we're going to get

an expression

in terms of x typically

it's going to look something like this 1

over r

absolute value

x minus c

which is less than 1.

so it's going to vary but

c could be four c could be zero so you

may just see an absolute value of x

r could be anything it could be one half

it could be two but usually in this

format and once you get to this point

you wanna multiply both sides by r

and keep in mind you want to set it less

than

one because you'll get this

but you need to introduce this

inequality because it only converges

when a ratio test gives you a value less

than one so this is something that you

have to add the less than one part and

then once you multiply both sides by r

for this expression

you're gonna get

x minus c

is less than r

so whatever number you see at this point

that is your radius of convergence

now because of space i'm going to clear

the board

so let's start with this

so this is where the series converges

now whenever

the absolute value of x minus c is

greater than r

then at that point the series

it diverges but since we're looking for

the radius of convergence

we're not going to use this

so we're just going to be focusing on

this expression

now to get rid of the absolute value

symbol

we could say that x minus c is less than

r but greater than negative r

now if you add c to both sides

you'll get that x

is between

negative r plus c

and r plus c

so the interval of convergence becomes

this

it's negative r plus c

comma

r plus c

now the last thing you need to do is

check the endpoints

because even though you have parentheses

here

sometimes you could have two brackets

two parentheses a bracket and a

parenthesis or parenthesis in a bracket

so there's four possibilities here

and you have to check your endpoints

which

i need to show you by example

but that's how you could find the

interval of convergence

so now you know the three scenarios so

let's work on some practice problems

so let's say if we have the power series

from zero to infinity of x to the n

over n factorial

go ahead and determine the radius and

the interval of convergence

for this power series

so let's start with the ratio test

the limit as n goes to infinity

of u sub n plus 1

divided by u sub n

and we need to see where

it's less than one where the series

converges

so u sub n plus one

all you gotta do is in this expression

replace n with n plus one so it's

x to the n plus one

divided by

n plus one factorial

and you said that that's just the

original expression x to the n over n

factorial

so let's take the limit

as n goes to infinity

of u sub n plus one

so that's this

and then divided by

u sub n

so let's simplify the expression

x to the n plus one we could say that's

x to the n

times x to the first power n plus 1

factorial

is n plus 1

times n factorial

and then using the expression keep

change flip

we're going to change division

to multiplication

and we're going to flip the second

fraction so it's n factorial

over x to the n

so we can cancel an n factorial

and we can also cancel x to the n

and so we're going to be left over with

so just i'm continuing from here just

keep that in mind

i'm not saying this is equal to

this here

so this is equal to the limit

as n goes to infinity

and then we have absolute value 1 over n

plus 1

times the absolute value of x so

basically separated x

over n plus one

into those two parts

now as n goes to infinity

what happens to one over n plus one

that's going to go to zero

and zero times x

is zero

now what do we say will happen

if we get a limit of zero

if the limit is equal to zero which is

always less than one

then that means that the series it

converges for all

x values

regardless of what x is x is a fixed

number by the way so if you choose x to

be 2 or 8 or 25 or negative 4

all of those numbers all of those fixed

values

multiplied by zero is equal to zero

so for any x value that you choose

the series will converge

and so what does this mean

so for this situation i believe we

defined it to be

what was a scenario number one i think

that was scenario number one

and so when a limit equals zero

the radius of convergence is infinity

and the interval of convergence

is negative infinity to infinity because

x could be anything and the series will

converge x can be a thousand it will

still converge

and so this is the answer

anytime you get a limit

that's equal to zero

that's all you need to write

now let's work on another example

this time we're going to have the series

of n factorial

times x to the n

feel free to pause the video if you want

to try

so let's start with the ratio test

now u to the n plus one

that's going to be n plus one factorial

times x

to the n plus one

and then we're going to divide it by u

to the n

which is the original expression

that we see here

so that's u to the n

so now let's simplify so now we have the

limit

as n goes to infinity

and then n plus 1 factorial that's n

plus 1

times n factorial and x to the n plus 1

that's x to the n

times x to the first power

divided by these two

so now we can cancel x to the n

and we can cancel

n factorial

and then i'm going to separate n plus 1

and

x so what i now have is that the limit

as n goes to infinity

of

n plus one

my absolute value doesn't need to be so

big

and then times the absolute value of x

so you can move this to the front

because it's not affected by n

so you can say that this is

the absolute value of x

times the limit

as n goes to infinity of n plus one and

n is always positive so we really don't

need the absolute value symbol there

but as n goes to infinity what happens

to n plus one

n plus one also

goes to infinity

now what does this mean

the absolute value of x times infinity

that means that the limit

goes to infinity

and what do we know when the limit goes

to infinity

this is the second scenario

that means that the radius of

convergence is zero

and the interval of convergence

is our c value

so what is c

so at what number

is the function

centered at

there's no x minus c to the n it's x

minus 0 to the n which is x to the m

so we can say that is centered

at zero

so when x is zero it's going to converge

now let's understand why

so if x is zero

we're going to have the limit

as n approaches infinity

n plus one times zero

n plus one times zero is zero

so then the whole limit will equal zero

if x is zero

now if x is anything else but zero let's

say if x is a half

one half times the limit as n goes to

infinity of n plus one

well this is going to go to infinity and

if you multiply that by a half

then

the whole thing is going to go it's

still going to be infinity half of

infinity is infinity

so if x is any other value then 0

the limit will go to infinity and so

it's going to diverge but when x is zero

then the limit goes to zero

and by the ratio test that's less than

one it converges

so anytime you do the ratio test

if you get infinity as your output that

means that

it's only going to converge

when

x

is equal to c

in this case your c value is zero so it

converges when x

is equal to zero

because the limit will go to zero

and so your answer for this situation

is the radius of convergence will be

zero

and your interval of convergence

is your c value it's when x is equal to

c so in this case

your interval of convergence is simply

zero

because c is zero

now let's try another similar but

different

example so consider the series from one

to infinity

of n factorial

times two x minus one

raised to the n factorial

so go ahead and determine

the radius of convergence

and the interval of convergence

so as always we're going to start with

the ratio test

so u to the n plus 1 that's going to be

n plus 1 factorial

times

2x minus 1

to the n plus 1 factorial

divided by

n factorial

times 2x minus 1 to the n so that's our

use of n

so this is

u sub n plus 1

and this is the original u sub n

expression that we see here

so go ahead and simplify it

so n plus one factorial we can write

that as n plus one

times n factorial

and two x minus one to the n plus one

that's two x minus one to the n

times 2x minus 1 to the first power

and so we could cancel

2x minus 1 to the n

and

n factorial

and so now

what we have left over

is the limit

as n goes to infinity

and then we have

n plus one

and also two x minus one

so what happens when n goes to infinity

n plus one

also

goes to infinity so we're gonna have

infinity times two x minus one

now where

is the power series centered at what is

our c value

to find our c value

we need to set the inside equal to zero

so if you set two x minus one equal to

zero

x is one half

and so c

is one half

now let's think about what we have

if x is anything

but one half

it's going to diverge let's say if x is

five

if we plug it in here two times five

minus one that's nine

nine times infinity is infinity

so whenever x

let's say if x does not equal one half

infinity times whatever that value is is

going to be infinity

and so the series is going to diverge

whenever x

is anything but one half but what

happens when x is a half

well if you plug in one half into this

expression

we know we're going to get zero

two times a half is one minus one is

zero

and so

zero times n plus one is zero

thus the whole limit is going to go to

zero which means that the power series

converges

so anytime you get infinity as a result

it's only going to converge

when x is equal to c

that is where the power series is

centered at

so for this problem the radius of

convergence

is zero because it's only one number

where

this series converges one x value

so the interval of convergence

is x equals c which is just one half

so it's not always zero

it depends on where the power series is

centered at

so this is the answer for this problem

so anytime you get infinity the radius

of convergence is zero

and the interval of convergence

is where the power series is centered at

that's where x equals c

so you just put in your c value

now let's work on another example

problem

so let's say we have the power series

from 0 to infinity

of x raised to the 2n

divided by 2n factorial

so go ahead and work on this problem

as always let's start with the ratio

test

so u sub n plus 1 that's going to be x

to the two

times n plus one

so replace n with m plus one and then

two n factorial is going to become two

times

n plus one

factorial

and then we're going to divide it by

u sub n

so u sub n is just x to the 2n

over

2 n factorial

so right now what we have is x

to the two n plus two

and on the bottom

we also have

two n plus 2

factorial

and then we're going to multiply by the

reciprocal

of the other fraction so this is going

to be 2n

factorial

over x to the 2n

now x to the two n plus two

we can separate that into x to the two n

times x squared

two n plus two factorial

so let's see if you have eight factorial

that's eight

times seven times six and if you want to

you can stop at five factorial

so you have to keep subtracting one to

get the next number

so if we start with the first number two

n plus two

to get to the next number

we need to take two n plus two and

subtract it by one

so it's going to be two n plus one and

then if we subtract two n plus one by

one

then we'll get two n

we wanna stop here because

we can cancel it

with the two n factorial on top

so now we can cancel x to the 2n

and 2n factorial

so what we now have is the limit

as n goes to infinity

and then

we have

since n is positive

i don't need an absolute value

expression anymore

so this is going to be 2n plus 2

times 2n minus 1

and then times x squared which is also

always positive

so what happens to this expression

when n

becomes very large when it goes to

infinity

as n goes to infinity that fraction

goes to zero

and so it's gonna be zero times x

squared

which is zero for all x so regardless of

what x is if it's 20 1000 negative 50 if

you multiply by zero it's going to be

zero

so this means that the series converges

for all

x values so anytime you get an answer of

zero for the ratio test

that means that the radius of

convergence

is going to be infinity

and the interval of convergence

is negative infinity to infinity

and so that's it for this problem

now let's try another one

so consider the power series

the square root n

times x minus 1 raised to the n

go ahead and try that one

so once again let's start with the ratio

test as always

u to the n plus 1 that's going to be the

square root of n plus one

times x minus one

raised to the n plus one

divided by the original expression u sub

n

so this is equal to the limit

as n goes to infinity now both of these

expressions they're inside the square

root so i can write it as

one expression

the square root of

n plus 1

over n

now

x minus 1 to the n plus 1 that's going

to be x minus

1

to the n power

times x minus 1

to the first power

and so i can cancel these two

so what i have left over is the limit

as n goes to infinity

square root n plus 1

over the square root of n or

just i'm going to write like this

and then times

x minus 1.

so what happens to

n plus 1 over n

when n goes to infinity

when n becomes very large

the one becomes insignificant

so you're gonna get one n over one n

and so it's going to become one and the

square root of one is one

so we're gonna have one

times x minus one

now this is all

still inside the absolute value

expression

so we have the absolute value of x minus

one

and whenever you have let's say

an answer in terms of x after the ratio

test

set that answer less than one because

you want to find out for what values of

x it converges

so we can rewrite this expression we can

get rid of the absolute value expression

by saying that x minus one is less than

one

but greater than negative one

so now if we add one

to all three sides

we can see that x

is less than one plus one which is two

but greater than negative one plus one

which is zero

so therefore the interval of convergence

is from zero

to two or at least it appears that way

we still need to check the endpoints

so we can adjust it shortly

now what is the radius of convergence

if you recall

once you have this form

if there's no number in front of the

absolute value then whatever number you

see here

is the radius of convergence in this

case it's one

and you could see that

we have x minus one or x minus c

so c

is one for this example it's centered at

one

and you can see it here too

that's x and minus c

now let's check the end points

so let's start with zero

we're gonna see if it converges

when x is equal to zero

so starting with the power series

replace x with zero so we're gonna have

the square root of n

times zero minus one to the n power or

negative one to the n power

so

does the series converge in this form

what would you say

well a simple way is to try the

divergence test

the limit as n goes to infinity

for the square root of n times negative

1 to the n power

as n goes to infinity

the square root of n what's going to

happen to that

that's going to be the square root of

infinity which is basically infinity

and then the only difference is

it's going to alternate in sign

it's going to switch from negative 1 to

1 so

our answer is going to be

positive infinity negative infinity and

so forth it's always going to alternate

so we could say that the limit doesn't

exist

because it doesn't equal one value

sometimes it will equal positive

infinity sometimes it will equal

negative infinity

now if the limit doesn't exist

then we say that according to the

divergence test

the limit diverges

so if it diverges

then

it doesn't converge when x is equal to

zero so we're not going to include zero

as an endpoint

so we're going to leave it

with parentheses it doesn't include zero

we're not going to use brackets if it

converged

then we would use a bracket at that

endpoint

so now let's test the other endpoint

when x is two

so we're going to have the square root

of n times 2 minus 1 to the n power

which is 1 to the n and 1 to the n is

just 1 so this is what we have

now using the divergence test

this is going to go to infinity so it

still diverges

so 2 should not be included in the

interval of convergence

now let's plot the interval of

convergence for this problem

so it starts at zero

and it goes to two

now it doesn't include zero or two so

we're gonna have an open circle

at those points

so anywhere in this region

where x is between 0 and 2

the power series will converge

now the power series is centered at one

so this is the c value

and the radius of convergence

is basically the distance from the

center

to the end of the interval

so this distance is r and so we can see

why r is one

because between zero and one

that distance is approximately one

so r is typically one half of the length

of the interval

so if it goes from zero to two

then if you take the difference between

zero and two which is two divided by two

that's your radius of convergence

and so that's it for this problem

hopefully

this gives you a better understanding of

it

let's try this power series

let's say it's x minus 3 to the n

divided by n

so let's start with the ratio test

so u sub n plus 1 that's going to be

x minus 3 raised to the n plus 1

divided by

n plus 1

and then we're going to divide that by u

sub n

which is

x minus 3 to the n over n

so then we can write x minus 3 to the n

plus 1

as x minus 3 to the n

times x minus 3 to the first

power and n plus one

that's not going to change we're just

going to leave that as n plus one

now we're going to change division to

multiplication and then flip the second

fraction

so we have this

the only thing we can cancel right now

is x minus 3 to the n power

and so we're left with the limit

as n goes to infinity

of

the absolute value or we could just say

1 over n plus 1

times the absolute value of x minus 3.

so as n goes to infinity actually wait

i'm forgetting something

i can't forget this n so let's put that

on top

that would have been a very big mistake

so as n goes to infinity what happens

to

n over n plus one

the 1 is insignificant so it becomes n

over n

which becomes 1.

so this is 1

times the absolute value of x minus 3.

and so that's less than 1.

so we can say that x minus 3 is between

1 and negative 1.

now if we add 3 to both sides

then x is between 4

and 2.

and the midpoint of two and four is

three so we can see that it's centered

athering

so c is three

now what is the radius of convergence

what is that equal to well if you take

the distance between

the midpoint of the interval which is

three and one of the endpoints the

difference between three and two is one

so the radius of convergence is one

or

once we had it in this form the absolute

value of x minus three

is less than one

we have it in the form of x minus c

the absolute value of that is less than

r

so you can see at this point r is one

so now we need to check the endpoints

so let's start with two

so when x is two

will the power series converge

or will it diverge

so 2 minus 3 that's negative 1. so we're

going to have negative 1 to the n power

over n

so notice that this is the alternating

harmonic series

so we need to employ the alternating

series test

ast

so let's start with the divergence test

so we're going to take the limit as n

goes to infinity of a sub n a sub n is

everything except the negative 1 to the

n power

so we'll just focus on one over n

and that goes to zero so it passes the

divergence test

this tells us that it may converge or it

may diverge

now for the alternating series test

the second thing we need to show

is that a sub n plus 1

is less than or equal to a sub n

so in this case for the alternating

series test

a sub n

is one over n

so just keep that in mind it's you take

out the negative one to the n power

now a sub n plus one that's one over n

plus one

one over n is always greater than one

over n plus one

so the two conditions of the alternating

series test has been met

so we could say that the series

converges

by the alternate series

tests so we can say that x is equal to

or greater than 2

so 2 is included

now we need to see if 4 is going to be

included in the interval of convergence

now let's substitute x with four

so we're gonna get the series

four minus three to the n

which is one to the n and one to the n

is just one

so we're gonna have 1 over n

so this is the divergent harmonic series

and based on a p-series test you can see

that p is equal to 1.

so if p is equal to 1

it's going to diverge

in order for the p series to converge

p has to be greater than one

then it will converge

if it's less than or equal to one it

diverges

so since p is exactly one

this particular series

diverges

so 4

is not included

so we can write the interval of

convergence

like this so it's going to be a bracket

at 2

and a parenthesis at 4.

so plotting it on a number line

we have the center at c

and the endpoints are 2 and 4.

so we're going to have a closed circle

at two

and an open circle at four

and here is our c value the power series

is centered at three

and so this

is the radius of convergence

which is one

and so that's it for this problem

so anywhere between two and four

if you plug it in

the series will converge it doesn't

converge at 4 but it converges at 2 and

anywhere between 2

and up to 4 but not including 4.

so let's work on one more example

problem

consider the power series

from 1 to infinity

of negative 1 raised to the n plus 1

times

x minus 4 raised to the n

divided by

n times 9 raised to the n

so go ahead and determine the radius and

the interval of convergence for this

power series

so let's start with the ratio test

u sub n plus one

that's gonna be negative one

to the n plus one

plus one so everywhere you see an n

replace it with n plus one

and then it's going to be divided by u

sub n

the original expression

negative 1 to the n plus 1.

i'm going to separate it like this

negative 1 to the n plus 1. well this is

m plus 2 really

i'm going to separate it as negative 1

to the n plus 1 times

negative 1 to the first power

and then x minus 4 to the n plus 1. i'm

going to write that as x minus 4 to the

n

times x minus

4

to the first power

n plus one i can't really change that

i'm just gonna leave it the way it is

nine to the n plus one i could write

that as nine to the n

times nine to the one

then change division to multiplication

and then flip

the second fraction

okay so now let's simplify what we have

these two we can cancel

we can cancel x minus four to the n

and also

nine to the n power

now let's write these two together

so this is the limit

as n goes to infinity

n over n plus

one and then we have

absolute value

negative one times x minus four

now as n goes to infinity

or let's not forget about the nine on

the bottom

so let's put this

over nine

as n goes to infinity

what happens

to n over n plus one

this becomes one

a thousand over a thousand and one

that's about one

and then the absolute value of negative

one over nine

we can take that out of the absolute

value expression and write it as one

over nine

so what we have now is

one over nine times the absolute value

of x minus four

and

this is going to converge

when the ratio test is less than one

so now what we need to do is multiply

both sides

by nine

and so these will cancel

and so now we have x minus four

is less than nine so notice that this is

the form

x minus c

is less than r

so therefore we could see that r

is 9 that's the radius of convergence

we could also see that c

the series

is centered at four

so c is four

so now let's determine the interval of

convergence

so we can say that x minus four

is less than nine but greater than

negative nine

now let's add four to both sides

nine plus four is thirteen

negative nine plus four is negative five

so this is what we now have

now let's rewrite the original series

which is negative 1 to the n plus 1

times x minus 4 to the n

divided by

n times nine to the n

so just by looking at it you could see

that the power series

is centered at c equals four because of

the x minus four in it

now let's check the end points so if we

plug in negative five

will it converge or diverge so let's

replace x with negative five

so we're going to get the series

negative

minus 4

that's negative 9 to the n

so we're going to have negative 1 to the

n plus 1

and then

negative 9 raised to the n

over n

times 9 to the n

now negative 9

to the n divided by 9 to the n we can

write that as

negative 9 over 9 to the n

which becomes negative 1 to the n

and so if we multiply negative one to

the n with negative one to the n plus

one

then that becomes negative one

to the two n plus one

so this gives us the series

negative one to the two n plus one

divided by n

so what can we do with

negative one to the two n plus one

so when n is one

two times one plus one that's going to

be three

and negative one to the third power

that's negative one so we're gonna have

negative one over one

now the next term when n is two

two times two plus one is five

negative one to the fifth power

is still negative one

and then when n is three

two times three plus one is seven

negative one to the seven is still

negative 1.

so as you can see

we don't have alternating signs

so we can reduce the series

to this we could say that

it's simply negative 1

to the n

and we can move the negative to the

front

so this becomes negative

and then we have the divergent

harmonic series

so according to the p series

p is one and because p is one

we could say that the series

is divergent

so negative five is not included

so we're going to have

a parenthesis at negative 5.

so the interval of convergence is

negative 5

to 13.

now let's check the endpoint 413.

so when x is 13

what's going to happen

so this is going to be 13 minus 4 which

is 9.

so we're going to have negative 1

to the n plus 1 one

times

nine to the n

over n times nine to the n

so these will simply cancel

and now

we have the alternating

harmonic

series we need to perform the

alternating series test

so let's start with

the divergence test

so the limit as n goes to infinity for a

sub n which is 1 over n

that's going to go to zero remember to

ignore this

so it passes the divergence test

at this point the series may converge or

it may diverge

if it doesn't equal zero then it

diverges

now we need to show that

a sub n plus one is less than or equal

to a sub n so a sub n is one over n

and a sub n plus one is one over n plus

one so that's true

so by the alternating series test we say

that the series

converges

which means that 13 is included

so the interval of convergence includes

13.

so we're going to use a bracket

for that

now let's plot the solution

on a number line

so it's going to start from negative 5

and it's going to end at 13

and it's centered at 4.

so 4 is right in the middle

so we're going to have an open circle

at negative 5

and a closed circle at 13.

so this is the interval of convergence

whenever x is between negative 5 and 13

the power series will converge

and so in the middle we have our c value

it's centered at x equals 4

and the radius of convergence

is nine

so 13 minus four is nine

and four minus negative five that's also

nine

and so that's it for this video

hopefully it gave you a good

understanding of power series

and how to find the interval of

convergence and the radius of

convergence thanks for watching