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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
