Original subtitles

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Today we're going to be talking about how to find the power series representation of a function and

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the associated radius an interval of convergence for the power series.

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And then this particular problem we've been given the function f of x is equal to x divided by the quantity

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9 plus x squared.

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So the first thing we need to do is find a power series representation of this function and the way

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that we're going to do it is by comparing it to this well-known power series The Power series acts to

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the power.

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We know that the sum of that power series is one divided by the quantity one minus X so what we're going

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to do is compare our original function r f of x function here to the sum one divided by 1 minus X because

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they're already somewhat similar.

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They're both rational functions we've got two terms here in the denominator of each one term in the

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numerator of each.

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We're going to compare those two together.

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Try to make our function f of x look more similar to this some here 1 divided by 1 minus X and in that

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way hope to find a power series representation for our function f of x.

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So in order to make our function look more like this some one divided by 1 minus x.

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The first thing we want to do is factor an X out of the numerator of our original function.

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So if we take an x out and we say that this is X times 1 divided by 9 plus x squared we haven't changed

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the function at all we just factor the X out of the numerator.

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But now what we have inside the parentheses here we have one in the numerator just like we have 1 in

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the numerator of this some 1 over 1 minus.

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So we're already closer.

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Now what we want to try to do is get a one a value of 1 for the first term of our denominator and the

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way we're going to do that is by factoring out and 9 from our denominator.

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So this is going to be come here now x times and then inside the parentheses we'll still have 1 in our

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numerator.

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But in our denominator will factor out a 9.

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And we'll get nine times one plus.

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Now in order to make this still true what we have to do is call this x squared divided by nine.

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That way when we multiply x squared divided by nine times nine are nine cancel and we're just left with

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x squared which is what we had originally here just this x squared term.

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So that's the way that we keep that still true.

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But now if we factor this nine outside of the parentheses what you can see we have is X divided by nine

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times what's inside of parentheses here 1 over 1 plus x squared divided by 9 like that.

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Now we're really close to this.

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We have one in the numerator like are some over here 1 over 1 minus x.

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We have our first term in the denominator as a 1.

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Just like this some over here the only difference now is the second term and the denominator and the

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fact that this sum has a negative sign here and the denominator verses are sum which has a positive

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sign here in the middle of our denominator.

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So what we're going to do is instead of this positive sign here we're going to turn this into a double

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negative we're going to say that this is equal to x over nine times 1 divided by 1 minus.

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And now we're going to call this negative x squared over 9 like this.

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What that does for us is it has this negative sign here.

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Now what we can do is say that this x value right here the x value in this particular spot in this sum

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is equal to this x value that we have here the x value in the same spot.

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And if we just say that x is equal to negative x squared divided by 9.

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Everything else is the same.

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So we can substitute this negative x squared plus 9 in for x right here because in this sum this infinite

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sum here.

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Pulling this x value from this sum right here this x value.

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So when ours is negative x squared over 9 we just plug that in right there.

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We multiply by x divided by 9.

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And now we have a power series representation.

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So our power series representation is going to be the sum from N equals zero to infinity and we're taking

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this directly from the well-known power series here.

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And then for X we're substituting negative x squared divided by nine so we're going to say negative

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x squared divided by nine.

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That's all raised to the end power according to our well-known power series here.

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So now we've represented this part right here this one over one minus X which we had right here inside

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our parentheses but we haven't accounted for this.

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X divided by 9.

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So we have to multiply this by X overnight and we'll just put that out in front.

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So this is our power series representation.

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All we have to do is simplify it.

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And the way that we're going to do that is by bringing the X divided by nine inside the infinite sum

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here so we're going to say that this is going to be the some from an equals zero to infinity.

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We're going to bring the X divided by nine inside.

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We're going to basically have X to the first power divided by 9:00 to the first power right both of

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these are raised to the first power and then here for our negative x squared divided by nine all race

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to the end power.

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The first thing we want to do with that is pull out the negative one value we can pull out this negative

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here as a negative one race to the end whenever we have a negative value inside parentheses like that.

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We can pull it out.

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We just want to make sure it's raised to this xponent here so we have negative one to the end.

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What that leaves us with is just everything left inside our parentheses x squared divided by nine.

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So we get X squared divided by nine.

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We just need to make sure that we keep it raised to the power so to the end power like that.

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Now if we simplify further we'll say we have some from an equal zero to infinity will bring the negative

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one to the an out in front.

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That's always a good idea a negative one to the power.

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Now notice here we have x rays to the to race to the nth power.

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Whenever you have two exponents like that something raised to an exponent raise to another xponent you

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can multiply those exponent together.

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So this is going to become X to the to end power.

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Same thing here you basically have nine to the first power race to the end you multiply one times and

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and you get in.

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So this is nine to the end instead of nine.

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The first race to the end.

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So now we have things with like bases that we can combine.

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So we have here to the first power and we have X to the two and power.

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Whenever you have two terms with like Bass's they both have a base of x.

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You can combine them as long as you add the exponent together.

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So we have an exponent here to N and an exponent 1 so we can combine these and say X to the two and

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plus one we just add them together.

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Same thing here with our denominator we have nine to the first power and we have nine to the end power.

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So we can combine them and we can say this is divided by nine to the end plus one we just add those

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exponents together.

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Now we've simplified this as much as we can.

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This is our power series representation of the original function f of x.

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We've got the first half of the problem done.

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Now we need to do is find the associated radius an interval of convergence for this power series.

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The easiest way to do that is to find the radius of convergence first and then use that radius to find

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the interval of convergence.

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The way that we're going to find the radius of convergence is we're going to use the ratio test and

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what the ratio test tells us we're going to call this L.

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We're going to set that equal to the limit as an goes to infinity of the absolute value of a seven plus

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one divided by a seven.

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This is the ratio test right here.

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What it tells you is that if you evaluate this right hand side based on your power series up here and

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you find a value for L you get some value of L if you set that value of L to be less than 1.

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That tells you where your series will converge.

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So this is an extremely useful convergence test for us to use to find the radius of convergence.

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All we need to do.

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We're going to say L is equal to the limit as and goes to infinity of the absolute value here for a

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seven plus one.

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We just plug and plus one into our original power series here.

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Every hour we have N.

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So we say negative 1 raise 2 instead of the n and plus 1.

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Then we multiply that by x to the two times the quantity and plus one plus one.

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So here we're going to get 2 times.

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And plus one plus one.

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When we play.

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And plus 1 and for this value right here.

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We get two and plus two plus one or two and plus three.

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So we have X to the two and plus three.

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Then we divide that by nine to the end plus 1 plus 1 or 9 to the end plus 2.

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That's a sub and plus one right here that we got.

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We divide that then by a sub N which is just our original series here.

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This whole thing right here we divide that by this part right here.

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But because we have a fraction divided by a fraction instead of just dividing we can kind of combine

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two steps here.

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And instead of dividing by this value we can multiply by the reciprocal.

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So then instead of a big division problem we get a multiplication problem instead.

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So we just take their supercool in other words we flip this upside down our denominator 9 to the end

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plus one becomes our numerator nine to the end plus 1 and our numerator negative 1 to the n negative

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1 to the end times x to the two and plus 1 which was our numerator becomes our denominator.

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This is the limit that we're going to be evaluating from here we just need to combine like terms so

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we're looking for things with like Bass's we're going to say L is equal to the limit as and goes to

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infinity of the absolute value.

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Now from here we're looking for like bases like a so we have this value here.

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Negative One race to the end plus one in our numerator and we have negative one race to the end in our

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denominator because we have like Bass's we can consolidate these terms.

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We just need to subtract the exponent in the denominator from the exponent in the numerator so the exponent

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in our numerator is this and plus one right here.

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So we're going to get an A plus one from the numerator the exponent and the nominator is n.

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So we're going to subtract that X point in the denominator so minus an and plus 1 minus.

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And our ends are going to cancel.

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And we're just going to be left with positive 1.

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So we're going to get negative 1 race to the positive one power in our numerator or just negative ones

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so we'll get a negative sign here to represent that in our numerator.

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When we take X to the two n plus three minus X to the two n minus 1 we subtract those exponents are

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two ends are going to cancel.

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We're going to get three minus one which is just two.

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We're going to be left with positive x squared here in our numerator then same thing here 9 to the end

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plus 1 over at to the end plus 2.

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When we subtract the exponent and the denominator from the exponent in the numerator our ends will cancel.

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We'll be left with 1 minus 2 or just negative 1.

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So we have nine to the negative 1.

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We know because we have a negative exponent that that value is going to be in a denominator and we have

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then 9 to the positive one in the denominator.

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So this is our simplified absolute value here now because we were able to cancel all of our values out

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of this function.

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This limit as and goes to infinity becomes irrelevant.

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It's only relevant if we have an end value left over but we don't have any values of x values so this

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limit is n goes to infinity really just goes away and what we are left with is that L is equal to the

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absolute value of negative x squared divided by 9.

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Because we have these absolute value brackets we can cancel out this positive sign and just say the

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absolute value of x squared over 9.

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Remember that we said at the beginning when we started this ratio test process that once we found a

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value for L we would set it less than 1 and that would give us our radius of convergence.

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And it does give us our radius of convergence.

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We just need to solve this inequality for a value of x and right now we have x squared divided by 9.

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So we're going to do is multiply both sides by nine and we'll be left with the absolute value of x squared.

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Less than nine we get nine times 1 which is nine over here in order to get rid of our absolute value

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bars and take the square root of this x squared value just to get X on its own.

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We need to say that x is greater than negative 3 and less than positive 3 right when we take the square

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root of x squared we get x we get positive 3 over here taking the square means we can also get negative

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3.

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We have to say x is greater than negative 3.

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This value now that we just found this X less than 3 right here.

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This gives us our radius of convergence it tells us that our radius of convergence is our equals 3 because

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once we solve for x whatever value we get right here is our radius of convergence.

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So now our radius of convergence is three we can say that our interval of convergence is negative 3

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to positive 3 we just take both sides of this inequality here.

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And this is fine except that we have to test the endpoints of the interval of convergence to see whether

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or not the value negative 3 itself is included in the interval of convergence and whether or not the

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value positive 3 itself is included in the interval of convergence or if the values between them are

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the only values for which the series converges and the end points actually aren't included in order

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to test the endpoints of our interval of convergence.

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We just want to plug the endpoints into the power series representation that we found earlier our function.

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We're going to be plugging in those end points for x.

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Not for N but for x.

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So let's start with negative 3 let's plug negative 3 in for x so with X equals negative 3.

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Here's what we're going to get the some from and equals zero to infinity of negative 1 to the nth power

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usually plug in negative 3 so you get negative 3 raised to the two and plus 1 divided by 9 to the end

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plus 1.

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And really this is just going to test our skills with algebra and exponents.

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We just need to simplify this function and determine whether or not this new series right here converges

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or diverges and we can use any convergence test that we want to figure out whether or not this series

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converges or diverges.

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But before we do that we want to simplify it as much as we can.

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So in order to simplify it the first thing you need to realize is that we can reverse the process we

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used earlier with exponents.

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Remember that we said when we had it terms with like Bass's we can combine them by just the exponents

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together well we can separate them by just separating the exponent so here we have negative 3 race to

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the two and plus one we can separate these and say negative three to the two N times negative 3 to the

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first power.

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Here we can do the same thing in the denominator we have nine to the end plus 1.

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We can call this nine to the N times 9 to the first power so times nine to the first power like this

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similar thing here we have negative three raised to the two and power.

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Remember how we said if we had some base here raised to an exponent raise to another exponent we could

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multiply those x points together.

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Well here we have the product of two xponent two.

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And and we can separate those and instead of saying negative three to the two n we can say negative

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three squared raised to the power and we probably put parentheses around this.

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But now essentially what we have is negative 3 squared which is 9 raised to the power.

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So this is going to become nine to the positive and from here we can cancel We have nine to the end

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in the denominator and 9 to the end in the numerator.

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So those all go away like this.

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And what we're left with is really just the sum from an equals zero to infinity.

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Here we have negative three to the first and nine to the first power so we're just left with negative

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3 over 9 or negative one third times negative 1 to the n.

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What we should be able to see here is that this is just a geometric series remember geometric series

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comes in the form a times x rays to the end.

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Power will here are value of a is negative one third that coefficient r x value is this negative one

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race to the end power.

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Remember that the geometric series Test says that the series converges if the absolute value of x is

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less than 1.

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What are x value is this negative one right here.

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So we say the absolute value of negative 1 less than 1.

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We take the absolute value of this we get positive 1 less than 1.

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Well this is not true.

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So what we know is that at the end point negative 3 the series diverges by the geometric series test.

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So we're just going to say for this endpoint the series diverges by geometric series test.

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So what that means because the series diverges at this endpoint it means that we leave this interval

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of convergence with a parentheses on this left side here.

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If we found that the series converged at this negative 3 and point then we would replace this parentheses

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with a hard bracket like this.

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But because it diverges we just leave this parentheses here.

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Now we have to test the other end point positive three.

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So we say at X equals positive 3.

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What do we get.

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Well we have the sum from N equals zero to infinity of negative 1 to the power.

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Here's where we plug in our positive 3 so positive 3 goes in here for x positive 3 rays to the two and

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plus 1 divided by 9 to the end plus 1 and we have the same process here with algebra.

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We're going to call our denominator instead of 9 to the end plus 1.

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We're going to call it 9 to the nth power times 9 to the first power like this in our numerator instead

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of three to the two and plus 1 we're going to get three to the two N times 3 to the first power.

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We're going to instead of three to the two and we're going to call this three squared raised to the

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power and then instead of three squared we're going to call this nine.

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So this is going to become nine to the end.

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We're going to get nine to the end and nine to the end to cancel from our numerator and denominator

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and you can see that all we're left with is the sum from N equals zero to infinity negative 1 to the

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nth power.

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We're just left with three divided by nine or one third we can put that one third out in front there

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it's the same as what we had before except positive ONE-THIRD instead of negative one third.

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But the fact remains that this value here that we had for X because this is a geometric series is unchanged.

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We're going to get this same inequality here.

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Yes the value of X less than 1.

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We're going to plug in this negative 1 value for x and we're going to find that one is not less than

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1.

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They're equal to each other.

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And so by the same test we're going to say diverges by the geometric series test.

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So both end points the series diverges at both end points which means we leave the interval of convergence

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written this way with these sort of parentheses instead of hard brackets like this.

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So this is our interval of convergence.

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This is our radius of convergence and this is our power series representation here that we found for

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the function f of x.