If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

# Power series of ln(1+x³)

AP.CALC:
LIM‑8 (EU)
,
LIM‑8.E (LO)
,
LIM‑8.E.1 (EK)
,
LIM‑8.F (LO)
,
LIM‑8.F.1 (EK)

## Video transcript

we have an infinite series here and the first thing I'd like you to try is to pause this video and see if you can express this as an infinite geometric series and if you can express it as an infinite geometric series see what it's sum would be given an interval of convergence so figure out what over what interval of X's would your infinite geometric series conversion what would that sum actually be so I'm assuming you've given a go at it so let's try to work through this together so the first thing I want to do is let me just factor out a common factor this might simplify it in terms of trying to express it so let's see if i factor out it looks like all of these are divisible by 3x squared so I can rewrite this as 3x squared times 1 minus X to the third power plus X to the sixth power minus X to the minus X to the ninth power and a pattern is starting to emerge and let me actually close the brackets with the same color with that pink right over there and let's see this looks like we are taking powers of X to the third so let me write it that way this is the same thing as 3x squared times we could write this first term or I guess a it could see the zeroth term this is X to the third to the zeroth power then minus this is X to the third to the first power and then plus this is X to the third to the second power and then I think you see what's going on this is X to the third to the third power and of course we can keep going but now we have to worry about this the switching of signs that we keep having so this would be negative one this is positive which is the same thing as negative one of the zero power this is negative which is negative one to the first power so let's actually write it this way we can write it as 3x squared times this first term we could write it as negative one or we could just write it as negative x to the third to the 0th power and then you're gonna say plus with plus we could say negative x to the third to the first power negative 1 to the first power is negative one to the third to the first powers next to the 3rd plus negative x to the third to the second power plus negative x to the third to the third power that's this term right over here negative one of the third powers negative one and of course x to the third to the third is X to the ninth and it keeps going and so this makes it a lot clearer what our common ratio is our common ratio here is is negative x to the third and over what interval would this converge well it's going to converge if our Comment if the absolute value of our common ratio is less than one so we're going to converge converge if the absolute value of our common ratio the absolute value of our common ratio which is negative x to the third is less than one or another way of saying this is the same thing the absolute value of a negative is going to be the same thing as the absolute value of a positive so that's the same way of saying the absolute value of x to the third is less than one or saying that X to the third is less than 1 and is greater than negative one and the way that's going to happen if you take if you take the cube roots of both sides of this or all the sides of this inequality you're going to get that X is going to be between negative 1 and 1 this right over here is our interval of convergence interval of convergence and if we restrict our X's to that what is this going to sum to was infinite geometric series our common ratio its absolute value is less than 1 and so this is going to sum to this is going to be equal to our first term I guess we could say or the thing that's multiplying by this whole thing but if you're multiplying it out this would be our first term it's going to be 3x squared all of that over 1 minus our common ratio so 1 minus negative x to the third well that's just going to be 1 plus x to the third so everything we've done so far is we've shown we've shown that this let me actually write it this way that this is equal to this thing over the interval of convergence so let me write the copy and paste it so write like that over the interval of convergence with X is between negative 1 and 1 these two things are the same now we can start to put our calculus hat on here because this looks interesting this you might remember this looks like the derivative of something that's familiar 1 plus X to the third what's the derivative of that well that's 3x squared so it looks like this is the this right over here is the derivative of the natural log of 1 plus X to the 3rd or the absolute value of 1 plus X to the third and if you don't believe me let's take the antiderivative of this thing right over here in fact for fun let's take the antiderivative of both sides of this and if we do that then we'll we will have shown essentially a geometric series representation of whatever the antiderivative of this thing is so I encourage you to pause the video again and try to take the antiderivative of both sides of this equation so let's take so we're going to take the antiderivative of the left hand side and we're going to take the antiderivative the antiderivative of the right hand side now on the left hand side I mentioned that hey looks like we have some expression and stir if ative that just calls out for u substitution so if we say that u is equal to 1 plus X to the third let me write this down so U is equal to 1 plus X to the third then what's D u going to be so then D U is going to be equal to 3x squared DX so notice we have U and then D u D U is this right over here so this this expression right over here could be rewritten as so let me go over here this could be re-written as the integral the integral of D u D u over u over u or I could say 1 actually I'm right it's 1 over u D u one over you d you which of course is equal to which is going to be equal to the natural log of the absolute value of U the natural log of the absolute value of U plus plus some constant plus sub constant and we of course know that U is 1 plus x to the third so this is going to be equal to the natural log of the absolute value of 1 plus x to the third one plus x to the third plus C plus C now we're restricting our domain for X being between negative 1 and 1 so for that domain this thing is always going to be this thing is always going to be non Celtic actually it's always going to be positive so let's so what we can do we can we don't have to write the absolute value sign so this is going to be equal to the natural log write it the natural log of 1 plus X to the 3rd 1 plus X to the 3rd plus C plus C so that's this left-hand side and the right-hand sides actually a lot more straightforward this is just a straightforward polynomial now as you can imagine we're gonna get a plus some type of constant there so let me differentiate them a little bit let me call this 1 C 1 and then on the right hand side what do we get the antiderivative of this is going to be let's see a the antiderivative of x squared is X to the third divided by 3 so this first term the antiderivative is just going to be X to the third power the derivative of x to the third is 3x squared now this term right over here negative 3x to the fifth the antiderivative of x to the fifth is X to the sixth X to the sixth over 6 X to the sixth over six but then we have that 3 over here 3 over 6 is 2 so it's negative X to the sixth over 2 let me do that in a different color just so we can keep track of it so this one right over here is negative the anti-derivatives negative X to the sixth over two and then let's see I'm running out of colors the antiderivative of X to the eighth is X to the ninth over nine so it's going to be plus X to the ninth and then we have this 3 3 over 9 is 3 and I think you see a pattern happening and then let's just do one more of these for fun X to the 12th over 12 but we have this 3 so negative X to the 12th over 4 and then we keep going and then we're both course going to have we're going to have some constant and actually let me put the constants in the front so let me copy and paste that or cut and paste that so I have some space so let me let me write over here I'll put some other constant C - it doesn't have to be the same one plus all of this now to simplify this I could subtract C 1 from both sides or essentially from C 2 and then I'm going to have I'm gonna have the natural log of 1 plus X to the 3rd power 1 plus X to the 3rd power this is kind of neat what we've just done with a little bit of integration is equal to C 2 minus C 1 well this is some constant minus some other constant so that's just going to be some arbitrary constant some arbitrary constant plus all of this business plus all of this business and we can even figure out what the constant is going to be by trying out some values of X that's in our that that's in our restricted domain well x equals 0 is between negative 1 and 1 so let's see what happens what X is equal to 0 to solve for C if X is equal to 0 if X is equal to 0 we get natural log natural log of 1 is equal to C c plus well all of these terms are going to be 0 0 to the third power minus 0 to the six on and on and on plus 0 plus 0 or another way and natural log of 1 of course e to the what power is 1 well that's 0 so C must be 0 C is equal to 0 so this thing right years equal to zero so what we've just done using a little bit of integration is starting with a let's just appreciate what went on starting with a an arbitrary infinite series we showed it could be represented as a geometric series we define an interval of convergence over which this would converge over which the common ratios absolute value is less than 1 and then using that we we expressed its sum and then we took the antiderivative of both sides to figure out to figure out an expansion for the natural log of 1 plus 1 plus X to the 3rd power which so at least in my mind that was pretty neat this is what we is a natural log of 1 plus X to the 3rd power is X to the 3rd power minus X to the sixth over 2 plus X to the 9th over 3 so on and on and on it actually let's just just to kind of give ourselves some closure here let's write it in Sigma notation so we could write the natural log of 1 plus X to the 3rd 1 plus X to the 3rd over our restricted domain where the absolute value of x is less than 1 is equal to is equal to the sum this sum from let's say let's say N equals 1 to infinity of X to the 3rd X to the 3rd to the nth power so to the first power a second power third power over n this is X to the third over 1 X to the X to the third squared over 2 O and I of course have to throw in let's see this first one is we're gonna have to care about the sign so let me throw in a negative 1 I see negative 1 to the first power should be negative but here it's positive so I'll say negative 1 to the n plus 1 negative 1 to the n plus 1 power does that work I think it does when N is equal to 1 this thing just becomes 1 this is X to the 3rd over 1 when n is equal to 2 this becomes negative which it needs to be and then this becomes X to the 6 and we're over 2 and so there you go we are done I found that pretty fine
AP® is a registered trademark of the College Board, which has not reviewed this resource.