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### Course: Integral Calculus > Unit 5

Lesson 14: Function as a geometric series# Power series of ln(1+x³)

We can represent ln(1+x³) with a power series by representing its derivative as a power series and then integrating that series. You have to admit this is pretty neat. Created by Sal Khan.

## Want to join the conversation?

- Why did we do the integration bit ? As in what was the aim of it ?(47 votes)
- It's not a well motivated video.

I believe expanded forms are useful for computational approximations where accuracy is controlled by however many terms you choose to include.

In the real world you'd start with ln(1+x^3) and try to find its expanded form (approximation) by taking the derivative, expanding, and then integrating -- you wouldn't start with a geometric series and integrate its closed form because you're just looking for something "neat" to do.(59 votes)

- At12:20, how do we know to start the summation at n=1 and not n=0? would we have to change the summation if we started at n=0?(7 votes)
- You have to create the summation so that it represents the series you have. You can start the summation at an
`n`

equal to whatever you want, but you would have to adjust the body of the summation to keep things the same.

The first term in the series is`x³`

, the second is`-x⁶/2`

, so that is what guides you.

If you decided to start the summation at`n=0`

, then the body of the summation would have to be:`(-1)ⁿ(x³ⁿ⁺³)/(n+1)`

If you decided to start the summation at`n=-1`

, then the body of the summation would have to be:`(-1)ⁿ⁺¹(x³ⁿ⁺⁶)/(n+2)`

Sal decided to start the summation at`n=1`

since that made the body of the summation quite simple.(16 votes)

- This pretty mathmatical art but i did not get what this can be useful for .....(14 votes)
- in the very last formula Is it okay to put (-1)^(n-1) ? instead of (-1)^(n+1) ?(6 votes)
- That is correct. Those expressions will have the same value for any value of n. So if the situation requires it, you could make that substitution.(5 votes)

- @5:00I don't think anybody will realize it's the derivative of natrual log of something lol(6 votes)
- What happens to the interval of convergence after we integrate the polynomial expression to find a series for log(1+x^3)? Are we still constrained by the same interval or does it not apply anymore, and if it doesn't, why is that?(5 votes)
- dont you have to check the limits of the interval of convergence?(3 votes)
- No, because you express the result of the left hand side integral in "x" terms,(1 vote)

- when trying to find the value of the constant of integration 'C', how was the value of x assumed to be 0 ? .... although -1 < x < 1 is our interval of convergence, is x = 0 a part of the interval?... wouldn't x = 0 make (-x^3) = 0 ? and if ( - x ^3) = 0 , wouldn't our common ratio become zero? and wouldn't the initial series cease to be a geometric series ?(0 votes)
- Sal wanted to use a value for x that would
`knock out all the power terms with x in them`

in order to solve for the C. He had removed all the unspecified terms and combined them into C2 to set the next step up.

A common strategy is to choose zero whenever you want to knock out a variable, as long as that was within the restricted domain of -1 < x < 1, and he mentioned he was checking, even though he knew very well that 0 is in that domain.

You asked whether it was: the domain -1 < x < 1 includes all numbers between -1 and 1. If we are counting by integers, the only integer in that domain is`zero`

. If we are counting by smaller divisions than integers, we could use -.99999 -.889653 -.1111111 -.000002`0`

0.000008, 0.55556, .999999 and an infinite number of other decimals or we could use

-½, -¼ ,`0`

, ¼, ½ or any combination just larger than -1 and just smaller than +1. The only really useful choice for this purpose, to avoid dealing with an infinite number of terms, is to use 0.

So

He started with his equation

ln (1 + x³) + C1 = C2 + x³ - ½x⁶ + ⅓x⁹ - ¼x¹²

He simplified it by subtracting C1 from both sides to get C3 or you can just call it C because you no longer have to differentiate between multiple unknown constants:

ln (1 + x³) = C + x³ - ½x⁶ + ⅓x⁹ - ¼x¹²

So, when x = 0, we have

ln (1 + 0) = C + 0 - 0 + 0 - 0

which is the same as

ln (1 ) = C + 0 - 0 + 0 - 0

Now, because the natural log of 1 is 0, this equation becomes

0 = C + 0 - 0 + 0 - 0

So C = 0

Your other question was wouldn't 0 make x³ = 0, the answer is`yes it would for that particular calculation`

, but the value of your common ratio does not change simply because you plugged in a value for one iteration of the calculations.

Hope that helps.(7 votes)

- Why did we factor out the 3x^2? I did it differently at the beginning of the video and more simpler.(3 votes)
- By factoring out 3x^2 it makes it a little bit easier for us to see the common ratio in the geometric series(1 vote)

## Video transcript

Voiceover: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 its sum would be given
an interval of convergence. Figure out over what
interval of xs would your infinite geometric series converge and what would that sum actually be. I'm assuming you've given a go at it, so let's try to work
through this together. 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. Let's see. If I factor out, it
looks like all of these are divisible by three x squared. I can rewrite this as
three x squared times one minus x to the third
power plus x to sixth power minus x to the ninth power, and a pattern is starting to emerge. Let me actually close the
brackets with the same color, with that pink right over there. 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 three x squared times, we could write this first term, or I guess I could say 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 the switching of signs
that we keep having. This would be negative one. This is positive, which is the same thing as negative one in 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 three x squared times this first term we could
write as negative one, or we could just write as
negative x to the third to the zeroth power, and then you're going to say plus. With plus, we could say
negative x to the third to the first power. Negative one to the first
power is negative one. x to the third to the first
power is x to the third 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 to the third
power is negative one, and of course, x to the third
to the third is x to the ninth and it keeps going. This makes it a lot clearer
what our common ratio is. Our common ratio here is
negative x to the third. Over what interval would this converge? It's going to converge if our common, if the absolute value of our common ratio is less than one. We are going to 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 one and is greater than negative one. The way that's going to happen, 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 one and one. This right over here is our
interval of convergence, interval of convergence. If we restrict our xs to that, what is this going to sum to? This infinite geometric
series, our common ratio, its absolute value is less than one, 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
them by this whole thing, but if you multiply it out, this would be our first term. It's going to be three x squared, all of that over one
minus our common ratio, so one minus negative x to the third, that's just going to be
one plus x to the third. Everything we've done so far is we've shown that this, let me actually write it this way, that this thing is equal to this thing over the interval of convergence. Let me write, copy and paste
it, so write like that, over the interval of convergence. If x is between negative one and one, 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. One plus x to the third,
what's the derivative of that? That's three x squared. It looks like this right over
here is the derivative of the natural log of one plus x to the third or the absolute value of
one plus x to the third. If you don't believe me, let's take the
anti-derivative of this thing right over here. In fact, for fun, let's take the anti-derivative
of both sides of this, and if we do that, then we
will have shown essentially a geometric series representation of whatever the anti-derivative
of this thing is. I encourage you to pause the video again and try to take the anti-derivative of both sides of this equation. We're going to take the anti-derivative of the left-hand side, and we're going to take
the anti-derivative, the anti-derivative of
the right-hand side. Now on the left-hand side, I mentioned that it looks like we have an expression and its derivative. That just calls out for u-substitution. If we say that u is equal
to one plus x to the third, let me write this down, u is equal to one plus x to the third, then what's du going to be? Then du is going to be
equal to three x squared dx. Notice, we have u and then du. du is this right over here. This expression right over
here could be rewritten as, let me go over here, this could be rewritten as
the integral of du over u, or I could say, actually, let me write this, one over u du, 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 some constant. We of course know that u
is one plus x to the third, so this is going to be
equal to the natural log of the absolute value of
one 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 one and one. So for that domain, this
thing is always going to be, this whole thing actually is
always going to be positive, so what we can do, we don't have to write
the absolute value sign, so this is going to be
equal to the natural log, let me write it, the natural log of one
plus x to the third, one plus x to the third plus c, plus c. That's this left-hand side, and the right-hand side is actually a lot more straightforward. This is just a straightforward polynomial. Now, as you can imagine,
we're going to get some type of constant there, so let me differentiate them a little bit. Let me call this one c one, and then on the right-hand
side, what do we get? The anti-derivative of
this is going to be, let's see, the anti-derivative of x
squared is x to the third divided by three. This first term, the anti-derivative, is just going to be x to the third power. The derivative of x to the
third is three x squared. Now, this term right over here, negative three x to the fifth, the anti-derivative of x to the fifth is x to the sixth over six, x to the sixth over six, but then we have that three over here. Three over six is two, so it's negative x to the sixth over two. Actually, let me do that
in a different color, just so we can keep track of it. This one right over here is negative, the anti-derivative is negative
x to the sixth over two, and then, let's see, I'm running out of colors. The anti-derivative 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 three. Three over nine is three. 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 three, so negative x to the 12th over four, and then we keep going, and then we're of course going to have, we're going to have some constant. Actually, let me put the
constant in the front. Let me copy and paste that, or cut and paste that
so I have some space. Let me write over here. I'll put some other constant, c two, it doesn't have to be the same one, plus all of this. Now to simplify this, I could subtract c one from both sides or essentially from c two, and then I'm going to
have the natural log of one plus x to the third power. One plus x to the third power, this is kind of neat what we've just done, with a little bit of integration, is equal to c two minus c one. This is some constant
minus some other constant, so that's just going to be
some arbitrary constant, plus all of this business. We can even figure out what
the constant is going to be by trying out some values of x that's in our restricted domain. x equals zero is between
negative one and one, so let's see what happens
when x is equal to zero to solve for c. If x is equal to zero, we get natural log of
one is equal to c plus, all of these terms are going to be zero, zero to the third power
minus zero to the sixth on and on and on, plus zero plus zero, or another way, and natural log of one, of course, [either what] power is one, well, that's zero, so c must be zero. C is equal to zero. This thing right over
here is equal to zero. What we have just done, using
a little bit of integration, is starting with a ... let's just appreciate what went on. Starting with an
arbitrary infinite series, we showed it could be represented
as a geometric series. We defined it into rule of convergence over which this would converge, over which the common
ratio's absolute value is less than one, and then using that, we expressed its sum, and then we took the
anti-derivative of both sides to figure out an expansion
for the natural log of one plus x to the third power, which, at least in my
mind, that was pretty neat. Natural log of one plus
x to the third power is x to the third power
minus x to the sixth over two plus x to the ninth over three, so on and on and on. Actually, just to give
ourselves some closure here, let's write it in sigma notation. We could write the natural
log of one plus x to the third over our restricted domain, where the absolute value
of x is less than one, is equal to the sum from, let's say, n equals one to infinity of x to the third to the nth power, so to the first power, the
second power, third power, over n. This is x to the third over one, x to the third squared over two, oh, and I of course have
to throw in, let's see, this first one is, we're going to have to
care about the sign, so let me throw in a negative one. Let's see. Negative one to the first
power should be negative, but here, it's positive, so I'll say negative
one to the n plus one, negative one to the n plus one power. Does that work? I think it does. When n is equal to one, this thing just becomes one. This is x to the third over one. When n is equal to two, this becomes negative,
which it needs to be, and then this becomes x to
the sixth, and we're over two, and so there you go. We are done. I found that pretty satisfying.