AP®︎/College Calculus BC
- Function as a geometric series
- Geometric series as a function
- Power series of arctan(2x)
- Power series of ln(1+x³)
- Function as a geometric series
- Maclaurin series of cos(x)
- Maclaurin series of sin(x)
- Maclaurin series of eˣ
- Worked example: power series from cos(x)
- Worked example: cosine function from power series
- Worked example: recognizing function from Taylor series
- Maclaurin series of sin(x), cos(x), and eˣ
- Visualizing Taylor series approximations
- Euler's formula & Euler's identity
- Geometric series interval of convergence
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 ?(46 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.(56 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
nequal 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:
If you decided to start the summation at
n=-1, then the body of the summation would have to be:
Sal decided to start the summation at
n=1since that made the body of the summation quite simple.(16 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?(4 votes)
- dont you have to check the limits of the interval of convergence?(3 votes)
- 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 themin 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
00.000008, 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.
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)
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.