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So the nice thing about math, if we're doing things that make logical sense, we should get to the same end point. But the point here is that there's multiple strategies. You could use the chain rule first and then the product rule or you could use the product rule first and then the chain rule. In this case, you could debate which one is faster. It looks like the one on the right might be a little bit faster. But sometimes, but these two are pretty close, but sometimes it'll be more clear than not which one is preferable. You really wanna minimize the amount of hairiness, the number of steps, the chances for careless mistakes you might have. | Applying the chain rule and product rule Advanced derivatives AP Calculus AB Khan Academy.mp3 |
And as you can imagine, based on the context of where this video shows up on Khan Academy, that maybe we will do it using the comparison test. And at any point, if you feel like you can kind of take this to the finish line, feel free to pause the video and do so. So in order to kind of figure out or get a sense for this series right over here, it never hurts to kind of expand it out a little bit. So let's do that. So this would be equal to, when n equals one, this is going to be one over two to the one plus one. So it's gonna be one over two plus one, it's gonna be 1 3rd, plus, that's n equals one, when n equals two, it's gonna be one over two squared, which is four plus two, plus one over six, plus, let's see, when we go to three, n equals one, n equals two, n equals three, it's gonna be one over two to the third, which is eight plus three is 11. So one over 11, maybe I'll do one more term. | Worked example direct comparison test Series AP Calculus BC Khan Academy.mp3 |
So let's do that. So this would be equal to, when n equals one, this is going to be one over two to the one plus one. So it's gonna be one over two plus one, it's gonna be 1 3rd, plus, that's n equals one, when n equals two, it's gonna be one over two squared, which is four plus two, plus one over six, plus, let's see, when we go to three, n equals one, n equals two, n equals three, it's gonna be one over two to the third, which is eight plus three is 11. So one over 11, maybe I'll do one more term. Two to the fourth power is going to be 16, plus four is 20, plus one over 20, and obviously we just keep going on and on and on. So in order to use, so it looks, it feels like this thing could converge. All of our terms are positive, but they are going, they're getting smaller and smaller quite fast, and if we really look at the, if we look at the behavior of this, the terms as n gets larger and larger, we see that the two to the n in the denominator will grow much, much faster than the n will. | Worked example direct comparison test Series AP Calculus BC Khan Academy.mp3 |
So one over 11, maybe I'll do one more term. Two to the fourth power is going to be 16, plus four is 20, plus one over 20, and obviously we just keep going on and on and on. So in order to use, so it looks, it feels like this thing could converge. All of our terms are positive, but they are going, they're getting smaller and smaller quite fast, and if we really look at the, if we look at the behavior of this, the terms as n gets larger and larger, we see that the two to the n in the denominator will grow much, much faster than the n will. So this will start to behave, this kind of behaves like one over two to the n, which is a clue of something that we might be able to use for the comparison test. So let me just write that down. So we have one over, so the infinite series from n equals one to infinity of one over two to the n. And so when n equals one, this is going to be equal to, this is going to be equal to 1 1⁄2. | Worked example direct comparison test Series AP Calculus BC Khan Academy.mp3 |
All of our terms are positive, but they are going, they're getting smaller and smaller quite fast, and if we really look at the, if we look at the behavior of this, the terms as n gets larger and larger, we see that the two to the n in the denominator will grow much, much faster than the n will. So this will start to behave, this kind of behaves like one over two to the n, which is a clue of something that we might be able to use for the comparison test. So let me just write that down. So we have one over, so the infinite series from n equals one to infinity of one over two to the n. And so when n equals one, this is going to be equal to, this is going to be equal to 1 1⁄2. When n is equal to two, this is going to be equal to 1⁄4. When n is equal to three, this is equal to 1⁸. When n is equal to four, this is equal to 1 16th. | Worked example direct comparison test Series AP Calculus BC Khan Academy.mp3 |
So we have one over, so the infinite series from n equals one to infinity of one over two to the n. And so when n equals one, this is going to be equal to, this is going to be equal to 1 1⁄2. When n is equal to two, this is going to be equal to 1⁄4. When n is equal to three, this is equal to 1⁸. When n is equal to four, this is equal to 1 16th. And you go on and on and on and on. And what's interesting about this is we recognize it. This is a geometric series, so let me be clear. | Worked example direct comparison test Series AP Calculus BC Khan Academy.mp3 |
When n is equal to four, this is equal to 1 16th. And you go on and on and on and on. And what's interesting about this is we recognize it. This is a geometric series, so let me be clear. This thing right over here, that is the same thing as the sum from n equals one to infinity of 1⁄2 to the n power. Just writing it in a different way. And since the absolute value of 1⁄2, which is just 1⁄2, so because the absolute value of 1⁄2 is less than one, we know that this geometric series converges. | Worked example direct comparison test Series AP Calculus BC Khan Academy.mp3 |
This is a geometric series, so let me be clear. This thing right over here, that is the same thing as the sum from n equals one to infinity of 1⁄2 to the n power. Just writing it in a different way. And since the absolute value of 1⁄2, which is just 1⁄2, so because the absolute value of 1⁄2 is less than one, we know that this geometric series converges. We know that it converges. And actually, we even have formulas for finding the exact sum of, or to find this, figure out what it converges to. And so we know this thing converges. | Worked example direct comparison test Series AP Calculus BC Khan Academy.mp3 |
And since the absolute value of 1⁄2, which is just 1⁄2, so because the absolute value of 1⁄2 is less than one, we know that this geometric series converges. We know that it converges. And actually, we even have formulas for finding the exact sum of, or to find this, figure out what it converges to. And so we know this thing converges. And we see that actually these two series combined meet all of the constraints we need for the comparison test. So let's go back to what we wrote about the comparison test. So the comparison test, we have two series. | Worked example direct comparison test Series AP Calculus BC Khan Academy.mp3 |
And so we know this thing converges. And we see that actually these two series combined meet all of the constraints we need for the comparison test. So let's go back to what we wrote about the comparison test. So the comparison test, we have two series. All of their terms are greater than or equal to zero. All of these terms are greater than or equal to zero. And then for the corresponding terms in one series, all of them are going to be less than or equal to the corresponding terms in the next one. | Worked example direct comparison test Series AP Calculus BC Khan Academy.mp3 |
So the comparison test, we have two series. All of their terms are greater than or equal to zero. All of these terms are greater than or equal to zero. And then for the corresponding terms in one series, all of them are going to be less than or equal to the corresponding terms in the next one. And so if we look over here, we could consider this one, the magenta series. This is kind of our infinite series of dealing with a sub n. And that this right over here is, well, I already did it in blue. This is kind of the blue series. | Worked example direct comparison test Series AP Calculus BC Khan Academy.mp3 |
And then for the corresponding terms in one series, all of them are going to be less than or equal to the corresponding terms in the next one. And so if we look over here, we could consider this one, the magenta series. This is kind of our infinite series of dealing with a sub n. And that this right over here is, well, I already did it in blue. This is kind of the blue series. And notice all their terms are non-negative. And the corresponding terms, 1⁄2 is greater than 1⁄3. 1⁄4 is greater than 1⁶. | Worked example direct comparison test Series AP Calculus BC Khan Academy.mp3 |
This is kind of the blue series. And notice all their terms are non-negative. And the corresponding terms, 1⁄2 is greater than 1⁄3. 1⁄4 is greater than 1⁶. 1⁸ is greater than 1⁹. 1 over 2 to the n is always going to be greater than 1 over 2 to the n plus n for the n's that we care about here. And since we know that this converges, since we know that the larger one converges, it's a geometric series where the common ratio, the absolute value of the common ratio is less than one. | Worked example direct comparison test Series AP Calculus BC Khan Academy.mp3 |
What we want to do in this video is figure out what the limit as x approaches zero of one minus cosine of x over x is equal to. And we're going to assume we know one thing ahead of time. We're going to assume we know that the limit as x approaches zero of sine of x over x, that this is equal to one. And I'm not gonna reprove this in this video, but we have a whole other video dedicated to proving this famous limit. And we do it using the squeeze or the sandwich theorem. So let's see if we can work this out. So the first thing we're going to do is algebraically manipulate this expression a little bit. | Limit of (1-cos(x)) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy.mp3 |
And I'm not gonna reprove this in this video, but we have a whole other video dedicated to proving this famous limit. And we do it using the squeeze or the sandwich theorem. So let's see if we can work this out. So the first thing we're going to do is algebraically manipulate this expression a little bit. What I'm going to do is I'm gonna multiply both the numerator and the denominator by one plus cosine of x. So times the denominator, I have to do the same thing, one plus cosine of x. I'm not changing the value of the expression. This is just multiplying it by one. | Limit of (1-cos(x)) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy.mp3 |
So the first thing we're going to do is algebraically manipulate this expression a little bit. What I'm going to do is I'm gonna multiply both the numerator and the denominator by one plus cosine of x. So times the denominator, I have to do the same thing, one plus cosine of x. I'm not changing the value of the expression. This is just multiplying it by one. But what does that do for us? Well, I can rewrite the whole thing as the limit as x approaches zero. So one minus cosine of x times one plus cosine of x. | Limit of (1-cos(x)) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy.mp3 |
This is just multiplying it by one. But what does that do for us? Well, I can rewrite the whole thing as the limit as x approaches zero. So one minus cosine of x times one plus cosine of x. Well, that is just going to be, let me do this in another color. That is going to be one squared, which is just one, minus cosine squared of x. Cosine squared of x, difference of squares. And then in the denominator, I am going to have these, which is just x times one plus cosine of x. | Limit of (1-cos(x)) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy.mp3 |
So one minus cosine of x times one plus cosine of x. Well, that is just going to be, let me do this in another color. That is going to be one squared, which is just one, minus cosine squared of x. Cosine squared of x, difference of squares. And then in the denominator, I am going to have these, which is just x times one plus cosine of x. Now what is one minus cosine squared of x? Well, this comes straight out of the Pythagorean identity, trig identity. This is the same thing as sine squared of x. | Limit of (1-cos(x)) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy.mp3 |
And then in the denominator, I am going to have these, which is just x times one plus cosine of x. Now what is one minus cosine squared of x? Well, this comes straight out of the Pythagorean identity, trig identity. This is the same thing as sine squared of x. So sine squared of x. And so I can rewrite all of this as being equal to the limit as x approaches zero. And let me rewrite this as, instead of sine squared of x, that's the same thing as sine of x times sine of x. | Limit of (1-cos(x)) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy.mp3 |
This is the same thing as sine squared of x. So sine squared of x. And so I can rewrite all of this as being equal to the limit as x approaches zero. And let me rewrite this as, instead of sine squared of x, that's the same thing as sine of x times sine of x. Let me write it that way. Sine x times sine x. So I'll take the first sine of x. | Limit of (1-cos(x)) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy.mp3 |
And let me rewrite this as, instead of sine squared of x, that's the same thing as sine of x times sine of x. Let me write it that way. Sine x times sine x. So I'll take the first sine of x. So I'll do, I'll take this one right over here and put it over this x. So sine of x over x times the second sine of x, let's say this one, over one plus cosine of x. Times sine of x over one plus cosine of x. | Limit of (1-cos(x)) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy.mp3 |
So I'll take the first sine of x. So I'll do, I'll take this one right over here and put it over this x. So sine of x over x times the second sine of x, let's say this one, over one plus cosine of x. Times sine of x over one plus cosine of x. All I've done is I've leveraged a trigonometric identity and I've done a little bit of algebraic manipulation. Well here, the limit of the product of these two expressions is going to be the same thing as the product of the limits. So I can rewrite this as being equal to the limit as x approaches zero of sine of x over x times the limit as x approaches zero of sine of x over one plus cosine of x. | Limit of (1-cos(x)) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy.mp3 |
Times sine of x over one plus cosine of x. All I've done is I've leveraged a trigonometric identity and I've done a little bit of algebraic manipulation. Well here, the limit of the product of these two expressions is going to be the same thing as the product of the limits. So I can rewrite this as being equal to the limit as x approaches zero of sine of x over x times the limit as x approaches zero of sine of x over one plus cosine of x. Now, we said going into this video that we're going to assume that we know what this is. We've proven it in other videos. What is the limit as x approaches zero of sine of x over x? | Limit of (1-cos(x)) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy.mp3 |
So I can rewrite this as being equal to the limit as x approaches zero of sine of x over x times the limit as x approaches zero of sine of x over one plus cosine of x. Now, we said going into this video that we're going to assume that we know what this is. We've proven it in other videos. What is the limit as x approaches zero of sine of x over x? Well, that is equal to one. So this whole limit is just going to be dependent on whatever this is equal to. Well, this is pretty straightforward here. | Limit of (1-cos(x)) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy.mp3 |
What is the limit as x approaches zero of sine of x over x? Well, that is equal to one. So this whole limit is just going to be dependent on whatever this is equal to. Well, this is pretty straightforward here. As x approaches zero, the numerator's approaching zero, sine of zero is zero. The denominator is approaching, cosine of zero is one, so the denominator is approaching two. So this is approaching zero over two or just zero. | Limit of (1-cos(x)) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy.mp3 |
Well, this is pretty straightforward here. As x approaches zero, the numerator's approaching zero, sine of zero is zero. The denominator is approaching, cosine of zero is one, so the denominator is approaching two. So this is approaching zero over two or just zero. So that's approaching zero. One times zero, well, this is just going to be equal to zero and we're done. Using that fact and a little bit of trig identities and a little bit of algebraic manipulation, we were able to show that our original limit, the limit as x approaches zero of one minus cosine of x over x is equal to zero. | Limit of (1-cos(x)) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy.mp3 |
In the last video, we took the Maclaurin expansion of e to the x, and we saw that it looked like it was some type of a combination of the polynomial approximations of cosine of x and of sine of x. But it's not quite, because there were a couple of negatives in there, if we were to really add these two together, that we did not have when we took the representation of e to the x. But to reconcile these, I'll do a little bit of a, I don't know if you can even call it a trick. Let's see, if we take this polynomial expansion of e to the x, this approximation, what happens, and if we say e to the x is equal to this, especially as this becomes an infinite number of terms, then it becomes less of an approximation and more of an equality. What happens if I take e to the ix? And before, that might have been kind of a weird thing to do. Let me write it down, e to the ix. | Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3 |
Let's see, if we take this polynomial expansion of e to the x, this approximation, what happens, and if we say e to the x is equal to this, especially as this becomes an infinite number of terms, then it becomes less of an approximation and more of an equality. What happens if I take e to the ix? And before, that might have been kind of a weird thing to do. Let me write it down, e to the ix. Because before, it's like, how do you define e to the ith power? That's a very bizarre thing to do, to take something to the xi power. How do you even comprehend some type of a function like that? | Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3 |
Let me write it down, e to the ix. Because before, it's like, how do you define e to the ith power? That's a very bizarre thing to do, to take something to the xi power. How do you even comprehend some type of a function like that? But now that we can have a polynomial expansion of e to the x, we can maybe make some sense of it, because we can take i to different amounts, to different powers, and we know what that gives. i squared is negative 1, i to the third is negative i, so on and so forth. What happens if we take e to the ix? | Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3 |
How do you even comprehend some type of a function like that? But now that we can have a polynomial expansion of e to the x, we can maybe make some sense of it, because we can take i to different amounts, to different powers, and we know what that gives. i squared is negative 1, i to the third is negative i, so on and so forth. What happens if we take e to the ix? Once again, it's just like taking the x up here and replacing it with an ix. Everywhere we see the x in its polynomial approximation, we would write an ix. Let's do that. | Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3 |
What happens if we take e to the ix? Once again, it's just like taking the x up here and replacing it with an ix. Everywhere we see the x in its polynomial approximation, we would write an ix. Let's do that. e to the ix should be approximately equal to, and it will become more and more equal. This is more to give you an intuition. I'm not doing a rigorous proof here, but it's still profound. | Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3 |
Let's do that. e to the ix should be approximately equal to, and it will become more and more equal. This is more to give you an intuition. I'm not doing a rigorous proof here, but it's still profound. Not to oversell it, but I don't think I can oversell what is about to be discovered or seen in this video. It would be equal to 1 plus, instead of an x, we'll have an ix, plus ix, plus, so what's ix squared? It's going to be, so let me write this down. | Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3 |
I'm not doing a rigorous proof here, but it's still profound. Not to oversell it, but I don't think I can oversell what is about to be discovered or seen in this video. It would be equal to 1 plus, instead of an x, we'll have an ix, plus ix, plus, so what's ix squared? It's going to be, so let me write this down. What is ix squared over 2 factorial? Well, i squared is going to be negative 1, and then you have x squared over 2 factorial. It's going to be minus x squared over 2 factorial. | Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3 |
It's going to be, so let me write this down. What is ix squared over 2 factorial? Well, i squared is going to be negative 1, and then you have x squared over 2 factorial. It's going to be minus x squared over 2 factorial. I think you might see where this is going to go. Then what is ix? Remember, everywhere we saw an x, we're going to replace it with an ix. | Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3 |
It's going to be minus x squared over 2 factorial. I think you might see where this is going to go. Then what is ix? Remember, everywhere we saw an x, we're going to replace it with an ix. What is ix to the third power? Actually, let me write this out. Let me not skip some steps over here. | Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3 |
Remember, everywhere we saw an x, we're going to replace it with an ix. What is ix to the third power? Actually, let me write this out. Let me not skip some steps over here. This is going to be ix squared over 2 factorial. Actually, let me, I want to do it just the way. Plus, plus ix squared over 2 factorial, plus ix to the third over 3 factorial, plus ix to the fourth over 4 factorial, and we can keep going, plus ix to the fifth over 5 factorial, and we could just keep going, so on and so forth. | Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3 |
Let me not skip some steps over here. This is going to be ix squared over 2 factorial. Actually, let me, I want to do it just the way. Plus, plus ix squared over 2 factorial, plus ix to the third over 3 factorial, plus ix to the fourth over 4 factorial, and we can keep going, plus ix to the fifth over 5 factorial, and we could just keep going, so on and so forth. Let's evaluate these ix's raised to these different powers. This will be equal to 1 plus ix. ix squared, that's the same thing as i squared times x squared. | Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3 |
Plus, plus ix squared over 2 factorial, plus ix to the third over 3 factorial, plus ix to the fourth over 4 factorial, and we can keep going, plus ix to the fifth over 5 factorial, and we could just keep going, so on and so forth. Let's evaluate these ix's raised to these different powers. This will be equal to 1 plus ix. ix squared, that's the same thing as i squared times x squared. i squared is negative 1. This is negative x squared over 2 factorial. Then this is going to be the same thing as i to the third times x to the third. | Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3 |
ix squared, that's the same thing as i squared times x squared. i squared is negative 1. This is negative x squared over 2 factorial. Then this is going to be the same thing as i to the third times x to the third. i to the third is the same thing as i squared times i. It's going to be negative i. This is going to be minus i times x to the third over 3 factorial. | Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3 |
Then this is going to be the same thing as i to the third times x to the third. i to the third is the same thing as i squared times i. It's going to be negative i. This is going to be minus i times x to the third over 3 factorial. Then plus, you're going to have, what's i to the fourth power? That's i squared squared. That's negative 1 squared. | Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3 |
This is going to be minus i times x to the third over 3 factorial. Then plus, you're going to have, what's i to the fourth power? That's i squared squared. That's negative 1 squared. That's just going to be 1. i to the fourth is 1, and then you have x to the fourth. Plus x to the fourth over 4 factorial. Then you're going to have plus, i to the fifth. | Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3 |
That's negative 1 squared. That's just going to be 1. i to the fourth is 1, and then you have x to the fourth. Plus x to the fourth over 4 factorial. Then you're going to have plus, i to the fifth. i to the fifth is going to be 1 times i. It's going to be i times x to the fifth over 5 factorial. Plus i times x to the fifth over 5 factorial. | Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3 |
Then you're going to have plus, i to the fifth. i to the fifth is going to be 1 times i. It's going to be i times x to the fifth over 5 factorial. Plus i times x to the fifth over 5 factorial. I think you might see a pattern here. Coefficient is 1, then i, then negative 1, then negative i, then 1, then i, then negative 1. x to the sixth over 6 factorial. Then negative i, x to the seventh over 7 factorial. | Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3 |
Plus i times x to the fifth over 5 factorial. I think you might see a pattern here. Coefficient is 1, then i, then negative 1, then negative i, then 1, then i, then negative 1. x to the sixth over 6 factorial. Then negative i, x to the seventh over 7 factorial. We have some terms. Some of them are imaginary. They have an i. | Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3 |
Then negative i, x to the seventh over 7 factorial. We have some terms. Some of them are imaginary. They have an i. They're being multiplied by i. Some of them are real. Why don't we separate them out? | Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3 |
They have an i. They're being multiplied by i. Some of them are real. Why don't we separate them out? Once again, e to the ix is going to be equal to this thing, especially as we add an infinite number of terms. Let's separate out the real and the non-real terms. Or the real and the imaginary terms, I should say. | Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3 |
Why don't we separate them out? Once again, e to the ix is going to be equal to this thing, especially as we add an infinite number of terms. Let's separate out the real and the non-real terms. Or the real and the imaginary terms, I should say. This is real. This is real. This is real. | Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3 |
Or the real and the imaginary terms, I should say. This is real. This is real. This is real. This right over here is real. Obviously, we could keep going on with that. The real terms here are 1 minus x squared over 2 factorial plus x to the fourth over 4 factorial. | Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3 |
This is real. This right over here is real. Obviously, we could keep going on with that. The real terms here are 1 minus x squared over 2 factorial plus x to the fourth over 4 factorial. You might be getting excited now. Minus x to the sixth over 6 factorial. That's all I've done here. | Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3 |
The real terms here are 1 minus x squared over 2 factorial plus x to the fourth over 4 factorial. You might be getting excited now. Minus x to the sixth over 6 factorial. That's all I've done here. They would keep going. Plus, so on and so forth. That's all of the real terms. | Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3 |
That's all I've done here. They would keep going. Plus, so on and so forth. That's all of the real terms. What are the imaginary terms here? I'll just factor out the i over here. Actually, let me just factor out. | Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3 |
That's all of the real terms. What are the imaginary terms here? I'll just factor out the i over here. Actually, let me just factor out. It's going to be plus i times, well this is ix, so this will be x. That's an imaginary term. This is an imaginary term. | Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3 |
Actually, let me just factor out. It's going to be plus i times, well this is ix, so this will be x. That's an imaginary term. This is an imaginary term. We're factoring out the i. Minus x to the third over 3 factorial. The next imaginary term is right over there. | Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3 |
This is an imaginary term. We're factoring out the i. Minus x to the third over 3 factorial. The next imaginary term is right over there. We factored out the i. Plus x to the fifth over 5 factorial. The next imaginary term is right there. | Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3 |
The next imaginary term is right over there. We factored out the i. Plus x to the fifth over 5 factorial. The next imaginary term is right there. We factored out the i. It's minus x to the seventh over 7 factorial. Then we obviously would keep going. | Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3 |
The next imaginary term is right there. We factored out the i. It's minus x to the seventh over 7 factorial. Then we obviously would keep going. Plus, minus, keep going, so on and so forth. Preferably to infinity so that we get as good of an approximation as possible. We have a situation where e to the ix is equal to all of this business here. | Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3 |
Then we obviously would keep going. Plus, minus, keep going, so on and so forth. Preferably to infinity so that we get as good of an approximation as possible. We have a situation where e to the ix is equal to all of this business here. You probably remember from the last few videos, the real part, this was the polynomial, this was the Maclaurin approximation of cosine of x around 0. I should say the Taylor approximation around 0, or we could also call it the Maclaurin approximation. This and this are the same thing. | Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3 |
We have a situation where e to the ix is equal to all of this business here. You probably remember from the last few videos, the real part, this was the polynomial, this was the Maclaurin approximation of cosine of x around 0. I should say the Taylor approximation around 0, or we could also call it the Maclaurin approximation. This and this are the same thing. This is cosine of x, especially when you add an infinite number of terms. Cosine of x. This over here is sine of x, the exact same thing. | Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3 |
This and this are the same thing. This is cosine of x, especially when you add an infinite number of terms. Cosine of x. This over here is sine of x, the exact same thing. It looks like we're able to reconcile how you can add up cosine of x and sine of x to get something that's like e to the x. This right here is sine of x. If we take it for granted, I'm not rigorously proving it to you, that if you were to take an infinite number of terms here, that this will essentially become cosine of x. | Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3 |
This over here is sine of x, the exact same thing. It looks like we're able to reconcile how you can add up cosine of x and sine of x to get something that's like e to the x. This right here is sine of x. If we take it for granted, I'm not rigorously proving it to you, that if you were to take an infinite number of terms here, that this will essentially become cosine of x. If you take an infinite number of terms here, this will become sine of x. It leads to a fascinating, fascinating formula. We could say that e to the ix is the same thing as cosine of x. | Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3 |
If we take it for granted, I'm not rigorously proving it to you, that if you were to take an infinite number of terms here, that this will essentially become cosine of x. If you take an infinite number of terms here, this will become sine of x. It leads to a fascinating, fascinating formula. We could say that e to the ix is the same thing as cosine of x. It is cosine of x, and you should be getting goose pimples right around now, is equal to cosine of x plus i times sine of x. This is Euler's formula. I always pronounce them right over here. | Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3 |
We could say that e to the ix is the same thing as cosine of x. It is cosine of x, and you should be getting goose pimples right around now, is equal to cosine of x plus i times sine of x. This is Euler's formula. I always pronounce them right over here. This right here is Euler's formula. If that by itself isn't exciting and crazy enough for you, because it really should be, because we've already done some pretty cool things. We're involving e, which we get from continuous compounding interest. | Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3 |
I always pronounce them right over here. This right here is Euler's formula. If that by itself isn't exciting and crazy enough for you, because it really should be, because we've already done some pretty cool things. We're involving e, which we get from continuous compounding interest. We have cosine and sine of x, which are ratios of right triangles. It comes out of the unit circle. Somehow we've thrown in the square root of negative 1. | Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3 |
We're involving e, which we get from continuous compounding interest. We have cosine and sine of x, which are ratios of right triangles. It comes out of the unit circle. Somehow we've thrown in the square root of negative 1. There seems to be this cool relationship here. It becomes extra cool, and we're going to assume we're operating in radians here, is if we assume Euler's formula, what happens when x is equal to pi? Just to throw in another wacky number in there, the ratio between the circumference and the diameter of a circle. | Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3 |
Somehow we've thrown in the square root of negative 1. There seems to be this cool relationship here. It becomes extra cool, and we're going to assume we're operating in radians here, is if we assume Euler's formula, what happens when x is equal to pi? Just to throw in another wacky number in there, the ratio between the circumference and the diameter of a circle. What happens when we throw in pi? We get e to the i pi is equal to cosine of pi. Cosine of pi is what? | Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3 |
Just to throw in another wacky number in there, the ratio between the circumference and the diameter of a circle. What happens when we throw in pi? We get e to the i pi is equal to cosine of pi. Cosine of pi is what? Cosine of pi is half way around the unit circle. Cosine of pi is negative 1. Then sine of pi is 0. | Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3 |
Cosine of pi is what? Cosine of pi is half way around the unit circle. Cosine of pi is negative 1. Then sine of pi is 0. This term goes away. If you evaluate it at pi, you get something amazing. This is called Euler's identity. | Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3 |
Then sine of pi is 0. This term goes away. If you evaluate it at pi, you get something amazing. This is called Euler's identity. I always have trouble pronouncing Euler. Euler's identity, which we could write like this, or we could add 1 to both sides, and we could write it like this. I'll write it in different colors for emphasis. | Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3 |
This is called Euler's identity. I always have trouble pronouncing Euler. Euler's identity, which we could write like this, or we could add 1 to both sides, and we could write it like this. I'll write it in different colors for emphasis. E to the i times pi plus 1 is equal to, I'll do that in a neutral color, is equal to, I'm just adding 1 to both sides of this thing right over here, is equal to 0. This is thought provoking. Here we have, just so you see, this tells you that there's some connectedness to the universe that we don't fully understand, or at least I don't fully understand. | Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3 |
I'll write it in different colors for emphasis. E to the i times pi plus 1 is equal to, I'll do that in a neutral color, is equal to, I'm just adding 1 to both sides of this thing right over here, is equal to 0. This is thought provoking. Here we have, just so you see, this tells you that there's some connectedness to the universe that we don't fully understand, or at least I don't fully understand. I is defined by engineers for simplicity so that they can find the roots of all sorts of polynomials as you could say the square root of negative 1. Pi is the ratio between the circumference of a circle and its diameter. Once again, another interesting number, but it seems like it comes from a different place as i. E comes from a bunch of different places. | Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3 |
Here we have, just so you see, this tells you that there's some connectedness to the universe that we don't fully understand, or at least I don't fully understand. I is defined by engineers for simplicity so that they can find the roots of all sorts of polynomials as you could say the square root of negative 1. Pi is the ratio between the circumference of a circle and its diameter. Once again, another interesting number, but it seems like it comes from a different place as i. E comes from a bunch of different places. E, you can either think of it, it comes out of continuous compounding interest, super valuable for finance. It also comes from the notion that the derivative of e to the x is also e to the x, so another fascinating number, but once again, seemingly unrelated to how we came up with i and seemingly unrelated to how we came up with pi. Then of course, you have some of the most profound basic numbers right over here. | Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3 |
Once again, another interesting number, but it seems like it comes from a different place as i. E comes from a bunch of different places. E, you can either think of it, it comes out of continuous compounding interest, super valuable for finance. It also comes from the notion that the derivative of e to the x is also e to the x, so another fascinating number, but once again, seemingly unrelated to how we came up with i and seemingly unrelated to how we came up with pi. Then of course, you have some of the most profound basic numbers right over here. You have 1. I don't have to explain why 1 is a cool number, and I shouldn't have to explain why 0 is a cool number. This right here connects all of these fundamental numbers in some mystical way that shows that there's some connectedness to the universe. | Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3 |
So let's say we have the indefinite integral of natural log of x to the 10th power, all of that over x, dx. Does u-substitution apply? And if so, how would we make that substitution? Well, the key for u-substitution is to see do I have some function and its derivative? And you might immediately recognize that the derivative of natural log of x is equal to one over x. To make it a little bit clearer, I could write this as the integral of natural log of x to the 10th power times one over x, dx. Now, it's clear. | _-substitution defining _ (more examples) AP Calculus AB Khan Academy.mp3 |
Well, the key for u-substitution is to see do I have some function and its derivative? And you might immediately recognize that the derivative of natural log of x is equal to one over x. To make it a little bit clearer, I could write this as the integral of natural log of x to the 10th power times one over x, dx. Now, it's clear. We have some function, natural log of x, being raised to the 10th power, but we also have its derivative right over here, one over x. So we could make the substitution. We could say that u is equal to the natural log of x. | _-substitution defining _ (more examples) AP Calculus AB Khan Academy.mp3 |
Now, it's clear. We have some function, natural log of x, being raised to the 10th power, but we also have its derivative right over here, one over x. So we could make the substitution. We could say that u is equal to the natural log of x. And the reason why I picked natural log of x is because I see something, I see its exact derivative here, something close to its derivative. In this case, it's its exact derivative. And so then I could say du, dx, du, dx is equal to one over x, which means that du is equal to one over x, dx. | _-substitution defining _ (more examples) AP Calculus AB Khan Academy.mp3 |
We could say that u is equal to the natural log of x. And the reason why I picked natural log of x is because I see something, I see its exact derivative here, something close to its derivative. In this case, it's its exact derivative. And so then I could say du, dx, du, dx is equal to one over x, which means that du is equal to one over x, dx. And so here you have it. This right over here is du, and then this right over here is our u. And so this nicely simplifies to the integral of u to the 10th power, u to the 10th power, du. | _-substitution defining _ (more examples) AP Calculus AB Khan Academy.mp3 |
And so then I could say du, dx, du, dx is equal to one over x, which means that du is equal to one over x, dx. And so here you have it. This right over here is du, and then this right over here is our u. And so this nicely simplifies to the integral of u to the 10th power, u to the 10th power, du. And so you would evaluate what this is, find the antiderivative here, and then you would back-substitute the natural log of x for u, and to actually evaluate this indefinite integral. Let's do another one. Let's say that we have the integral of, let's do something interesting here. | _-substitution defining _ (more examples) AP Calculus AB Khan Academy.mp3 |
And so this nicely simplifies to the integral of u to the 10th power, u to the 10th power, du. And so you would evaluate what this is, find the antiderivative here, and then you would back-substitute the natural log of x for u, and to actually evaluate this indefinite integral. Let's do another one. Let's say that we have the integral of, let's do something interesting here. Let's say the integral of tangent x, dx. Does u-substitution apply here? And at first you say, well, I just have a tangent of x. | _-substitution defining _ (more examples) AP Calculus AB Khan Academy.mp3 |
Let's say that we have the integral of, let's do something interesting here. Let's say the integral of tangent x, dx. Does u-substitution apply here? And at first you say, well, I just have a tangent of x. Where is its derivative? But one interesting thing to do is, well, we could rewrite tangent in terms of sine and cosine. So we could write this as the integral of sine of x over cosine of x, dx. | _-substitution defining _ (more examples) AP Calculus AB Khan Academy.mp3 |
And at first you say, well, I just have a tangent of x. Where is its derivative? But one interesting thing to do is, well, we could rewrite tangent in terms of sine and cosine. So we could write this as the integral of sine of x over cosine of x, dx. And now you might say, well, where does u-substitution apply here? Well, there's a couple of ways to think about it. You could say the derivative of sine of x is cosine of x, but you're now dividing by the derivative as opposed to multiplying by it. | _-substitution defining _ (more examples) AP Calculus AB Khan Academy.mp3 |
So we could write this as the integral of sine of x over cosine of x, dx. And now you might say, well, where does u-substitution apply here? Well, there's a couple of ways to think about it. You could say the derivative of sine of x is cosine of x, but you're now dividing by the derivative as opposed to multiplying by it. But more interesting, you could say the derivative of cosine of x is negative sine of x. We don't have a negative sine of x, but we can do a little bit of engineering. We can multiply by negative one twice. | _-substitution defining _ (more examples) AP Calculus AB Khan Academy.mp3 |
You could say the derivative of sine of x is cosine of x, but you're now dividing by the derivative as opposed to multiplying by it. But more interesting, you could say the derivative of cosine of x is negative sine of x. We don't have a negative sine of x, but we can do a little bit of engineering. We can multiply by negative one twice. So we could say the negative of the negative sine of x. And I stuck one of the, you could say, negative ones outside of the integral, which comes straight from our integration properties. This is equivalent. | _-substitution defining _ (more examples) AP Calculus AB Khan Academy.mp3 |
We can multiply by negative one twice. So we could say the negative of the negative sine of x. And I stuck one of the, you could say, negative ones outside of the integral, which comes straight from our integration properties. This is equivalent. I can put a negative on the outside and a negative on the inside so that this is the derivative of cosine of x. And so now this is interesting. In fact, let me rewrite this. | _-substitution defining _ (more examples) AP Calculus AB Khan Academy.mp3 |
This is equivalent. I can put a negative on the outside and a negative on the inside so that this is the derivative of cosine of x. And so now this is interesting. In fact, let me rewrite this. This is going to be equal to negative, the negative integral of one over cosine of x times negative sine of x dx. Now, does it jump out at you what our u might be? Well, I have a cosine of x in a denominator and I have its derivative. | _-substitution defining _ (more examples) AP Calculus AB Khan Academy.mp3 |
In fact, let me rewrite this. This is going to be equal to negative, the negative integral of one over cosine of x times negative sine of x dx. Now, does it jump out at you what our u might be? Well, I have a cosine of x in a denominator and I have its derivative. So what if I made u equal to cosine of x? u is equal to cosine of x and then du dx would be equal to negative sine of x. Or I could say that du is equal to negative sine of x dx. | _-substitution defining _ (more examples) AP Calculus AB Khan Academy.mp3 |
Well, I have a cosine of x in a denominator and I have its derivative. So what if I made u equal to cosine of x? u is equal to cosine of x and then du dx would be equal to negative sine of x. Or I could say that du is equal to negative sine of x dx. And just like that, I have my du here. And this, of course, is my u. And so my whole thing is now simplified to, it's equal to the negative indefinite integral of one over u, one over u du, which is a much easier integral to evaluate. | _-substitution defining _ (more examples) AP Calculus AB Khan Academy.mp3 |
We approximated it using this polynomial. And we saw this pretty interesting pattern. Let's see if we can find a similar pattern if we try to approximate sine of x using a Maclaurin series. And once again, a Maclaurin series is really the same thing as a Taylor series, where we are centering our approximation around x is equal to 0. So it's just a special case of a Taylor series. So let's take f of x in this situation to be equal to sine of x. f of x is now equal to sine of x. And let's do the same thing that we did with cosine of x. | Maclaurin series of sin(x) Series AP Calculus BC Khan Academy.mp3 |
And once again, a Maclaurin series is really the same thing as a Taylor series, where we are centering our approximation around x is equal to 0. So it's just a special case of a Taylor series. So let's take f of x in this situation to be equal to sine of x. f of x is now equal to sine of x. And let's do the same thing that we did with cosine of x. Let's just take the different derivatives of sine of x really fast. So if you have the first derivative of sine of x is just cosine of x. The second derivative of sine of x is derivative of cosine of x, which is negative sine of x. | Maclaurin series of sin(x) Series AP Calculus BC Khan Academy.mp3 |
And let's do the same thing that we did with cosine of x. Let's just take the different derivatives of sine of x really fast. So if you have the first derivative of sine of x is just cosine of x. The second derivative of sine of x is derivative of cosine of x, which is negative sine of x. The third derivative is going to be the derivative of this. So I'll just write a 3 in parentheses there instead of doing all the prime, prime, prime. So the third derivative is the derivative of this, which is negative cosine of x. | Maclaurin series of sin(x) Series AP Calculus BC Khan Academy.mp3 |
The second derivative of sine of x is derivative of cosine of x, which is negative sine of x. The third derivative is going to be the derivative of this. So I'll just write a 3 in parentheses there instead of doing all the prime, prime, prime. So the third derivative is the derivative of this, which is negative cosine of x. The fourth derivative is the derivative of this, which is positive sine of x again. So you see, just like cosine of x, it kind of cycles after you take the derivative enough time. And in order to do the Maclaurin series, we care about evaluating the function and each of these derivatives at x is equal to 0. | Maclaurin series of sin(x) Series AP Calculus BC Khan Academy.mp3 |
So the third derivative is the derivative of this, which is negative cosine of x. The fourth derivative is the derivative of this, which is positive sine of x again. So you see, just like cosine of x, it kind of cycles after you take the derivative enough time. And in order to do the Maclaurin series, we care about evaluating the function and each of these derivatives at x is equal to 0. So let's do that. So for this, let me do this in a different color, not that same blue. So I'll do it in this purple color. | Maclaurin series of sin(x) Series AP Calculus BC Khan Academy.mp3 |
And in order to do the Maclaurin series, we care about evaluating the function and each of these derivatives at x is equal to 0. So let's do that. So for this, let me do this in a different color, not that same blue. So I'll do it in this purple color. So f, that's hard to see, I think. So let's do this other blue color. So f of 0 in this situation is 0. | Maclaurin series of sin(x) Series AP Calculus BC Khan Academy.mp3 |
So I'll do it in this purple color. So f, that's hard to see, I think. So let's do this other blue color. So f of 0 in this situation is 0. And f, the first derivative evaluated at 0, is 1. Cosine of 0 is 1. Negative sine of 0 is going to be 0. | Maclaurin series of sin(x) Series AP Calculus BC Khan Academy.mp3 |
So f of 0 in this situation is 0. And f, the first derivative evaluated at 0, is 1. Cosine of 0 is 1. Negative sine of 0 is going to be 0. So f prime prime, the second derivative evaluated at 0 is 0. The third derivative evaluated at 0 is negative 1. Cosine of 0 is 1. | Maclaurin series of sin(x) Series AP Calculus BC Khan Academy.mp3 |
Negative sine of 0 is going to be 0. So f prime prime, the second derivative evaluated at 0 is 0. The third derivative evaluated at 0 is negative 1. Cosine of 0 is 1. You have a negative out there. It is negative 1. And then the fourth derivative evaluated at 0 is going to be 0 again. | Maclaurin series of sin(x) Series AP Calculus BC Khan Academy.mp3 |
Cosine of 0 is 1. You have a negative out there. It is negative 1. And then the fourth derivative evaluated at 0 is going to be 0 again. And we could keep going, but once again, it seems like there's a pattern. 0, 1, 0, negative 1, 0. Then you're going to go back to positive 1, so on and so forth. | Maclaurin series of sin(x) Series AP Calculus BC Khan Academy.mp3 |
And then the fourth derivative evaluated at 0 is going to be 0 again. And we could keep going, but once again, it seems like there's a pattern. 0, 1, 0, negative 1, 0. Then you're going to go back to positive 1, so on and so forth. So let's find its polynomial representation using the Maclaurin series. And just a reminder, this one up here, this was approximately cosine of x. And you'll get closer and closer to cosine of x. I'm not rigorously showing you how close and that it's definitely the exact same thing as cosine of x. | Maclaurin series of sin(x) Series AP Calculus BC Khan Academy.mp3 |
Then you're going to go back to positive 1, so on and so forth. So let's find its polynomial representation using the Maclaurin series. And just a reminder, this one up here, this was approximately cosine of x. And you'll get closer and closer to cosine of x. I'm not rigorously showing you how close and that it's definitely the exact same thing as cosine of x. But you get closer and closer and closer to cosine of x as you keep adding terms here. And if you go to infinity, you're going to be pretty much at cosine of x. Now, let's do the same thing for sine of x. | Maclaurin series of sin(x) Series AP Calculus BC Khan Academy.mp3 |
And you'll get closer and closer to cosine of x. I'm not rigorously showing you how close and that it's definitely the exact same thing as cosine of x. But you get closer and closer and closer to cosine of x as you keep adding terms here. And if you go to infinity, you're going to be pretty much at cosine of x. Now, let's do the same thing for sine of x. So I'll pick a new color. This green should be nice. So this is our new p of x. | Maclaurin series of sin(x) Series AP Calculus BC Khan Academy.mp3 |
Now, let's do the same thing for sine of x. So I'll pick a new color. This green should be nice. So this is our new p of x. So this is approximately going to be sine of x as we add more and more terms. And so the first term here, f of 0, that's just going to be 0. So we're not even going to need to include that. | Maclaurin series of sin(x) Series AP Calculus BC Khan Academy.mp3 |