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
Euler's formula is eⁱˣ=cos(x)+i⋅sin(x), and Euler's Identity is e^(iπ)+1=0. See how these are obtained from the Maclaurin series of cos(x), sin(x), and eˣ. This is one of the most amazing things in all of mathematics! Created by Sal Khan.
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- If e^(i*pi)=-1, then (e^(i*pi))^2=(-1)^2=1=e^(i*2*pi). It follows that e^(i*2*pi)=e^0. Taking the natural logarithm of both sides: i*2*pi=0, so either i=0, 2=0, or pi=0, all of which are false. Does this make sense?(107 votes)
- actually the problem is that the complex logarithm has an infinite amount of solutions so the complex log of 1 is not only 0 but rather i*2*pi*n where n is an arbitrary integer (for n=0 the term results in 0).(115 votes)
- Why is it assumed that d/dx[e^ix] = e^ix, instead of i(e^ix)?(31 votes)
- The only way this could be is if we were to take the derivative with respect to 'ix', rather than 'x'(26 votes)
- sal, you keep saying that you're proof is not rigorous, but didn't euler himself prove this by proving that the taylor series are equivalent?(13 votes)
- I think that the proof can be make rigorous by doing two things Sal didn't:
1) Show that the expansion isn't just an approximation, it's /exactly/ eˣ
2) Show that it's not just true for real numbers centered at zero, it's true for every single number in ℂ.(45 votes)
e^(ix) = cos(x) + i*sin(x)
only true around x = 0 since you based this on Maclaurin Series? If not how do you generalize this proof for any value of x?(15 votes)
- This proof is accurate for all real numbers, noted in the video by x. In the video Khan keeps mentioning that this proof isn't general. The proof is only non-gendral in the sense that it is an approximation as accurate as the number of terms included. (ref, the ellipses used in the polynomials) As the number of terms increases, the proof becomes more accurate. It hasn't been shown here, but it is known that the taylor expansion of sine and cosine approach perfect accuracy as the number of terms increases, and therefore Euler's identity is correct.(20 votes)
- Would you say it makes sense that the use of "i" is related to trigonometry in the sense that trig functions, sin and cos cycle 4 times before they return to their original values (i.e. the 4th derivative of cosine is also cosine) ; whereas 4 additional iterations of "i" will also return it back to it's original value (i.e. i_^5 = _i) ?? What do you guys think, coincidence?(20 votes)
- Perhaps that is why Euler's formula works! And when you look into it actually does explain why it works because since both the derivatives of trig functions and powers of i have a "cycle" of 4, only the powers of x and the factorials don't cycle, which is exactly like the Maclaurin expansion of trig functions so you can factor out the cos(x) and i*sin(x) to get Euler's formula! And by the way, if you want italics, you can't put symbols like ^ and = inside the underscores.(7 votes)
- When we add i in the exponent of e^x, how does it appear with all of the x terms?(12 votes)
- he replaced 'x' with 'ix'. he plugged it in to the polynomial approximation formula that he devised earlier. He could have replaced 'x' with '2x' or 'y'. Remember that x is the input to this function, e is just a number somewhere between 2 and 3.(9 votes)
- Help! I think I proved 0=i!
log base anything of 1 = 0, since x^0 always =1
Please show what I did wrong! I don't wanna be the person who broke algebra!(5 votes)
- You are actually assuming that, since ln(1) is 0. No, because, you see, that is saying e^0 = 1, which is true, but, you see, e^i2pi also is 1. Therefore, you are already assuming that e^i2pi = e^0 , which is true, but in this cas ln(1) is indeterminable, with multiple values!
I don't wanna be the person who broke algebra... LOL, cute in a way. I really want to be that guy.(3 votes)
- why the i^2 is -1...?(3 votes)
- Because i is the square root of -1, when you square it the operations of squaring and square rooting cancel each other out.(8 votes)
- How is e^(i*pi) equal to -1?
I mean, I know that it's cos(pi) + i sin(pi), but I thought sin(pi) = 1 and cos(pi) = 0! (not 0's factorial)
Therefore, -1 = i?
- You got your trigonometry mixed up! π is one half turn around the unit circle. Thus sin(π)=0 and cos(π)=-1(18 votes)
- I think Euler's Formula is very impressive, but I don't hold the same opinion for Euler's Identity. I mean, it just comes from the way radians are defined. After all, we could have defined radians in such a way that instead of 2*pi radians being a full circle, we could say 2*golden ratio radians, or 2*Avogadro's number radians is a full circle, and then someone would be claiming that there's a "profound relationship" between the golden ratio and e and i, or between chemistry and e and i, etc. So I do think the relationship between e and i is authentic, but I can't say I perceive any genuine relationship to pi.(4 votes)
- That is an interesting argument you have got there.
I think the thing with π is that it is fundamentally related to circles, since π is the ratio of circumference of the circle to its diameter. The radian is not defined in terms of golden ratios, avogadro's number nor π. It is defined as the ratio between the arc length and the radius of the circle. If your arc length is equal to 2π and your radius is equal to 1, the angle at the center of the circle is 2π radians. If your radius is equal to 2 instead, you have got an angle of π radians and you would have only gotten a quarter around the circle because now you need an arc length of 4π to go fully around the circle. My point is that radians are not defined using π, but since circles have a fundamental connection to π it shows up when you use radians.
Moreover, you need π as argument to
sinevaluate to 0 and
-1(to get the other gem that is -1 into the formula).
Maybe I am not seeing things exactly like you do... But that was just my argument in favor of the greatness of π in the formula.(4 votes)
Voiceover: 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 a combination of the polynomial approximations of cosine of X and of sine of X. But it's not quite, because there's a couple of negatives in there. If we were to 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 if we say E to the X is equal to this, especially as this becomes an infinite number of terms and 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, I said how do you define E to the Ith power? That's a very bizarre thing to do, 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 sense of it. Because we can take I to different amounts to different powers and we know what that gives, I squared is negative one, I to the third is negative I, and so on and so forth. So what happens if we take E to the IX. So once again, just like taking the X up here and replacing it with an IX, so every where we see the X in it's polynomial approximation, we would write an IX. So let's do that. So E to the IX should be approximately equal to, and it will become more and more equal, and 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 one plus instead of an X will have an IX, plus IX, plus, so what's IX squared? So let me write this down. What is IX squared over two factorial? Well I squared is going to be negative one, and then you'd have X squared over two factorials. It's going to be minus X squared over two factorial, I think you might see where this is going to go. And then what is IX? Remember everywhere we saw an X, we're going to replace it with an IX. So what is IX to the third power? Actually let me write this out, let me not skip some steps over here. So this is going to be IX squared over two factorial, actually let me, I want to do it just the way. So plus IX squared over two factorial, plus IX to the third over three factorial, plus IX to the fourth over four factorial, and we can keep going, plus IX to the fifth over five factorial, and we can just keep going so on and so forth. Let's evaluate if these IXs raised to these different powers. So this will be equal to one plus IX, IX squared, that's the same thing as I squared times X squared, I squared is negative one. So this is negative X squared over two factorial, and 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, so it's going to be negative I. So this is going to be minus I times X to the third over three factorial. So then plus, you're going to have, what's I to the fourth power? So that's I squared squared. So that's negative one squared, that's just going to be one. So I to the fourth is one and then you have X to the fourth. Plus X to the fourth over four factorial. And then you're going to have, I don't even write the plus yet, I to the fifth. So I to the fifth is going to be one times I, so it's going to be I times X to the fifth over five factorials plus I times X to the fifth over five factorial. I think you might see a pattern here. Coefficient is one, then I, then negative one, then negative I, then one, then I, then negative one, X to the sixth over six factorial, and then negative IX to the seventh over seven factorial. So we have some terms, some of them are imaginary, they're being multiplied by I. Some of them are real. Why don't we separate them out? So 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. So this is real, this is real, this is real, and this right over here is real. And obviously we could keep going on with that. So the real terms here are one minus X squared over two factorial plus X to the fourth over four factorial, you might be getting excited now, minus X to the sixth over six factorial and that's all I've done here, but they would keep going so plus so on and so forth. So that's all of the real terms. And what are the imaginary terms here. And I'll just factor out the I over here. Actually, let me just factor out. So it's going to be plus I times, well this is IX, so this will be X. And then the next, so that's an imaginary term, this is an imaginary term. We're factoring out the I, so minus X to the third over three factorial, and then the next imaginary term is right over there. We factor out the I, plus X to the fifth over five factorial and then the next imaginary term is right there, we factored out the I. So it's minus X to the seventh over seven factorial. And then we would obviously keep going. So plus minus keep going, so on and so forth, preferably to infinite so that we get as good of a approximation as possible. So we have a situation where E to the IX is equal to all of this business here. But you probably remember from the last two videos the real part, this was the polynomial, this was the Maclaurin approximation of cosine of X around I should say the Taylor approximation around zero or you could also call it the Maclaurin approximation. So this and this are the same thing. So this is cosine of X, especially when you added an infinite power of terms, cosine of X. This over here is sine of X, the exact same thing. So 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 of the X. This right here is sine of X. And so if we take it for granted, I'm not rigorously proving it to you, and if you were to take an infinite number of terms here that this will essentially become cosine of X and if you take an infinite number of terms here this will become sine of X. It leads to a fascinating formula. We could say that E to the IX, is the same thing as 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. And this right here is Euler's Formula. And if that by itself isn't exciting and crazy enough for you, because is 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. And some how we've thrown in the square root of negative one. There seems to be this cool relationship here. But it becomes extra cool, and we're going to assume we're operating in radians here. 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 cirle. What happens when we throw in pi? We get E to the Ipi is equal to cosine of pi. Cosine of pi is what? Pi is half way around the unit circle, so cosine of pi is negative one and then sine of pi is zero, so this term goes away. So if you evaluated at pi you get something amazing, it's called Euler's Identity. I always have trouble pronouncing Euler's. Euler's Identity, which we could write like this, or we could add one to both sides and we could write it like this. And I'll write it in different colors for emphasis. E to the I times pi plus one is equal to , I'll do that in a neutral color, is equal to, I'm just adding one to both sides of this thing right over here, is equal to zero. And this is thought provoking. I mean here we have, 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 one. Pi is the ratio between the circumference of a circle and it's diameter. Once again another interesting number but seems like it comes from a different place as I. E comes from a bunch of different places. E you could either think of it comes out of continuing 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, another fascinating number. But once again seemingly unrelated to how we came up with I and seemingly unrelated with how we came up with pi. And then of course you have some of the most profound basic numbers right over here. You have one, I don't have to explain why one is a cool number. And I shouldn't have to explain why zero is a cool number. And so this right here connects all of these fundamental numbers in some mystical way that shows that there's some connectedness to the universe. So frankly, if this does not blow your mind, you really have no emotion.