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

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

## Integral Calculus

### Course: Integral Calculus>Unit 5

Lesson 15: Maclaurin series of eˣ, sin(x), and cos(x)

# Euler's formula & Euler's identity

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.

## Want to join the conversation?

• 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?
• 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).
• Why is it assumed that d/dx[e^ix] = e^ix, instead of i(e^ix)?
• The only way this could be is if we were to take the derivative with respect to 'ix', rather than 'x'
• 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?
• 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.
• is
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?
• 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.
• 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?
• 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.
• When we add i in the exponent of e^x, how does it appear with all of the x terms?
• 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.
• Help! I think I proved 0=i!

e^(i2π )=1
ln(1)=i2π
(ln(1))/2π =i
log base anything of 1 = 0, since x^0 always =1
0/2π =i
0=i

Please show what I did wrong! I don't wanna be the person who broke algebra!
• 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.
• why the i^2 is -1...?
• Because i is the square root of -1, when you square it the operations of squaring and square rooting cancel each other out.
• 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?
Help!
• You got your trigonometry mixed up! π is one half turn around the unit circle. Thus sin(π)=0 and cos(π)=-1
Moreover, you need π as argument to `sin` and `cos` to make `sin` evaluate to 0 and `cos` evaluate to `-1` (to get the other gem that is -1 into the formula).