Deriving Euler's formula four ways!

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[Bodyknock]

Fun video. 🙂 One small thing, in method 3 there's a substitution of x=tan(θ/2), which is fine but note that there's a restricted domain on it since tan(π/2) is undefined. So unlike the first two methods you might need to be a little careful handling the domain.

[ClaraDeLemon]

Personally, my favourite method is as follows:
The diff equation y'' = -y has cosine and sine as obvious solutions, and obviously any linear combination of them as well. But we could also have taken exp(c•x), whose second derivative is c²exp(cx), with c some constant. If c²=-1, i.e. if c=i, this is also a solution. But how do we compute exp(ix)? Luckily, we don't need to, because by Picard-Lindelöf we know there is a two parameter family of solutions for this equation, already covered by the sines and cosines from before, so e^ix must be some combination of the two. Adjusting constants and noticing e^i0 = 1 and its derivative at zero is i, we reach the famous result.

I love this method, because it is both intuitive and really highlights the connection between the trigonometric functions and the exponential. In fact, if you're more knowledgable in complex analysis, you'll know the relationship is actually backwards, and sine and cosine are actually combinations of complex exponentials and not the other way around

[girianshiido]

The problems of these proofs is how do you define each functions? What is assumed about each functions? For example, in the second proof, you are supposed to know that the derivative of exp(iθ) wrt θ is equal to iexp(iθ), but you first need to define the exponential of a complex number.

[coreyyanofsky]

here's another way: set f(θ) = cos(θ) + i sin(θ) and take the derivative
f'(θ) = -sin(θ) + i cos(θ) = i (cos(θ) + i sin(θ)) = i f(θ)
which is a differential equation whose solution (with initial condition f(0) = 1) is f(θ) = exp(iθ)

[gregsarnecki]

So good to see the return of the magic chalk squares as you progress through the problem. It's what got me into your videos in the first place. Long may they continue...

[zkxxutp]

There is a way with the definition through the limit. e^(ix)=lim (1+ix/n)^n=lim [sqrt(1+x^2/n^2)] ^n * [cos(atg(x/n))+i sin (atg(x/n))]^n=lim [sqrt(1+x^2/n^2)]^n * [cos(n atg(x/n))+i sin (n atg(x/n))] =1 [cos(x)+i sin(x)]

[mathhack]

Amazing as usual. Thanks for this weekend mental refreshment.
@20:42 cos(φ) *First equation + sin(φ)* second equation

[routemath]

The Beatles of math formulas. Nicely done! 🙂

[dominicellis]

I love the complex logarithm way. You can derive the formula from the harmonic oscillator equation. The eigenvalues are complex and when you diagonalize you end up with a rotation matrix. This proves Euler formula.

[BbNn]

Fantastic!!!...& very nice!..Thank you.

[idolgin]

I love all patterns that bring together exponential and trig functions.

[geekwhoeatsrice]

I know of a 5th way!

It uses the close approximation of theta and sine (and by extension tangent) and the limit form of e^(r). Pretty lengthy but decent enough for a PreCalculus class

[youtubeaureus]

I'm a math teacher here in France, and I love your videos! Very well explained, a pure pleasure! Thanks 💜

[dougrife]

The derivations using calculus are probably circular. That is, the derivative of e raised to an imaginary power is undefined until and unless Euler’s formula is derived. For example, taking the derivative of both sides of Euler’s formula yields Euler’s formula. But that only works if you already know that the derivative of e^i*theta = i*e^i*theta. The power series for the sine, cosine and exponential function are really fundamental to proving the formula without any circularity. I think that’s how Euler discovered his famous formula.

[sniperwolf]

I derived it in a similar way as the last one, but without making any initial assumptions about what the polar form should be.

Take some complex number 𝑧 = 𝑎 + 𝑖∙𝑏. From the interpretation that a complex number is a point in the complex plane, we can verify that 𝑧 can be equivalently defined as the point that is at a distance 𝑟 from the origin 𝑂 and which has a line segment connecting it to the origin forming an angle 𝜃 with the positive real axis measured in the counter-clockwise direction. 𝑟 and 𝜃 will be called the polar coordinates of 𝑧 and it follows that 𝑧 = 𝑟(cos 𝜃 + 𝑖∙sin 𝜃) from trigonometry. Similarly, we can take a second complex number 𝑤 = 𝑠(cos 𝜑 + 𝑖∙sin 𝜑), where 𝑠 and 𝜑 are the polar coordinates of 𝑤.

Then we can calculate the product 𝑧𝑤 = 𝑟𝑠[cos 𝜃 cos 𝜑 - sin 𝜃 sin 𝜑 + 𝑖∙(cos 𝜃 sin 𝜑 + sin 𝜃 cos 𝜑)] = 𝑟𝑠[cos (𝜃 + 𝜑) + 𝑖∙sin(𝜃 + 𝜑)], by using trigonometric identities, which has polar coordinates 𝑟𝑠 and 𝜃 + 𝜑. This fact motivates us to represent a complex number in polar form as 𝑧 = 𝑟∙𝑢^𝜃, where 𝑢 is some yet to be determined complex number, such that 𝑧𝑤 = 𝑟𝑠∙𝑢^(𝜃 + 𝜑).

To determine 𝑢, we can take the derivative of 𝑧 w.r.t. 𝜃 in both the rectangular and polar forms:
∂𝑧/∂𝜃 = 𝑟(-sin 𝜃 + 𝑖∙cos 𝜃) = 𝑟∙𝑖∙(cos 𝜃 + 𝑖∙sin 𝜃) = 𝑖∙𝑧;
∂𝑧/∂𝜃 = 𝑟∙ln 𝑢∙𝑢^𝜃 = ln 𝑢∙𝑧.
Therefore, ln 𝑢 = 𝑖 ⟺ 𝑢 = 𝑒^𝑖.

In conclusion, it was shown that a complex number 𝑎 + 𝑖∙𝑏 can be equivalently represented in polar form as 𝑟∙𝑒^(𝑖∙𝜃) and Euler's formula was automatically proven along the way.

[jamesbaugh]

Try using rotation matrices: Look at R(theta) = R(theta/n)^n in the limit as n->inf. Show that in this limit you get the limit of (I+J*theta/n)^n =exp(J*theta). Where Z=xI+yJ is the matrix representation of the complex numbers. (I = diag(1, 1) J = ((0, -1), (1, 0))^T. R(theta)= ((c, -s), (s, c))^T c =cos(theta), s = sin(theta). [I'm using column of rows to represent square matrix, and column = row^T ... 'cause, YT comments don't allow LaTex input. ]

[physicsatroeper]

Here's a method I learned that is very similar to your METHOD 2: Let f_1(x) = e^{ix} and f_2(x) = \cos(x) + i\sin(x). Both of these functions satisfy the differential equation: F' = iF, and the initial condition F(0) = 1. Therefore, by the existence-uniqueness theorem of first-order, homogeneous, linear ordinary differential equations, the two functions must be identical.

[gawater]

z = cos(x) + i sin(x) = cos(xN/N) + i sin(xN/N) = (cos(x/N) + 1 sin(x/N))^N by de Moivre's formula which is proved by indiction and intuitively clear from the fact that the angles of complex numbers add up when multiplied. Letting N -> infinity: cos(x)+i sin(x) = (1 + ix/N)^N = e^(ix).

[Jack_Callcott_AU]

IMHO, the second proof is the best...it's the most elegant and simple.

[smxlong]

I like to do it by showing the following:

Let f(x) = exp(i*x)
Let g(x) = cos(x) + i sin(x)

Proceed to show that f'''' = f, = f', = f'', = f''.
Show that the same holds for g.

Then show that f(0) = g(0), f'(0) = g'(0), f''(0) = g''(0), and f'''(0) = g'''(0).

Then observe that this means that f(x) and g(x) are not only equal at x=0, but all of their infinite derivatives are also equal. Thus f and g must be equal everywhere.