Please explain this to me. My calculator confirms but I sense some deeper meaning here

...

there's nothing *le Deep biuitufol matt* about it

That identity comes from e^ix = cosx + isinx
Let x = pi then solve. If you don't know what radians are, don't bother trying to understand this until you do

It's beautiful but it isn't deep at all.

It has to do with rect to polar conversions on the imaginary number line. Look at the works of Euler and you will find stuff like this.

Look

[math] e = lim_{n \to \infty} (1 + 1/n)^{n} [\math]

When we construct an exponential function with this value, explicitly f(x) = e^x, we get a function whose "rate of change" at any given x-value is itself.

Maybe you should review what exponential functions are and plot a few graphs so you get a feel for them.

Now for i. First consider the real numbers. Imagine a line that extends to +/- infinity, no gaps on the line, and a measure defined on the line. This is your geometric intuition of the real numbers; every point on the line corresponds to a real number. No gaps on the line ensures you can zoom in as much as you want and you will never see any gaps, and it ensures irrational numbers like pi, e, and sqrt(2) are there. If this isn't clear you should brush up on the different number systems and their properties, especially the real numbers and the motivation for its construction.

Now this line is inadequate for finding solutions to some polynomial equations. Primarily because some solutions require sqrt(-1). (Not sure how much basic algebra you know, but of course some solutions are of the form sqrt(-K), K a real number, but it can be shown that for all K, sqrt(-K) = sqrt(-1) * sqrt(K), so it always boils down to dealing with sqrt(-1)). Instead of always writing sqrt(-1) all the time, we just call this value i. This is where the field of complex numbers comes in; by the fundamental theorem of algebra it is this field that is algebraically closed.

More on this field: it is essentially an extension of the real numbers. The numbers in it are of the form z = a + ib, where a and b are real numbers. So every real number is also a complex number with imaginary part 0 (I.e. b = 0 in our notation above). As mentioned this extension forms a field, which means it shares a lot of nice properties with other familiar number systems and we can do many of our beloved mathematics with them, including algebra and calculus. More shortly

Just as we captured our visual intuition of the real numbers with an infinitely extending line with a metric imposed on it, we can do the same with the complex numbers with a 2d plane. Imagine the regular real line and then at 0 (the origin) construct a second real line running perpendicular to it. This is the complex plane, and each z = a + ib can be found by going along the real line a units and then up the second line b units. If this sounds eerily familiar to the Cartesian plane, well, it certainly is. In fact, it is isomorphic to what is called R^2, which is the Cartesian plane in vector coordinates. Now I'm not going talk anymore about this other than that it is common to treat complex numbers similar to 2d vectors (directed line segments with computable magnitude). In fact it is common notation to write z as a point; I.e write z = a + ib = (a, b).

I unfortunately don't have enough time to continue, but I hope this has put you on the right track. Never stop searching for intuition.

Cos pi =-1
Sin pi=0
-1+0+1=0

math] e = lim_{n \to \infty} (1 + 1/n)^{n} [/math]

What book is this? It looks nice and structured.

that's Stewart buddy

Consider the differential equation [eqn] \ddot y = k^2 y [/eqn]It has as a solution [math] y(x) = e^{kx} [/math]. Now consider [eqn] \ddot y = -y [/eqn][math] \sin [/math] and [math] \cos [/math] are solutions to this equation.

If you let [math] k = i [/math] in the first equation, you get the second. Using [math] i [/math] you've related [math] e^{ix} [/math] and the trigonometric functions: both are solutions to the same DE.
As a corollary, you've also linked [math] e, i, \pi [/math].

The best understanding I've gotten for how it works comes from these two sites:

First you need to understand the exponential function as the process of continously return:
betterexplained.com/articles/an-intuitive-guide-to-exponential-functions-e/

Then you can understand how a continous return with an imagine number as the interest works :
betterexplained.com/articles/intuitive-understanding-of-eulers-formula/

Draw the complex plane somewhere and pick a random complex number, now draw a line from the origin to the point that represents the chosen number. Note that this number can be represented by

[math]z=\cos \theta + i \sin \theta[/math]

Where theta is the angle between the line you draw and the real axis. Now it's a know fact that

[math]e^{i \theta} = \cos \theta + i \sin \theta [/math]

Now draw a line from the origin where the number -1 is located in the plane, what is the angle between the line you've drawn and the "positive" axis of the real numbers? Its pi, so when theta=pi

[math] e^{i \pi} =-1[/math]u

that is really nice thx !

You should try to understand understand Euler's formula. If you get that, plugging in pi for x is trivial and gives you -1.

Basically what 90% of every college calc student reads

Euler' identity, faggot. Ain't u ever taken calc 2??

This video explains it in an approachable way

youtube.com/watch?v=F_0yfvm0UoU

Another thing this channel taught me is how you can interpret operations on numbers as movements

for example, multiplication of complex numbers adds their angles

I would recommend watching all his videos

My point is:

complex exponents are also easily interpreted geometrically.

This video is good:
But I personally like this series more for understanding complex numbers, which is kind of prerequisite to getting Euler's identity:

Imaginary Numbers are Real: youtube.com/playlist?list=PLiaHhY2iBX9g6KIvZ_703G3KJXapKkNaF

Interesting argument, but if it were valid then sin = exp + a constant. Can it be salvaged to say something true?

err, not a constant but a linear function.

If you define a whole bunch of different numbers based on their relationship with the number one (or negative one), it turns out that sometimes a random set of operations involving those numbers will spit out one (or negative one).

Whoa! Amazing!

Whoa, also, did you realize that cos^2(x) + sin^2(x) = 1?! It's unbelievable!

I fucking hate Euler.

Not sure it's relevant but both relate to differential equations in one way or another.
f'(x) = f(x)
>f(x) = e^x
g"(x) = -g(x)
>g(x) = sin(x)
>Periodic function with period = 2*pi

>When a random person on Veeky Forums can explain this better than my differential equations professor
There is a reason I failed that class even though I can solve linear differential equations just fine.

There is some theorem that states (lim n -> infinity of (1 + 1/n)^n)^z = lim n-> infinity (1 + z/n)^n, so e^z = lim n -> infinity (1 + z/n)^n, for all real z. Because of that, we can generalize this formula to all complex z. Using the geometric interpretation of complex multiplication, and the fact that sin(t) is approximately equal to t for very small t, we conclude that e^z = x*(cos(y) + i*sin(y)), where y is the imaginary part of z.

You want to know what's funny? I got a tattoo of this on my wrist my senior year in high and boy what a wonderful legacy I left after that *full blown sarcasm* anyways a year later I got that shit removed never make the mistake I made math tattoos on visible body parts will buy you a good ticket to no pussy getting land. Since I got it removed I never regretted, hey, we all make autistic mistakes every now and the in life.

he forgot to include the boundary conditions y(0)=1, y'(0)=i. then it goes through.

If you've been worth while to raise into society you can proof it quite easily using the definition of complex numbers and e

thank you

Don't you mean "through" instead of "to" ?

>muh imaginary numbers so we get an arbitrary formula that says -1+1=0

wow so biutiful