12:00 AM
specifically, those are the taylor series for sin(x), cos(x), and e^x
now i have a question for you - what happens if you take e^(ix)

Welp those are the only things I know so :)
Urm

raising to a complex power - seems weird, doesn't it

(have to write this out on paper hold on)
Does it become (xi)^n?
Or (xi)^n/n!?

well, the sum of that over all natural n, but yes
so, what happens when you expand that out

It was messing with my head a bit but from what I'm getting it looks like
1+(xi)-(x^2/2!)-(x^3i/3!?)+(x^4/2!), etc?

12:06 AM
yep!
exp(ix) =
1
+ (i) x/1!
+ (-1) x²/2!
+ (-i) x³/3!
+ (1) x⁴/4!
+ (i) x⁵/5!
+ (-1) x⁶/6!
+ ...

now... what happens if you look at the terms with an i and the terms without an i

bobble is confused about how this relates to trig

(it's ok bobble magic will soon be happening)

Hold on

12:07 AM
to start, just look at the terms without an i

Oooh the ones with an i looks like the sin series and the ones without the i looks like the cosine series

not just "looks like" - it is!

so, if you collect them together and factor out the i... what do you get?

cos(x)+sin(x)?

12:08 AM
well what happened to the i

Oh. Urm i(cos(x)+sin(x))?

not every term had an i though

Hm

12:10 AM
so what is the result?

so...

cos(x)+isin(x)?
Wait. Oh cos(x)+sin(ix)?

which one?

The latter one

explain?

12:12 AM
So, the cosine is 1-x^2/2!+x^4/4! etc.

yes, that's cos(x)

Sine series is x-x^3/3!+x^5/5! etc. If you insert "i" into sin(x) so it's expressed as sin(ix), it would be xi/1!-x^3i/3!+x^5i/5! etc. right?
So cos(x)+sin(x) would combine the two series, right?

sin(ix) is (ix)/1! - (ix)³/3! + (ix)⁵ / 5! - ...

And.. oh, I see the problem
I would flip the order of the signs

here, you're just seeing that every term is multiplied by i
so you have something that looks like "ia+ib+ic+id+..."
(where a,b,c,d are their own things)

12:16 AM
Hm

hm?

Just thinking
I'm tempted to say sin(-xi), but that wouldn't match the series above either

so, where do you "see" the power series of sine in the terms?
what would you have to do to each term to get to the power series of sine?

(thinking)
Oh, does it need to be mupltiplied by i?
Wait, nope
I'm thinking -i
Well, let me clarify. If you multiply every term in the e^(xi) series by -i, you get the taylor series for sine

sure
so, we have cos(x) + sin(x)/(-i), right?

12:25 AM
Divided by -i?

well, if you multiplied it by -i, you would get the sine
is that not what division is?

(processing)

okay let's back up

I guess so

you're familiar with the idea of factoring out a common term, correct?

12:26 AM
Yes

so, rearrange the terms
so you have:
e^(ix) = (1 + x²/2! + x⁴/4! + ...) + (ix - ix³/3! + ix⁵/5! - ...)

Right

what happens when you factor out an i
from the right side

i(x-x^3/3+x^5/5!-....)

and what is that in the parentheses now

12:30 AM
The sin taylor series

so...
you can write e^ix in terms of sine and cosine; what is the result

cos(x)+i(sin(x))?

yep!
in other words: e^(ix) = cos(x) + i sin(x)

So why divided -i?

has the magic happened yet?

12:31 AM
dividing by -i is the same as multiplying by i

Oh it is?

was trying a different approach, didn't work

(Oh, I was right the first time around)

this formula is important enough to be called Euler's formula

(Ohhhh you're right about the i thingy. Just had to convert -i as i^3)
Okay

12:33 AM
euler invented like 70 things that have his name attached, but with some qualifier like "Euler characteristic formula"
In mathematics and physics, many topics are named in honor of Swiss mathematician Leonhard Euler (1707–1783), who made many important discoveries and innovations. Many of these items named after Euler include their own unique function, equation, formula, identity, number (single or sequence), or other mathematical entity. Many of these entities have been given simple and ambiguous names such as Euler's function, Euler's equation, and Euler's formula. Euler's work touched upon so many fields that he is often the earliest written reference on a given matter. In an effort to avoid naming everything...
this one is just called "Euler's Formula". this is The Important One

i mean... basically yeah i guess
but now you can do this:
e^i(a+b) = e^(ia) · e^(ib)
that's just a normal thing from exponentiation - do you see why that's true?

now use euler's formula on that and Magic will happen

Oooh okay
urm cos(a)+isin(a)*cos(b)+isin(b)

12:37 AM
parentheses!
don't forget them!
and apply it to both sides
@bobble the bold formula was us gathering reagents for the magic, and now the actual magic is happening

e ^ix = cos x + i sin x. Just typing that so I can see it to reference it
I'm confused on the magic :P
Do I have to expand the left side as well?

yep, do that

Ok, I'll try
cos(a+b)+isin(a+b)??

yep!

and the right hand side?

12:41 AM
(^)

cos(a+b)+isin(a+b)=(cos(a)+isin(a))*(cos(b)+isin(b))

parentheses on the right side!
you're missing them

What
Ohhh

(cos(a)+isin(a)) · (cos(b)+isin(b))

I just say what you were talking About
Forgot PEMDAS :P

12:42 AM
So, now, expand that out. What do you get?

Ack hold on. I'm messing myself up. Gonna do some substitutions

cos(a+b)+isin(a+b)=cos(a)cos(b)+isin(a)cos(b)+icos(a)sin(b)-sin(a)sin(b)

now, when are two complex numbers equal?
that's a pretty general question
but what do you look at to check if two complex numbers are equal?

Oh darn complex number
Is it the what was that word... conjugate?

12:50 AM
nah, you're overthinking it

Hm
Oh do I just set it as an equation?

what's the general form of a complex number?

explain?

Like the general form is Ai+b
So isolate "i" to one side?

well, not to one side of the equation, but yes - break it down into its real and imaginary components

12:53 AM
(if z = x + iy then we say that x is the real part and iy is the imaginary part. Or component. Sometimes we say that y is the imaginary part instead, just to make life more interesting.)

Right

(two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal)

Oh. I did over think this

yes, the way i phrased it was probably unclear

So cos(a+b)+isin(a+b) is like the complex form
(or already in its "complex" form of a sort")

12:57 AM
"standard form" is the usual term
of course standard form has several meanings in different contexts, like most mathematical terms involving some synonym of "normal" or "regular"

Right. So what do I do with the right side?

the same thing!
group all the things without i together, and group all the things with i together, and then factor
to get (stuff) + i(other stuff)

Oh I see
cos(a)cos(b)-sin(a)sin(b)+i(sin(a)cos(b)+cos(a)sin(b))? That looks wrong. Nope, I think that's correct

looks right to me
now, as gareth said, two complex numbers are equal only when their real and imaginary components are both equal

1:02 AM
( cos(a+b) ) + i( sin(a+b) ) = ( cos(a)cos(b)-sin(a)sin(b) )+i( sin(a)cos(b)+cos(a)sin(b) )

i'm trying, one sec

No problem

there we go
this is the equation you had above, just with both converted to standard form (with some suggestive bolding)

Looks like the cosine and sine addition formulas to me with the i :P. But how do we prove it?
(Well from there)

1:04 AM
that is the proof

we already know those are equal

Ahhhhh
That is cool

because we started with two equal numbers - e^(i(a+b)) and e^ia · e^ib

Ohhhhh

1:04 AM
and everything else has just been us changing their forms -- same number, just rewriting and rewriting

Neat!

you can do... basically all trig identities this way
or at least a lot of them that you'll have to deal with

So like sin^2+cos^2=1?

sure
e^0 = e^(ia) · e^(-ia)
expand those out

hold on, I gotta call someone real quick

1:13 AM
PARENTHESES

you forgot the is
also what gareth said

(cos(a)+isin(a))*(cos(a)-isin(a)

right. Now expand that out and do what you need to do with i times i...

Which is (cos^2(a))-(-sin^2(a)) which is (cos^2(a))+(sin^2(a))
Is this also where people derive the "most beautiful" equation in math or whatnot? The e^(i)(pi)=1 or e^(i)(pi)-1=0?

yep
(personally i think it's overrated - the more general version is better)

1:18 AM
Hold on I gotta walk my dogs but I'm going to try expanding it out

1:38 AM
Ok so I was thinking about it and it's uh fairly straightforward
e^[(pi)(i)]=cos(pi)+isin(pi). Cosine of pi=-1, and sin of pi is 0. -1+0=-1. So e^[(pi)(i)]=-1, or e^[(pi)(i)]+1=0

1:54 AM
yep