Using your imagination =d

So, I've learned quite a bit during my first quarter here at irvine, but I think the most powerful idea I've picked up is Eulers Formula.

I'm sure you've all seen it before. It's

E^(ix) = cos(x) + isin(x)

Which comes from the taylor expansion of E^(x) and then plugging in (ix), and magically the cos(x) expansion appears, and all the terms with i as a multiple are the expansion for sin(x).

Anyway, I didn't think much about it until I got into this diffy Q class. We pretty much ALWAYS use this formula to solve anything that has to do with sin(ax) or cos(ax).
Which got me to thinking that maybe you could use it for anything that has to do with sin's or cos's.

I tried it out, and yes it works for everything, I'll give an example.

If you want to know the derivative of sin(x) . You think of sin(x) as the IMMAGINARY part of E^(ix).

So, the derivative of E^(ix) is i*E^(ix) which, if you substitute back to cos's and sin's you get

d/dx E^(ix) = i* (cos(x) + i*sin(x) ). If you distribute the (i) to both terms, and remember that i^2 = -1 then you get

d/dx E^(ix) = icos(x) - sin(x). Since we wanted to know the IMMAGINARY part of this solution, we take cos(x) to be the d/dx of sin(x), which it is. Notice that if you wanted the REAL part of the solution (which would be the same as wanting to know the d/dx of cos(x) you would have found the real part to be -sin(x) which is the derivative of cos(x).


Anyway, that's just easy shit.

If you remember problems like this (integral) (E^(x) cos(x) ) which is one of those integrals that you have to use integration by parts twice, and then add the original integral to both sides to solve for the integral. It turns out to be MUCH easier using Eulers equation.

replacing cos(x) with E^(ix) in the integral, and remembering that since it's cos(x) we want the REAL part of the final solution that we come up with. we get this

(integral) (E^(x) * E^(ix) )
which we can add the exponents to get
(intelgral) ( E^(i+1)x )

that integral is super easy to do, since E^(whatever) is just E^(whatever) / (whatever)

so we get this for the integral

E^(i+1)x / (i+1)

but that part does not have the immaginary and real portions seperated . so you have to rationalize the denominator by multiplying the top and bottom by the conjugate (i - 1)

doing that. you get this

E^(i + 1)x *(1-i) / 2

distributing in E^(i+1)x to (i - 1) you get this

( E^(i+1)x *i - E^(i+1)x ) / 2

converting all E^(ix) back to (cosx + isinx) you get this

( i( cosx + i sinx ) E^(x) - ( cosx + isinx ) E^(x) ) / 2
then distributing everything and remembering that you only wanted the REAL solutions,

you get this

- (sinx*E^(x) + cosx*E^(x) ) / 2

which is solution to the integral.


I know that looks like a LOT of work, but when I do it on paper it turns out this method saves me alot more time than doing two integration by parts.

also you only have to do one integration and that integral uses only E^(constant * x)
which is super easy to integrate.


ANYWAY lol, sorry to write all this out, but seriously I think it's the coolest shit in the world.
praise be the Eulers formula.

Seriously guys, try this sometime. It really amazes me that this stuff works.

Maf works.