e^(j*pi) = sinusoidal function of course.... but why!

Euler. He is dead, but his formula - and its proof - live on as one of the jewels of mathematics (to quote Ricard Feynman)

-- Bill Sloman, Nijmegen

e^(j*pi) = -1, that doesn't in itself produce a sinusoidal function. Perhaps you mean the more general case of e^(j*x) = cos(x) + i*sin(x)? There are a couple of ways to go about showing that this equality is the proper choice. Perhaps the easiest way to "prove" it is to try computing the Taylor series expansion for both e^(j*x) and cos(x) + i*sin(x), or more specifically the expression for the remainder term of the Taylor series.

You should find that the expression for the remainder of the Taylor series for both sides of the equation is the same, and if you take the limit as "n" goes to infinity in the expression for the remainder the remainder term goes to zero, showing that the two functions are equivalent for all x.

Wow.... if I would have realized Wikipedia had proofs, I would have looked there first! Thanks

oh wow, yeah, I meant to type e^(j*x), but pi flew out instead of x. oops

And just a correction of my own, you'd want to use the "ratio test" on the Taylor series remainder term, not just take the limit.

Tim

-- Deep Friar: a very philosophical monk. Website:

I dislike that approach, because we don't know (a priori) that the series equals the function until we know the function has no poles. By definition, we know the differential equation

d/dx(e^(a*x) ) =3D a * e^(a*x)

as well as e^0 =3D 1

So, it's a no-brainer to take the second derivative

d2/dx2( e^(j*x)) =3D j*j * e^(j*x) =3D -e^(j*x)

and solve.

You might consider looking at my website

there are some flash programs that illustrate how complex numbers work.

One thing to bear in mind... when you see e^(j*pi) and ask why it produces a sinusoidal you must understand that this is a short had notation. There is an implied e^(-j*pi) that is paired with it to produce the sinusoidal as you would understand it. This "never explicitly mentioned but always there" sister is really important to understand.

In the end the e^(j*pi)+e^(-j*pi) is an alternate way of expressing Acosx+Bsinx. It is short hand and it is easier to mathematically manipulate. It also a way to express a sinusoidal waveform that can take on any starting phase position.

But always remember: If e^(j*pi) is stated in the context of real sinusoidal signals there is always an implied e^(-j*pi) that may seem to be invisible or that you are assumed to undertand.

My website also goes over the transfer function of an rlc circuit and shows how the complex numbers in the transfer function flush out.

You're referring to phasors here?

It was quite explicitly mentioned when I went to school -- not only in the first circuits class, but also later on in signals & systems classes where there's discussion of phasors being a specific instance of a Fourier transform, the Fourier transform being a specific instance of the Laplace transform, etc.

Yes, I am refering to phasors, but I think that because the OP is only mentioning one side he may be a bit confused on this point. But also in the fourier transform it is a little confusing because the implied pair is generated by running the integral from - infinity to + infinity so the pair gets created by the mirror image nature of the integral.

That's a good point; it's easy to start thinking that negative frequencies are "something special" (i.e., you can go back in time! -- or something like that :-) ) when that isn't really the case. I do like the plots on a phasor diagram of e^(jw) and e^(-jw) and how, via their counter-rotation, it's clear that you end up with a real signal (...I expect you have this on your web site but I haven't visited in quite awhile).

The other thing that can be a little tricky is realizing that after all this effort you've gone to to end up with real signals, if you're planning to modulate the signal onto a carrier anyway well, hey!, cos() and sin() are orthogonal and all, so -- poof! -- you can now transmit your complex data directly "for free." Sure, the opposite sidebands are no longer complex conjugates, but that's OK...

---Joel

To be precise, e^(j*pi)is -1. I think you are talking about y =3D e^(j*x).

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I came to peace about negative frequencies when I realized that they only exist in the frequency domain. In the time domain you just have frequencies.

I think I follow you - showing, for example, that the Taylor series of e^ix on an interval (x-a) represents the function only shows definitively that e^ix converges absolutely in the neighborhood of that interval. It could hypothetically have poles elsewhere where the Taylor series does not converge. Tricky.

Oh, I dunno, x(t) = cos(-2*pi*1000*t) can most certainly exist in the time domain, right? :-)

ime

I know you know, but cos(-x) =3D cosx so its the same thing. and sin(-x) =3D -sin(x) which means a 180 degree phase reversal.

And that is why thinking of negative frequencies in the time domain will blow your mind. At least it blew my mind until I discovered the joy of accepting negative frequencies in the frequency domain but not worrying about negative frequencies in the time domain.

If you use the exponential form, it's just an artifact of not having true complex signals. In fact, every real signal you're looking at is merely a superposition of complex conjugates. This isn't strange at all; the equations are the same for standing waves. Something going each way, you just can't observe the components seperately.

Tim

e^(x*j) = cos x + j*sin x by Euler

cos x = {1 - sin^2 x}^(1/2)

e^(x*j) = {1 - sin^2 x}^(1/2) + j*sin x

cos x = sin( x + pi/2)

e^(x*j) = sin( x + pi/2) + j*sin x

...unless that wave itself is made of multiple orthogonal parts, such as the electric and magnetic field components that a directional coupler uses to separate the waves travelling in each direction...