Well known identity

The identity exp(iwt)=cos(wt)+i*sin(wt).
I've always thought of it as natural outcome of polynomial series expansion. Is there a different way to derive this relationship?

Quote: pacomartin

The identity exp(iwt)=cos(wt)+i*sin(wt).
I've always thought of it as natural outcome of polynomial series expansion. Is there a different way to derive this relationship?

Hyperbolics?

EDIT: Took a minute to pencil this out, is this what you're asking?

e^n = cosh(n) + sinh(n)
cosh(n) = cos(in) ; sinh(n) = -i sin(in)
e^n = cos(in) - i sin(in)
n = iwt
e^(iwt) = cos(i^2*wt) -i sin(i^2*wt)
e^(iwt) = cos(-wt) - i sin(-wt)

e^(iwt) = cos(wt) + i sin(wt), but not sure if this is what you're asking ...

<Hyperbolics is the same way as Taylor series (power expansion).

What about differential equations ?

d/dt exp(i w t) = i w exp(i w t)

d/dt [ cos(w t) + i sin(w t)] = - w sin(w t) + i w cos(w t) = i w [cos(w t) + i sin(w t)]

Both solve the same differential equation df/dt = i w f.
Since exp(0) = 1 = cos(0) + i sin(0), both functions must be equal.

Quote: pacomartin

Is there a different way to derive this relationship?

Don't Drink and Derive.
That tee-shirt slogan from Snorgtees is about all I can contribute.

Quote: MangoJ

<Hyperbolics is the same way as Taylor series (power expansion).
What about differential equations ?
d/dt exp(i w t) = i w exp(i w t)
d/dt [ cos(w t) + i sin(w t)] = - w sin(w t) + i w cos(w t) = i w [cos(w t) + i sin(w t)]
Both solve the same differential equation df/dt = i w f.
Since exp(0) = 1 = cos(0) + i sin(0), both functions must be equal.

Now I just feel stupid. Thank you, very logical.

I was reviewing the landmark paper below, and I could only remember the relationship being derived via polynomial expansion.
The Quefrency Alanysis of Time Series for Echoes: Cepstrum, Pseudo Autocovariance, Cross-Cepstrum and Saphe Cracking
Proceedings of the Symposium on Time Series Analysis (M. Rosenblatt, Ed) Chapter 15, 209-243. New York: Wiley, 1963.
B. P. Bogert, M. J. R. Healy, and J. W. Tukey:

Now be honest fellow forum readers. Am I the only one who has no idea what the hell this post is about ??????????????

Quote: buzzpaff

Now be honest fellow forum readers. Am I the only one who has no idea what the hell this post is about ??????????????

It's mathspeak. Impossible to understand unless you're trained in it, much like classical music scores or World of Warcraft raid strategy. It's also paco continuing to confirm my suspicions that he actually IS the internet. =)

Quote: Face

It's mathspeak. Impossible to understand unless you're trained in it, much like classical music scores or World of Warcraft raid strategy. It's also paco continuing to confirm my suspicions that he actually IS the internet. =)

Is there actually a formalized WoW raid strategy?

I thought so, but then again I thought Pacomartin was wrong about Robert Duvall being in " To Kill a Mockingbird"
Just happened to have it on VHS so I watched it from beginning to end. Paco got me again. Probably would have remembered Robert if he had a speaking role.

Quote: MathExtremist

Is there actually a formalized WoW raid strategy?

Ha! Well, I'd often make some crude calculations when gold and equipment farming, changing tactics and specialties based on the mob to maximize DPS efficiency while minimizing downtime. I imagine if you could quanitify the strength, resistance, weakness, damage mitigation and DPS of a specific raid section and/or end boss, as well as quantify the values of the healers, DPS, crowd control and tankers in your group, you very well could make a formalized, math based strategy for a raid. Of course, you have to figure out standard deviation, which could vary greatly depending on experience, how many beers your healer had, and whether your tank is being yelled at to finish his homework.

Ok, I'll stop derailing. That concept was too funny to pass up =).

Quote: MangoJ

<Hyperbolics is the same way as Taylor series (power expansion).

What about differential equations ?

d/dt exp(i w t) = i w exp(i w t)

d/dt [ cos(w t) + i sin(w t)] = - w sin(w t) + i w cos(w t) = i w [cos(w t) + i sin(w t)]

Both solve the same differential equation df/dt = i w f.
Since exp(0) = 1 = cos(0) + i sin(0), both functions must be equal.

Same song, second verse. All these functions are pretty incestuous. You can differentiate series as well.

Aren't you using the identity to prove the identity here? My proofs are admittedly a little rusty, but ...

df/dt = iwf
df/f = iw dt
ln f = iwt + C
f = Ce^(iwt)
cos(wt) + i sin (wt) = Ce^(iwt), which only works if C = 1, so I'm not sure the diffy q does what you're wanting, and I don't think it answers paco's question (but I'm still not sure what he's asking ...)

Backing up:

df/dt = iCwe^(iwt) which not only does not = the above but uses the identity to prove itself. So I'm not sure that's the way to go here ...

Quote: ItsCalledSoccer

Same song, second verse. All these functions are pretty incestuous. You can differentiate series as well.

Depending on what you know about the functions exp, cos and sin. There are many many ways to formulate a true statement.
You don't need a power series expansion to formulate derivatives. In fact log(x) doesn't have a power series expansion (at x=0), but still has a derivative (1/x).

Quote:

Aren't you using the identity to prove the identity here?

Nope I'm not. I only use the knowledge of the derivatives of exp, cos and sin. The derivatives of exp, cos and sin can be obtained by other ways than just power series.
Then I use the theorem that a differential equation with same starting values has a unique solution (with certain technical assumptions, but they are met here).

The proof of the original identity is simple: Show that both functions obey the same differential equation with same starting values.
From the theorem above, you conclude that both functions are equal.

No use of power series in here, just using knowledge of derivatives and differential equations.

Still not sure how ...

iw e^(iwt) = iw Ce^(iwt) for any C ...

Quote: ItsCalledSoccer

...

f = Ce^(iwt)
cos(wt) + i sin (wt) = Ce^(iwt), which only works if C = 1, so I'm not sure the diffy q does what you're wanting ...

My math is rusty, but I think the value of the constant of integration is typically established by a boundary condition. A convenient one might be f=1 at t=0.

f(0) = cos(w0) + i sin(w0) = 1 + i 0 = 1

f(0) = Ce^(iw0) = C e^(0) = C * 1 = 1 ---> C = 1 ---> f(t) = e^(iwt) = cos(wt) + i sin(wt)

But some mathematician should probably comment on whether this settles the issue.

Maybe you missed a shared start value, which is a requirement of the identity theorem for differential equations.

Say f(t) is the solution of df/dt = i w f(t).

One family of solution you already found by integrating the differential equation, that is f(t) = C exp(iwt) for any C.
Another solution is "cos(wt) + i sin(wt)", which you can verify by direct differentiation.

Both functions are equal, only if they share a same starting value t0. You can choose any t0 you like.I like t0=0.
Then f(t0) = C
and
cos(w t0) + i sin(w t0) = 1 + i *0 = 1.

Hence both solutions are equal only if C = 1.

And that is the proof of
exp(i w t) = cos(w t) + i sin(w t)