Any oscillating function can be written as $$ f(t)=k+\Re(\sum_{n=0}^{\infty}\check{c}_ne^{inwt} ) $$
One can find t by
$$ \check{c}_n=\frac{2\int_{0}^{T}f(t)e^{-inwt}}{T} $$
I understood finding real Fourier coefficients for the real, but here why is there a 2 in the denominator?

Question

Any oscillating function can be written as $$ f(t)=k+\Re(\sum_{n=0}^{\infty}\check{c}_ne^{inwt} ) $$
One can find t by
$$ \check{c}_n=\frac{2\int_{0}^{T}f(t)e^{-inwt}}{T} $$
I understood finding real Fourier coefficients for the real, but here why is there a 2 in the denominator?

anonymous · Answer

$$\check{c}_n=a_n-ib_n$$ Where a is the cosine coefficient and b is the sine coefficient for the real fourier series, to clarify.

experimentx · Answer

can you give me the link were you got this?

anonymous · Answer

Feynman lectures

anonymous · Answer

For example, imagine a function$$f(t)=3\cos(6t)+12sint(6t)=\Re((3-12i)e^{i6t})$$
We want to find 3-12i.
$$\check{c}_n=\frac{2\int\limits_{0}^{2\pi}(3-12i)1dt}{2\pi}=\frac{2(3-12i)(2\pi)}{2\pi}=2(3-12i)$$

experimentx · Answer

it's pretty weird eq 26 http://mathworld.wolfram.com/FourierSeries.html

anonymous · Answer

So it seems the 2 was incorrect.

experimentx · Answer

$$  f(t)=3\cos(6t)+12\sin (6t)= 3 \; \Re (e^{i6t}) + 12 \Im ( e^{i6t}) $$

anonymous · Answer

Left hand side of that equation is only real- it can't have an imaginary part. Did you read a nonexistent i?

experimentx · Answer

Seems you eluded Imaginary part by putting complex coefficient.

experimentx · Answer

it would work out too.

anonymous · Answer

Sorry, I forgot taking the imaginary part takes the size of the imaginary part only without adding an I afterwards.

experimentx · Answer

3 - 12i <-- this is your coefficient. Your function is $ 3cos(6t)+12\sin(6t) $
try getting this out of that relation. I haven't tried this yet though.

anonymous · Answer

I tried it above, was I incorrect?

experimentx · Answer

your case, n=6 ... putting this on 
try this ... I'm not sure though
$$ {3 \over \pi }\int_0^{\pi \over 3} (3\cos(6t)+12\sin(6t))e^{-i6t}dt = ?$$

experimentx · Answer

If that value = (3−12i) then your claim is correct.

anonymous · Answer

It should work also with 2pi also, which would make it a little easier.

It is, just convert f(t) into ke^pt, $$(3-12i)e^{6t}$$, as  above, and the exponentials multiply to 1.

anonymous · Answer

I assume that if you 'complexify' a real function, operate on it, and take the real part of the result, it's the same as not complexifying it in the first place and operating just on reals for real result.

experimentx · Answer

$$ {3 \over \pi }\int_0^{\pi \over 3} (3\cos(6t)+12\sin(6t))e^{-i12t}dt =$$

anonymous · Answer

$$(3-12i)e^{i6t}e^{-i6t}=(3-12i)e^0=(3-12i)$$Am I missing something?

experimentx · Answer

http://www.wolframalpha.com/input/?i=Integrate [%283+Cos[6+t]+%2B+12+Sin[6+t]%29%2FE^%28%286+I%29+t%29%2C+{t%2C+0%2C+Pi%2F3}]*3%2Fpi

experimentx · Answer

2 seems to be right!! or somehow i must get some other 2

anonymous · Answer

I don't see how...

experimentx · Answer

go to that wolf link.

anonymous · Answer

I saw that. Is my complexfying of 3cos(6t)+12sin(6t) right?

experimentx · Answer

that should be your coefficinet according to your formula. if you multiply by that by 2 ... you exactly get your coefficient and hence your claim is correct.

experimentx · Answer

since your function is very nice, you easily got this form
$$ f(t)=3\cos(6t)+12sint(6t)=\Re((3-12i)e^{i6t})  ---- (1)$$

if your function were not nice, you should have gone through more rigorous processes 
which is to do this
$$ \check{c}_n=\frac{2\int_{0}^{T}f(t)e^{-inwt}}{T} ------ (2)$$

the coefficient from (1) and (2) must match each other.

experimentx · Answer

and f(t) is not 3 - 12 i

anonymous · Answer

I wasn't proposing that. Why should the 2 exist?

experimentx · Answer

It's petty weird!!

experimentx · Answer

do you have mathematical derivation?

anonymous · Answer

Here's my logic why it shouldn't:$$F(t)=(3-12i)e^{i6t}$$$$\large \frac{2\int\limits_{0}^{\pi/3}(3-12i)e^{i6t}e^{-i6t}dt}{\pi/3}= \frac{2\int\limits_{0}^{\pi/3}(3-12i)dt}{\pi/3}=\frac{2(3-12i)\int\limits_{0}^{\pi/3}dt}{\pi/3}$$$$=\frac{2(3-12i)\pi/3}{\pi/3}=2(3-12i)$$

experimentx · Answer

All fourier coefficients have|dw:1349031061922:dw| while not expressing into complex form. May be due to the fact that you are using Re(....) there.

experimentx · Answer

you are on wrong track

anonymous · Answer

You may be correct.I think my logic rests on this, is it incorrect:$$g(t)=\Re(h(t))$$$$F(g(t))=\Re(F(h(t))$$?

experimentx · Answer

http://www.wolframalpha.com/input/?i=Re [%283-12i%29e^%28i6t%29] these two are just two different form of same equation.

experimentx · Answer

you already have done this stuff ... the only thing you need to prove is 
|dw:1349031300459:dw|