Question

(PDE)(Fourier Series) I don't understand why the Fourier Series in this document is an *exponential* Fourier Series, or what information we look to to deduce that we should express it as an Exponential FS.

Answer 1

http://see.stanford.edu/materials/lsoftaee261/Midterm-2006-Solutions.pdf

It's the same question as last time, just part B. If I was given something like this on an exam, how should I know to express the FS as an exponential FS? @Kainui

Answer 2

What would be the alternative way of expressing the FS if not exponential?

Answer 3

Lol God, I know it's obviously not intentional, but you guys happen to phrase your questions by complete chance in possibly the worst way for me to respond to them from my misunderstanding, hahahahah, give me just a minute

Answer 4

Hahaha XD hey that's probably a good thing! Better have the hard questions now than on the test.

Answer 5

The exponential series is not, to my understanding, the "default" or "typical" way to express the FS of some given function, it's only supposed to be generally used when complex numbers are involved in the function or something like that. This is why I would think of trying to express it using a full range Fourier Sine & Cosine Series or something like that.

Answer 6

That's what my mindset and viewpoint is, so the alternative way of expressing it would by default be as a Fourier Sine & Cosine Series. I haven't dealt with an integral as a function *in* a fourier series, though, so I am not familiar with it.

Answer 7

Well you can sorta convert from one to the other \[\ f(t)=\sum_{n=-\infty}^\infty c_n e^{2\pi nit} = \sum_{n=-\infty}^\infty (a_n+b_n i)(\cos(2\pi nt) +i \sin(2\pi n t))\] Then from here you can combine them to go from 0 to infinity instead of from negative infinity. The reason the e^ix version is better imo is that it lets you use exponent rules to play around with stuff inside of sines and cosines which is hard cause who wants to play with trig identities?

Answer 8

What I'm trying to find is a strongly definitive answer in why one approach was taken and one was not. Is there? Or is it a matter of choice, and will pursuing either method give you an identical end result. This problem only has one right answer, so I highly doubt that they are equivalent, and thus there has to be some systematic way to determine whether to use an exponential FS or a Fourier Sine & Cosine Series

Answer 9

I just don't want to go into an exam, try to express one function in the fundamentally wrong form of FS and get a paper back like "lel 0/10 you chose the wrong way to express this FS"

Answer 10

I think that's really the answer, exponential expresses the same thing but allows manipulation of exponents instead of trig identities.

Answer 11

So if I took a function that met the general conditions to be expressed in a FS and chose either method, what you are saying is both end results are algebraically the exact same?

Answer 12

Yep. Ask your teacher if you don't believe me. I guess maybe I'll show it this way: \[\Large \sum_{n=0}^\infty a_n \cos (2\pi n t) + b_n \sin(2\pi n t)\] Start there then use these identities: \[\Large \cos(x) = \frac{e^{ix}+e^{-ix}}{2} \\ \Large \sin(x) = \frac{e^{ix}-e^{-ix}}{2i}\] \[\Large \sum_{n=0}^\infty a_n \frac{e^{i 2 \pi n t}+e^{-i2\pi n t}}{2}+b_n \frac{e^{i 2 \pi n t}-e^{-i2\pi n t}}{i2}\] \[\Large \sum_{n=0}^\infty \frac{a_n-i b_n}{2} e^{i 2\pi n t} + \frac{a_n + i b_n}{2} e^{-i 2\pi n t}\] Notice that the only difference in the powers of the exponent is a negative sign, so we can remove the negative term and make the summation from negative infinity to positive infinity. Since these are all arbitrary constants, we can call it something else as an arbitrary complex number, and here we go: \[\Large \sum_{n=- \infty }^\infty c_n e^{i 2\pi n t} \]

Answer 13

Also if you're not used to complex numbers, keep in mind that \[\Large \frac{1}{i} = -i\] It's pretty simple to prove, just multiply both sides of the equation by i.