# How can an exponential Fourier Series model a real function f(x) if the series contains imaginary numbers??

I'm an trying to model an arch using Fourier series. since I don't have the exact function of the arch, I want to take a sequence of points along the arch at fixed intervals and find the Fourier series using those points. However, I'm having difficulty understanding the difference between a Discrete Time Fourier Series (DTFS) and Discrete Fourier Transform (DFT). Some sources say that the DFT is used to find the complex Fourier Coefficients $$C_k$$ in the exponential Fourier series $$f(x) = \sum_{k=-\infty}^\infty C_k e^{i2\pi kx\over T}$$

wouldn't that mean that DTFS and DFT are the same thing?

On the other hand, also I dont understand how I can model my arch using a complex exponential series, since the series contains the imaginary number $$i$$, yet the original function was real. I understand how the formula above is derived from the original trigonometric series using Euler's identites, but I can't grasp how the exponential series is converted back to the trigonometric series. Do the imaginary terms cancel out? if so how/why does that happen?

context: I'm a highschool student so there are some big gaps in mathematical abilities. I would appreciate it if someone could answer my 2 questions in a way that does not include rigorous mathematical proof or advanced mathematical frameworks and tools. Thanks in advance!

• A quick thought: it is not unusual that a formula contains coefficients from some set while producing values from a smaller/different set. E.g. Fibonacci numbers are all natural numbers, but the explicit formula involves square root of $5$. Commented Oct 12, 2023 at 11:45

Yes indeed, the imaginary terms cancel out if $$f$$ is real.
For example, look that the formula $$\cos(x) = \frac{e^{ix}+e^{-ix}}{2}.$$ If you come back to the definition of the exponential as a series, that is $$e^x = \sum_{n=0}^\infty \frac{x^n}{n!}$$, then you get $$\cos(x) = \frac{1}{2} \sum_{n=0}^\infty \frac{(ix)^n+(-ix)^n}{n!} \\ = \frac{1}{2} \sum_{n=0}^\infty \frac{(i^n+(-i)^n)\,x^n}{n!}$$ and since $$i^n+(-i)^n = (-1)^{n/2}$$ if $$n$$ is even (because $$i^2=-1$$, so $$i^{2k} = (-1)^k$$) while $$i^n+(-i)^n = 0$$ if $$n$$ is odd (because $$(-i)^{2k+1} = (-1)^{2k+1}i^{2k} = -i^{2k}$$), so the above sum becomes a sum of positive terms.