Question about the start of a cosine fourier series

[deckoff9]

Question about the "start" of a cosine Fourier series

Hey. I was just looking through Paul's Online Notes http://tutorial.math.lamar.edu/Classes/DE/FourierCosineSeries.aspx to teach myself Fourier Series and I had a question about the a[itex]_{0}[/itex] term of the cosine series.

In the online lesson, it says assume an even function has the series f(x) = [itex]\Sigma[/itex]a[itex]_{n}[/itex]cos(n[itex]\pi[/itex]x/L) where -L≤x≤L. The series starts at 0, and the way Paul gave a prove of it was to multiply the series by cos(m[itex]\pi[/itex]x/L) and then integrated and used the fact that cos(m[itex]\pi[/itex]x/L) and cos(n[itex]\pi[/itex]x/L) were orthogonal if m!=n.

So that for example, for the Fourier series of x[itex]^{2}[/itex], he got a[itex]_{0}[/itex] = L[itex]^{2}[/itex]/3, where -L≤x≤L.

However, my question is, why do we need to start at n = 0? The proof using orthogonality would work just as well if n were to start at 1 or 100, and the formula for the coefficients would remain the same. In addition, I'm not sure convergence explains it, since the beginning terms of a infinite series have no effect on the convergence of an infinite series. So I was hoping someone could clear this up for me.

Thanks in advance!

 

[HallsofIvy]

Changing or deleting a finite number of terms will not change whether or not the series converges but it surely changes what it converge to!

 

[Padfoot89]

You're making use of the fact that the set of [itex]{ \cos \frac{n \pi x}{L} } [/itex] for all integers n serves as a complete basis for even functions on the interval L. Without the [itex] n = 0 [/itex] term, you don't have a complete basis and thus can't span the space of even continuous on this interval (in reality, the space this set spans is a little bigger than just continuous even functions, but for brevity of the conversation, I've cut it down).

At its heart the Fourier expansion is based in applications of linear algebra, hence why my explanation is based in linear algebra terms. If you struggle with the explanation, you should look into some basic linear algebra concepts and definitions. These expansions are one of my favorite examples of "hidden" linear algebra.

 

[deckoff9]

Thanks for the answers guys! It helped clear up my confusion a lot.

 

[blue_raver22]

Hi there,

As a scientist, it is important to understand the reasoning behind mathematical concepts and not just blindly accept them. In this case, the reason why the cosine Fourier series starts at n = 0 is due to the nature of even functions.

An even function is symmetric about the y-axis, meaning that for every value of x, there is a corresponding negative value that gives the same y-value. Therefore, when we consider the Fourier series for an even function, we only need to consider the even terms (cosine terms) since the odd terms (sine terms) will cancel out due to the symmetry. This is why the series starts at n = 0, as the first term (a0) represents the constant term in the Fourier series for an even function.

You are correct that the proof using orthogonality would work for any starting value of n, but starting at n = 0 simplifies the calculations and makes the formula for the coefficients more concise. As for convergence, it is true that the beginning terms of an infinite series have no effect on its convergence. However, in this case, we are not just looking at the convergence of the series, but also the accuracy of the Fourier approximation. Starting at n = 0 gives the best approximation for an even function, as the odd terms would only add unnecessary error.

I hope this helps clarify the reasoning behind starting the cosine Fourier series at n = 0. Keep exploring and questioning mathematical concepts, it will only enhance your understanding as a scientist.