# Fourier series specific value

I am learning Fourier series, and I must compute specific values for the Fourier series of the function $$5x+7$$ on the interval [-8,8].

The fourier series of a function on an interval [-L, L] is given by: $$f(x)=\sum_{n=0}^{\infty} A_{n} \cos \left(\frac{n \pi x}{L}\right)+\sum_{n=1}^{\infty} B_{n} \sin \left(\frac{n \pi x}{L}\right)$$

• I find first $$A_0$$: $$A_{0}=\frac{1}{2 L} \int_{-L}^{L} 5x+7 d x$$ The first term disappears (odd), and integrating 7 on the interval gives $$A_0 = 7$$

• Finding $$A_m$$ $$A_{m}=\frac{1}{L} \int_{-L}^{L} (5x+7) \cos \left(\frac{m \pi x}{L}\right) d x \quad m=1,2,3, \ldots \\ = \int_{-L}^{L} 5x \cos \left(\frac{m \pi x}{L}\right) d x + \int_{-L}^{L} 7 \cos \left(\frac{m \pi x}{L}\right) d x$$ The first term is odd (odd * even = odd), the second is orthogonal, so $$A_m=0$$.

• Finding $$B_m$$ in the same manner:

$$B_{m}=\frac{1}{L} \int_{-L}^{L} (5x+7) \sin \left(\frac{m \pi x}{L}\right) d x \quad m=1,2,3, \ldots \\ = \int_{-L}^{L} 5x \sin \left(\frac{m \pi x}{L}\right) d x + \int_{-L}^{L} 7 \sin \left(\frac{m \pi x}{L}\right) d x$$ The second term is orthogonal, but the first is odd and can be integrated. By partial integration I find: $$B_{m}=\frac{-1^{n+1}}{n \pi}$$

ie my Fourier series is: $$7 + \sum_{n=1}^{\infty} \frac{-1^{n+1} 80}{n \pi} \sin \left(\frac{n \pi x}{8}\right)$$

EDIT: I asked about $$f(0)$$ and as @Ninad pointed in the comment, it was easy to calculate. I edited my question for clarity!

EDIT 2: added missing 80 in the fourier series.

My issue is when calculating special values for this series. The site I am following is not covering this part and focusing in computing Fourier series (if I am not blind!: https://tutorial.math.lamar.edu/Classes/DE/FourierSeries.aspx) I cannot find a good example to follow (I don't understand how the person comes to their conclusion here: Computing value of fourier series)

So I would be grateful for an explanation on how one computes positive and negative values in the interval of a Fourier series, or some resources to read with exemplified process!

For example, how do I take f(2), and f(-2) and deal with the infinite sum?

• $f(0)$ looks like $7$ on that graph, I'm not sure what you're confused about. Apr 23, 2021 at 8:10
• Sorry, you are right! Looks like I cannot read the graph properly, I was reading the x-axis! But I am still confused about computing specific values. I edited my question, please have a look! Apr 23, 2021 at 8:53

There is a number of useful results about the pointwise convergence of Fourier series.

Take $$f$$ to be integrable on $$[-L, L]$$ and have a Fourier series $$\sum a_n \cos (n\pi x/L) + b_n \sin (n\pi x /L)$$. Then two examples are

1. If $$f$$ has left and right derivative at $$x\in [-L, L]$$ then the Fourier series converges at $$x$$ to $$\tfrac{1}{2}(f(x+)+f(x-))$$ where $$f(x\pm)$$ indicate the limits of $$f(x+t)$$ as $$t\to x$$ from above and below.
2. If $$f$$ is continuous and the sum of the Fourier coefficients converges absolutely (i.e. $$\sum |a_n| + |b_n|$$ converges) then the Fourier series converges to $$f(x)$$.

Use example 1 applied to your case, (assuming the coefficients are OK) we see that $$17 = f(2) = 7 + \sum_{n=1}^\infty \frac{(-1)^{n+1}80}{n\pi} \sin \left( \tfrac{1}{4}n\pi \right)$$ where the infinite sum on the right will converge. You can calculate values $$\sin \left( \tfrac{1}{4} n\pi \right)= \left\{ \begin{array}{2} \sqrt{2}/2&n=1,3,9,11, \cdots\\ 1&n=2,10, \cdots \\ -\sqrt{2}/2&n=5,7,13,15, \cdots \\ -1&n=6,14,\cdots \\ 0&n=4,8,12,\cdots \end{array} \right.$$ and if you combine into groups of eight terms you have, $$\tfrac{1}{8}\pi = \sum_{n=0}^\infty \left\{ -\frac{\tfrac{1}{2}\sqrt{2}}{8n+1} + \frac{1}{8n+2} - \frac{\tfrac{1}{2}\sqrt{2}}{8n+3} + \frac{0}{8n+4} + \frac{\tfrac{1}{2}\sqrt{2}/2}{8n+5} -\frac{1}{8n+6}+\frac{\tfrac{1}{2}\sqrt{2}/2}{8n+7}-\frac{0}{8n+8} \right\}$$

• Thank you very much @WA Don, that was very informative! So the important is to generate a series of terms that are repeating (cyclic?) Apr 23, 2021 at 16:52
• No - that won't always happen as it did in this case. The important thing is that the Fourier series converges for the right type of $f$ so you can then say the $f(x)$ and the sum of the series are the same. For example, to find $$\sum_{n=1}^\infty \frac{1}{n^2}$$ you can develop a Fourier series for a simple function (Maybe $f(x) = |x|$ on $[-\pi,\pi]?)$ which gives the $1/n^2$ terms when $x=0$ and obtain the famous $$\sum \frac{1}{n^2} = \frac{\pi^2}{6}$$ Apr 24, 2021 at 8:40
• I will file this knowledge for later, we just started with Fourier and did not yet come ton convergence :) Apr 24, 2021 at 8:47
• Btw sorry there was a typo in my question, 80 is missing in B_m, let me fix that... Apr 24, 2021 at 8:47
• By the way I have a follow up question if you have time and energy :) math.stackexchange.com/questions/4114511/… Apr 24, 2021 at 8:50