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This is a follow-up question to this.

I now understand that when calculating a specific value in a Fourier series, the goal is to identify a pattern and recognize a known infinite series or Taylor expansion.

I have the function $g(x) = 5x+7$, and its Fourier series: $\displaystyle7 + \sum_{n=1}^{\infty} \frac{-1^{n+1} \cdot 80}{n \pi} \sin \left(\frac{n \pi x}{8}\right)$. I verified it was correct. I now need to find f(5).

As the teacher showed me, I took several values of $n$ for $\sin(\frac{5n\pi}{8})$ to try to identify my infinite sum.

$$ \begin{aligned} \displaystyle \begin{array}{lll} \displaystyle n=1 & \sin \left(\frac{5 \pi}{8}\right) & \frac{\sqrt{2+\sqrt{2}}}{2} \\ n=2 & \sin \left(\frac{5 \pi}{4} \right) & \frac{-1}{\sqrt{2}} \\ n=3 & \sin \left(\frac{15 \pi}{8} \pi\right)=\sin \left(\frac{- \pi}{8}\right) & -\frac{\sqrt{2-\sqrt{2}}}{2}\\ n=4 & \sin \left(\frac{5 \pi}{2} \pi\right)=\sin \left(\frac{\pi}{2}\right) & 1\\ n=5 & \sin \left(\frac{25 \pi}{8} \pi\right)=\sin \left(\frac{ \pi}{8} \pi\right) & \frac{\sqrt{2+\sqrt{2}}}{2} \\ n=6 & \operatorname{an}\left(\frac{30 \pi}{8}\right)=\sin\left(\frac{-\pi}{4}\right) & \frac{-1}{\sqrt{2}}\\ n=7 & \sin\left(\frac{35 \pi}{8}\right)=\sin\left(\frac{3 \pi}{8}\right) & \frac{\sqrt{2+\sqrt{2}}}{2}\\ n=8 & \sin\left(5 \pi\right) & 0\\ n=9 & \sin\left(\frac{45 \pi}{8}\right)=\sin\left(\frac{5 \pi}{8}\right) & \frac{\sqrt{2+\sqrt{2}}}{2}\\ n=10 & \sin\left(\frac{50 \pi}{8}\right)=\sin\left(\frac{\pi}{4}\right) & \frac{1}{\sqrt{2}}\\ n=11 & \sin \left(\frac{55 \pi}{8} \pi\right)=\sin \left(\frac{- \pi}{8}\right) & -\frac{\sqrt{2-\sqrt{2}}}{2}\\ \end{array}\\ ... \end{aligned} $$

When I try to put it with coefficients I get: $$ \begin{aligned} 7+\frac{80}{\pi} &\left(\frac{1}{1} \frac{\sqrt{2+\sqrt{2}}}{2}+\frac{1}{2} \frac{1}{\sqrt{2}}+\frac{1}{3} \frac{\sqrt{2-\sqrt{2}}}{2}-\frac{1}{4} \frac{\sqrt{2+\sqrt{2}}}{2}+\frac{1}{5} \frac{\sqrt{2+\sqrt{2}}}{2}\right.\\ &+\frac{1}{6} \frac{1}{\sqrt{2}}+\frac{1}{7} \frac{\sqrt{2+\sqrt{2}}}{2}-\frac{1}{8} \cdot 0+\frac{1}{9} \frac{\sqrt{2+\sqrt{2}}}{2}-\frac{1}{10} \frac{1}{\sqrt{2}}-\frac{1}{11} \frac{\sqrt{2-\sqrt{2}}}{2} ...) \end{aligned} $$

I am sorry in advance for typos, I have a tendency to miss them when typing latex!

Question: How do I see my series for f(5)? I can't identify any Taylor series or familiar sum. Are there particular techniques to distinguish a series?

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    $\begingroup$ Hi. I am not sure it is correct to say that the 'goal ' is to find a pattern. Sometimes you might find the Fourier series generates an interesting pattern of terms, $e.g. \sum 1/n^2$, and equating series and function value allows you to obtain the sum; but Fourier series have other uses. In the case above, I cannot see any pattern and terms do not obviously cancel, so not sure what the question intended. Moreover, I don't know of any guaranteed way to identify a pattern. $\endgroup$
    – WA Don
    Apr 24, 2021 at 9:03
  • $\begingroup$ Alright. This might mean I have the wrong approach again! I had f(0), f(8), f(-8) which were equal to 7 because the sin terms cancels out, and f(-4) = -13 because of a pattern of -1, 0, 1, 0, which, together with the $\frac{-1^{n+1}}{n}$ term, that was corresponding to $tan^{-1}(1)$. So I assumed f(5) would also hide a pattern. What would your approach be to solve this question? $\endgroup$
    – Dovendyr
    Apr 24, 2021 at 9:14
  • $\begingroup$ I am not sure what the objective of the question is. Are you expected to find a pattern? There might not be any other than the one you have derived. $\endgroup$
    – WA Don
    Apr 24, 2021 at 9:18
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    $\begingroup$ OK. I am not sure what you can do if no pattern is apparent. Maybe stop at the sum of $\sin$ terms. Note that the cycle must be $n=1,2,\cdots,16$ in this case to get a full period of $\sin (5n\pi/8)$. $\endgroup$
    – WA Don
    Apr 24, 2021 at 9:32
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    $\begingroup$ Thanks - as long as your happy. I won't cut and past because I'd like to answer more thoroughly to make it a good answer but not inclined to put extra work in! $\endgroup$
    – WA Don
    Apr 25, 2021 at 17:41

1 Answer 1

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The answer was clearly stated by @WA Don in my previous question and I wish I can select it as a solution for this question as well.

I did not understood the condition (2.) at he time, because I simply did not understood how one could take all Fourier coefficients (I still don't actually!).

From WA Don answer:

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 $f(x)$.

The same theorem was available in my book, Differential Equations with Boundary-Value Problems:

enter image description here

In conclusion, since the function $g(x) = 5(x) + 7$ on the interval $[-8,8]$ is continuous, its Fourier series $ 7 + \sum_{n=1}^\infty \frac{(-1)^{n+1}80}{n\pi} \sin \left( \tfrac{1}{4}n\pi \right)$ converges to $5(x) + 7$, that is f(5) = 32.

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    $\begingroup$ Hi. The second condition (see [wikipedia[(en.wikipedia.org/wiki/…) entry) means that the sum of the positive values $|a_n|, |b_n|$ of the coefficients - $80/(n\pi)$ - converges to a finite number. This is not the case because (ignoring $80/\pi$) you would sum terms of $1/n$ and this sum gows without limit (diverges to $\infty$). Thus you cannot use condition 2 but condition 1 works fine because $f$ has left and right derivatives and a discontinuity at the ends of the interval where $f(8)\neq f(-8)$. $\endgroup$
    – WA Don
    Apr 26, 2021 at 8:35
  • $\begingroup$ So you mean that I can use either of conditions, and both do not need to be true? $\endgroup$
    – Dovendyr
    Apr 26, 2021 at 9:48
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    $\begingroup$ Yes - see the page reference $\endgroup$
    – WA Don
    Apr 27, 2021 at 11:15

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