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If $x(t) = \{t\}$ (the fractional part of $t$), then the Fourier series of $x(t)$ is $$\frac{1}{2}-\frac{1}{\pi}\sum_{n=1}^{\infty} \frac{\sin(2\pi nt)}{n}$$

My question is why $x(1) = 0.5$ (using the Fourier series) and not $0$, as expected?

Notice that $x(t) = \{t\} = t - \lfloor t \rfloor$, so $x(1) = 1 - \lfloor 1 \rfloor = 1 - 1 = 0$.

However, with the Fourier series, it gives $0.5$ because

$$\begin{split} x(1) &= \frac{1}{2}-\frac{1}{\pi}\sum_{n=1}^{\infty} \frac{\sin(2\pi n\cdot1)}{n}\\ &= \frac{1}{2} - \frac{1}{\pi} \cdot 0\\ &= \frac{1}{2} - 0\\ &=\frac{1}{2} \end{split} $$

Is it only true for non-integer values, e.g., is it true for $\sqrt 2$, but not for the number $1$?

Joshua Ortiz
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    To recollection, the Fourier series of a suitably-nice function can differ from the original function on a set of measure zero, and commonly at discontinuities (as the integers are for the periodic extension of the fractional part), so this result shouldn't be unexpected. – PrincessEev Nov 06 '22 at 23:59
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    Because $\lim_{x\to1-}{x}=0$ and $\lim_{x\to0+}{x}=1.$ Remember, any finite Fourier series sum is continuous, but ${x}$ is discontinuous. – Thomas Andrews Nov 07 '22 at 00:01

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The fractional function is continuous everywhere but the integers on the real line. At a continuous point, Fourier series converge nicely to the value of the function, but at a discontinuous point, Fourier series converge to the average of the right and left limits at the point, which gives $\frac{0+1}{2}=0.5$ in your case.

In general, Fourier series is not necessarily convergent. Or Convergence might be in another sense. Pointwise convergence is rare.

enihcamemit
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