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My question is why do we define $f(\xi+0)$ as $\lim\limits_{\varepsilon \rightarrow 0} f(\xi + \varepsilon^2)$ and $f(\xi-0)$ as $\lim\limits_{\varepsilon \rightarrow 0} f(\xi - \varepsilon^2)$.

Thanks.

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    The mystery symbol there is actually called \xi.2010-08-26
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    We generally don't define $f(\xi+0)$ in this way: I've never seen this cheesy trick of defining one-sided limits using squares before. What is this book?2010-08-26
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    @Chapman: It is from _Introduction to Calculus and Analysis_ by R.Courant and F.John.2010-08-26
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    Concerning the title question, here is a quote from page 49 of Riemann's Zeta Function by H. Edwards: "In other words $\psi (x)=\sum_{p^{n}$x$ is a power; at the jumps $x=p^{n}$ the value of $\psi $\ is defined *as usual*, to be halfway between the new and old values $\psi (x)=\frac{1}{2}[\psi(x-\epsilon )+\psi (x+\epsilon )]$". Concerning your last question, I have never seen those definions in this way. Since $\epsilon ^2$ tends to zero fast than $\epsilon$ as $\epsilon$ approches zero, I guess that there is some connection with the derivative of function. – 2010-08-26
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    @Robin Chapman:But if we do not use squares then $\epsilon$ could be negative.2010-08-27

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It is not that we assign that value: Fourier series simply converge to that independently of our desires!

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    Yes, or even better if $f$ is not given as such then we can redefine $f$ to have the mean value at the point of discontinuity and then $S[f]$ (the Fourier series of $f$) will converge to $f$ at that point.2010-08-26