# About the existence of a periodic function $f$ with exactly one singularity and matches with a sinusoid of same period?

I want to know whether a function $f$ with the following properties can exist. $f$ should be periodic with period $2\pi$, $f$ should be smooth everywhere except at exactly one point per period, such that the function $f$ itself or one of its derivatives (of different order) should jump or blow-up at this point. Equally spaced sample points are chosen such that there are at least 3 points in one period. Now the values of this function $f$ at these sample points should be equal to those of a given sinusoidal function of period $2\pi$ i.e, $\sin(x)$. What I'd like to know is, can such a function $f$ exist ?

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For a "jump discontinuity", just take your sinusoidal function and alter it at that one point. –  uncookedfalcon Jun 2 '12 at 11:26
@uncookedfalcon : Thanks for the comment. I have made an attempt at an answer. Please comment any mistakes. –  Rajesh D Jun 2 '12 at 11:58

The answer is yes, for any ordered derivative jumping. Just consider a bump function having a support $(0,l)$, now modify it by replacing the second half of the bump function with a negative of the original. $b_m(x) = bump(x), x\in[0,l/2)$ and $b_m(x) = -bump(x), x \in [l/2,l]$. Now Let $$g = b_m\ast b_m\ast ...(k~times)..\ast b_m$$, where $\ast$ is a convolution operation. Now $g$ is smooth except at $x = l/2$ where $g^{(k)}(x)$ jumps. The support of $g(x)$ is $(0,(k+1)l)$.
Choose $l$ such that $(k+1)l \leq d$ where $d$ is the spacing between the samples where you want functions to be equal. Now form the desired function $f(x)$ by adding a shifted version of $g(x)$ i.e., $g(x-\tau)$ to $\sin(x)$ to form $f(x) = g(x-\tau) + \sin(x)$. Choose $\tau$ such that the support of $g(x-\tau)$ sits exactly between two sample points. Now $f(x)$ agrees with $\sin(x)$ at all the required sample points.