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periodic extension (Definition)

Let $ f$ be a function defined on some real interval $ [a,b]$. By a periodic extension of $ f$ to the real line we mean a function $ g$ such that

  1. $ g$ is defined on $ \mathbb{R}$ except perhaps at points $ a+n(b-a)$, where $ n\in\mathbb{Z}$;
  2. $ g(x)=f(x)$ for all $ x\in (a,b)$, and
  3. $ g(x+n(a-b))=g(x)$ for all $ x\in (a,b)$ and all integers $ n$.

The best way to understand periodic extensions of a function is to look the graph of a periodic extension of a real-valued function. For example, let $ f(x)=x$ be defined on $ [-1,1]$. The graph of $ f$ looks like


\begin{pspicture} % latex2html id marker 79 (-6,-2)(6,2) \psaxes[Dx=9,Dy=2]{->}(... ....2,1.6){$y$} \psline(-1,-1)(1,1) \psdots[dotscale=1](1,1)(-1,-1) \end{pspicture}

Then the graph a periodic extension $ g$ of $ f$ may look like


\begin{pspicture} % latex2html id marker 86 (-6,-2)(6,2) \psaxes[Dx=9,Dy=2]{->}(... ...3,-1)(5,-1) \psdots[dotscale=1](1,1)(-1,1)(3,1)(-3,1)(5,1)(-5,1) \end{pspicture}

or look like


\begin{pspicture} % latex2html id marker 93 (-5.5,-2)(5.5,2) \psaxes[Dx=9,Dy=2]{... ...-5,1)(5,-1) \psdots[dotscale=1](-1,0)(1,0)(-3,0)(3,0)(-5,0)(5,0) \end{pspicture}

Notice the two periodic extensions of $ f$ are identical except at odd integer points on the $ x$-axis. The reason why we do not require $ g$ to agree with $ f$ on the end points of $ [a,b]$ is because we do not know if $ f(a)=f(b)$. If they do not agree, requiring that $ g=f$ on all of $ [a,b]$ may result in points $ a+n(b-a)$ getting mapped to two distinct values $ f(a)$ and $ f(b)$, rendering $ g$ not well-defined. In fact, if $ f$ does not agree on its endpoints, no periodic extensions of $ f$ are continuous.

Notice, also, that the domain of function $ f$ does not have to be the entire closed interval $ [a,b]$. The domain of $ f$ may very well be a subset $ S\subseteq [a,b]$. For example, $ f(x)=x$ may be a function defined on the open interval $ (-1,1)$. The two graphs above are again graphs of periodic extensions of $ f$.

However, if $ S$ is a proper subset of $ [a,b]$ that is not the open interval $ (a,b)$, then the definition of a periodic extension needs to be modified: $ g$ is a periodic extension of $ f$ defined on $ S\subseteq [a,b]$ if

  1. $ g$ is defined on a subset $ T\subseteq \mathbb{R}$ except perhaps at points $ a+n(b-a)$, where $ T=\lbrace x+n(a-b)\mid x\in S\rbrace$ and $ n\in\mathbb{Z}$;
  2. $ g(x)=f(x)$ for all $ x\in S-\lbrace a,b\rbrace$, and
  3. $ g(x+n(a-b))=g(x)$ for all $ x\in S-\lbrace a,b\rbrace$ and all integers $ n$.
We generally assume that $ a=\inf S$ and $ b=\sup S$.

For example, if $ f(x)=x$ for all rational numbers $ x\in [-1,1]$, then a periodic extension of $ f$ has its domain the set of all rational numbers except perhaps at are odd integers.

Remarks.

  • Trigonometric functions defined on $ \mathbb{R}$ are periodic extensions of the trigonometric functions defined for angles in the interval $ [0,2\pi]$.
  • Suppose $ f$ is defined either on a closed interval $ [a,b]$ or an open interval $ (a,b)$, $ f$ has a continuous periodic extension (defined on all of $ \mathbb{R}$) iff $ f$ is continuous and that
    1. either $ f(a)=f(b)$ when $ f$ is defined on a closed interval, or
    2. $ f(a+)=f(b-)$ when $ f$ is defined on an open interval, where $ f(a+)$ is the one-sided limit approaching $ a$ from the right, and $ f(b-)$ is the one-sided limit approaching $ b$ from the left.
    Simply define the periodic extension $ g$ so that either $ g(a)=f(a)$, or $ g(a)=f(a+)$. For example, the following graph

    \begin{pspicture} % latex2html id marker 144 (-6,-0.5)(6,2) \psaxes[Dx=9,Dy=2]{-... ...)(3,1) \psline(3,1)(4,0) \psline(4,0)(5,1) \psline(5,1)(5.5,0.5) \end{pspicture}
    is the graph of the continuous periodic extension of a function $ f_1$ given by $ f_1(x)=\vert x\vert$ defined on $ (-1,1)$, or a function $ f_2$ defined on $ (0,2)$, given by $ f_2(x)=x$ for $ 0<x\le 1$ and $ f_2(x)=2-x$ for $ 1\le x<2$. With $ f_1$, we see that $ f_1(-1+)=f_1(1-)=1$, while with $ f_2$, we have $ f_2(0+)=f_2(2-)=0$.

    It is easy to see that if a continuous periodic extension of a function exists, then it is unique.

  • Higher dimensional periodic extensions may also be defined for functions defined on a parallelepiped ($ n$-dimensional analog of a parallelogram). A periodic extension $ g$ of a function $ f$ defined on a parallelepiped is a function such that its projection $ p_i(g)$ onto axis $ i$ in $ \mathbb{R}^n$ is a periodic extension of the projection $ p_i(f)$ of $ f$ onto axis $ i$.

Bibliography

1
G.P. Tolstov, Fourier Series, Prentice-Hall, 1962.



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See Also: periodic functions, triangular wave function

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This is version 14 of periodic extension, born on 2007-09-21, modified 2007-09-26.
Object id is 9955, canonical name is PeriodicExtension.
Accessed 467 times total.

Classification:
AMS MSC42A99 (Fourier analysis :: Fourier analysis in one variable :: Miscellaneous)

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