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Let $f$ be the function $f(x) = x^2 + 2 $, where $ 0<x<1 $.
Extend the function $f(x)$
(1) As an odd periodic function with period $2$
(2) As an even periodic function with period $2$
(3) As a periodic function with period $1$

I know what exactly are odd and even functions. But have no idea how to extend as such. Please explain with a correct method.

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  • $\begingroup$ How are you defining an even or odd function? $\endgroup$ – Daniel Pietrobon Nov 17 '13 at 13:37
  • $\begingroup$ are you sure you can extend that way.... $\endgroup$ – user87543 Nov 17 '13 at 13:38
  • $\begingroup$ @01000100 simply even f(x) = f(-x) and odd -f(x) = f(-x) $\endgroup$ – Marlon Abeykoon Nov 17 '13 at 13:47
  • $\begingroup$ i am actually talking about periodicity... $\endgroup$ – user87543 Nov 17 '13 at 13:48
  • $\begingroup$ Yes thats what my issue is, Question is correct since i obtained it from a trustworthy source.. I have no idea in extending it :) $\endgroup$ – Marlon Abeykoon Nov 17 '13 at 13:53
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  • Even case

Due to symmetry we need to define $f(x)=x^2+2$ on $(-1,0)$. It remains to discuss the $2$-periodicity. One starts with "special points", for example $x=0$. Then

$$f(0)=0^2+2=\text{(2-periodicity)}=f(0+2)=f(2),$$

i.e. $f(2):=2$. To find the extension of $f$ on $(2,3)$ one continues with all $\delta\in (0,1)$, i.e.

$$f(\delta)=\delta^2+2=\text{(2-periodicity)}=f(\delta+2),$$

where $\delta+2\in (2,3)$. Similarly, considering all points $-\delta\in(-1,0)$, using $2$-periodicity one finds the extension of $f$ on the interval $(1,2)$ via $-\delta+2\in (1,2)$, as we did above. The extension of $f$ for all other points easily follow by drawing.

  • Odd case

Due to symmetry we need to define $f(x)=-x^2-2$ on $(-1,0)$. One selects once again the "special points", for example $x=0$ as in the even case. The analysis is completely similar, with due changes.

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Given a function $f:\ A\to{\mathbb R}$ and a superset $\tilde A\supset A$ extending $f$ to $\tilde A$ means defining a function $\tilde f:\ \tilde A\to{\mathbb R}$ such that the restriction of $\tilde f$ to $A$ coincides with the given $f$. Usually the extended function is denoted by $f$ again.

In your case $A=\ ]0,1[\ $. As the extended functions are required to be periodic I shall assume that $\tilde A={\mathbb R}$ is intended; but $\tilde A={\mathbb R}\setminus{\mathbb Z}$ would also be okay.

ad 1: The requirement $f(-x)=f(x)$ enforces $f(x)=x^2+1$ $\>(-1<x<0)$. The value $f(0)$ may be chosen arbitrarily; and the same holds for $f(1)=f(-1)$. Now $f$ is specified on the full interval $[-1,1]$. Since $[-1,1]$ is a fundamental domain for the envisaged $2$-periodicity it follows that the extended $f$ is now determined on all of ${\mathbb R}$. The value of the extended function at an arbitrary point $x\in{\mathbb R}$ is obtained as follows: Write $x$ in the form $x=2k+x'$ with $x'\in[-1,1]$ and put $$f(x):=f(x')\ .$$ When we have chosen $f(0)=0$ and $f(1)=f(-1)=2$ then the extended function will be continuous on all of ${\mathbb R}$.

ad 2: The requirement $f(-x)=-f(x)$ enforces $f(x)=-x^2-1$ $\>(-1<x<0)$, as well as $f(0)=0$. Now $f$ is specified on the full interval $\ ]-1,1[\ $. Since $[-1,1]$ is a fundamental domain for the envisaged $2$-periodicity it follows that the extended $f$ is now determined on all of ${\mathbb R}$, apart from the odd integers. Now the condition $f(-1)=-f(1)$ together with $f(1)=f(-1)$ enforces $f(-1)=f(1)=0$. The value of the extended function at an arbitrary point $x\in{\mathbb R}$ is therefore obtained as follows: Write $x$ in the form $x=2k+x'$ with $x'\in[-1,1]$ and put $$f(x):=f(x')\ .$$ The extended function, which was uniquely determined, is not continuous at the odd integers, since it has a jump discontinuity there.

I leave case 3. to you.

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