Odd/even extensions

Nov 2019
Good evening all,

I have this question,

f(t) is defined on the interval 0 ≤ t < 2 by f(t) = t(2 − t).
a) sketch the odd extension of f for -6≤ t ≤ 6, and state the fundamental period of this extension.
b) sketch the even extension for -6≤ t ≤ 6 and state the fundamental period of this extension.

The graph I have ended up with doesn't seem right, as a t=0, it is 0, t=1 it is 1, t=2 it is 0, and then as I plot up to t=6, i end up getting -24, is this completely wrong? should I just repeat the 0,1,0 pattern? I also can't figure out my fundamental periods.

Hope someone can help.

Thank you.


Math Team
Sep 2015
$f^+(t) = \begin{cases}0 &t<0\\t(2-t)&0\leq t \leq 2\\0 &2< t\end{cases}$

The odd extension mirrors the domain and inverts the sign, i.e. and repeats periodically

$f^-(t) = \begin{cases}0 &t < 2\\-t(2-t) &-2 \leq t \leq 0\\0 &0 < t\end{cases}$

And then typically you'll have these repeat with a period of the non-zero domain length.
Twice the originally given domain length as we've mirrored the domain.

$f(t) = \begin{cases}0 &t< 2\\-t(2-t) &-2 \leq t < 0\\t(2-t) &0 \leq t \leq 2\\0 &2 < t\end{cases}$

$f_o(t) = \sum \limits_{k=-\infty}^\infty f(t - 4k)$

The even extension is identical except that you don't invert the sign. I leave that to you.
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