# Show these functions are 2 pi periodic

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g(t)=½( f(t)+f(-t) ) h(t)=½( f(t)-f(-t) )
show its 2π periodic so: g(t+2π) = ½( f(t+2π)+f(t-2π) ) why does -t become t-2π ?
½( f(t)+f(-t) ) = g(t)
h(t+2π)=½( f(t+2π)-f(t-2π) )
½( f(t)-f(-t) ) = h(t) is this correct?
can anybody show me some similar examples please?
this is from a fourier series question paper.
thanks

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LCKurtz
Homework Helper
Gold Member
Of course, this problem is false unless $f(t) = f(t+2\pi)$, which you haven't told us.

g(t)=½( f(t)+f(-t) ) h(t)=½( f(t)-f(-t) )
show its 2π periodic so: g(t+2π) = ½( f(t+2π)+f(t-2π) ) why does -t become t-2π ?
It doesn't. If you put $t+2\pi$ in for $t$ in $f(-t)$ you get $f(-(t+2\pi)) = f(-t -2\pi)$. Do you see why the numerator becomes $f(t)+f(-t)$?

• Delta2
RUber
Homework Helper
An example would be the exponential definition of the cosine function.
$\cos(x) = \frac{1}{2} (e^{ix} + e^{-ix})$
This is analogous to your problem for $g(x)$ with $f(x) = e^{ix}$.
$\cos(x+2\pi) = \frac{1}{2} (e^{i(x+2\pi)} + e^{-i(x+2\pi)})\\ = \frac{1}{2} (e^{ix}e^{i2\pi} +e^{-ix}e^{-i2\pi})\\ = \frac{1}{2} (e^{ix} + e^{-ix})\\ = \cos(x)$

Delta2
Homework Helper
Gold Member
Hint: prove that if a function has period T, then $f(x-T)=f(x)$ as long as f is defined in x-T.

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haruspex
Homework Helper
Gold Member
Of course, this problem is false unless $f(t) = f(t+2\pi)$,
or maybe the sum of a periodic function and an odd function.

LCKurtz
Homework Helper
Gold Member
Of course, this problem is false unless $f(t) = f(t+2\pi)$, which you haven't told us.

or maybe the sum of a periodic function and an odd function.
Well, I assumed, which was also unstated, that problem was to prove that if $f$ has period $2\pi$, then when you write $f = g + h$ with $g$ even and $h$ odd, that $g$ and $h$ have period $2\pi$. In that context I'm not sure what you are suggesting.

And, annoyingly enough and a pet peeve of mine, the OP hasn't returned to the thread to clarify anything

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