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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  • #2
LCKurtz
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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)##?
 
  • #3
RUber
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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) ##
 
  • #4
Delta2
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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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  • #5
haruspex
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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.
 
  • #6
LCKurtz
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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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