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I thought I am supposed to do check $f(t)=-1$ and compare with $f(-t)$

$f(-t)=-1$

If $f(t)=f(-t)$ the function is even. But this function is odd.

mangekyou
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2 Answers2

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It is for sure not odd function as f(0)=-1. For odd f(0)= 0 only.

mathsdiscussion.com
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You should specify that you are adopting the convention that a discontinuous function of the first kind assumes in the discontinuity point the average of the left and right limits;

In that way, left limit is -1, right limit is +1, the average is 0 and the function then is odd as

$$f(x)=1\quad(x>0)$$ $$f(-x)=-1=-f(x)$$ $$f(0)=0$$

Anyway, the definition of odd function only wants $$f(-x)=-f(x)$$ So if the function is undefined at x=0 there’s no point in insisting on f being 0 there

See for reference: Do odd functions pass through the origin?

Francesco Bernardini
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  • I'm asked to find the Fourier series of the function. There is nothing else specified in the problem. – mangekyou Dec 02 '19 at 18:04
  • Isn't it $f(0)=-1$ ? – mangekyou Dec 02 '19 at 18:16
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    I just meant that you should specify the value for x=0 if you want to put the cherry on the pie for the oddness; the function you mentioned is a combination of Heaviside functions, and the value in x=0 is not univocal, should be specified either by you or by the problem, depending on the application; see for example https://en.wikipedia.org/wiki/Heaviside_step_function#Zero_argument In your case I would say that it’s not harmful to assume f(x)=0 – Francesco Bernardini Dec 02 '19 at 18:17
  • I'm starting to understand better. We know $f(-π)=1$ from $f(t)=1, -π ≤ t < 0$

    What throws me off is that $f(t)=-1$ for $0≤t<π$ Less than π not equal to π.

    How can we then assume $f(π)=-1$?

    As I write this I realize that if we multiply $f(t)=1, -π ≤ t < 0$ by $(-1)$ we get $f(t)=-1, π≥ t > 0$ which means $f(π)=-1$.

    Now we have $f(-π)=1$ and $f(π)=-1 => f(-x)=1$ and $f(π)=-1$

    $$ f(-x)= 1 = -f(x)=-(-1)=1$$

    Am I correct?

     – mangekyou Dec 02 '19 at 18:41
  • You are not assuming that, you are defining your function to be -1 at $\pi$, as well as you define it to be 1 at $-\pi$. They are boundary points and the behaviour you must expect depends to some extent on what you are going to do next. For example, if you extend the Fourier expansion outside of $[-\pi,\pi]$, you might have a limit for $x\rightarrow\pi^-$ of $f(x)$ different from the limit for $x\rightarrow\pi^+$. At boundary points, clearly, the function is undefined and you need to apply some convention to define what its value is (for example, again, the average of left and right limits) – Francesco Bernardini Dec 02 '19 at 20:25