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I am a bit confused with the Fourier series. The first step should be to determine if my function is odd or even, then find the coefficients (with eventually the shortcut for odd or even function) and finally I can compute the series.

My confusion comes from this exercise, where the function f with periodicity 2$\pi$ is given in the interval $0\le x\le2\pi $ as:

$f(x)= \left\{ \begin{array}{ll} x\ \text{for} \ 0\le x\le\pi\\ 0 \ \text{for}\ \pi\le x < 2\pi \end{array} \right.$

I thought that the function was odd, at least for the first interval, so I computed the Fourier series only by looking at the coefficient $b_n$ (I assumed that $a_n$ was 0).

I got the answer back from the TA saying that it's wrong, because $f(x)$ is neither odd nor even and therefore, the Fourier series should be computed with no shortcuts.

So my questions are:

1) Is possible for a function to be neither odd nor even?

2) Has this something to do with the discontinuity? How do you evaluate if a function is odd or even if the function is piecewise?

I feel like I must have missed some important concept about that.

Captain Lama
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Kundera
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    Regarding the first question, consider the function $e^x$: $e^{-x} \not= -e^x$ and $e^{-x} \not= e^x$ (i.e. yes, a function can be neither odd nor even). – Chill2Macht Apr 28 '16 at 15:45
  • But how can you conclude that $f(x)$ is neither odd nor even? – Kundera Apr 28 '16 at 16:07
  • The definition of an odd function: $f(-x)=-f(x)$. The definition of an even function: $f(-x)=f(x)$. If you know how to solve for/write $f(-x)$, you can test to see whether or not it equals either $-f(x)$ or $f(x)$. If it equals neither of the two, then the function must be neither odd nor even. I don't know if that helps you though... – Chill2Macht Apr 28 '16 at 16:15
  • Thank you for answering. It is clear to me the difference between odd and even, and that $f(x)=x$ is an odd function. But the teacher assistant said that the function was neither odd nor even, and I am not sure how you conclude that. – Kundera Apr 28 '16 at 16:33

1 Answers1

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1) Yes it is possible to have a function which is neither odd or even, for example, observe function $f(x)$=$x^2+x$, now $f(-x)$=$(-x)^2+(-x)$=$x^2-x$, now for the function to be odd the answer should either be $-f(x)$ i.e. $-x^2-x$ or for it to be even the answer should be $f(x)$ i.e. $x^2+x$, but it is neither, hence the function is neither even nor odd because it does not satisfy the condition for even function nor does it satisfy the condition for odd function.

hope this helps :)

user374502
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