Is the function odd, even or neither?

[cabellos]

The function f(x) 1, -pi < x < 0

0, 0 < x < pi

now after sketching the function i believe i am correct in saying it is neither symmetrical about the x-axis or the origin and therefore is neither odd nor even?

It is a square wave?

Am i correct? Also how should i now go about finding its Fourier series?

Thankyou

 

[HallsofIvy]

Assuming you mean f(x)= , then, yes, it is neither even nor odd.

You could find its Fourier series by using the usual formulas:
A_n= \frac{1}{\pi}\int_{-\pi}^0 sin(nx) dx[/itex]<br /> and<br /> B_n= \frac{1}{\pi}\int_{-\pi}^0 cos(nx)dx[/itex]&lt;br /&gt; where A&lt;sub&gt;n&lt;/sub&gt; and B&lt;sub&gt;n&lt;/sub&gt; are the coefficients of sin(nx) and cos(nx) respectively for n&amp;gt; 0. &lt;br /&gt; B_0= \frac{1}{2\pi}\int_{-\pi}^0 dx= \frac{1}{2}&lt;br /&gt; is the constant term.&lt;br /&gt; &lt;br /&gt; Or you could write f(x)= 1/2 + g(x) where g(x)= 1/2 for -\pi \le x\le 0 and g(x)= -1/2 for 0&amp;amp;lt; x \le \pi. g(x) is an odd function so the calculations are little simpler.

 

[matt grime]

I'm going to be more abrupt than the lovely mentors here. What is the bloody definition? Sorry, cabellos, but I'm truly fed up with the number of questions which are answerable with 'look at the definition'.

 

[HallsofIvy]

"Lovely", moi? Oh, how sweet of you!

Actually, I think the first sentence of just about every response should be "look up the definition"!