How to show a function is even/odd

  • Thread starter Thread starter mcfc
  • Start date Start date
  • Tags Tags
    Function
mcfc
Messages
16
Reaction score
0

Homework Statement


Hi,
I'm new to this site, I've had a look around and there are a lot of useful sections, particularly the section with math and science learning materials.
Anyway, I need to show that the following function is odd
f(x)=\left\{\begin{array}{ccc}<br /> -\sin x&amp;\mbox{ for }-\pi \leq x&lt; \frac{-\pi}{ 2}\\<br /> \sin x &amp;\mbox{ for } \frac{-\pi}{2} \leq x \leq \frac{\pi}{2}\\<br /> -\sin x &amp;\mbox{ for } \frac{\pi}{2}&lt;x&lt;\frac{\pi}{2}<br /> \end{array}\right.

\mbox{ and }f(x + 2 \pi) = f(x) \mbox{<br /> for all other values of x, is an odd function.}

Homework Equations



I know an odd function is definded as f(-x) = -f(x)

The Attempt at a Solution


In the interval
-\pi\leq x &lt; {-\pi \over 2} \mbox{ if I substiture } -\pi \mbox{ it becomes }-\sin(-x) = -\sin[-(-{\pi \over 2})] = -\sin({\pi \over 2})

Is that the correct way to solve it?
But I'm not sure how to show it's odd in the other intervals!
 
Last edited:
Physics news on Phys.org
If f(-x)=f(x), then the function is even. If f(-x)=-f(x), then the function is odd.
 
This might be a useful variation of the definitions:

If the function f(x) is even, then f(x)-f(-x)=0 for all x.
If the function f(x) is odd, then f(x)+f(-x)=0 for all x.
 
robphy said:
This might be a useful variation of the definitions:

If the function f(x) is even, then f(x)-f(-x)=0 for all x.
If the function f(x) is odd, then f(x)+f(-x)=0 for all x.

To show it's odd:
look at values in the intervals?
-\sin(-\pi) - \sin({\pi }) = 0

\sin({-\pi \over 2}) + \sin({\pi \over 2}) = 0

-\sin({3 \pi \over 4}) - \sin({-3 \pi \over 4}) = 0

do I need to show anything else?
 
mcfc said:
To show it's odd:
look at values in the intervals?
-\sin(-\pi) - \sin({\pi }) = 0

\sin({-\pi \over 2}) + \sin({\pi \over 2}) = 0

-\sin({3 \pi \over 4}) - \sin({-3 \pi \over 4}) = 0

do I need to show anything else?

You would have to show that it's true for every value in the interval, not just at a few random points. So you'd have to let a be a random value in each interval, and then look at f(a) and f(-a). Since the intervals are symmetric, once you've assigned an interval for a, it will be obvious what interval -a is in and therefore which definition of the function you need to use.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top