How to show a function is even/odd


by mcfc
Tags: even or odd, function
mcfc
mcfc is offline
#1
Aug22-09, 03:49 PM
P: 18
1. The problem statement, all variables and given/known data
Hi,
I'm new to this site, I've had a look around and there are alot of useful sections, particularly the section with math and science learning materials.
Anyway, I need to show that the following function is odd
[tex]f(x)=\left\{\begin{array}{ccc}
-\sin x&\mbox{ for }-\pi \leq x< \frac{-\pi}{ 2}\\
\sin x &\mbox{ for } \frac{-\pi}{2} \leq x \leq \frac{\pi}{2}\\
-\sin x &\mbox{ for } \frac{\pi}{2}<x<\frac{\pi}{2}
\end{array}\right.[/tex]

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


2. Relevant equations

I know an odd function is definded as [tex] f(-x) = -f(x)[/tex]

3. The attempt at a solution
In the interval
[tex]-\pi\leq x < {-\pi \over 2} \mbox{ if I substiture } -\pi \mbox{ it becomes }-\sin(-x) = -\sin[-(-{\pi \over 2})] = -\sin({\pi \over 2})[/tex]

Is that the correct way to solve it?
But I'm not sure how to show it's odd in the other intervals!
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zcd
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#2
Aug22-09, 05:24 PM
P: 200
If f(-x)=f(x), then the function is even. If f(-x)=-f(x), then the function is odd.
robphy
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#3
Aug22-09, 05:38 PM
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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.

mcfc
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#4
Aug22-09, 06:16 PM
P: 18

How to show a function is even/odd


Quote Quote by robphy View Post
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?
[tex]-\sin(-\pi) - \sin({\pi }) = 0 [/tex]

[tex] \sin({-\pi \over 2}) + \sin({\pi \over 2}) = 0[/tex]

[tex]-\sin({3 \pi \over 4}) - \sin({-3 \pi \over 4}) = 0[/tex]

do I need to show anything else?
mathie.girl
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#5
Aug22-09, 06:46 PM
P: 32
Quote Quote by mcfc View Post
To show it's odd:
look at values in the intervals?
[tex]-\sin(-\pi) - \sin({\pi }) = 0 [/tex]

[tex] \sin({-\pi \over 2}) + \sin({\pi \over 2}) = 0[/tex]

[tex]-\sin({3 \pi \over 4}) - \sin({-3 \pi \over 4}) = 0[/tex]

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 [tex]a[/tex] be a random value in each interval, and then look at [tex]f(a)[/tex] and [tex]f(-a)[/tex]. Since the intervals are symmetric, once you've assigned an interval for [tex]a[/tex], it will be obvious what interval [tex]-a[/tex] is in and therefore which definition of the function you need to use.


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