How to show a function is even/odd

  • Thread starter Thread starter mcfc
  • Start date Start date
  • Tags Tags
    Function
Click For Summary

Homework Help Overview

The discussion revolves around demonstrating whether a given piecewise function is odd. The function is defined in specific intervals involving sine functions and periodicity. Participants are exploring the definitions and properties of odd functions in the context of this problem.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the definition of odd functions and explore specific values to test the function's oddness. There are attempts to substitute values from the defined intervals to verify the property.

Discussion Status

Some participants have provided variations of the definitions of even and odd functions. Others suggest that while specific values have been tested, a more general approach is needed to confirm the function's oddness across all intervals.

Contextual Notes

There is an emphasis on needing to show the property holds for all values in the defined intervals, not just selected points. The discussion also highlights the importance of understanding the symmetry of the intervals in relation to the function's definition.

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.
 

Similar threads

  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 40 ·
2
Replies
40
Views
5K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 12 ·
Replies
12
Views
2K
Replies
3
Views
2K