How to show a function is even/odd

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To demonstrate that the function f(x) is odd, it is essential to verify that f(-x) = -f(x) across all defined intervals. The function is defined piecewise, and examples provided show that values at specific points satisfy the odd function condition. However, it is crucial to generalize this by selecting a variable 'a' within each interval to prove the relationship holds for all values in those intervals. The symmetry of the intervals simplifies the process, allowing for a straightforward application of the function's definitions. Ultimately, a comprehensive proof requires confirming the odd function property for every value in the specified ranges.
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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:
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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.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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