show that if f is any function, then the function O defined by

O(x)=f(x)-f(-x)all over 2 is odd

Aren't there some restrictions on the sort of function f(x) is? Suppose f(x)=4. Then f(-x)=4 also, and (f(x)-f(-x))/2 = (4-4)/2 = 0, and that ISN'T odd. Another one: put f(x)=x², then f(-x)=x², and (f(x)-f(-x))/2=0 again. Same problem: something's wrong here.

Also consider f(x)=sin(x). f(-x)=sin(-x)=-sin(x), so O(x)=(sin(x)+sin(x))/2=sin(x) - and for most values of x that isn't even an integer, let alone an odd one.

To show that the function O(x) = (f(x) - f(-x))/2 is odd, we need to verify that O(-x) = -O(x) for any input value x.

Let's substitute -x into the function O(x):

O(-x) = (f(-x) - f(--x))/2
= (f(-x) - f(x))/2
= -(f(x) - f(-x))/2
= -O(x)

As we can see, O(-x) = -O(x) for all values of x. This means that the function O(x) is odd.

To understand this result, we can analyze the properties of odd functions. An odd function is symmetric with respect to the origin, which means that if we reflect the graph of the function across the y-axis, it remains unchanged. Mathematically, if f(x) is odd, then f(-x) = -f(x) for all x.

In our case, O(x) = (f(x) - f(-x))/2. By substituting -x into the function, we obtain O(-x) = (f(-x) - f(x))/2. As we can see, when we compare O(x) and O(-x), they both have the form (f(x) - f(-x))/2, but with opposite signs, indicating the symmetry around the origin. Therefore, O(x) is an odd function.

You can use this information to prove the oddness of the function O(x) for any given function f(x), by following the steps explained above.