Even plus odd functions

[{~}]

I have been looking at my old calculus textbook because to my dismay I seem to have forgotten most of the calculus I learned. I am given 3 cases of $(f+g)(x)$.

Case 1 both f and g are even:
I know $f(x) = f(-x)$ and $g(x)=g(-x)$ for the domain of the function. I can reason by substitution that
$f(x)+g(x)=f(-x)+g(-x)$
$(f+g)(x)=(f+g)(-x)$
$(f+g)(x)$ is even. So far so good.

Case 2 both f and g are odd:
I found that if $f(x)=-g(x)+c$ then
$(f+g)(x)=c$ which is even.
Otherwise I think that $(f+g)(x)$ would be odd though I don't know how to assert that.

Case 3 f is even and g is odd:
I think that other than is special case where one or both of our functions are zero for all x in the domain $(f+g)(x)$ would neither be even or odd. I don't know how to prove this.

I know that the notions of even an odd is defined in terms of sets rather than algebraically like I did here. I think if I understood sets better I might have more of a handle on this, I don't know. Hints?

 

[DarthMatter]

If $f$ and $g$ are odd:

$(f+g)(-x)=f(-x)+g(-x)=-f(x)-g(x)=-(f+g)(x)$.

 

[{~}]

If $f$ and $g$ are odd:

$(f+g)(-x)=f(-x)+g(-x)=-f(x)-g(x)=-(f+g)(x)$.

I had something similar, but why does this look like it's always true when it isn't? Or is my special case wrong?

 

[DarthMatter]

What special case? $-g(x)+c$ is not odd in general.

 

[{~}]

What special case? $-g(x)+c$ is not odd in general.

Thanks for pointing out my error. That made things less confusing.

 

[jbunniii]

Case 3 f is even and g is odd:
I think that other than is special case where one or both of our functions are zero for all x in the domain $(f+g)(x)$ would neither be even or odd. I don't know how to prove this.

Suppose that $f$ is even and $g$ is odd.

If $f+g$ is even then $f(x) + g(x) = f(-x) + g(-x)$ for all $x$. Since $f(x) = f(-x)$, it follows that $g(x) = g(-x)$. But $g$ is odd, so also $g(x) = -g(-x)$. Therefore, $g(-x) = -g(-x)$, or equivalently $2g(-x) = 0$. This is true for all $x$, so $g$ is the zero function.

If $f+g$ is odd then $f(x) + g(x) = -f(-x) - g(-x)$ for all $x$. Since $g(x) = -g(-x)$, this is equivalent to $f(x) = -f(-x)$. But $f$ is even, so also $f(x) = f(-x)$. Therefore, $f(-x) = -f(-x)$, or $2f(-x) = 0$. This is true for all $x$, so $f$ is the zero function.

 

[FactChecker]

Case 3 f is even and g is odd:
I think that other than is special case where one or both of our functions are zero for all x in the domain (f+g)(x) would neither be even or odd. I don't know how to prove this.

In Case 3 you can not say anything about the sum of an even + odd function. Any function, h(x), can be represented as the sum of an even function and an odd function. So you can not state any conclusions about h.
( Define even(x) = (h(x) + h(-x))/2 and odd(x) = (h(x) - h(-x))/2. Prove that h(x) = even(x) + odd(x), that even(x) is even, and that odd(x) is odd.)