# Even plus odd functions

• {~}
In summary, it seems that in Case 3, f is even and g is odd, but you can't say anything else about the sum of the function.f

#### {~}

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?

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?

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.

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.

Last edited:
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.)

Last edited: