Even and odd Functions question

[Tido611]

My prof said "every function is the sum of an even and an odd function, explain."

ive spent about 2 hours off and on thinking about this and i haven't come up with anything really.

is it because f(g(x)) is an even function if either f(x) or g(x) is even, and you can just split every function into simpler functions , one being even and the other odd? or is that just proving that every function is a product of an even and an odd?

 

[LeBrad]

Suppose you write f(x) as
f(x) = \frac{f(x)+f(-x)}{2} + \frac{f(x)-f(-x)}{2}
Can you say anything about the evenness or oddness of (f(x)+f(-x)) and (f(x)-f(-x))?

 

[Tido611]

is (f(x) + f(-x)) an even function because even function, f(x)= f(-x) and
is (f(x) - f(-x)) an odd function because even functions, f(x) = -f(x)?


im not really sure

 

[LeBrad]

First, do you agree that I can write
f(x) = \frac{f(x)+f(-x)}{2} + \frac{f(x)-f(-x)}{2}
for any f(x)?

Second, do you know the definition of even and odd functions?
For an even function g(x): g(x) = g(-x)
For an odd function g(x): g(-x) = -g(x)

Now let h(x) = f(x)+f(-x)
and let k(x) = f(x) - f(-x)

Check to see if h or k are even or odd.

 

[Tido611]

umm i agree with your second point but I am still not understanding the first one.

 

[LeBrad]

Ok, the point I'm trying to make is that for any f(x), I can always write

f(x) = \frac{f(x)+f(-x)}{2} + \frac{f(x)-f(-x)}{2}

Now I'm going to define

h(x) = \frac{f(x)+f(-x)}{2} \;,\;\; k(x) = \frac{f(x)-f(-x)}{2}

By our definitions of h and k, it is clearly true that
f(x) = h(x) + k(x)
So if you can show that h(x) is always even and k(x) is always odd, then you are done. To do this, apply the definition of even and odd functions.
i.e.
h(x) = 1/2(f(x) + f(-x))
h(-x) = 1/2(f(-x) + f(--x)) = 1/2(f(-x)+f(x)) = ?

Then do the same for k(x). If you find that h(x)=h(-x) or h(x)=-h(-x), then you can say h is even or odd.

 

[Tido611]

Very Nice, now every thing makes sense thank you soo much but the only thing I am still wondering is where did the the 1/2 come from?

h(x) is the even function and k(x) is the odd function right?


but that doesn't really show that all functions are a sum of...

 

[LeBrad]

A lot of times it helps to rewrite things by cleverly adding 0 or multiplying by 1.
Consider
f(x) = \frac{2}{2}f(x) = \frac{1}{2}(f(x)+f(x)) = \frac{f(x)}{2}+\frac{f(x)}{2} = \frac{f(x)}{2}+\frac{f(x)}{2} + 0 = \frac{f(x)}{2}+\frac{f(x)}{2} + \left( f(-x) - f(-x) \right) =

\frac{f(x)}{2}+\frac{f(x)}{2} + \frac{1}{2}\left( f(-x) - f(-x) \right) =

\frac{f(x)}{2}+\frac{f(x)}{2} + \frac{f(-x)}{2}-\frac{f(-x)}{2} = \left(\frac{f(x)}{2}+\frac{f(-x)}{2} \right) + \left(\frac{f(x)}{2} -\frac{f(-x)}{2} \right)

 

[Tido611]

HaHa(giddy laugh)

LeBrad you are a King, thank you very much for you help.

 

[HallsofIvy]

By the way, sine and cosine are already odd and even functions (respectively) but e^x is not. It's even and odd "parts"
are
\frac{e^x+ e^{-x}}{2}= cosh(x)
and
\frac{e^x- e^{-x}}{2}= sinh(x)