Proof of an odd function plus an even function

I was looking at the wiki page http://en.wikipedia.org/wiki/Even_and_odd_functions#The_sum_of_even_and_odd_functions and it says that to prove an even function plus an odd function, we first have to re-write the f(x) as

$f(x)$ as $f(x)/2 + f(x)/2 + f(-x)/2 - f(-x)/2$

Why has f(x) been written like this? For example, where on earth did "+ f(-x)/2" and "- f(-x)/2" come from? I have my own idea as to where these functions have come from, and you can read my understanding of it below (and tell me if my thinking is correct or incorrect). Secondly, I don't see why they have re-written the functions as being divided by two. Why have they done that? I feel that I could happily re-write the function f(x) as
$f(x)$ as $f(x)/ + f(x)/ + f(-x)/ - f(-x)/$ But would that be incorrect? If so, why?

My thoughts to my questions:

Is each function f(x) divided by 2 because f(x) is a function for all real numbers? If not, why is each function divided by 2 in this proof?

Secondly, why is the function of f(x) re-written in that form? Is it because:

an even function is $f(x)/2 = f(-x)/2$ which is $0 = f(x)/2 - f(-x)/2$

and an odd function is $f(x)/2 = -f(-x)/2$ which is $0 = f(x)/2 + f(-x)/2$

and these added together makes the 're-written f(x) function' $f(x)$ as $f(x)/2 + f(x)/2 + f(-x)/2 - f(-x)/2$ ?

If not, please do try to explain so I can understand this proof. Thanks

• Proof of what?${}$ Commented Sep 28, 2013 at 16:21
• Proof of an even function plus an odd function? Commented Sep 28, 2013 at 16:28
• Do you mean that any function can be expressed as the sum of an even function and an odd function? Commented Sep 28, 2013 at 16:30
• @GitGud I interpreted it as "Explain the above proof of why every function can be written as odd + even". Commented Sep 28, 2013 at 16:38
• Also, you should care for the definition domain. For example, $\large{\rm f}\left(x\right) = \ln\left(x\right)$. Commented Sep 28, 2013 at 16:38

If $f(x)$ is not defined in a symmetric domain, we extend $f(x)$ such that it takes on values of 0 otherwise.

Suppose that $f(x) = g(x) + h(x)$, where $g(x)$ is an even function and $h(x)$ is an odd function. Then,

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

Hence, we may solve for $g(x), h(x)$ to obtain

$$g(x) = \frac{f(x)}{2} + \frac{ f(-x) } { 2} , h(x) = \frac{f(x)}{2} - \frac{ f(-x) } { 2}$$

This approach motivates the function, and it remains to check that they satisfy the conditions. We can easily check that $g(x)$ is indeed an even function, and $h(x)$ is an odd function, and $f(x) = g(x) + h(x)$, in the (extended) domain of $f(x)$, and in particular, the original domain of $f(x)$.

The idea of the proof you presented is that -

1) $f(x)$ can be rewritten in the following way

$$f(x) = [\frac{f(x)}{2} + \frac{ f(-x) } { 2} ] + [ \frac{f(x)}{2} - \frac{f(-x)} { 2} ].$$

This follows by just cancelling terms on the RHS. $_\square$

2) Show that the first term is an even function, and the second term is an odd function.

Let $g(x) = \frac{f(x)}{2} + \frac{ f(-x) } { 2}$. Then $g(-x) = \frac{f(-x)}{2} + \frac{ f(-(-x) ) } { 2} = \frac{f(x)}{2} + \frac{ f(-x) } { 2} = g(x)$. Hence it is even.
Do the same for $h(x) = \frac{f(x)}{2} - \frac{ f(-x) } { 2}$ to show that it is odd.$_\square$

Of course, this doesn't explain where we plucked those functions from.

• @GitGud Ah, I didn't see that interpretation. Let me add it in. Commented Sep 28, 2013 at 16:33