Expressing $f(x) = \frac{x-1}{x+1}$ as a sum of an even and odd function

Question

I am trying to write $$f(x) = \frac{1-x}{x+1}$$ as a sum of an even and odd function. One solution, though messy, is to use the following derivation:

We work backwards. Let $f = f_e + f_0$, where $f_e$ is even and $f_0$ odd. Then $f(x) = f_e (x) + f_0 (x)$ for all $x$ in the domain of $f$. But then $$ f(-x) = f_e (-x) + f_0 (-x) = f_e (x) - f_0 (x). $$ Adding, we eliminate $f_0 (x)$: $$ f(x) + f(-x) = 2f_e (x), $$ so $$ f_e (x) = \frac{f(x) + f(-x)}{2}. $$ Now, $f_0 = f(x) - f_e$. Then: $$ f_0 (x) = f(x) - \frac{f(x) + f(-x)}{2} = \frac{2f(x) - f(x) - f(-x)}{2} = \frac{f(x) -f(-x)}{2}. $$ One easily checks that $f_e$ is even and $f_0$ odd.

Using this derivation, I get: \begin{align*} \frac{1-x}{1+x} = \frac{1}{2}\left(\frac{1-x}{1+x} + \frac{1 + x}{1 - x} \right) + \frac{1}{2} \left(\frac{1-x}{1+x} - \frac{1+x}{1-x} \right). \end{align*} Assuming I haven't made an algebra mistake, this seems to work, but it's not elegant. I did some research into this and found another proposed solution, but I believe it to be faulty. First, I'll just present it. \begin{align*} \frac{1-x}{1+x} & = \frac{1-x}{1+x} \cdot \frac{1-x}{1-x} \\ & = \frac{1 - 2x + x^2}{1 - x^2} \\ & = \frac{x^2 + 1}{1 - x^2} + \frac{2x}{1 - x^2}. \end{align*} One then checks that the first function is even and the second is odd.

My problem with this solution is that it is not valid for every $x$ in the domain of $f$. By definition, the domain of $f$ is $\mathbb{R} \setminus \{-1\}$. But this solution, in multiplying by $\frac{1-x}{1-x}$, presupposes that $x \neq 1$, which does not seem to me to be allowed.

I'm back at square one, then, because the solution I found does not seem in any way elegant or natural, and I'm assuming there is some kind of a trick that I am missing. I tried polynomial long division, partial fractions, and so forth, and nothing, other than the above derivation, brought me any progress.

Answer 1

You do have a sign error in your derivation. Your calculation $$ f(x) \ = \ \frac{1-x}{x+1} \ \ \Rightarrow \ \ f(-x) \ = \ \frac{1 \ - \ (-x)}{(-x) \ + \ 1} \ = \ \frac{1 \ + \ x}{1 \ - \ x} $$ is correct, but one of your symmetric functions is affected: $ \ f_{odd} \ $ should be $ \frac{2x}{x^2-1} \ = \ -\frac{2x}{1-x^2} \ \ . $ (Interestingly, $ \ f(-x) \ = \ \frac{1}{f(x)} \ \ . $)

[Also, the function expression in your title does not match the one in your discussion; the even and odd functions for that would be $ \ f_{even} \ = \ \frac{ x^2 + 1}{ x^2 - 1} \ $ and $ \ f_{odd} \ = \ \frac{ 2x }{ 1 - x^2} \ $ . I am assuming you want the one under discussion here.]

We can check this by $$ \frac{1 \ + \ x^2}{1 \ - \ x^2} \ + \ \frac{-2x}{1 - x^2} \ \ = \ \ \frac{(1 \ - \ x)^2}{1 \ - \ x^2} \ \ = \ \ \frac{1 \ - \ x}{1 \ + \ x} \ \ . $$

I noticed the error when I initially made a plot of your functions; the graph here shows $ \ \frac{1 \ - \ x}{1 \ + \ x} \ $ in blue, and the even and (corrected) odd functions in green and red, respectively. As José Carlos Santos remarks, we cannot expect an asymmetric function with a vertical asymptote to be "reconstructed" perfectly by symmetric functions, since vertical asymptotes are arranged symmetrically about the $ \ y-$axis in both even and odd functions. So there is actually a "hole" in the domain of $ \ f_e \ + \ f_o \ $ at $ \ x = 1 \ $ that is not present in the original function. The best that can be managed in such a sum then is to have the limit of the sum be equal to $ \ f(1) \ . $ Indeed, while the two symmetric functions are not defined at $ \ x = 1 \ $ , and moreover, the limits of the individual functions do not exist as $ \ x \ $ approaches $ \ 1 \ , $ we do have $$ \ \lim_{x \ \rightarrow \ 1} \ \left[ \ \frac{1 \ + \ x^2}{1 \ - \ x^2} \ \ + \ \ \frac{-2x}{1 - x^2} \ \right] \ \ = \ \ \lim_{x \ \rightarrow \ 1} \ \ \frac{1 \ - \ x}{1 \ + \ x} \ \ = \ \ 0 \ \ . $$

This sort of situation occurs in other kinds of summing of functions where there is a discontinuity in either the "component" functions or in the function to be constructed. An example that comes to mind is when a Fourier series (an infinite sum of continuous trigonometric functions) must "mimic" a jump discontinuity.

Answer 2

In both cases, your solutions are undefined when $x=\pm1$. And that's natural. After all, $f(x)$ is undefined when $x=-1$ and, if you express $f(x)$ as $f_o(x)+f_e(x)$, with $f_o$ odd and $f_e$ even, then it is natural that at least one of the functions $f_o$ and $f_e$ is undefined when $x=-1$. But then, since $f_o(-x)=-f_o(x)$ and $f_e(-x)=f_e(x)$, if $f_o(x)$ is undefined when $x=-1$, then it is also undefined when $x=1$, and the same thing applies to $f_e(x)$.

Answer 3

Just to add a little more perspective. Others have explained why you should not be surprised that one or both of $f_o$ and $f_e$ are undefined at $+1$ as well as $-1$. However, this is not always the case.

Consider the function $f(x) = \frac{x^2-1}{x + 1}$. Like your function, this is defined on $\mathbb{R} \setminus \{-1\}$. However, it can be expressed as the sun of an odd function and an even function which are defined at $\{+1\}$ and $\{-1\}$. $f_o(x) = x$ and $f_e(x) = -1$.

In a sense, I have cheated since my function is $f(x) = x - 1$ in disguise except that it is not defined at $-1$. We could extend and clean up my function by saying that it is defined at $-1$ and that the value is $-2$. Borrowing a term that is more commonly used in complex analysis, this is a "removable singularity". We can get a nice function by giving it at a value at the missing point. "Nice" in this context means "continuous".

Your function is different in that the singularity is "essential". We can extend your function by defining a value at $-1$ but we cannot pick a value that will make it nice (continuous). In this case, one or both of $f_o$ and $f_e$ will be undefined at $-1$ and also $+1$.