# Express a function as a sum of even and odd functions

• chwala
In summary, the first line of the summary is f(x)=g(x)+ h(x), which is the equation for the even and odd functions. The second line is f(x)=\left[\dfrac{1}{2}\left[\dfrac{1}{2x-x^2}+\dfrac{1}{-2x-x^2}\right]\right] which is the equation for the sum of the even and odd functions. The last line is h(x)=\dfrac{1}{2}\left[\dfrac{1}{x(2-x)}+\dfrac{1}{x(2+x)}\right] which

#### chwala

Gold Member
Homework Statement
See attached questions (highlighted in Red)
Relevant Equations
Even and odd number concept
I am refreshing on this; of course i may need your insight where necessary...I intend to attempt the highlighted...this is a relatively new area to me...

For part (a),

We shall let ##f(x)=\dfrac{1}{x(2-x)}##, let ##g(x)## be the even function and ##h(x)## be the odd function. It follows that,

##f(x)=g(x)+ h(x)##

##g(x)=\dfrac{1}{2}\left[\dfrac{1}{2x-x^2}+\dfrac{1}{-2x-x^2}\right]##
##g(x) = \dfrac{1}{2}\left[\dfrac{2x+x^2-2x+x^2}{(2x-x^2)(2x+x^2}\right]##
##g(x)=\dfrac{1}{2}\left[\dfrac{2x^2}{(x^2(2-x)(2+x)}\right]##
##g(x)=\left[\dfrac{1}{(2-x)(2+x}\right]##

##h(x)=\dfrac{1}{2}\left[\dfrac{1}{x(2-x)}-\dfrac{1}{-x(2+x)}\right]##
##h(x)=\dfrac{1}{2}\left[\dfrac{1}{x(2-x)}+\dfrac{1}{x(2+x)}\right]##
##h(x)=\dfrac{1}{2}\left[\dfrac{1(2+x)+1(2-x)}{x(2-x)(2+x)}\right]##
##h(x)=\dfrac{1}{2}\left[\dfrac{2+x+2-x)}{x(2-x)(2+x)}\right]##
##h(x)=\dfrac{1}{2}\left[\dfrac{4}{x(2-x)(2+x)}\right]##
##h(x)=\left[\dfrac{2}{x(2-x)(2+x)}\right]##

therefore,

##\dfrac{1}{x(2-x)}=\dfrac{1}{(2-x)(2+x)}+\dfrac{2}{x(2-x)(2+x)} ##

Bingo! I will attempt the others later...

Do you even need to do all that simplification? Why not just let h and g be the first lines of what you wrote?

Office_Shredder said:
Do you even need to do all that simplification? Why not just let h and g be the first lines of what you wrote?
...you mean all under one denominator?

As these are all quotients, bear in mind that multiplying numerator and denominator by a factor chosen to make the denominator even can be more efficient than calculating $(f(x) \pm f(-x))/2$. So for (a), $$\frac{1}{x(2-x)} = \frac{x(2+x)}{x^2(4-x^2)} = \frac{2}{x(4-x^2)} + \frac{1}{4-x^2}.$$ To complete the answer, note that in addition to the points where the left hand side is not defined, the right hand side is not defined at $x = -2$.

chwala
pasmith said:
As these are all quotients, bear in mind that multiplying numerator and denominator by a factor chosen to make the denominator even can be more efficient than calculating $(f(x) \pm f(-x))/2$. So for (a), $$\frac{1}{x(2-x)} = \frac{x(2+x)}{x^2(4-x^2)} = \frac{2}{x(4-x^2)} + \frac{1}{4-x^2}.$$ To complete the answer, note that in addition to the points where the left hand side is not defined, the right hand side is not defined at $x = -2$.
Thanks...let me check on this simplified approach...

chwala said:
...you mean all under one denominator?

i mean

##f(x)=\left(\dfrac{1}{2}\left[\dfrac{1}{2x-x^2}+\dfrac{1}{-2x-x^2}\right]\right) +\left(\dfrac{1}{2}\left[\dfrac{1}{x(2-x)}-\dfrac{1}{-x(2+x)}\right]\right)##

chwala
Office_Shredder said:
i mean

##f(x)=\left(\dfrac{1}{2}\left[\dfrac{1}{2x-x^2}+\dfrac{1}{-2x-x^2}\right]\right) +\left(\dfrac{1}{2}\left[\dfrac{1}{x(2-x)}-\dfrac{1}{-x(2+x)}\right]\right)##
Thanks @Office_Shredder .

Actually, I can now see your point of having all in one line...pretty easy stuff...point is to just just deal with some algebra...I have the solutions, so I guess no need of continuing with the thread. Cheers guys!

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