Express a function as a sum of even and odd functions

[chwala]

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...

 

[Office_Shredder]

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

 

[chwala]

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?

 

[pasmith]

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]

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...

 

[Office_Shredder]

...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]

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 .

 

[chwala]

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!