## Generalizing to an arbitrary function.

Limits, differentiation, related rates, integration, trig integrals, etc.

### Generalizing to an arbitrary function.

$\mbox{Exercise 1A-4:$
$\mbox{a) Show that every polynomial is the sum of}$
$\mbox{an even and an odd function.}$

$\mbox{b) Generalize part (a) to an arbitrary function }\, f(x)$
$\mbox{by writing}$

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

$\mbox{Verify this equation, and then show that the two}$
$\mbox{functions on the right are respectively even and odd.}$

$\mbox{c) How would you write }\, \frac{1}{x\, +\, a}\, \mbox{ as the sum of}$
$\mbox{an even and an odd function?}$

$\mbox{Solution to 1A-4:}$
$\mbox{a) }\, p(x)\, =\, p_e(x)\, +\, p_o(x),\, \mbox{ where }\, p_e(x)\, \mbox{ is the sum of the}$
$\mbox{even powers and }\, p_o(x)\, \mbox{ is the sum of the odd powers.}$

$\mbox{b) }\, f(x)\, =\, \frac{f(x)\, +\, f(-x)}{2}\, +\, \frac{f(x)\, +\, f(-x)}{2}$

$F(x)\, =\, \frac{f(x)\, +\, f(-x)}{2}\, \mbox{ is even and }\, G(x)\, =\, \frac{f(x)\, -\, f(-x)}{2}$

$\mbox{is odd because}$

$F(-x)\, =\, \frac{f(-x)\, +\, f(-(-x))}{2}\, =\, F(x);$

. . . . .$G(-x)\, =\, \frac{f(-x)\, -\, f(-(-x))}{2}\, =\, -G(-x)$

$\mbox{c) Use part (b):}$

$\frac{1}{x\, +\, a}\, +\, \frac{1}{-x\, +\, a}\, =\, \frac{2a}{(x\, +\, a)(-x\, +\, a)}\, =\, \frac{2a}{a^2\, -\, x^2}\, \mbox{ even}$

$\frac{1}{x\, +\, a}\, -\, \frac{1}{-x\, +\, a}\, =\, \frac{-2x}{(x\, +\, a)(-x\, +\, a)}\, =\, \frac{-2x}{a^2\, -\, x^2}\, \mbox{ odd}$

$\Rightarrow \, \frac{1}{x\, +\, a}\, =\, \frac{a}{a^2\, -\, x^2}\, -\, \frac{x}{a^2\, -\, x^2}$

In the two above snippets I am wondering how "b)" was arrived at. In the question both parts of the equation show a plus and at "b)" one of the signs has changed to a minus symbol. I'm really not sure what is meant by generalizing to an arbitrary function as I am being asked to do.

original images:
https://skydrive.live.com/redir?resid=D ... gpF0ko&v=3
https://skydrive.live.com/redir?resid=D ... yVVJws&v=3
sepoto

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Joined: Thu Sep 19, 2013 3:38 am