Method of Partial Fractions - simplifying by formula

Consider http://www.mathhelpforum.com/math-he...a2c0c21206.png

The fraction can be broken down into http://www.mathhelpforum.com/math-he...704334880a.png + http://www.mathhelpforum.com/math-he...789ed62982.png with A and B to be determined.

You can solve this using algebra as it's normally taught. However since the denominator consists of linear factors, then there's a formula that says A = 1/(2-1) = 1 and B = 1/(1 - 2) = -1. More generally if you have http://www.mathhelpforum.com/math-he...242801c288.png , then A = (b - a) and B = (a - b).

Your first puzzle is given an expression http://www.mathhelpforum.com/math-he...362b8241b2.png, break it up into the individual fractions and determine what A, B and C are *without first using algebra* (you might see a pattern in connection with the previous example).

Your second puzzle is given the expression http://www.mathhelpforum.com/math-he...55d7c55a93.png, again break it up into the individual fractions and determine what A, B and C are in this case (please note I take exception with *Schaum's* that says some quadratics are irreducible since they all

are reducible as the last problem illustrates).

The point to this exercise is that it's far simpler to do these problems by formula than by algebra (also note that this particular method works with fractions with one as the numerator and only linear factors in the denominator).

Okay sharpen your pencils and go to work.

Partial Fractions: a pattern

Here's a stunning pattern in Partial Fractions.

I "discovered" it many years ago and a recent posting reminded me of it.

$\displaystyle \begin{array}{c}\;\;\dfrac{1}{n(n+1)} \;=\;\dfrac{1}{1!}\left(\dfrac{1}{n} - \dfrac{1}{n+1}\right) \qquad\;\; \\ \\[-3mm]

\dfrac{1}{n(n+1)(n+2)} \;=\; \dfrac{1}{2!}\left(\dfrac{1}{n} - \dfrac{2}{n+1} + \dfrac{1}{n+2}\right);\;\;\; \\ \\[-3mm]

\dfrac{1}{n(n+1)(n+2)(n+3)} \;=\;\dfrac{1}{3!}\left(\dfrac{1}{n} - \dfrac{3}{n+1} + \dfrac{3}{n+2} - \dfrac{1}{n+3}\right)\;\; \\ \\[-3mm]

\dfrac{1}{n(n+1)(n+2)(n+3)(n+4)} \;=\; \dfrac{1}{4!}\left(\dfrac{1}{n} - \dfrac{4}{n+1} + \dfrac{6}{n+2} - \dfrac{4}{n+3} + \dfrac{1}{n+4}\right)\\ \\[-3mm]

\vdots \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\q quad \vdots \qquad

\end{array}$

Another partial fractions pattern

Here's another pattern . . .

. . $\displaystyle \begin{array}{ccc}

\dfrac{x}{(x-a)^2} \;=\;\dfrac{1}{x-a} + \dfrac{a}{(x-a)^2} \\ \\

\dfrac{x^2}{(x-a)^3} \;=\;\dfrac{1}{x-a} + \dfrac{2a}{(x-a)^2} + \dfrac{a^2}{(x-a)^3} \\ \\

\dfrac{x^3}{(x-a)^4} \;=\;\dfrac{1}{x-a} + \dfrac{3a}{(x-a)^2} + \dfrac{3a^2}{(x-a)^3} + \dfrac{a^4}{(x-a)^4} \\ \\

\dfrac{x^4}{(x-a)^5} \;=\;\dfrac{1}{x-a} + \dfrac{4a}{(x-a)^2} + \dfrac{6a^2}{(x-a)^3} + \dfrac{4a^3}{(x-a)^4} + \dfrac{a^4}{(x-a)^5} \\ \\

\vdots \qquad\qquad\qquad\qquad\qquad\qquad\qquad \vdots \qquad\qquad\qquad\qquad\qquad\end{array}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Let's check the fourth equation.

On the right, get a common denominator:

$\displaystyle \displaystyle \frac{1}{x-a} + \frac{4a}{(x-a)^2} + \frac{6a^2}{(x-a)^3} + \frac{4a^3}{(x-a)^4} + \frac{a^4}{(x-a)^5} $

. . $\displaystyle \displaystyle =\;\frac{(x-a)^4 + 4a(x-a)^3 + 6a^2(x-a)^2 + 4a^3(x-a) + a^4}{(x-a)^5} $

Do *not* muiltiply it out!

Don't you recognize the numerator?

. . It is: .$\displaystyle \bigg[(x-a) + a\bigg]^4 \;=\;x^4$ . . . . . *Ha!*