# Method of Partial Fractions - simplifying by formula

• Dec 29th 2010, 02:53 PM
wonderboy1953
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.
• Dec 29th 2010, 03:02 PM
DrSteve
Schaum's probably means that some quadratics are irreducible over the reals. Of course, as you've pointed out, all quadratics are reducible over the complexes.
• Dec 29th 2010, 03:08 PM
wonderboy1953
Quote:

Originally Posted by DrSteve
Schaum's probably means that some quadratics are irreducible over the reals. Of course, as you've pointed out, all quadratics are reducible over the complexes.

Schaum's should have been more explicit. Still they're a good study aid.
• Jan 2nd 2011, 12:43 PM
Soroban
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.

$\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]

\end{array}$

• Jan 3rd 2011, 01:19 PM
wonderboy1953
Nice
Quote:

Originally Posted by Soroban

Here's a stunning pattern in Partial Fractions.

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

$\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]

\end{array}$

Beautiful pattern. A combo of Pascal's triangle (with alternating signs) plus an inverted factorial (I'm wondering if there's a deeper significance to this). Great discovery Soroban.
• Jan 3rd 2011, 01:39 PM
wonderboy1953
I guess it's time for me to solve the puzzles but first a correction. The general formula for a two-termed linear denominator is A = 1/(b - a) and B = 1/(a - b).

Regarding the puzzle, the general formula in both cases, is the same which is A = 1/(b - a)(c - a), B = 1/(a - b)(c - b), C = 1/(a - c)(b - c) so letting a = 1, b = 2 and c = 5, then the answer is A = 1/(2 - 1)(5 - 1) = 1/8, B = 1/(1 - 2)(5 - 2) = -1/3 and C = 1/(1 - 5)(2 - 5) = 1/12.(Clapping)
• Jan 3rd 2011, 05:41 PM
Soroban
Another partial fractions pattern

Here's another pattern . . .

. . $\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} \\ \\

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

Let's check the fourth equation.

On the right, get a common denominator:

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

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

Do not muiltiply it out!

Don't you recognize the numerator?

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

• Jan 4th 2011, 12:33 AM
Drexel28
Quote:

Originally Posted by Soroban

Here's a stunning pattern in Partial Fractions.

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

$\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]

\end{array}$

This is fairly well-known actually. Let

$\displaystyle P(x)=\sum_{k=1}^{m}(-1)^k{m\choose k}\prod_{\substack{j=0\\ j\ne k}}^m (x+j)$

It's evident that $\deg P\leqslant m$ yet it's easy to see that $P(0)=P(-1)=\cdots=P(-m)=m!$ from where it follows that $P(x)=m!$. Thus, dividing both sides by $\displaystyle m!\prod_{j=0}^{m}(x+j)$ gives $\displaystyle \left(\prod_{j=0}^{m}(x+j)\right)^{-1}=\frac{1}{m!}\sum_{k=1}^{m}{m\choose k}\frac{(-1)^k}{x+k}$ just as you noticed.
• Jan 4th 2011, 02:46 AM
Ackbeet
The OP is referring to what the EE's call the Heaviside cover-up method, and what the mathematicians call the residue method.
• Jan 4th 2011, 09:28 AM
TheCoffeeMachine
I think it's covered in most books (or rather at least those which I've seen) and if it's omitted it's probably
because it's judged to be rather limiting -- the cases where it works are only those which could be done
without any particular methods anyway; it doesn't work in where the method of partial fractions matters!
• Jan 4th 2011, 01:23 PM
wonderboy1953
You did it again
Quote:

Originally Posted by Soroban

Here's another pattern . . .

. . $\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} \\ \\

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

Let's check the fourth equation.

On the right, get a common denominator:

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

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

Do not muiltiply it out!

Don't you recognize the numerator?

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

This part: $\bigg[(x-a) + a\bigg]^4 \;=\;x^4$, is of special interest to me which I'll explain at another time.