Results 1 to 11 of 11

Math Help - Method of Partial Fractions - simplifying by formula

  1. #1
    Banned
    Joined
    Oct 2009
    Posts
    769

    Method of Partial Fractions - simplifying by formula

    Consider

    The fraction can be broken down into + 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 , then A = (b - a) and B = (a - b).

    Your first puzzle is given an expression , 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 , 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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Nov 2010
    From
    Staten Island, NY
    Posts
    451
    Thanks
    2
    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.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Banned
    Joined
    Oct 2009
    Posts
    769
    Quote Originally Posted by DrSteve View Post
    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.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,861
    Thanks
    742

    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]<br /> <br />
\dfrac{1}{n(n+1)(n+2)} \;=\; \dfrac{1}{2!}\left(\dfrac{1}{n} - \dfrac{2}{n+1} + \dfrac{1}{n+2}\right);\;\;\; \\ \\[-3mm]<br /> <br />
\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]<br /> <br />
\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]<br /> <br />
\vdots \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\q  quad \vdots \qquad<br />
 \end{array}

    Follow Math Help Forum on Facebook and Google+

  5. #5
    Banned
    Joined
    Oct 2009
    Posts
    769

    Nice

    Quote Originally Posted by Soroban View Post

    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]<br /> <br />
\dfrac{1}{n(n+1)(n+2)} \;=\; \dfrac{1}{2!}\left(\dfrac{1}{n} - \dfrac{2}{n+1} + \dfrac{1}{n+2}\right);\;\;\; \\ \\[-3mm]<br /> <br />
\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]<br /> <br />
\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]<br /> <br />
\vdots \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\q  quad \vdots \qquad<br />
 \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.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Banned
    Joined
    Oct 2009
    Posts
    769

    The answer

    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.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,861
    Thanks
    742

    Another partial fractions pattern


    Here's another pattern . . .


    . . \begin{array}{ccc}<br />
\dfrac{x}{(x-a)^2} \;=\;\dfrac{1}{x-a} + \dfrac{a}{(x-a)^2} \\ \\<br />
\dfrac{x^2}{(x-a)^3} \;=\;\dfrac{1}{x-a} + \dfrac{2a}{(x-a)^2} + \dfrac{a^2}{(x-a)^3} \\ \\<br />
\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} \\ \\<br />
\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} \\ \\<br />
\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 \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!

    Last edited by Soroban; January 4th 2011 at 05:43 AM.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by Soroban View Post

    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]<br /> <br />
\dfrac{1}{n(n+1)(n+2)} \;=\; \dfrac{1}{2!}\left(\dfrac{1}{n} - \dfrac{2}{n+1} + \dfrac{1}{n+2}\right);\;\;\; \\ \\[-3mm]<br /> <br />
\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]<br /> <br />
\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]<br /> <br />
\vdots \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\q  quad \vdots \qquad<br />
 \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.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    The OP is referring to what the EE's call the Heaviside cover-up method, and what the mathematicians call the residue method.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Super Member
    Joined
    Mar 2010
    Posts
    715
    Thanks
    2
    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!
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Banned
    Joined
    Oct 2009
    Posts
    769

    You did it again

    Quote Originally Posted by Soroban View Post

    Here's another pattern . . .


    . . \begin{array}{ccc}<br />
\dfrac{x}{(x-a)^2} \;=\;\dfrac{1}{x-a} + \dfrac{a}{(x-a)^2} \\ \\<br />
\dfrac{x^2}{(x-a)^3} \;=\;\dfrac{1}{x-a} + \dfrac{2a}{(x-a)^2} + \dfrac{a^2}{(x-a)^3} \\ \\<br />
\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} \\ \\<br />
\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} \\ \\<br />
\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 \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.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Simplifying fractions.
    Posted in the Algebra Forum
    Replies: 3
    Last Post: September 4th 2010, 05:08 PM
  2. Simplifying fractions.
    Posted in the Algebra Forum
    Replies: 5
    Last Post: May 14th 2010, 10:59 PM
  3. Replies: 0
    Last Post: April 28th 2010, 10:53 AM
  4. simplifying fractions :P
    Posted in the Calculus Forum
    Replies: 0
    Last Post: February 24th 2010, 04:49 PM
  5. Simplifying fractions
    Posted in the Algebra Forum
    Replies: 2
    Last Post: November 9th 2009, 06:34 PM

Search Tags


/mathhelpforum @mathhelpforum