Partial fraction, a shorter method need if possible

1. Apr 19, 2007

Mathgician

1. The problem statement, all variables and given/known data
1/[(x^2)(x-3)^2]

I know that this can be separated by the long way of finding A,B,C,D constants and etc... But, I was curious if there is a shorter way to separate it by inspection like how you would do if you were using "heavyside" method for distinctive first order multiples.

2. Apr 19, 2007

Dick

I don't see why you should think that there is an easy way. Partial fractions IS the 'easy way'. If you were given an unfactored quartic, you probably couldn't do it at all. Feel grateful.

Last edited: Apr 19, 2007
3. Apr 20, 2007

HallsofIvy

Staff Emeritus
It's hard to say if there is a "simpler" method when you don't say what method you are using!

"Partial fractions" says there exist numbers A, B, C such that
$$\frac{1}{(x-2)(x-3)^2}= \frac{A}{x-2}+ \frac{B}{x-3}+ \frac{C}{(x-3)^2}[/itex] One method of finding finding A, B, and C is to add the fractions on the right, then equate coefficients of like powers of x in the numerators. I would call that the "hard" way. Better is to get rid of the fractions by multiplying both sides of the equation by (x-2)(x-3)2. That gives 1= A(x-3)2+ B(x-2)(x-3)+ C(x-2). Now you could multiply out that right side and equate coefficients of like powers. I would call that the "medium" way. Since this is equation, 1= A(x-3)2+ B(x-2)(x-3)+ C(x-2), is to be true for all x, choosing three values of x will give you three equations for A, B, and C. Do you see that choosing x= 2 and then x= 3 will give particularly easy equations? Since you don't have three distinct linear factors, there is no third value of x that will reduce so easily but you already know two of the values and taking x to be any simple number, say x= 0 or x= 1, will give you a fairly easy equation for the third number. 4. Apr 20, 2007 Gib Z I think you read the OPs original question wrong Halls :( It was $x^2$, not x-2. 5. Apr 20, 2007 Mathgician I know what you mean, I am not saying I am looking for something other than separating by partial fractions. What I am saying is, if there is a way of solving by partial fractions in a quicker way by inspection as you can do that for the "heavyside" method. Are you familiar with the heavyside method? 6. Apr 20, 2007 Dick No. I guess that's why I'm not getting your question. 7. Apr 20, 2007 Mathgician For the heavyside method, If you plug in for a value of x for each factor that would make it a zero, and then cover that factor up only, then plug in for the part that is not covered. You will get the constant for that factor fraction. Then you do the same for the other. It is very useful method if you already know how to do it the long way (you are getting the fractions practically by inspection), but you do not want to take the time. But, the only downside is that it only works for distinct first order factors. 8. Apr 20, 2007 Mathgician What do you mean couldn't do it at all? Are you insulting me because I don't have years of math experience? I will tell you something buddy, I've been out of it for 8 years when I got back to taking college algebra, took calculus I in the summer, never dropped any of my math classes, I am finishing my calc III right now. Don't insult me because you have more experience than I do, I can learn quick! Are you a teacher? I know many of the people are teachers in here. Thats is funny because if you are, all you do is the same crap all over again and again, you got to be retarded not to know stuff. Is Dick your real name? Don't come in here thinking you are some brilliant bla bla bla... If you want to be brilliant grow some imagination. Doing the same thing over and over and blabbing about what they know is for people that don't have one. Last edited: Apr 20, 2007 9. Apr 20, 2007 HallsofIvy Staff Emeritus I believe that is what I did, missing the square on the x, of course. For higher powers, or irreducible quadratics, there is no equally simple method but you can, of find some of the values that way. Then just plug in simple number to get equations for the rest. I saw no insult. Dick asked why you thought there might be a simpler method. There is no harm in asking for a simpler method nor is there any is suggesting, as Dick did, that there was no simpler method (complicated by the fact that you didn't say what method you ARE using). Certainly, when he said "If you were given an unfactored quartic, you probably couldn't do it at all. " he was not insulting you. I probably couldn't factor a general quartic- I don't feel insulted by that. 10. Apr 20, 2007 Dick I apologize for not anticipating how you would take my flip answer. The correct way to phrase that would have been "if it were an unfactored quartic then WE probably couldn't do it at all. WE should feel grateful that it is factored and there is something as easy as partial fractions." I had no intention of questioning your capabilities. Sorry again. In fact your reply got me curious about this "heavyside" method. Which I hadn't heard of. 11. Apr 20, 2007 Mathgician I made a blunder, its called "Heaviside cover-up method," popularized by an electrical engineer Oliver Heaviside (1850-1925) 12. Apr 20, 2007 Mathgician I apologize for taking your message personally. I am still learning right now, as you probably are too, but I am not very experienced. I was just curious because the Heaviside method probably didn't come from saying it is something is impossible, I was thinking with a little creativity someone has come up with a method for separating complex partial fraction that can be factored in a much efficient way. 13. Apr 20, 2007 Päällikkö I think by the heaviside Mathgician means something like eg. (I'm in a hurry and just throw an example without much mathematical rigor or generality) [tex]F := \frac{1}{(x-2)(x-3)^2}= \frac{A}{x-2}+ \frac{B}{x-3}+ \frac{C}{(x-3)^2}$$

Then:
$$A = \lim_{x \to 2}F(x-2)$$

$$B = \lim_{x \to 3}\left( \frac{d}{dx} \left(F(x-3)^2\right) \right)$$

$$C = \lim_{x \to 3}F(x-3)^2$$

EDIT: Damn I'm slow. Now that the method's name has come up, I suppose it's easy to google. Well, off I go, I'll leave the above undeleted although it's a bit unnecessary.

14. Apr 20, 2007

Dick

Hey, that is a bit faster than the brute force approach. And that's how you extend it past distinct first order factors. Very clever. Learn something new every day...