Posted by Soroban on September 13, 2002 at 16:45:34:
In Reply to: Yes I have used that method.... posted by Subhotosh Khan on September 13, 2002 at 15:10:57:
: That method is simple as long as we do not have repeated roots for the denominators (I suppose that's why chaste-in-heart-mathematicians do not like this method).Suppose you had:
: f(n)/[n*(n-a)^2] this has to be "broken up" as
: A/n + B/(n-a) + (Cn + D)/(n-a)^2
: Now problem gets complicated (for n=a)......
Now, we are different worlds, Mr. S!
For "repeated linear factors", I know this method...
f(n)/n(n-a)^2 = A/n + B/(n-a) + C/(n-a)^2
That is, for (n-a)^k, we have a fraction "for each power" (1 to k).
I agree that there are "problems" when n = a, but I maintain that
it's still simpler to use the "illegal" values of n for solving.
Let n = a, and we instantly have the value of A.
Let n = 0 and let n = any (-a?), and we have a system of two equations.
And we have the two equations immediately., without expanding expressions,
then collecting coefficients to set up the system.
I have no wish to start a rivalry or a war, Mr. S ~ I learned the
"Equal Coefficients" method in college. It was tedious, but correct.
While teaching, I saw the "Zero Method" in several texts.
I was stunned by its simplicity and speed. I taught it that way
ever since. However, I did warn my students that we were breaking
some Sacred Laws ~ and I showed them one example of the Equal
Coefficients method, in case they later had a "purist" professor.
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