| Let F(s) be your origional fraction with a partial fraction expansion
1 A B C
F(s) = --------------- = ----- + ----- + -----
(s-1)(s+2)(s+4) (s-1) (s+2) (s+4)
Consider the limit of (s-1)F(s) as s -> 1
1 (s-1) (s-1)
(s-1)F(s) = ---------- = A + B ----- + C -----
(s+2)(s+4) (s+2) (s+4)
This knocks off the B and C terms, leaving only the constant A, so
we find
A = 1/((1+2)(1+4)) = 1/15
If there is a repeated pole, then it can be handled by taking
derivatives.
This can also be justified by a contour integral.
- Jim
|
| Easy,
It is an old trick used to avoid listing of all the rational fraction in
catalog of functions. It is possible to write any rational fraction
-- i.e ratio of polynomials -- as a sum of fraction in which the
denominator is of degree at most 1. Furthermore those fraction can be
taken of the type 1/(x-a). If a transformation is linear -- such as
integral, derivative, and many other -- it is possible to deal with all
the rational fraction using only the case of x^n, the case 1/(x-a), and
some simple computaion rules. Usefull for the editor.
Take the fraction P(x)/Q(x) where P(x) and Q(x) are polynomials. The first
step is eliminate high degrees of P. A division provides two polynomials S(x)
and R(x) such that P = S*Q + R, and deg(R)<deg(Q). Therefore write
P/Q = (S*Q+R)/Q = S + R/Q, with deg(R)<deg(Q)
The next step is to factorise Q(x). Working in the complex plane will gives
you Q(x) = (x-x1)(x-x2)...(x-xq), where x1,x2,...xq are the roots of Q(x).
Then you can consider arbitrary numbers a,b,...c and compute
a/(x-x1) + b/(x-x2)+ ...+ c/(x-xq)
It happend that this quantity -- which is a rational fraction that it would
be of few interest to write in the general case, but is usually simple to
compute on practical applications -- can take any value of the form R/Q as
above. Thus it is allways possible to write a fraction in the form:
(*) P/Q = (S*Q+R)/Q = S + a/(x-x1) + b/(x-x2)+ ...+ c/(x-xq)
To compute the practical values of a,b,..c start from the result and finds
the values of a,b,...c. Reduce (*) to the same denominator. The denominator
is simply (x-x1)(x-x2)...(x-xq) = Q(x). And the numerator will look like
a(x-x2)...(x-xq) + b(x-x1)...(x-xq) + ... + c(x-x1)(x-x2)..(x-x[q-1])
Which is "easy" to compute and gives a polynomial in x whose coefficients
depend on a,b,..c. This polynomial must be equal to S(x). Therefore the
coefficients of the two polynomials must be equal. This gives a system of
equations in a,b,..c that will determine the values of a,b,..c. Note that
the system is linear -- which is NOT by chance.
Well, to tell the true, things are not so simple. The above system is not
solvable in the case where Q(x) has multiplie roots. If, for example, x1
is a double root we get
a/(x-x1) + b/(x-x1) = (a+b)/(x-x1)
and loose a degree of freedom. Rather consider
a/(x-x1) + b/(x-x1)^2 = (ax+(b-a*x1))/(x-x1)^2
Or more generaly
a/(x-x1) + b/(x-x1)^2 + ... + c/(x-x1)^m
if x1 is a root of multiplicity m.
Did I answer the question?
Alain
|
| Sorry, I think I answered the wong question.
Let restart from the moment you have to find a,b,..c from the equation:
a(x-x2)...(x-xq) + b(x-x1)...(x-xq) + ... + c(x-x1)(x-x2)..(x-x[q-1]) = R(x),
where x1,x2,..xq are the roots of Q(x)=(x-x1)(x-x2)...(x-xq).
The products in the above sum are the product of q-1 of the factors (x-xi).
Exactlyone is missing. Therefore, putting x=xi will zero all but one of
the terms. Doing this q times, once for each xi, will provide:
a = R(x1), b = R(x2),... c = R(xq)
Goodie, goodie. Unless you have silly multiple roots.
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