Partial Fraction Expansion

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  • Thread starter nikozm
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Hi,

I would like to expand the following expression:

1/[((a+s)*(1+b/s)^m)], where a, b, and s are real nonnegative values and m is an arbitrary positive integer.

Particularly, according to partial fraction expansion, it becomes:

Sum[A_j/[(1+b/s)^j],{j,1,m}]+B/(a+s). I look for a closed-form expression of A_j and B.

Any help could be useful.

Thank you in advance.
 

Answers and Replies

  • #2
34,688
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Hi,

I would like to expand the following expression:

1/[((a+s)*(1+b/s)^m)], where a, b, and s are real nonnegative values and m is an arbitrary positive integer.

Particularly, according to partial fraction expansion, it becomes:

Sum[A_j/[(1+b/s)^j],{j,1,m}]+B/(a+s). I look for a closed-form expression of A_j and B.

Any help could be useful.
I don't know if there is such an expansion. Partial fraction decomposition is used when the denominator is the factorization of some polynomial. For your problem, the denominator is ##(a + s)(1 + \frac b s)^m## which is not a polynomial in s.

Consider a much simpler example. Can we decompose ##\frac 1 {(a + x)(b + 1/x)}##
That is, can we write ##\frac 1 {(a + x)(b + 1/x)}## as ##\frac A {a + x} + \frac B {b + 1/x}##?
If so, multiplying on both sides by (a + x)(b + 1/x) yields the equation ##1 = A(b + 1/x) + B(a + x)##, which must hold for all x for which the original denominator isn't zero.

Multiplying out the right side, we have ##1 = Ab + Ba + \frac A x + Bx##
In order for this equation to be identcally true, we must have 1 = Ab + Ba, ##\frac A x = 0##, and ##Bx = 0##, implying that A = B = 0, and Ab + Ba = 1.
 

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