What Is the Partial Fraction Expansion of 1/[((a+s)*(1+b/s)^m)]?

[nikozm]

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.

 

[Mark44]

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.