# partial fraction did wrongly...

• Dec 2nd 2009, 05:32 AM
xcluded
partial fraction did wrongly...
Hi pals , possible to check where did i went wrong ?

2 / s(1 + s^2) = A/s + B/(1 + s^2)

then , 2 = A(1 + s^2) + Bs

when s = 0 , A = 2
when s = 1 and A = 2 , B = -2

however i could not get B = -2s

(Crying)
• Dec 2nd 2009, 05:50 AM
Jester
Quote:

Originally Posted by xcluded
Hi pals , possible to check where did i went wrong ?

2 / s(1 + s^2) = A/s + B/(1 + s^2)

then , 2 = A(1 + s^2) + Bs

when s = 0 , A = 2
when s = 1 and A = 2 , B = -2

however i could not get B = -2s

(Crying)

Tt should be

$\frac{2}{s(s^2+1)} = \frac{A}{s} + \frac{Bs+C}{s^2+1}$.
• Dec 2nd 2009, 05:54 AM
xcluded
Quote:

Originally Posted by Danny
Tt should be

$\frac{2}{s(s^2+1)} = \frac{A}{s} + \frac{Bs+C}{s^2+1}$.

hmm why Bs + C ?
Sorry , i'm pretty poor in these. (Worried)
• Dec 2nd 2009, 06:03 AM
HallsofIvy
Quote:

Originally Posted by xcluded
hmm why Bs + C ?
Sorry , i'm pretty poor in these. (Worried)

Because the denominator is an "irreducible" (unfactorable in terms of real numbers) quadratic. If x- a is a factor of the denominator, the partial fractions expansion will include $\frac{A}{x- a}$. If $x^2+ cx+ d= (x-a)^2+ b$, so that it is irreducible, then the partial fractions expansion will include $\frac{Ax+ B}{(x-a)^2+ b}$. If the denominator has $(x-a)^n$ as a factor, the partial fraction expansion will include $\frac{A_1}{x-a}+ \frac{A_2}{(x-a)^2}+ \cdot\cdot\cdot+ \frac{A_n}{(x-a)^n}$. Finally, if the denominator has $(x^2+ a)^n$ as a factor, the partial fraction expansion will include $\frac{A_1x+ B_1}{x^2+ a}+ \frac{A_2x+ B_2}{(x^2+a)^2}+$ $\cdot\cdot\cdot+$ $\frac{A_nx+ B_n}{(x^2+a)^n}$.

Since any polynomial, with real coefficients, can be factored (in the real numbers) to powers of linear or powers of quadratic terms, those are all the terms you need.