How to break it into simple fractures

[electron2]

[tex]\frac{1}{s(s^2-1)}=\frac{A}{s}+\frac{b}{s-1}+\frac{C}{s+1}[/tex]

i get A=-1
B=-C
A+B+C=0

where is the mistake

 

[Cyosis]

We cannot find your mistake if you don't show us your entire solution.

 

[Mark44]

How do you know you made a mistake? What did you get for B and C? After you find these values, you should be able to check that the expression on the right side (with numbers substituted for A, B, and C) is equal to 1/(s(s^2 - 1)). If the two expressions are identically equal then your values for A, B, and C are correct. If not, then you made a mistake.

 

[Cyosis]

The problem lies in the three expressions he has gotten so far. If A=-1 and B=-C then A+B+C=A=0=-1.

 

[Count Iblis]

You can do it faster, without solving any equations, by expanding around the poles. A rational function f(z) with a pole at z = p can be expanded as:

a/(z-p)^n + b/(z-p)^(n-1) + ...

If you add up all the singular terms of all the expansions around all the poles, and call that function you get g(z), then the difference:

h(z) = f(z) - g(z)

doesn't have any poles anymore, therefore it must be a polynomial. If
f(z) tends to zero at inifinty, then so does h(z), because g(z) also tends to zero to infinity. But because h(z) is a polynomial, that means that h(z) = 0, therefore f(z) = g(z).

In the given case, the expansions around the three poles is very easy to obtain. Around s = 0, we have:

f(s) = 1/s [expansion of 1/(s^2 - 1) around s = 0] =

1/s [-1 + O(s)] = -1/s + nonsingular terms.

Around s = 1:

f(s) = 1/(s-1) [expansion of 1/s 1/(s+1) around s = 1] =

1/2 1/(s-1) + nonsingular terms

Around s = -1:

f(s) = 1/(s+1) [expansion of 1/s 1/(s-1) around s = -1] =

1/2 1/(s+1) + nonsingular terms

So, we have:

f(s) = -1/s + 1/2 [1/(s-1) + 1/(s+1)]

 

[nickmai123]

[tex]\frac{1}{s(s^2-1)}=\frac{A}{s}+\frac{b}{s-1}+\frac{C}{s+1}[/tex]

i get A=-1
B=-C
A+B+C=0

where is the mistake

Your mistake isn't where you solved for A. It's that you wrote your equations incorrectly. Check here:

[tex]\frac{1}{s(s+1)(s-1)}=\frac{A(s^{2}-1)+B(s^{2}+s)+C(s^{2}-s)}{s(s+1)(s-1)}[/tex]

Now redo your system and you will get A = -1. However, your 2nd and 3rd equations will be different.

Hint: [tex]B \neq -C[/tex]

 

[nickmai123]

You can do it faster, without solving any equations, by expanding around the poles. A rational function f(z) with a pole at z = p can be expanded as:

a/(z-p)^n + b/(z-p)^(n-1) + ...

If you add up all the singular terms of all the expansions around all the poles, and call that function you get g(z), then the difference:

h(z) = f(z) - g(z)

doesn't have any poles anymore, therefore it must be a polynomial. If
f(z) tends to zero at inifinty, then so does h(z), because g(z) also tends to zero to infinity. But because h(z) is a polynomial, that means that h(z) = 0, therefore f(z) = g(z).

In the given case, the expansions around the three poles is very easy to obtain. Around s = 0, we have:

f(s) = 1/s [expansion of 1/(s^2 - 1) around s = 0] =

1/s [-1 + O(s)] = -1/s + nonsingular terms.

Around s = 1:

f(s) = 1/(s-1) [expansion of 1/s 1/(s+1) around s = 1] =

1/2 1/(s-1) + nonsingular terms

Around s = -1:

f(s) = 1/(s+1) [expansion of 1/s 1/(s-1) around s = -1] =

1/2 1/(s+1) + nonsingular terms

So, we have:

f(s) = -1/s + 1/2 [1/(s-1) + 1/(s+1)]

Haha I think you might have lost him here, as you lost me for a second. The equations become trivial in this case, so it is a bit easier to just set up the system and stare at it. I've never seen that before, but now I'll remember it as an alternative to doing stupid systems.