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I need to decompose (1/((s - 2)^(2) + 1^(2))) into partial fractions, but I am not sure how exactly.

Here are my attempts:

Attempt 1

(1/((s - 2)^(2) + 1^(2))) =

(A/(s - 2)) + (B/(s - 2)^(2)) + (C/(1)) + (D/(1^(2))) =

(A/(s - 2)) + (B/(s - 2)^(2)) + C + D

I only get (1/((s - 2)^(2))) from this, I am missing the "+ 1^(2)" part.

Attempt 2

(1/((s - 2)^(2) + 1^(2))) =

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

(A/(s - 2)) + (B/(s - 2)^(2)) + C

I'm obviously messing up somewhere. My original denominator is s^(2) - 4s + 5 in which I completed the square and converted it to (s - 2)^(2) + 1^(2). I am unable to find any partial fraction examples using the case of having the extra "+ 1^(2)" remaining.

Question Summarized: How can I correctly decompose a completed square like the one in my example into partial fractions?

Edit: In my case, I didn't have to use partial fractions after all in this problem as it was in a Laplace transform form. However, I marked the answer as correct in case it helps anyone in the future.

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If you are working only with real numbers, you can't decompose that expression.

Are you trying to find the inverse Laplace transform of that expression? If so, completing squares is the first step, so you're on the way. Then you need to apply the Laplace translation theorem. You can read about it here: http://mathfaculty.fullerton.edu/mathews/c2003/LaplaceShiftingMod.html .

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  • $\begingroup$ This is the correct solution. My fellow classmate pointed this out as well. I'll mark this as the accepted answer. $\endgroup$ – Khaltazar Jun 24 '15 at 22:35
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$$\frac 1 {(s - 2)^2 + 1^2}$$ I wonder why you're trying to put this into partial fractions form? If it's for any purpose that should not involve complex numbers, then you're done: it's already in partial fractions form and has only one fraction. Remember the forms:

$$ \frac A {\text{1st-degree polynomial}},\quad \frac{B}{(\text{1st-degree polynomial})^2},\quad \frac{C}{(\text{1st-degree polynomial})^3},\quad\ldots $$ $$ \frac {Ax+B}{\text{irreducible 2nd-degree polynomial}}, \qquad \frac {Cx+D}{(\text{irreducible 2nd-degree polynomial})^2}, \qquad \ldots $$ Etc.

You already have an irreducible 2nd-degree polynomial as your denominator and $0x+1$ as your numberator. You're done.

If you had $(s-2)^2 - 5^2$ then it would not be irreducible, since it would be $\Big((s-2)-5\Big)\Big((s-2)+5\Big) = (s-7)(s+3)$. But if you have $(s-2)^2 + 5^2$ that's irreducible unless you bring in complex numbers, in which case you'd get $(s-2-5i)(s-2+5i)$ and then you could go on to partial fractions, which might involve imaginary numbers in the numerator.

If this is for the purpose of finding an indefinite integral, then you have $$ \int\frac{ds}{(s-2)^2+1} = \int\frac{dt}{t^2+1} = \arctan t+C = \arctan(s-2)+C. $$

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are you working with complex numbers? If not then I suggest that you had a typo and should be using $s^2-4s-5$ which has an easier factorization.

but if you using complex numbers then we can do this $$ \frac{1}{(s-a)^2+b^2} = \frac{A}{(s-2)+ib} + \frac{B}{(s-2)-ib} $$ solving the above we find $$ A(s-2-ib) + B(s-2+ib) = 1 $$ this leads to $$ A(-2ib) = 1\to A = \frac{1}{2b}i\\ B(2ib) = 1 \to B = -\frac{1}{2b}i $$ we get $$ \frac{1}{(s-a)^2+b^2} = \frac{i}{2b}\left[\frac{1}{(s-2)+ib} - \frac{1}{(s-2)-ib}\right] $$

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  • $\begingroup$ Okay, that is good to know in the future, but I asked another student in my class that happened to stop by and he said we can skip the partial fraction step in this problem as it is already in a Laplace transform form. The result ended up being e^(2t) * sin(t). Thanks for your answer. I'll give it a thumbs up so it can potentially assist others and note this in the question. $\endgroup$ – Khaltazar Jun 24 '15 at 22:31
  • $\begingroup$ @Khaltazar : You might also want to look at the other answer posted here. $\endgroup$ – Michael Hardy Jun 24 '15 at 22:41

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