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I want to do partial fractions on: $1/[(s^2+1)(s+1)]$

$(as+b)/(s^2+1) + c/(s+1)$

Multiplying both sides by $(s^2+1)(s+1)$, get $a=-1/2, b=1/2, c=1/2$

$(-s/2+1/2)/(s^2+1) + (1/2)/(s+1) = 1/[(s^2+1)(s+1)]$

Unfortunately when I add $(-s/2+1/2)/(s^2+1) + (1/2)/(s+1)$ I get $(1/4)/[(s^2+1)(s+1)]$.

Where did I go wrong?

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  • $\begingroup$ Can you explain how you did your last step of "when I add"? $\endgroup$ – Calvin Lin Feb 18 '14 at 9:34
  • $\begingroup$ [(-s/2+1/2)(s+1)+(1/2)(s^2+1)]/[(s^2+1)(s+1)] $\endgroup$ – user122415 Feb 18 '14 at 9:43
  • $\begingroup$ Yes, and I'd ask you to justify that expansion. For example, expand the numerator : $( - \frac{s}{2} + \frac{1}{2} ) ( s+1) + \frac{1}{2} (s^2 + 1)$. What do you get? $\endgroup$ – Calvin Lin Feb 18 '14 at 9:47
  • $\begingroup$ -s^2/2-s/2+s/2+1/2+s^2/2+1/2 = 1/4 $\endgroup$ – user122415 Feb 18 '14 at 9:51
  • $\begingroup$ Please triple check. For example, when you substitute in $s=0$, what do you get? $\endgroup$ – Calvin Lin Feb 18 '14 at 9:53
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No,when you add $$ \frac{-s/2+1/2}{s^2+1} + \frac{1/2}{s+1} $$ you get $$ \frac{1}{2}\frac{1-s^2+1}{(s^2+1)(s+1)} $$ or $$ \frac{1}{(s+1)(s^2+1)}. $$

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