Posted by David on Tuesday, March 25, 2008 at 9:43pm.
I made this question, just wanted to make sure it didn't drown under the tide...
I've set a problem up, something like this.
-It's easier to substitute for s the values that make the denominators zero and then demand equalty. You then get simpler equations than comparing the coefficients of equal powers of s on both sides.-
Would anybody mind showing it out? I tried what Count Iblis said, but it didn't really work out well...
Then, Count Iblis said this...
If I give you the answer (I just verified that it is correct):
1/(s-1)^3 - 1/2 1/(s-1)^2 +
Can you verify that adding up the partial fractions yields the original fraction?
Can you extract your A, B, C, etc. from the solution and verify that it satisfies your equations?
Now, you have to admit that just writing down the equations by equating equal powers of s and solving them is not an effient way to solve the problem. It is perhaps a good algebra exercise, but I can assure you that people like me who have to do this for their work usually do not use this awkward method. only in somple cases when you have just two terms is the method useful.
There is a slight shortcut when you solve partial fractions by introducing the unknown constants: You add up the partial fractions so that you have a common denominator and then equate the two numerators as you've done.
You can insert s = 1, s = i and s = -i and demand that the two numerators are the same for these three cases. If you do that then you lose some terms so the equations becomes simpler Also, you can demand that the first and second derivatives of both sides are the same at s = 1.-
I'm sorta confused at this point. I just can't find the A, B, C, D, and E.
Calculus - Integrals - mclovin', Tuesday, March 25, 2008 at 9:53pm
O_O OH DEAR GOD.
Calculus - Integrals - Count Iblis, Tuesday, March 25, 2008 at 10:57pm
If we forget the overall factor of 4, we have:
A/(s-1)^3 + B/(s-1)^2 + C/(s-1) +
(D + E s)/(s^2+1)
Finding A,...E is just a matter of solving the linear equations.
What I ask you to do is to verify the answe I gave. So, if you already know that:
1/(s-1)^3 - 1/2 1/(s-1)^2 +
A = 1,
B = -1/2,
C = 0,
D = 1/2,
E = 0
Your A,B etc. probably differ by a factor of 4 from these numbers.
Now, to practice doing partal fractions by writing down the expression with unknown constants, you should practice simpler problems.
If you practice the more complicated problems you are actually practicing linear algebra, which may distract you from learning integration.
Also, if you want to learn to do partial fraction that are more complicated like this one, you should learn the methods people actually use to do that: Series expansions around the singular points.
If you expand around s = 1 the singular terms are:
1/(s-1)^3 - 1/2 1/(s-1)^2 + nonsingular terms
In this case we are basically done, because the contribution of the form
(a + bs)/(s^2+1)
to the partial fraction expansion is now completely fixed by the large s asymptotic behavior. For large s:
(s+1)/[(s^2+1)(s-1)^3] becomes 1/s^4
But the contribution form the expansion arounf s = 1 yields -1/2 1/s^2 as the dominant asymptotic behavior. This must thus be cancelled by (a + bs)/(s^2+1) therfore a = 1/2 and b = 0 (nonzero b would introduce a net 1/s asymptotic behavior).
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