Unformatted text preview: artial fraction decomposition.
Theorem 8.11. Suppose
an xn + an−1 xn−1 + · · · + a2 x2 + a1 x + a0 = bm xm + mm−1 xm−1 + · · · + b2 x2 + b1 x + b0
for all x in an open interval I . Then n = m and ai = bi for all i = 1 . . . n.
Believe it or not, the proof of Theorem 8.11 is a consequence of Theorem 3.14. Deﬁne p(x) to be
the diﬀerence of the left hand side of the equation in Theorem 8.11 and the right hand side. Then
p(x) = 0 for all x in the open interval I . If p(x) were a nonzero polynomial of degree k , then, by
Theorem 3.14, p could have at most k zeros in I , and k is a ﬁnite number. Since p(x) = 0 for all
the x in I , p has inﬁnitely many zeros, and hence, p is the zero polynomial. This means there can
be no nonzero terms in p(x) and the theorem follows. Arguably, the best way to make sense of
either of the two preceding theorems is to work some examples.
Example 8.6.1. Resolve the following rational functions into partial fractions.
1. R(x) =
2. R(x) = x+5
−x−1 3. R(...
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