In this video I go over a pretty extensive proof of the simplest case of partial fraction decomposition. The simplest case is for the case of a proper fraction without repeating or non-linear factors in the denominator of the rational function. The proof involves first proving a simpler theorem which allows us to use the proof of the simpler theorem to prove the overall partial fraction decomposition proof.
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Partial Fraction Decomposition: Simplest Case Proof
In my earlier videos on partial fraction decomposition I went over the techniques in decomposing rational functions but I did not explain WHY those techniques worked.
In this video I go over the proof of the simplest case for partial fraction decomposition.
Some examples of these fractions are shown below:
Proof of the simplest case:
where:
- The ai's are all different (i.e. non-repeating factors)
- N(x) is a polynomial of degree at most d - 1
In proving that the above holds true we first need to show the following theorem holds true first.
Lemma 1:
(Note that lemma just means that it's a subsidiary theorem which will help in the overall proof)
Let N(x) and D(x) be polynomials of degree n and d respectively, with n ≤ d. Suppose that a is NOT a zero of D(x).
Then there is a polynomial P(x) of degree p < d and a number A such that:
Now we have to prove lemma 1, and then we can apply it repeatingly to prove the overall partial fraction decomposition theorem for the simplest case.
Proof of Lemma 1:
Thus we have proved Lemma 1.
Now we can apply Lemma 1 repeatingly to prove the overall theorem:
