partial fractions
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Partial Fractions. Splitting Fractions into Component fractions Component-> part of whole One into many smaller!!. - PowerPoint PPT PresentationTRANSCRIPT
Partial Fractions
122
782
x
B
x
A
xx
x
Splitting Fractions intoComponent fractions
Component-> part of whole One into many smaller!!
In Calculus, there are several procedures that are much easier if we can take a rather large fraction and break it up into pieces. The procedure that can decompose larger fractions is called Partial Fraction Decompostition. We will proceed as if we are working backwards through an addition of fractions with LCD.
EXAMPLE 1: For our first example we will work an LCD problem frontwards and backwards. Use an LCD to complete the following addition.
21
78
21
10533
2
2
1
5
2
3
1
11
5
2
3
xx
x
xx
xx
x
x
xxx
xxx
1
5
2
3
xx
The LCD is (x + 2)(x – 1). We now convert each fraction to LCD status.On the next slide we will work this problem backwards
2
782
xx
xFind the partial fraction decomposition for:
As we saw in the previous slide the denominator factors as (x + 2)(x – 1). We want to find numbers A and B so that:
122
782
x
B
x
A
xx
x
The bad news is that we have to do this without peeking at the previous slide to see the answer. What do you think will be our first move?
Congratulations if you chose multiplying both sides of the equation by the LCD. The good news is that, since we are solving an equation, we can get rid of fractions by multiplying both sides by the LCD.
So we multiply both sides of the equation by (x + 2)(x – 1).
Now we expand and compare the left side to the right side.
If the left side and the right side are going to be equal then:
A+B has to be 8 and
-A+2B has to be 7.
BAxBAx
BBxAAxx
xBxAx
xxx
Bxx
x
A
xx
xxx
xxx
B
x
A
xx
xxx
278
278
2178
121
12212
7812
12122
7812
2
A + B = 8
-A + 2B = 7
This gives us two equations in two unknowns. We can add the two equations and finish it off with back substitution.
3B = 15
B = 5
If B = 5 and A + B = 8 then A = 3.
Cool!! But what does this mean?
Remember that our original mission was to break a big fraction into a couple of pieces. In particular to find A and B so that:
122
782
x
B
x
A
xx
x
We now know that A = 3 and B = 5 which means that
1
5
2
3
2
782
xxxx
x
Now we will look at this same strategy applied to an LCD with one linear factor and one quadratic factor in the denominator.
And that is partial fraction decomposition!
EXAMPLE 2: Find the partial fraction decomposition for
1243
101123
2
xxx
xx
First we will see if the denominator factors. (If it doesn’t we are doomed.)
The denominator has four terms so we will try to factor by grouping.
43
34312432
223
xx
xxxxxx
Since the denominator is factorable we can pursue the decomposition.
3434
1011
1243
101122
2
23
2
x
C
x
BAx
xx
xx
xxx
xx
Because one of the factors in the denominator is quadratic, it is quite possible that its numerator could have an x term and a constant term—thus the use of Ax + B in the numerator.
3434
101122
2
x
C
x
BAx
xx
xx As in the first example, we multiply both sides of this equation by the LCD.
)43()3(1011
4331011
343
344
1011
343434
101134
22
222
222
2
222
22
CBxBAxCAxx
CCxBBxAxAxxx
xxx
Cxx
x
BAxxx
xxx
C
x
BAx
xx
xxxx
If the two sides of this equation are indeed equal, then the corresponding coefficients will have to agree:
-1 = A + C
11 = 3A + B
-10 = 3B + 4C
On the next slide, we solve this system. We will start by combining the first two equations to eliminate A.
-1 = A + C
11 = 3A + B
-10 = 3B + 4C
3 = -3A - 3C
11 = 3A + B
Multiply both sides by -3
Add these two equations to eliminate A.
14 = B – 3C
Multiply both sides of this equation by –3.
Add this equation to eliminate B.
-42 = -3B + 9C-10 = 3B + 4C
-52 = 13C
-4 = C
We now have two equations in B and C. Compare the B coefficients.
We can finish by back substitution.
-1 = A + C -1 = A - 4 A = 3
-10 = 3B + 4C -10 = 3B + 4(-4) -10 + 16 = 3B 2 = B
We have now discovered that A = 3, B = 2 and C = -4.
OK, but I forgot what this means.
Fair enough. We began with the idea that we could break the following fraction up into smaller pieces (partial fraction decomposition).
3443
1011
1243
1011
22
2
23
2
x
C
x
BAx
xx
xx
xxx
xx
Substitute for A, B and C and we are done.
3
4
4
232
xx
x
EXAMPLE 3: For our next example, we are going to consider what happens when one of the factors in the denominator is raised to a power. Consider the following for partial fraction decomposition:
22
2
2
23
2
3
724813
96
724813
96
724813
xx
xx
xxx
xx
xx
xx
There are two setups that we could use to begin:
Setup A proceeds along the same lines as the previous example.
22
2
)3(3
724813
x
CBx
x
A
xx
xx
Setup B considers that the second fraction could have come from two pieces.
22
2
)3(33
724813
x
C
x
B
x
A
xx
xx
AxCBAxBAxx
CxBxBxAAxAxxx
CxBxBxxxAxx
CxxBxxAxx
xxx
Cxx
x
Bxx
x
Axx
xxx
C
x
B
x
A
xx
xxxx
936724813
396724813
396724813
33724813
33
33
3724813
3)3(33
7248133
22
222
222
22
2
2
222
2
22
22
Since we have already done an example with Setup A, this example will proceed with Setup B. Step 1 will be to multiply both sides by the LCD and simplify.
Expand.
Group like terms and factor.
We now compare the coefficients of the two sides.
AxCBAxBAxx 936724813 22 The last line of the previous slide left us here.
If we compare the coefficients on each side, we have:
A + B = 13
6A + 3B + C = 48
9A = 72From the third equation A = 8. Substituting into the first equation:
A + B = 13 so 8 + B = 13 and B = 5.
Substituting back into the second equation:
6A + 3B + C = 48 so 6(8) + 3(5) + C = 48
48 + 15 + C = 48 63 + C = 48 and C = -15
22
2
)3(33
724813
x
C
x
B
x
A
xx
xx
To refresh your memory, we were looking for values of of A, B and C that would satisfy the partial fraction decomposition below and we did find that A= 8, B=5 and C=-15.
So…..
22
2
)3(
15
3
58
3
724813
xxxxx
xx
Our last example considers the possibility that the polynomial in the denominator has a smaller degree than the polynomial in the numerator.
EXAMPLE 4: Find the partial fraction decomposition for
82
515422
23
xx
xxx
Since the order of the numerator is larger than the order of the denominator, the first step is division.
5
1642
82
52
5154282
23
2
232
x
xxx
xx
xx
xxxxx
By long division we have discovered that:
82
52
82
5154222
23
xx
xx
xx
xxx
We will now do partial fraction decomposition on the remainder.
BAxBAx
BBxAAxx
xBxAx
xxx
Bxx
x
Ax
xxx
B
x
A
xx
xxx
x
B
x
A
xx
x
xx
x
425
425
425
242
244
5
242424
524
2424
5
82
52
Multiply both sides by the LCD.
Distribute
Group like terms
Compare coefficients
From the previous slide we have that:
BAxBAx 425
If these two sides are equal then:
1 = A + B and 5 = 2A – 4B
To eliminate A multiply both sides of the first equation by –2 and add.
2A – 4B = 5
-2A – 2B = -2
-6B = 3 so B = -1/2
If A + B = 1 and B = -1/2 then
A –1/2 = 2/2 and A = 3/2
22
1
42
32
2
2/1
4
2/32
242
82
52
82
5154222
23
xxx
xxx
x
B
x
Ax
xx
xx
xx
xxx
In summary then:
You should now check out the companion piece to this tutorial, which contains practice problems, their answers and several complete solutions.
Tips for partial fraction decomposition of N(x)/D(x):
1. If N(x) has a larger order than D(x), begin by long division. Then examine the remainder for decomposition.
2. Factor D(x) into factors of (ax + b) and cbxax 2
3. If the factor (ax + b) repeats then the decomposition must include:
2bax
B
bax
A
4. If the factor
decomposition must include:
cbxax 2 repeats then the
222cbxax
DCx
cbxax
BAx