Partial Fractions
Partial fractions though not directly in the syllabus of IIT and state boards finds its application in the chapters namely integration and sequences and series, so here it is a brief overview of partial fractions:
here we split an algebraic expression in proper fractions (proper fraction means degree of numerator is smaller than that of denominator).
i.e Writing:
6 1 1
------------- = ------- - -------
x² + 2x - 8 x - 2 x + 4
6
means that you have expressed ----------- in partial fractions...
x² + 2x - 8
Now, there are three main "methods" of doing this:
(1) Substitution of strategic values
(2) Solving with co-efficiants
(3) Cover up method (can only be used on fractions without powers in the bottom bit)
Method 1
Substitution of values.
=======================
Express as a partial fraction.
x - 1
---------------
(3x - 5)(x - 3)
=======================
First write the fraction as:
x - 1 A B
--------------- = -------- + -------
(3x - 5)(x - 3) (3x - 5) (x - 3)
Notice that I have taken the two terms that are in brackets and placed them on there own,
in there own fraction (the A and B are what we need to find out)
next, multiply denominator (the bottom bit of the big fraction) by both sides:
so we now have:
x - 1 = A(x - 3) + B(3x - 5)
to get rid of one term, substitute a 'strategic' value - example: to get rid of (x - 3), make x = 3, so (3 - 3) ends up as 0 - its gone! Remember that you have to do this to all the x's in the equation though.
3 - 1 = B(3*3 - 5)
2 = 4B
so, 2/4 = B
or, 1/2 = B
now, do the same except this time get rid of B leaving A behind.
- make x = 5/3
so, (5/3 * 3) = 5
so, (5 - 5) = 0
- the (3x - 5) term is gone!
5/3 -1 = A(5/3 - 3)
2/3 = A * -4/3
2/3
------ = A
-4/3
-1/2 = A
so, we now have A and B - the answer!
x - 1 -1 1
--------------- = --------- + --------
(3x - 5)(x - 3) 2(3x - 5) 2(x - 3)
#########################################################################################
Method 2
Solving by co-efficiants
========================
Usually people use this method in conjunction with another method ie substitution of values, there is however nothing wrong with using this method on its own if you prefer it - as below...
=======================
Express as a partial fraction.
x - 1
---------------
(3x - 5)(x - 3)
=======================
First write the fraction as:
x - 1 A B
--------------- = -------- + -------
(3x - 5)(x - 3) (3x - 5) (x - 3)
The first step (as in the first method) is to multiply the denominator by both sides
x - 1 = A(x - 3) + B(3x - 5)
It may help if the A and B parts are expanded but this step can usually be missed.
x - 1 = A * x - A * 3 + B * 3 * x - 5B
x - 1 = Ax - 3A + 3Bx - 5B
now look at this and try to equate co-efficiants for 'x'
Ok, we've now done the first step, now do the same agian but with somthing else (like x² or
constants (the constants are the actual number's - i.e not x's which can be anything)
Lets do, co-efficiant for constants.
-1 = -3A -5B
we now have a simultanious equation...
1 = A + 3B --- (1)
-1 = -3A + (-3B) --- (2)
Solve.
multiply (1) by -3
-3 = -3A + (-9B) --- (1a)
subtract (1a) from (2)
-2 = -4B
-2
------- = B
-4
0.5 = B
plug into (1)
1 = A + 3 * 0.5
1 - 3 * 0.5 = A
-0.5 = A
we now know A and B
x - 1 0.5 -0.5
--------------- = --------- + --------
(3x - 5)(x - 3) (3x - 5) (x - 3)
this can be written with "1/2's" instead of the "0.5's"
x - 1 -1 1
--------------- = --------- + --------
(3x - 5)(x - 3) 2(3x - 5) 2(x - 3)
#########################################################################################
Method 3
cover up method
================
This method is in my opinion the easiest of all the three methods, but it can be misleading if you follow it like a cookbook recipe, not knowing what you are really doing. It also cannot be used with non-linear fractions such as explained after this section...I basically just done a partial fraction not explaing it much - since you can pretty much see how you do it from example.
=======================
Express as a partial fraction.
x - 1
---------------
(3x - 5)(x - 3)
=======================
delete first term.
3x - 5
3x - 5 = 0
3x = 5
x = 5/3
plug 5/3 as x into what is left.
5/3 - 1 2/3 -1
------------- = ------------ = ---
5/3 - 3 -4/3 2
this is the answer for the (3x - 5) expression
repeat for other expression.
delete second term.
x - 3
x - 3 = 0
x = 3
plug in.
3 - 1 2 1
----------- = --- = ---
3*3 - 5 4 2
this is the answer for the (x - 3) expression.
so,
x - 1 -1 1
--------------- = --------- + --------
(3x - 5)(x - 3) 2(3x - 5) 2(x - 3)
done.
pretty quick huh?
#########################################################################################
Strange exceptions (repeated-linear fractions)
for a repeated linear fraction, i.e. (1 + x)2 the format will look somthing like:
A B C
fraction = ------------ + --------- + ---------
expression (1 + x) (1 + x)²
notice that becuase (1 + x)
2 is in the 'expression' it also has an (1 + x)
not squared to go with it (obviously there will not be a 1 + x in every expression and the number is merely a representation).
example:
=======================
Express as a partial fraction.
1
---------------
(x - 3)(x + 1)²
=======================
1 A B C
--------------- = ------- + ------- + --------
(x - 3)(x + 1)² (x - 3) (x + 1) (x + 1)²
multiply by the denominator.
1 = A(x + 1)² + B(x - 3)(x + 1) + C(x - 3)
multiply out (in this case its only the 'A' term).
1 = A(x +1)(x +1) + B(x - 3)(x + 1) + C(x - 3)
solve by making x = -1 (this is an 'inspection method' part)
1 = C(-1 - 3)
1 = C*4
-1/4 = C
make x = 3
1 = A(3 + 1)(3 + 1)
1 = A*16
1/16 = A
solve coefficients of x2 for B
0 = A + B
0 = 1/16 + B
-1/16 = B
solved.
1 1 1 1
--------------- = ----------- - ----------- - ------------
(x - 3)(x + 1)² 16(x - 3) 16(x + 1) 4(x + 1)²
#########################################################################################
Strange exceptions (quadratics in the denominator - bottom of the fraction - )
If you have got a fraction in the denominator i.e. x² + 3x + 2 the format should look like this:
A Bx + C
fraction = --------- + ------------
factor quadractic
example:
=======================
Express as a partial fraction.
x - 1
--------------------
(x + 3)(x² + 3x + 2)
=======================
x - 1 A Bx + C
-------------------- = --------- + -------------
(x + 3)(x² + 3x + 2) x + 3 x² + 3x + 2
multiply by denominator.
x - 1 = A(x² + 3x + 2) + (Bx + C)(X + 3)
By Inspection
make x = -3 (at this point be careful that your selected value doesn't also make the
quadratic equal 0 as well!)
-3 - 1 = A((-3)² + 3*(-3) + 2)
-4 = A(9 - 9 + 2)
-4 = 2A
-2 = A
solve co-efficiant of x²
0 = A + B
0 = -2 + B
2 = B
solve co-efficiants of x
1 = 3A + 3B + C
1 = 3*(-2) + 3*(2) + C
1 = C
done.
x - 1 -2 2x + 1
-------------------- = --------- + ---------------
(x + 3)(x² + 3x + 2) (x + 3) x² + 3x + 2
Hope all of that helped!
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