Why Include Different Powers in Partial Fraction Expansion?

[Swapnil]

Why is it that when you have a repeated root in the denominator of a rational proper function, you include different powers of the same root in the function's partial fraction expansion?

For example,
$$\frac{x^2 + 4x + 7}{(x-3)^3} = \frac{k_1}{(x-3)} + \frac{k_2}{(x-3)^2} + \frac{k_3}{(x-3)^3}$$

why do you do this?

 

[Gib Z]

It let's you achieve the different powers on the numerator. Eg To get the same denominator, for the first part we must multiply by (x-3)^2, giving us an x^2, then for the 2nd part we must multiply by (x-3), giving us the x, and 3rd part gives us our constant. Subtracting and multiplying these in the end gives up our original expression.

Lets see how this works out in this example.

$$\frac{x^2 + 4x + 7}{(x-3)^3} = \frac{k_1}{(x-3)} + \frac{k_2}{(x-3)^2} + \frac{k_3}{(x-3)^3}$$

Multiply to get a common denominator.

$$\frac{x^2 + 4x + 7}{(x-3)^3} = \frac {k_1 \cdot(x-3)^2 + k_2\cdot(x-3) + k_3}{(x-3)^3}$$

Expand.

$$\frac{x^2 + 4x + 7}{(x-3)^3} = \frac {k_1 \cdot x^2 - k_1 \cdot 6x +9 \cdot k_1 +k_2 \cdot x -k_2 \cdot 3 + k_3}{(x-3)^3}$$

Simplify Like terms etc.

$$\frac{x^2 + 4x + 7}{(x-3)^3} = \frac {k_1\cdot x^2 + (k_2 -6k_1) \cdot x + (k_3 -3k_2 + 9)}{(x-3)^3}$$

Phew, that was a bit of confusing [itex]tex[/tex].

Anyway, The simplest way to solve is to equate co-efficients on both sides :D.

So we get:
$$k_1 = 1$$
$$(k_2 -6k_1)=4$$
$$(k_3 -3k_2 + 9) =7$$

YAY! Simultaneous Equations!

For the 2nd equation, since [itex]k_1=1[/tex], the equation simplifies to
$$k_2 -6=4$$
$$k_2 =10$$
Put that into equation 3.
$$k_3 -30 + 9 = 7$$
$$k_3=28$$

YAY we have our question solved!

$$\frac{x^2 + 4x + 7}{(x-3)^3} = \frac{1}{(x-3)} + \frac{10}{(x-3)^2} + \frac{28}{(x-3)^3}$$

HOORAH!

 

[murshid_islam]

thank you Gib Z. that was very well explained.

 

[Swapnil]

It let's you achieve the different powers on the numerator. Eg To get the same denominator, for the first part we must multiply by (x-3)^2, giving us an x^2, then for the 2nd part we must multiply by (x-3), giving us the x, and 3rd part gives us our constant. Subtracting and multiplying these in the end gives up our original expression.

Oh, I see. Makes sense. Thanks.

 

[Von Gastl]

It let's you achieve the different powers on the numerator. Eg To get the same denominator, for the first part we must multiply by (x-3)^2, giving us an x^2, then for the 2nd part we must multiply by (x-3), giving us the x, and 3rd part gives us our constant. Subtracting and multiplying these in the end gives up our original expression.

Lets see how this works out in this example.

$$\frac{x^2 + 4x + 7}{(x-3)^3} = \frac{k_1}{(x-3)} + \frac{k_2}{(x-3)^2} + \frac{k_3}{(x-3)^3}$$

Multiply to get a common denominator.

$$\frac{x^2 + 4x + 7}{(x-3)^3} = \frac {k_1 \cdot(x-3)^2 + k_2\cdot(x-3) + k_3}{(x-3)^3}$$

Expand.

$$\frac{x^2 + 4x + 7}{(x-3)^3} = \frac {k_1 \cdot x^2 - k_1 \cdot 6x +9 \cdot k_1 +k_2 \cdot x -k_2 \cdot 3 + k_3}{(x-3)^3}$$

Simplify Like terms etc.

$$\frac{x^2 + 4x + 7}{(x-3)^3} = \frac {k_1\cdot x^2 + (k_2 -6k_1) \cdot x + (k_3 -3k_2 + 9)}{(x-3)^3}$$

Phew, that was a bit of confusing [itex]tex[/tex].

Anyway, The simplest way to solve is to equate co-efficients on both sides :D.

So we get:
$$k_1 = 1$$
$$(k_2 -6k_1)=4$$
$$(k_3 -3k_2 + 9) =7$$

YAY! Simultaneous Equations!

For the 2nd equation, since [itex]k_1=1[/tex], the equation simplifies to
$$k_2 -6=4$$
$$k_2 =10$$
Put that into equation 3.
$$k_3 -30 + 9 = 7$$
$$k_3=28$$

YAY we have our question solved!

$$\frac{x^2 + 4x + 7}{(x-3)^3} = \frac{1}{(x-3)} + \frac{10}{(x-3)^2} + \frac{28}{(x-3)^3}$$

HOORAH!

When your rusty, it's best to peack over someone shoulders.
I use to remember how to that, 1st year Calculus.
Excellent job.

 

[Gib Z]

Lol thanks guys, Its fine. If you guys look over to my thread on Integrals, I am horrible :P