Why Include Different Powers in Partial Fraction Expansion?

  • Context: Undergrad 
  • Thread starter Thread starter Swapnil
  • Start date Start date
  • Tags Tags
    Expansion Partial
Click For Summary

Discussion Overview

The discussion revolves around the inclusion of different powers of a repeated root in the partial fraction expansion of a rational proper function. Participants explore the reasoning behind this approach, particularly in the context of achieving the necessary terms in the numerator.

Discussion Character

  • Technical explanation, Conceptual clarification

Main Points Raised

  • One participant questions why different powers of the same root are included in the partial fraction expansion, using a specific example of a rational function.
  • Another participant explains that including different powers allows for achieving the necessary terms in the numerator, detailing the process of multiplying to obtain a common denominator and expanding the expression.
  • Several participants express understanding and appreciation for the explanation provided, indicating that the reasoning makes sense to them.
  • One participant reiterates the explanation of achieving different powers in the numerator through the multiplication of terms, emphasizing the process of equating coefficients to solve for constants.
  • There is a light-hearted acknowledgment of past struggles with similar concepts, suggesting a shared experience among participants.

Areas of Agreement / Disagreement

Participants generally agree on the reasoning behind including different powers in the expansion, but the discussion does not resolve any deeper theoretical questions or alternative approaches.

Contextual Notes

The discussion does not delve into potential limitations or alternative methods for partial fraction decomposition, nor does it address any assumptions that may underlie the participants' reasoning.

Who May Find This Useful

This discussion may be useful for students or individuals seeking clarification on partial fraction decomposition, particularly those grappling with repeated roots in rational functions.

Swapnil
Messages
459
Reaction score
6
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?
 
Last edited:
Physics news on Phys.org
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 tex[/tex]. <br /> <br /> Anyway, The simplest way to solve is to equate co-efficients on both sides :D.<br /> <br /> So we get:<br /> k_1 = 1<br /> (k_2 -6k_1)=4<br /> (k_3 -3k_2 + 9) =7<br /> <br /> YAY! Simultaneous Equations! <br /> <br /> For the 2nd equation, since k_1=1[/tex], the equation simplifies to&lt;br /&gt; k_2 -6=4&lt;br /&gt; k_2 =10&lt;br /&gt; Put that into equation 3. &lt;br /&gt; k_3 -30 + 9 = 7&lt;br /&gt; k_3=28&lt;br /&gt; &lt;br /&gt; YAY we have our question solved!&lt;br /&gt; &lt;br /&gt; \frac{x^2 + 4x + 7}{(x-3)^3} = \frac{1}{(x-3)} + \frac{10}{(x-3)^2} + \frac{28}{(x-3)^3}&lt;br /&gt; &lt;br /&gt; HOORAH!
 
Last edited:
thank you Gib Z. that was very well explained.
 
Gib Z said:
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.
 
Gib Z said:
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 tex[/tex]. <br /> <br /> Anyway, The simplest way to solve is to equate co-efficients on both sides :D.<br /> <br /> So we get:<br /> k_1 = 1<br /> (k_2 -6k_1)=4<br /> (k_3 -3k_2 + 9) =7<br /> <br /> YAY! Simultaneous Equations! <br /> <br /> For the 2nd equation, since k_1=1[/tex], the equation simplifies to&lt;br /&gt; k_2 -6=4&lt;br /&gt; k_2 =10&lt;br /&gt; Put that into equation 3. &lt;br /&gt; k_3 -30 + 9 = 7&lt;br /&gt; k_3=28&lt;br /&gt; &lt;br /&gt; YAY we have our question solved!&lt;br /&gt; &lt;br /&gt; \frac{x^2 + 4x + 7}{(x-3)^3} = \frac{1}{(x-3)} + \frac{10}{(x-3)^2} + \frac{28}{(x-3)^3}&lt;br /&gt; &lt;br /&gt; HOORAH!
&lt;br /&gt; &lt;br /&gt; When your rusty, it&amp;#039;s best to peack over someone shoulders.&lt;img src=&quot;https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f644.png&quot; class=&quot;smilie smilie--emoji&quot; loading=&quot;lazy&quot; width=&quot;64&quot; height=&quot;64&quot; alt=&quot;:rolleyes:&quot; title=&quot;Roll Eyes :rolleyes:&quot; data-smilie=&quot;11&quot;data-shortname=&quot;:rolleyes:&quot; /&gt; &lt;br /&gt; I use to remember how to that, 1st year Calculus.&lt;br /&gt; Excellent job.
 
Lol thanks guys, Its fine. If you guys look over to my thread on Integrals, I am horrible :P
 

Similar threads

Graduate Function expansion in Cartesian and spherical tensors
  • · Replies 1 ·
Replies
1
Views
2K
MHB Would Partial Fractions Simplify This Integral?
  • · Replies 8 ·
Replies
8
Views
2K
Undergrad How to manipulate functions that are not explicitly given?
  • · Replies 16 ·
Replies
16
Views
3K
Undergrad Finite expansion of a fraction of functions
  • · Replies 11 ·
Replies
11
Views
3K
Undergrad Understanding Mixed Partial Derivatives: How Do You Solve Them?
  • · Replies 3 ·
Replies
3
Views
2K
MHB Partial fractions (5x^2+1)/[(3x+2)(x^2+3)]
  • · Replies 3 ·
Replies
3
Views
3K
MHB 206.08.05.44 partial fraction decomposition
  • · Replies 6 ·
Replies
6
Views
2K
Undergrad Why is the integral of ##\arcsin(\sin(x))## so divisive?
  • · Replies 3 ·
Replies
3
Views
3K
A doubt in Partial fraction decomposition
  • · Replies 8 ·
Replies
8
Views
2K
MHB 242t.8.5.9 expand the quotient by partial fractions
  • · Replies 3 ·
Replies
3
Views
2K