HELP on Partial Fraction Decomposition Problem - Heavy Side Technique

In summary, khatche4 is asking for help with a partial fraction decomposition problem using the Heavy Side Technique. They have attempted to split the fraction and set up equations, but have gotten stuck. Other users suggest using substitution to simplify the equations and find the values for the variables.
  • #1
khatche4
22
0
HELP! on Partial Fraction Decomposition Problem! - Heavy Side Technique

I am doing a partial fraction decomposition problem for my calc 2 class.
We use the Heavy Side Technique, but I will take help either way!

[tex]\frac{2x-1}{x(x^2+1)^2}[/tex]

Thank you!
 
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  • #2


Hi khatche4, welcome to PF!:smile:

What have you tried?...You need to show some attempt at the problem before we will assist you.
 
  • #3


I tried splitting it up as..

[tex]\frac{2x-1}{x(x^2+1)^2}[/tex] = [tex]\frac{A}{x}[/tex] + [tex]\frac{Bx+C}{x^2+1}[/tex] + [tex]\frac{Dx+E}{x^2+1}[/tex]

Then I used x=0, x=1, x=-1, x=2, and x=-2 to get...

1= -4 + 2B + 2C + 2D + 2E
3= 4 - 2B + 2C - 2D + 2E
3= -25 + 20B + 10C + 20D + 10E
5= 25 - 20B + 10C - 20D + 10E

And then I got stuck...


Should it be this, initially?:

[tex]\frac{2x-1}{x(x^2+1)^2}[/tex] = [tex]\frac{A}{x}[/tex] + [tex]\frac{Bx+C}{x^2+1}[/tex] + [tex]\frac{Dx+E}{(x^2+1)^2}[/tex] ?
 
  • #4


khatche4 said:
Should it be this, initially?:

[tex]\frac{2x-1}{x(x^2+1)^2}[/tex] = [tex]\frac{A}{x}[/tex] + [tex]\frac{Bx+C}{x^2+1}[/tex] + [tex]\frac{Dx+E}{(x^2+1)^2}[/tex] ?

Yes, it should.
 
  • #5


Ok! thank you! If I need anymore help, I'll come back!
=]
 
  • #6


Ok... I'm stuck again.

I went along with
[tex]\frac{2x-1}{x(x^2+1)^2}[/tex] = [tex]\frac{A}{x}[/tex] + [tex]\frac{Bx+C}{x^2+1}[/tex] + [tex]\frac{Dx+E}{(x^2+1)^2}[/tex]

From there, I used
x=0 to get A = -1
x=1
x=-1
x=2
x=-2
(Am I supposed to use 4?? We've only done up to Bx+C, and there we use 2.. So I assume to use 4)..

Then, from there, I got
1 = -4 + 2B + 2C + D + E
3 = 4 - 2B + 2C - D + E
3 = -25 + 20B + 10C + 4D + 2E
5 = 25 - 20B + 10C - 4D + 2E

So that boils down to..

1 = 2C + E
3 = 2C + E
3 = 10C + 2E
5 = 10C + 2E
right??

Will I be using substitution for this one? If so, do I use 2 different substitutions or just one?

Thanks, again!
 
  • #7


khatche4 said:
Ok... I'm stuck again.

I went along with
[tex]\frac{2x-1}{x(x^2+1)^2}[/tex] = [tex]\frac{A}{x}[/tex] + [tex]\frac{Bx+C}{x^2+1}[/tex] + [tex]\frac{Dx+E}{(x^2+1)^2}[/tex]

From there, I used
x=0 to get A = -1

good :approve:
x=1
x=-1
x=2
x=-2
(Am I supposed to use 4?? We've only done up to Bx+C, and there we use 2.. So I assume to use 4)..

You can get away with using only 3 values, if you choose those values wisely. What made x=0 such a useful choice was that it made one of the factors of [itex]x(x^2+1)^2[/itex] zero...The other factor is of course just [itex]x^2+1[/itex], which is zero at [itex]x=\pm i[/itex], so try using [itex]x=i[/itex] to find easy solutions for [itex]D[/itex] and [itex]E[/itex]; you'll have a much easier time finding [itex]B[/itex] and [itex]C[/itex] after that.
 

What is partial fraction decomposition?

Partial fraction decomposition is a method used in mathematics to break down a rational expression into simpler fractions. It involves rewriting an expression as the sum of its constituent parts, with each part having a single denominator.

What is the Heavy Side Technique?

The Heavy Side Technique is a method used in partial fraction decomposition to solve problems involving complex or repeated factors in the denominator. It involves multiplying both sides of the equation by the factor(s) in question to eliminate them from the denominator.

What are the steps for using the Heavy Side Technique?

The steps for using the Heavy Side Technique are as follows:1. Factor the denominator of the rational expression.2. Set up a system of equations using the factors from step 1.3. Multiply both sides of the equation by the factors in question to eliminate them from the denominator.4. Solve the resulting system of equations to find the coefficients of the partial fractions.5. Substitute the coefficients back into the original equation to get the final solution.

Can the Heavy Side Technique be used for any rational expression?

Yes, the Heavy Side Technique can be used for any rational expression, as long as the denominator can be factored into linear or quadratic factors.

What are some common mistakes to avoid when using the Heavy Side Technique?

Some common mistakes to avoid when using the Heavy Side Technique include:- Forgetting to factor the denominator before setting up the system of equations.- Not multiplying both sides of the equation by the factors in question.- Making errors when solving the resulting system of equations.- Forgetting to substitute the coefficients back into the original equation to get the final solution.

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