Partial Fraction Decomposition #2

[themadhatter1]

Homework Statement

Write the Partial Fraction decomposition for:

\frac{x}{16x^4-1}

Homework Equations

The Attempt at a Solution

\frac{x}{16x^4-1}
=\frac{x}{(2x-1)(2x+1)(4x^2+1)}
=\frac{x}{(2x-1)(2x+1)(4x^2+1)}=\frac{a}{(2x-1)}+\frac{b}{(2x+1)}+\frac{cx+d}{4x^2+1}

x=(2x+1)(4x^2+1)a+(2x-1)(4x^2+1)b+(2x-1)(2x+1)(cx+d)

From here I can find that A=1/8 and B=1/8

I can't find c and d because I would have to plug in i/2 into x to get (4x^2+1)=0

Now, I believe I need to set up a system of equations. I can foil and get what I had into polynomial form and I'd get.

x=(8a+8b+4c)x^3+(4a-4b+4d)x^2+(2a+2b-c)x+(a-b-d)

From here I'm not quite sure what to do to create the system of equations.

 

[vela]

Match the coefficients from the LHS and the RHS for each power of x. You can also simplify a bit by plugging in your results for a and b that you found earlier.

BTW, you're missing the x³ term.

Also, the sign is wrong on 4b in the coefficient of x².

And d shouldn't appear in the coefficient of x.

And you're missing a few contributions to the constant term as well.

 

[themadhatter1]

Yeah, I foiled incorrectly. but it's fixed now.

so I would have the system of equations8a+8b+4c+0d=0
4a+4b+0c+4d=0
2a+2b+0c+0d=1
1a-1b-1c-1d=0when the system is solved I don't get the right answers.

Also what does LHS and RHS stand for?

 

[vela]

You still have a few algebra mistakes. The -c should be part of the coefficient of x¹ and not x⁰.

 

[themadhatter1]

Ok, I'm certain that my algebra is correct now

8a+8b+4c+0d=0
4a+4b+0c+4d=0
2a+2b-1c+0d=1
1a-1b+0c-1d=0

I get a=0 b=.25 c=-.5 d=-.25

I know the a=1/8 and b=1/8 are correct because it's the answer in the back of my book. So how is this system wrong?

 

[Char. Limit]

Oh, since you asked, and I see no answer, LHS is left-hand side and RHS is right-hand side of an equation.

 

[Mark44]

There's a much easier way to do this that you have chosen to pursue.

\frac{x}{16x^4-1}=\frac{A}{(2x-1)}+\frac{B}{(2x+1)}+\frac{Cx+D}{4x^2+1}
\Rightarrow x =A(2x+1)(4x^2 + 1) + B(2x-1)(4x^2 + 1) + (Cx+D)(2x - 1)(2x + 1)

The bottom equation has to be true for all values of x, so substitute four values of x to get four equations in four unknowns. If you choose the values wisely you'll get equations that are easy to solve.

Two obvious values to choose are x = 1/2 and x = -1/2.

 

[themadhatter1]

I saw that and that was how I originally got my a=1/8 and b=1/8 but to get the (4x^2+1) to equal 0 you'd have to sub in i/2 for x. can you have an imaginary number in a partial fraction decomposition?

If you sub it in you wind up getting a 2 variable equation

\frac{i}{2}=(i-1)(i+1)c\frac{i}{2}+d

\frac{-1}{4}=c\frac{i}{2}+d

 

[Char. Limit]

Actually, I'm of the opinion that for any system of equations with more than three unknowns, Hatter's way is the best way. And Hatter, your system is correct. Your solutions are not. Try putting them into your third equation: it won't work. I just used row reduction to work it out, and I got the values for a and b in the back of your book. Are you sure you didn't make a mistake somewhere?

 

[themadhatter1]

The solutions I found for the system of equations I last posted up are correct. You can plug in the solutions for each equation and the values satisfy each equation. I used a matrix to solve it. Are you solving some different system of equations?

ps. I changed the third equation. It was wrong in the first post with the system. The correct system is the most recent one I posted

 

[Char. Limit]

Woops, sorry, after looking at your system again, I find that the second equation in the system is wrong. My bad.

Multiplying (2x-1)(4x^2+1) gives a -4 coefficient to the x^2 term. So your second equation should be

4a - 4b + 0c + 4d = 0

There's the mistake. Change the second equation and resolve it. If I'm right (not a guarantee, as you've just seen), you should get the correct solutions.

 

[themadhatter1]

hmm. aha.. ok let's see...8a+8b+4c+0d=0
4a-4b+0c+4d=0
2a+2b-1c+0d=1
1a-1b+0c-1d=0

Finally haha! a=1/8 b=1/8 c=-1/2 d=0

Thanks, I think we can say that small algebra errors kill. I'll try and be more careful since all my errors were with negatives and simple combining of like terms.

Thanks again!

 

[Char. Limit]

You're welcome and have a great day! Feel free to come back if you have any other questions.

 

[vela]

hmm. aha.. ok let's see...8a+8b+4c+0d=0
4a-4b+0c+4d=0
2a+2b-1c+0d=1
1a-1b+0c-1d=0

Finally haha! a=1/8 b=1/8 c=-1/2 d=0

Thanks, I think we can say that small algebra errors kill. I'll try and be more careful since all my errors were with negatives and simple combining of like terms.

Thanks again!

Glad to see you got it worked out. One thing you could have done is plugged your answers for a and b in that you found earlier using x=1/2 and x=-1/2 to get

<br /> \begin{align*}<br /> 2+4c &amp; = 0 \\<br /> 4d &amp; = 0 \\<br /> 1/2-c &amp; = 1 \\<br /> -d &amp; = 0<br /> \end{align*}<br />

You can easily see what c and d have to equal. Also, having four equations and two unknowns provides a check. If the system is inconsistent, you know you made an algebra mistake somewhere.

I saw that and that was how I originally got my a=1/8 and b=1/8 but to get the (4x^2+1) to equal 0 you'd have to sub in i/2 for x. can you have an imaginary number in a partial fraction decomposition?

If you sub it in you wind up getting a 2 variable equation

\frac{i}{2}=(i-1)(i+1)c\frac{i}{2}+d

\frac{-1}{4}=c\frac{i}{2}+d

You can use complex numbers. If you do that, you'd get

\begin{align*}<br /> \frac{i}{2} &amp; = (i-1)(i+1)(c\frac{i}{2}+d) \\<br /> 0 + \frac{1}{2} i&amp; = -ic - 2d<br /> \end{align*}<br />

Since c and d are real, if you equate the real parts of each side, you'd get -2d=0, and the equating the imaginary parts gives you -c=1/2.

The main problem with using complex numbers is that you're much more likely to make algebra mistakes, which, as you have discovered, are easy enough to make already. It's bad enough keeping track of signs, and now you're throwing in multiples of i.

 

[Mark44]

I saw that and that was how I originally got my a=1/8 and b=1/8 but to get the (4x^2+1) to equal 0 you'd have to sub in i/2 for x. can you have an imaginary number in a partial fraction decomposition?

If you sub it in you wind up getting a 2 variable equation

\frac{i}{2}=(i-1)(i+1)c\frac{i}{2}+d

\frac{-1}{4}=c\frac{i}{2}+d

I substituted x = 0 and x = 1. These values don't get rid of any terms, but that's OK, since you already know that A and B are 1/8.