Question

Partial Fractions

Answer 1

Find the exact values of A, B, C, and D in the following (see attachment).

I forgot how to derive these values. Can someone help?

Answer 2

you would multiply both sides by the \[z ^{4}-1\] then distribute the values. this is a long one to explain so i will go step by step with you if you need me to

Answer 3

okay, so if you multiplied z^4-1 on both sides, you would end up with:
z = A / (z +1) * z^4 - 1 + B / (z-1) *z^4 -1 + (Cz + D) / (z^2 + 1) * z^4 -1

z^4 - 1 = (z - 1) (z+1)(z^2 +1), so you'd end up with:

A(z-1)(z^2 +1) + B(z+1)(z^2-1) + (Cz + D)(z-1)(z+1)

Correct?

Answer 4

Oh yeah, all of that equal to z

Answer 5

How would I proceed from here? (assuming im correct up until this point)

Answer 6

so you have z = A(z-1)(z^2 +1) + B(z+1)(z^2-1) + (Cz + D)(z-1)(z+1)
then you distribute everything out- A(z-1)(z^2 +1) = (Az-a)(Z^2+1) = AZ^3+AZ+AZ^2-a
and so on through the rest of the values

Answer 7

Nevermind c: looks like you're on the right track.

Answer 8

Az^3 + Az - Az^2 - A + Bz^3 -Bz + Bz^2 - B + (Cz + D)(z^2-1)

= Az^3 + Az - Az^2 - A + Bz^3 -Bz + Bz^2 - B + Cz^3 - Cz + Dz^2 - D

Answer 9

Az^3 - Az^2 + Az - A + Bz^3 + Bz^2 - Bz - B + Cz^3 + Dz^2 - Cz - D = z

Answer 10

how am I doing

Answer 11

looks great!
so the next step is to figure out what a, b, c, and d equal. the left equation could read 0z^3+0z^2+z+0. you would take all of the like powers of Z and set them equal to the corresponding coeffiecent on the left side.
For power z^3, we have 0z^3 = Az^3 + Bz^3 +Cz^3.
then, because all of the z^3 are common, we have 0= a + B + C. you would do the same thing for 0z^2,z, and 0. does that make sense? i can try to explain it better if i need to

Answer 12

you'd have 4 equations?

Answer 13

yes.

Answer 14

0z^3 = Az^3 + Bz^3 +Cz^3
0z^2 = -Az^2 +Bz^2
0z = Az - Bz - Cz
0 = -A - B

Answer 15

?

Answer 16

you are missing the D variable in those. the second one and the last one should have a D

Answer 17

and it should be Z= Az+Bz+Cz. the coefficient for that was 1, not zero.

Answer 18

0z^3 = Az^3 + Bz^3 +Cz^3
0z^2 = -Az^2 +Bz^2 + Dz^2
z = Az - Bz - Cz
0 = -A - B - D

Im not seeing where the coefficient youre talking about is coming from

Answer 19

never mind about the coefficient. you got it right now. now you eliminate the z's from all the equations- effectively dividing each by the respective term, so you get
0 = A + B +C
0 = -A +B + D
1 = A - B - C
0 = -A - B - D
which is a system of equations you can use to solve for each of the variables. the end result is substituting the value in for each variable in the original equation:

Answer 20

So can I use the elimination method to simplify the first two equations, then do it again with the bottom two, and then again with the result from those 2?

Answer 21

you could. or use matrices or any method for solving.

Answer 22

1 = A - B - C
0 = -A - B - D

1 = -2B - (C - D)


0 = A + B +C
0 = -A +B + D

0 = 2B + (C + D)

1 = -2B - (C - D)
0 = 2B + (C + D)

Answer 23

If I eliminate again, that doesn't look like it will come out right.

Answer 24

Wait

Answer 25

I'm trying to isolate 1 letter. I don't know if that's possible.

Answer 26

thats why i think in this case matrices would be best. im trying to find a resource to help explain.

Answer 27

I did this, Matrix A =

1 1 1 0
-1 1 1 0
1 -1 -1 1
-1 -1 -1 0

rref(A) =

1 0 0 0
0 1 1 0
0 0 0 1
0 0 0 0

Answer 28

I don't know if that helps me though

Answer 29

\[\frac{z}{z^4-1}=-\frac{z}{2 \left(z^2+1\right)}+\frac{1}{4 (z-1)}+\frac{1}{4 (z+1)} \]

Answer 30

What, how do you even get those numbers?

Answer 31

ok. here's a link. its about halfway down the page- starts with 'consider to following example' https://www.boundless.com/algebra/50d49650e4b0965b097acfe2/50d49650e4b0965b097acfe9/50d49650e4b0965b097ad0a2/#.UQIj1yfC1qU

Answer 32

I did that morgan, but my calculator didnt give me any fractions like that guys post up there.

Answer 33

you shouldn't get fractions. you would get the values of A, B, C, and D. which you then plug into that. hold on im trying to write it out

Answer 34

If one had to solve this type of problem for a living, then one would buy a copy of the Mathematica program and enter the following and then press the Enter key on the numeric pad of a PC keyboard:\[\text{Apart}\left[\frac{z}{z^4-1}\right] \]

Answer 35

the B matrix would be
0
0
1
0
the a matrix would be
1 1 1 0
-1 1 0 1
1 -1 -1 0
-1 -1 0 -1

Answer 36

To model the equations
0 = A + B + c
0 = -A + B + D
1 = A - B - C
0 = -A - B - D

wouldnt matrix a be
1 1 1 0
-1 1 1 0
1 -1 -1 1
-1 -1 -1 0

Answer 37

Or wait, would it be 5 x 5 to account for the fact you're missing C and D in some cases? in which case you put a 0 there.

Answer 38

4 x 5

Answer 39

it would be a 4x4. you place a 0 where a variable is missing

Answer 40

but what about the = 0, doesn't that part go on the last column of each row?

Answer 41

you also need to keep the order of the variables the same.
we are doing two different types i think. are you doing row echelon?

Answer 42

the way i recall how to do it, you take the coefficients for the variables, set them up in a matrix A, the constant ( the =0 part) in another matrix B. multiply the inverse of A times B.

Answer 43

i apologize that this is so confusing and long.

Answer 44

http://mathbits.com/MathBits/TIsection/Precalculus/matrices.htm

If you look at example 2 when they use reduced row echelon, the equals part is on the far right. That's what I'm going by at the moment.

Answer 45

I appreciate your patience! I really want to understand this.

Answer 46

in that case yes it would be a 4 by 5.

Answer 47

the thing i noticed you were missing was the placeholders. in your matrix you put up earlier

Answer 48

Okay, the resulting matrix from rref(A) is:

1 1 1 0 0
0 1 .5 .5 0
0 0 1 -1 -1
0 0 0 0 1

You're right, I was missing them, but I've corrected that and put the variables in same order when I constructed the 4 x 5 matrix that produced that answer I just got above.

Answer 49

well darn. haha.

Answer 50

ok. hold on

Answer 51

yeah this one is a doozy

Answer 52

I thought I was on to something when we were back here:

A(z-1)(z^2 +1) + B(z+1)(z^2-1) + (Cz + D)(z-1)(z+1) = z

If we let z equal something like 1, you still can't isolate A, B, C, or D. you end up with 0 = 1, which is weird

Answer 53

or if you let z = 2

Answer 54

you get 5A + 11B + 6C + 3D = 2

Answer 55

sorry, 9B

Answer 56

well it makes sense that z couldnt be 1, because then the original equation would be undefined. so it has to be a value greater than 1, between 1 and -1, or less than -1.

Answer 57

so maybe that's the problem. its not diffrentiable on all x. but i also dont understand why they would give such a problem if that was unsolvable, really.
so. im not sure how we want to proceed. its possible there is an error somewhere in all of this that got perpetuated into the matrix- the matrix was close. i tried my matrix method and it came up with a error.

Answer 58

i guess the question becomes, do you understand now how to derive the values, though we can't exactly complete this problem?

Answer 59

Yes, I believe I have a much better understanding/approach to these types of problems.

Answer 60

Your patience is legendary :)

Answer 61

Thank you

Answer 62

thanks. and you're welcome!

Answer 63

ok.. im gonna go now. i wish you luck!