Question

i am having problems with integration of rational functions by partial fractions. the question is z+1/z^2(z-1)

Answer 1

\[\frac{z+1}{z^2(z-1)}=\frac{A}{z}+\frac{B}{z^2}+\frac{C}{z-1}\]
find A,B, and C such that equation is true

Answer 2

finding A, B and C is what i need help with

Answer 3

do you know how to combine fractions?

Answer 4

look at the first fraction A/z
what do we need on bottom so that eventually we will have the same denominators
well it is lacking another z and a (z-1)
so you will multiply top and bottom by z(z-1)

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look at the second fraction B/z^2
what do we need on bottom?
well it is lacking a (z-1)
so we will multiply top and bottom by (z-1)


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look at C/(z-1)
what do we need on bottom?
well it is lacking a z^2
so we will multiply top and bottom by z^2

Answer 5

\[\frac{z+1}{z^2(z-1)}=\frac{A}{z}+\frac{B}{z^2}+\frac{C}{z-1} \\ \frac{z+1}{z^2(z-1)}=\frac{A z(z-1)+B(z-1)+Cz^2}{z^2(z-1)}\]

Answer 6

bottoms are the same
just compare the tops

Answer 7

you have to satisfy the equation
\[z+1=A(z^2-z)+B(z-1)+Cz^2\]

Answer 8

collect your like terms together

Answer 9

You need to get the constants A,B,C shown in the next ecuation.
\[z+1/z ^{2}(z-1)= (A/z-1) + (B/z) + (C/z ^{2})\]
Multiplying each side by z^2(z-1)
\[z+1=z ^{2}(z-1)(A/z-1)+z ^{2}(z-1)(B/z)+z ^{2}(z-1)(C/z ^{2})\]
After that you can give z any value that could hel you to solve the ecuation and get A,B,C. Example: z=1, z=0. Both of them make the ecuation easier to solve and give you constant values.
After that evaluate with the constants and solve the 3 integers. You should obtamin solutions with logarithms.

Answer 10

\[z+1=(A+C)z^2+(-A+B)z+(-B)\]

Answer 11

notice you have 0 z^2 's on left hand side
notice you have (A+C) z^2 's on the right hand side
so you need A+C to be 0
So you have A+C=0

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notice you have 1 z 's on left side
notice you have (-A+B) z's on right side
so you need 1=-A+B

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notice you have constant 1 on left side
notice you have constant -B on right side
so you need -B=1

Answer 12

last equation solve it first
then use it to solve second equation
then use that result and solve the first equation

Answer 13

your wrong because the answer is -2/z + -1/z^2 + 2/z-1

Answer 14

who are you talking to?

Answer 15

you have to solve the equations...

Answer 16

you can do this
if -B=1 then B=?

Answer 17

if -B=1 then B=-1
now go to the second equation 1=-A+B
you have B=-1 solve for A

Answer 18

you and no, from my teacher, in class told me the equation is z+1=Az(z-1)+B(z-1)+ Cz^2, i have C=2, A=-1 but i cant find B

Answer 19

Your A value in incorrect

Answer 20

i mean C=2, A=-1 and i dont know B

Answer 21

anyways i don't know what you are calling wrong
but I have not said anything incorrect

I gave you the three equations to solve
all I did was compare both sides after combining like terms

and got the following three equations:
A+C=0
-A+B=1
-B=1

Answer 22

The last equation is the easiest equation to solve
if -B=1 then B=-1
then use the equation equation -A+B=1
replace B with -1
-A+(-1)=1 solve for A
-A=2
A=-2
then go to the first equation A+C=0
-2+C=0 which gives C=2

Answer 23

now you just go back to what were trying to split up as before
\[\frac{z+1}{z^2(z-1)}=\frac{A}{z}+\frac{B}{z^2}+\frac{C}{z-1}\]

Answer 24

yeah i have that, so there for \[you would have z+1=Az(z-1)+B(z-1)+cz ^{2}\]

Answer 25

yes same equation I have way above somewhere

Answer 26

you do know z(z-1) is the same as z^2-z
so that is not wrong

Answer 27

still don't know what you refer to as wrong

Answer 28

because you get a common denominator

Answer 29

saying your wrong helps me in no way to help you
you could say if you don't understand something I did

Answer 30

but I would like to know what it is

Answer 31

i quess my teacher does it a little differently than you, lets start over, i already have \[A/z+B/z ^{2}+C/(z-1)\]

Answer 32

ok and you said you were able to get \[z+1=A(z^2-z)+B(z-1)+Cz^2\]

Answer 33

now I said something about combining any like terms
...
\[z+a=(A+C)z^2+(-A+B)z+(-B)\]
do you understand how I rearranged things here?

Answer 34

that lowercase a is suppose to be 1

Answer 35

\[z+1 =(A+C)z^2+(-A+B)z+(-B)\]

Answer 36

yes i see what you did there, now in order to find c you, you put like a 1 in for z and solve,. and the same with the others, just with low numbers like -1 or 0

Answer 37

no

Answer 38

i didn't do that

Answer 39

i compared both sides of the equation

Answer 40

as i said before there are 0 z^2's on the right but (A+C) z^2's on the left
so 0=A+C

and there is 1 z on the right and (-A+B) z on the left
so 1=-A+B

and there there is 1 on the right and -B on the left
so 1=-B

Answer 41

but if you want to do it that other way we can try
normally i go for solving the system of equations

Answer 42

i just learned this last week so i am a little lost on your method

Answer 43

\[z+1=Az(z-1)+B(z-1)+Cz^2 \\ \\ z=1 \text{ gives } 1+1=A(1)(1-1)+B(1-1)+C(1)^2 \\ 2=A(0)+B0)+C \\ C=2 \text{ which is the same value I got with the previous method } \\ z=0 \text{ gives } 0+1=A(0)(0-1)+B(0-1)+2(0)^2 \\ \text{ so here you get } 1=B(-1) \\ \text{ or } -1 =B \text{ which is also what I have by previous method } \\ \text{ do you understand this so far }\]

Answer 44

We only have to find A now

Answer 45

we have C=2 and B=-1
so we have so far the following equation:
\[z+1=Az(z-1)-1(z-1)+2z^2 \]

Answer 46

use some value that is not 0 or 1
and solve for A

Answer 47

ok, i have been using -2, -1, -+.5 and getting bad answers

Answer 48

how did you plus or mins 1/2

Answer 49

those are the values i plugged in for z

Answer 50

i'm going to replace z with -1...
\[z+1=Az(z-1)-1(z-1)+2z^2 \\ -1+1=A(-1)(-1-1)-1(-1-1)+2(-1)^2 \\ 0=A(-1)(-2)-1(-2)+2(1) \\ 0=2A+2+2 \\ \text{ subtract 4 on both sides } -4=2A \text{ divide by 2 on both sides } -2=A\]
which is the result i got by previous method

Answer 51

i must of done my rhythmic wrong thank you for you patience, calc 2 is frustrating me

Answer 52

just a pointer
you may not want to refer to someone 's work as wrong without even knowing anything about that method
it just comes off as negative and kinda rude

Answer 53

im sorry

Answer 54

it's okay
i was just a tad bit annoyed

Answer 55

anyways try to go back through the problem
or go back to my work and please let me know if I'm can clarify anything I did
you can look at both methods or not
it is up to you

Answer 56

I did start with z=1 because it made two terms 0
then I when to z=0 because it made one term 0
then I could have chose any z for the last one but I thought z=-1 would make things easiest

Answer 57

ok, well thank you

Answer 58

np