Question

Hi everyone! Can anyone discuss with me partial fraction decomp? I know how to do it but I don't understand why it's okay to set up the problems in certain ways...Here is what I mean>>>
for example s+1/s^2(s+2)^3 is set up like this>>>A/s +B/s^2 + C/(s+2) + D/ (s+2)^2 + E/(s+2)^3
the reason I'm confused is because it seems like it should be set-up like this instead>>>
A/s +B/s + C/(s+2) + D/ (s+2) + E/(s+2)
Can anyone explain this to me? Thanks soooo much! :o)

Answer 1

Hi again amistre! aren't you tired of me yet? :o)

Answer 2

you already have a root of s defined

A/s + B/s = (A+B)/s which is just redundant

Answer 3

almost, bladders getting full :)

Answer 4

when you say root, I am thinking of where the function crosses the x-axis...what does a root of "s" actually mean?

Answer 5

notice this:

A/s + B/s^2 when we add fractions, which is what decomp is doing, we are trying to convert it into all the possible setups

A(s)/s(s) + B/s^2 = (As+B)/s^2

Answer 6

the roots are the factors of the setup

Answer 7

okay yes...roots are factors! totally get that

Answer 8

gimme just a sec to work that example on paper to prove to myself...one sec

Answer 9

\[\frac{5}{2*3^2*5^3}\]
notice the prime factorization of the denominator
\[\frac{5}{2*3^2*5^3}=\frac{A}{2}+\frac{B_0}{3}+\frac{B_1}{3^2}+\frac{C_0}{5}+\frac{C_1}{5^2}+\frac{C_2}{5^3}\]
these are all the possible formations that we can compost it into

Answer 10

lol, compost

Answer 11

so you are saying if you get a common denominator on the right side, and add, it will equal the left side then?

Answer 12

the decomposition is accounting for ALL possible combinations of the summation. its up to us to determine which ones are actually valid.

Answer 13

spose we ignore some of them
\[\frac{5}{2*3^2*5^3}=\frac{A}{2}+\frac{B}{3^2}+\frac{C}{5^3}\]
we havent accounted for all possible summations, have we?

Answer 14

well according to what you said, no...but my intuition wants to say, "I need to prove that to myself"

Answer 15

we dont get a truely decompsed set up if we do not account for ALL possible combinations.

Answer 16

okay...i get now that we have to "completely decompose" and write out all combination on the right...but can you answer this yes or no please...if we get a common denominator on the right side and add all those terms, will it equal the left side?

Answer 17

yes, that is after all how you add up fractions

Answer 18

okay...just making sure! :o)

Answer 19

because I tried to add up your example >A(s)/s(s) + B/s^2 = (As+B)/s^2
and I wasn't getting anywhere

Answer 20

you still have to solve for A and B

Answer 21

if both sides are to be equal, then they have to be equal for all values of s

Answer 22

so humor me for a sec...take your example>>>A(s)/s(s) + B/s^2 = (As+B)/s^2
and prove it for me

Answer 23

I can't get it

Answer 24

hi zepdrix...I see you spying on us!

Answer 25

wrote it all out but I can't prove itI

Answer 26

wait! got it!!!! :o) YAY!

Answer 27

okay amistre!, not that I ever doubted you, but I proved it to myself now! You have magical powers and I believe you 100%!

Answer 28

so bottom line, write out every facotr on the right to completely "compost" haha

Answer 29

(s+1)/s^2(s+2)^3
= A/s +B/s^2 + C/(s+2) + D/ (s+2)^2 + E/(s+2)^3

let s = 0, and cover up the bad parts

(0+1)/0^2(0+2)^3
= A/0 +B/0^2 + C/(0+2) + D/ (0+2)^2 + E/(0+2)^3


1/8 = C/2 + D/4 + E/8

1 = 4C + 2D + E
-------------------------------

let s = -2, and cover up the bad parts

(-2+1)/(-2)^2(-2+2)^3
= A/-2 +B/-2^2 + C/(-2+2) + D/ (-2+2)^2 + E/(-2+2)^3

-1/4 = -A/2 + B/4

-1 = -2A + B
-------------------------------

now we have some conditions defined that must hold

let B = 2A - 1

let E = 1 - 4C - 2D

and plug them in the setup, let s=-1 and see what the results are

eventually you get to a point where you can start defining all the tops

Answer 30

yay!!

Answer 31

question...I completely get what you did to solve for B and E...but what made you choose s=-1? I would have never done that

Answer 32

its something small, and it makes the top left go to 0, remember, in order for left and right to be equal, they must be true for ALL values of s, so when we narrow it down out of the simpler stuff we start branching out

(s+1)/s^2(s+2)^3
= A/s +(2A-1)/s^2 + C/(s+2) + D/ (s+2)^2 + (1 - 4C - 2D )/(s+2)^3


0= -A +(2A-1) + C + D + (1 - 4C - 2D )

0= A -1 + C + D + 1 - 4C - 2D

0= A - 3C - D

A = 3C + D must be true

Answer 33

notice that A in terms of C and D reduce us to 2 unknowns

Answer 34

makes the top left of "what" go to zero?

Answer 35

let s=1, and s=2 to get 2 equations in 2 unknowns

Answer 36

you really need to pay attention to what you post lol

(s+1)/s^2(s+2)^3 is our left side
^^^^
the top of it

-1+1 = 0

Answer 37

A = 3C + D

B = 2A - 1 = 6C + 2D - 1

E = 1 - 4C - 2D

and we have only C and D left to solve ...

Answer 38

the simpler stuff is used up, so we simply have to define 2 equations for a particular s value that hasnt been used and work the results

Answer 39

i was paying attention to what I posted...I just didn't understand, so I asked...not that you clarified, I get it...thanks

Answer 40

:)

Answer 41

I hate these systems of equations...can't I just distribute and make a matrix?

Answer 42

maybe, dunno how that would go tho :)

(s+1)/s^2(s+2)^3
= (3C+D)/s +(6C + 2D - 1)/s^2 + C/(s+2) + D/ (s+2)^2 + (1 - 4C - 2D )/(s+2)^3


let s=1

2/27 = (3C+D) +(6C + 2D - 1) + C/3 + D/9 + (1 - 4C - 2D )/27

2 = 3(27)C+ 27D + 6(27)C + 2(27)D - 27 + 9C + 3D + 1 - 4C - 2D

28 = [3(27)+6(27) +5]C+ [27 + 2(27) + 1]D

28 = 248 C+ 82 D
-----------------------------------

let s=2

3/16(6)^3 = (3C+D)/2 +(6C + 2D - 1)/4 + C/6 + D/ 6^2 + (1 - 4C - 2D )/6^3

3/16 = 108(3C+D) +54(6C + 2D - 1) + 36C + 6D + (1 - 4C - 2D )

3/16 = 108(3)C +108D +54(6)C + 54(2)D - 54 + 36C + 6D + 1 - 4C - 2D

oy vey this is long

Answer 43

amistre, i really do admire your work ethic but I honestly would rather gargle peanut butter! wouldn't a matrix be much easier?

Answer 44

(53(16) +3)/16 = [108(3) +54(6)+32] C + [108+54(2)+4] D

im not sure i have the mental capacity to figure out a matrix solution lol

Answer 45

okay...well I will do the matirx and then compare your answer...I will need like 5 minutes...brb

Answer 46

28 = 248 C+ 82 D, (53(16) +3)/16 = (108(3) +54(6)+32) C + (108+54(2)+4) D

http://www.wolframalpha.com/input/?i=28+%3D+248+C%2B+82+D%2C+%2853%2816%29+%2B3%29%2F16+%3D+%28108%283%29+%2B54%286%29%2B32%29+C+%2B+%28108%2B54%282%29%2B4%29+D

assuming i made absolutely no errors .....

C = -14389/9600, D = 11699/2400

Answer 47

yikes...gimme a sec

Answer 48

http://www.wolframalpha.com/input/?i=decompose+%28s%2B1%29%2F%28s%5E2%28s%2B2%29%5E3+%29

im sure i made some faux pauxes

Answer 49

apparently D = 0 so yeah, oh well

Answer 50

yeah, i made some errors along the way, but the method is sound regardless of my ineptitudity

Answer 51

this is really good practice for my test! :o)
getting there!

Answer 52

1 0 1 0 0 0
6 1 4 1 0 0
12 6 4 2 1 0
8 12 0 0 0 1
0 8 0 0 0 1
equals>>>

Answer 53

A=-.0625 = -1/16
B=.125 = 1/8
C=.0625 = 1/16
D=0 = 0
E=-.25 = -1/4

Answer 54

I can't get wolfram to answer the question...but I think it's correct

Answer 55

(s+1)/s^2(s+2)^3
= A/s +B/s^2 + C/(s+2) + D/ (s+2)^2 + E/(s+2)^3

(s+1)/(s+2)^3
= As +B + Cs^2/(s+2) + Ds^2/ (s+2)^2 + Es^2/(s+2)^3

(s+1) = As(s+2)^3 +B(s+2)^3 + Cs^2(s+2)^2 + Ds^2(s+2) + Es^2

let s=-2,-1,0,1,2

-1 = A0 +B0 + C0 + D0 -4E ; E = 1/4

-1/4 = -A +B + C + D

1 = 8B ; B = 1/8

-3/8 = -A + C + D

3 = 8A -8C -8D
-------------------------

2 = A(3)^3 +(3)^3/8 + C(3)^2 + D(3) + 1/4

-6.5 = 108A + 36C + 12D

---------------------------

-14 = 128A + 64C + 16D

rref{{8,-8,-8,3},{108,36,12,-6.5},{128,64,16,-14}}

bah!! i still messed it up long the way ...

Answer 56

http://www.wolframalpha.com/input/?i=decompose+%28s%2B1%29%2F%28s%5E2%28s%2B2%29%5E3+%29

this is the wolfs reply

Answer 57

@amistre64
Got it! YAY! My matrix worked! Take a peak...it says it's true!

http://www.wolframalpha.com/input/?i= [%28s%2B1%29%2F%28s^2%28s%2B2%29^3%29]%3D[%28-1%2F16%29%2Fs+%2B+%281%2F8%29%2Fs^2+%2B+%281%2F16%29%2F%28s%2B2%29+%2B+%280%29%2F%28s%2B2%29^2+%2B+%28-1%2F4%29%2F%28s%2B2%29^3]

Answer 58

I can't get the dumb link to post correctly

Answer 59

your not on chrome ... are you

Answer 60

firefox

Answer 61

just paste this into wolfram
[(s+1)/(s^2(s+2)^3)]=[(-1/16)/s + (1/8)/s^2 + (1/16)/(s+2) + (0)/(s+2)^2 + (-1/4)/(s+2)^3]

Answer 62

btw, your link didn't show anything

Answer 63

firefox is going bad i spose ....

Answer 64

so that's good news...I did it correctly, and I never took linear algebra yet either! :o)

Answer 65

bladders at its limit, so im heading towards bed. gnite :)

Answer 66

a couple questions...
1) this problem was like A/s + B/ s^2 etc
I have seen problem like Ax+B/(something) etc...
when do you know to combine letters instead of keeping the separate?

Answer 67

but yeah, to answer the question, we need to account for ALL the possible combinations of the linear roots.

Answer 68

well, the one simplifies to the other

A/s + B/s^2 adds up to (As+B)/s^2

so its really a matter of how you want to go about it

Answer 69

tis is why i said the top needs to be one degree less then the bottom

Answer 70

so when I have seen problems set up like Ax+B/(something) etc... in the past, they could have just separated it further? They just took a shortcut or something?

Answer 71

they shortcutted it yes

Answer 72

okay neat!
I forgot my second question and it was easy but important darn it! :O(

Answer 73

i remember!

Answer 74

now when we have a prime factor such as (s^3+3)

we can split it into imaginary roots, and the same thing happens in the sum, we get

(As^2 + Bs + C)/(s^3+3)

Answer 75

pretty sure thats not a prime factor but hey, its late lol

Answer 76

when solving these crazy problem with matrices and your calculator...
the calculator gives you answers in decimals, but what if you get an answer like .66666667...the calc won't give you a fraction for that...any suggestions?

Answer 77

s^2 + 3 has imaginary roots, thats a better example

Answer 78

repeating decimals can be "summed" to a fraction

S = .6666666....
10S = 6.6666666......

10S - S = 9S = 6

S = 6/9 = 2/3

Answer 79

so this way you can always use your calc to solve your systems of equations, even if the calculator won't give you a fraction right?

Answer 80

S = .5134823823823...

1000 S = 5134.823823823...
100000S = 513482.3823823...

(100000 - 1000) S = 513482 - 5134

S can then be solved as a fraction

Answer 81

ugh. forgot i did a triple repeat ... fill in the detials eh

Answer 82

so you start with S = .5134823823823...
do you have to just make up 2 new values like
1000 S = 5134.823823823...
100000S = 513482.3823823... ?
I mean, can you sum the steps in words?

Answer 83

the idea is to get it so that you can subtract the decimals parts and be left with whole numbers

Answer 84

okay, I will look into how to do it

Answer 85

S = .5134823823823...

1000S brings the nonrepeating stuff out front, moves the deci 4 places

1000S = 5134. 823823823... we have triples repeating in this case so another 3 zeros shoould pull out enough stuff


1000000S = 5134823. 823823823...

now the subtraction eliminates the decimal expansions

1000000S = 5134823. 823823823...
- 1000S = - 5134. 823823823...
---------------------------------
a S = b.0000000000

S = b/a

Answer 86

.777777... = 7/9

.23232323... = 23/99

.512512512512.. = 512/999

Answer 87

so the amount of zeros you choose is equal to how many repeating places there are?

Answer 88

depends on the number, if they are repeating from the start, then yeah

otherwise we need to adjust it so that we are subtracting out the decimal expansion

Answer 89

.n repeat ..... 10^n gets the first n parts out of the way, and then for the repetaing part we multiply out another 10^k for some k repeating terms

Answer 90

so if you have .777777...
you set it equal to like X so
x=.777777...
then 100x = 77.7777... and
1000x=777.777...
(1000-100)x=(777.777...-77.777...)
x=700/900
x=7/9

Answer 91

.666666666

10^0 pulls out the first 0 non repeating terms

10^0 * 10^1 pulls out the 1 repeating term

(10^1 - 10^0)S = 6.6666..... - 0.6666.....

9S = 6

Answer 92

thats fine too

Answer 93

.066666...
x=.066666...
(1000-100)x = 66.666...-6.666...
x=600/900
x=2/3

Answer 94

just as long as you have more zeros than like non repeating spaces right?

Answer 95

not quite

1000(.06666) = 66.6
100(.06666) = 6.6

900 x = 60

x = 60/900 = 6/90 - 2/30

Answer 96

=1/15

Answer 97

oh no! I got it wrong?