Question

partial fractions fail

Answer 1

ok i have 1/ (x (x^2-4)) = A/ x + (bx+c) / (x^2-4)

Answer 2

im breaking the rule of irreducible factoring, but it still works

Answer 3

you can factor x^2-4 = (x-2)(x+2)

Answer 4

oh i know

Answer 5

i don't think that will always work

Answer 6

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

Answer 7

myin, can you think of a counterexample

Answer 8

i will think one sec

Answer 9

here is another example int ( 1/ ( x ( x^2 - 15))

Answer 10

here we have reducible over reals, irreducible over rationals (integers)

Answer 11

so this form only right?
where you have 1/(a*(x^2-b))

Answer 12

i mean that a is an x

Answer 13

i mean you want me to find a counterexample of that form?

Answer 14

right,

Answer 15

( 1/ ( x ( x^2 - 15)) = A / x + (bx + c) / (x^2 - 15)

Answer 16

in general
1 / [(x-a)^n (x^2+bx+c)^n] = A1/(x-a) + A2/ ( x-a)^2 + ... An/(x-a)^n + (B1x+C1)/ (x^2+bx+c) + (B2x + C2) / (x^2+bx+c)^2 + ... + (Bn x + Cn ) / (x^2+bx+c)^n

Answer 17

i guess you may be right

Answer 18

still thinking

Answer 19

see my book keeps saying "irreducible"

Answer 20

yes it works
A = -1/b
B = 1/b

Answer 21

one must use quadratics that are irreducible, but im finding that is not the case

Answer 22

now i remember what a pain this is.
i got A = -1/4

Answer 23

right we have 1/ (x (x^2-4)) = (-1/4) / x + (1/4 * x ) / (x^2 - 4)

Answer 24

1/ (x (x^2-4)) = (-1/4) / x + (1/4 * x + 0 ) / (x^2 - 4)

Answer 25

B=C and A+B+C=0
-1/4+2B=0
making B = 1/8

Answer 26

but it is late and i could have made a mistake

Answer 27

no you do not always have to reduce quadratics, as long as there is a product in denominator

Answer 28

satellite, you did it out the normal way, we have same solution though

Answer 29

whew. glad to know i can still solve a simple system

Answer 30

i was just making a point, you can relax the rule about quadratics being irreducible

Answer 31

sure if you use complex you can even do it with \[x^2+1\] and even use it to integrate

Answer 32

right but i mean , sometimes you have, say this case
int 1 / ( x ( x^2 - 15)

Answer 33

\[\int \frac{1}{x^2+1} dx\] can use partial fractions.

Answer 34

technically, yes over the complex numbers

Answer 35

i think the issue boils down, what field you are restricting yourself to

Answer 36

restricting to rationals , to reals, or to complex

Answer 37

myinanay, are you in agrement?

Answer 38

yes i just made a proof lol i will scan it

Answer 39

for when we were talking about 1/[x(x^2-a)] form
sorry i doubted you cantorset :)

Answer 40

its alittle sloppy and i wanted it on one page so after i got to the end of the page i went back up and drew a square to finish showing the combination of fractions from i reduced the polynomial x^2-a would still give you the same combination of frations if you did not reduce

Answer 41

ok im back

Answer 42

can you read it?

Answer 43

so you got the same answer doing it both ways?

Answer 44

right! so since we proved we can make it a theorem lol

Answer 45

it is no longer a conjunction to us

Answer 46

how do you know they are equal though?

Answer 47

do you in the square i combined the fractions?

Answer 48

from the reduced way

Answer 49

it came out to be the same as the non reduced way

Answer 50

oh i see

Answer 51

you can combine the last two fractions ,

Answer 52

right

Answer 53

i can write it more neatly if you like

Answer 54

thats fine, just that last step i did on paper

Answer 55

so why does my book insist on reducing only the irreducible

Answer 56

irreducible actually depends on the field i think, since 1 / (x^2 +1) is reducible in C

Answer 57

well i only know this true so far for the form 1/[x*(x^2-a)]
which is reduceable

Answer 58

hmmmm

Answer 59

you dont think 1 / [(x-a)^n (x^2+bx+c)^n] = A1/(x-a) + A2/ ( x-a)^2 + ... An/(x-a)^n + (B1x+C1)/ (x^2+bx+c) + (B2x + C2) / (x^2+bx+c)^2 + ... + (Bn x + Cn ) / (x^2+bx+c)^n

Answer 60

maybe it is easier just to memorize just reduce just in case instead of remembering for certain forms it is akay not to reduce

Answer 61

thats ugly to me cantorset lol

Answer 62

no it has a nice pattern

Answer 63

The property of irreducibility depends on the field F; a polynomial may be irreducible over some fields but reducible over others

Answer 64

right, the field F, but we can just change the field to suit our problem

Answer 65

yes...we are dealing with real analysis here

Answer 66

so F is the field of R

Answer 67

like integral 1 / ( x^2 + 1) = int 1 / (x+ i ) + 1/ (x-i) i believe

Answer 68

tan inverse of x

Answer 69

1/ x^2 + 1 = A / x+i + B / x- i

Answer 70

+C

Answer 71

Well thats complex analysis cantorset!

Answer 72

what is e^(ix) in terms of sine and cosine again i cant remember
something like sin(x)+icos(x)

Answer 73

yes, ok i see your point

Answer 74

so you want to stay in field R, ok

Answer 75

by the way x+i and x-i are not polynomials in real analysis

Answer 76

also if we assume int(1/(x^2+1))=int(1/(x+i))+int(1/(x-i))=then we get ln(x^2+1)
but this gives us 2x/(x^2+1) when we take derivative which is not 1/(x^2+1)

Answer 77

i actually didnt do the A , and B thingy

Answer 78

that was a guess

Answer 79

1/ x^2 + 1 = A / x+i + B / x- i

Answer 80

so A=-1/2i and B=1/2i
lets see what happens

Answer 81

brb

Answer 82

i/2 *ln{(x-i)/(x+i)}
i don't know if we can do anything with this

Answer 83

right

Answer 84

youre right about the arctangent though

Answer 85

so my point was, can we relax our condition about being reducible , or irreducible?

Answer 86

so far we can relax it for some reducibles like 1/[x*(x^2-a)] but not for forms that are irreducible

Answer 87

is e^(arctanx)==((x-i)/(x+i))^(1/2i) true?

Answer 88

irreducible over the reals, you mean

Answer 89

i don't know saubihik i fail at complex analysis

Answer 90

or whatever that is

Answer 91

i can check my calculator

Answer 92

but it might be off by a constant

Answer 93

how do you plug imaginarys into your calculator?

Answer 94

can we just pretend since they are imaginary lol

Answer 95

i have TI 84

Answer 96

i think its true and may be by some constant and so yah u can do partial fracs with rational functions having COMPLEX VARIABLES

Answer 97

only the answer will come in complex terms but dont woory thats same!!