partial fractions fail

- anonymous

partial fractions fail

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- anonymous

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

- anonymous

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

- dumbcow

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

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## More answers

- anonymous

oh i know

- myininaya

i don't think that will always work

- anonymous

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

- anonymous

myin, can you think of a counterexample

- myininaya

i will think one sec

- anonymous

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

- anonymous

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

- myininaya

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

- myininaya

i mean that a is an x

- myininaya

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

- anonymous

right,

- anonymous

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

- anonymous

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

- myininaya

i guess you may be right

- myininaya

still thinking

- anonymous

see my book keeps saying "irreducible"

- dumbcow

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

but it is late and i could have made a mistake

- dumbcow

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

technically, yes over the complex numbers

- anonymous

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

- anonymous

restricting to rationals , to reals, or to complex

- anonymous

myinanay, are you in agrement?

- myininaya

yes i just made a proof lol i will scan it

- myininaya

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

##### 1 Attachment

- myininaya

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

- anonymous

ok im back

- myininaya

can you read it?

- anonymous

so you got the same answer doing it both ways?

- myininaya

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

- myininaya

it is no longer a conjunction to us

- anonymous

how do you know they are equal though?

- myininaya

do you in the square i combined the fractions?

- myininaya

from the reduced way

- myininaya

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

- anonymous

oh i see

- anonymous

you can combine the last two fractions ,

- myininaya

right

- myininaya

i can write it more neatly if you like

- anonymous

thats fine, just that last step i did on paper

- anonymous

so why does my book insist on reducing only the irreducible

- anonymous

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

- myininaya

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

- anonymous

hmmmm

- anonymous

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

- myininaya

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

- myininaya

thats ugly to me cantorset lol

- anonymous

no it has a nice pattern

- anonymous

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

- anonymous

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

- anonymous

yes...we are dealing with real analysis here

- anonymous

so F is the field of R

- anonymous

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

- myininaya

tan inverse of x

- anonymous

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

- myininaya

+C

- anonymous

Well thats complex analysis cantorset!

- myininaya

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

- anonymous

yes, ok i see your point

- anonymous

so you want to stay in field R, ok

- anonymous

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

- myininaya

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)

- anonymous

i actually didnt do the A , and B thingy

- anonymous

that was a guess

- anonymous

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

- myininaya

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

- anonymous

brb

- myininaya

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

- anonymous

right

- anonymous

youre right about the arctangent though

- anonymous

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

- myininaya

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

- anonymous

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

- anonymous

irreducible over the reals, you mean

- myininaya

i don't know saubihik i fail at complex analysis

- myininaya

or whatever that is

- anonymous

i can check my calculator

- anonymous

but it might be off by a constant

- myininaya

how do you plug imaginarys into your calculator?

- myininaya

can we just pretend since they are imaginary lol

- anonymous

i have TI 84

- anonymous

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

- anonymous

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

- anonymous

So our rule is:
partial fraction expansions provide an approach to integrating a general rational function. Any rational function of a real variable can be written as the sum of a polynomial function and a finite number of fractions. Each fraction in the expansion has as its denominator a polynomial function of degree 1 or 2, or some positive integer power of such a polynomial. (In the case of rational function of a complex variable, all denominators will have a polynomial of degree 1, or some positive integer power of such a polynomial.) If the denominator is a 1st-degree polynomial or a power of such a polynomial, then the numerator is a constant. If the denominator is a 2nd-degree polynomial or a power of such a polynomial, then the numerator is a 1st-degree polynomial.

- myininaya

really? thats neat

- myininaya

man math is deep

- anonymous

So we have a rule of partial fractions now with the prerequisite: ANY RATIONAL FUNCTION WITH REAL OR COMPLEX VARIABLES.
that is very good generalization!!

- anonymous

but whats the field?

- myininaya

saubhik i didn't realize i made a mistake until u posted the 1/(2i)
dont know i put the i on top

- anonymous

I think there is no field restrictions for partial fracs. its C

- anonymous

oh,

- anonymous

where did you get that quote from?

- anonymous

which one?

- anonymous

the big one

- anonymous

partial fraction expansions provide an approach to integrating a general rational function. Any rational function of a real variable can be written as the sum of a polynomial function and a finite number of fractions. Each fraction in the expansion has as its denominator a polynomial function of degree 1 or 2, or some positive integer power of such a polynomial. (In the case of rational function of a complex variable, all denominators will have a polynomial of degree 1, or some positive integer power of such a polynomial.) If the denominator is a 1st-degree polynomial or a power of such a polynomial, then the numerator is a constant. If the denominator is a 2nd-degree polynomial or a power of such a polynomial, then the numerator is a 1st-degree polynomial.

- anonymous

wikipedia rocks!!:)

- anonymous

But integration is limited to R.
Recall the fundamental theorems of calculus!!!

- anonymous

hmmmm

- anonymous

every function to be integrated/differentiated (if such operation is allowed by the function) has to be real-valued.
So we have a doubt here: Applying partial fracs if we get complex fracs. then how we are integrating??

- anonymous

for eg. A/(x-i) is not real valued so integral of this is not defined.!!

- anonymous

but it should come out the same

- anonymous

which type of integral are you using? U cannot use the general Newton-Leibniz integral!

- anonymous

I think if you want to avoid the complex analysis stuff (residue theorem,contour integration) avoid splitting to complex rational functions!

- anonymous

ok, but i dont have to reduce things like x^2 - 15

- anonymous

you may/maynot

- anonymous

reducing x^2-15=(x-rt(15))(x+rt(15)) is recommended and the PROPER procedure to partial fraction decomposition.
i.e. 1/(x^2-15)=A/(x-rt(15))+B/(x+rt(15))

- anonymous

actually its not

- anonymous

it makes it much more difficult

- anonymous

that is the PROPER way. See that x^2-15 is reducible. So why not use this advantage. But if u use cx+d thing then don't call this partial frac decomposition.

- anonymous

that may be "partial" partial frac decomposition :D

- anonymous

well try to do partial fract, it will be harder

- anonymous

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

- anonymous

yah, but that's not partial frac decompositon

- anonymous

the degree of the rational function (i.e. the degree of both the numerator and denominator polynomials) must be reduced to the utmost extent

- anonymous

the outcome of a full partial fraction expansion expresses that function as a sum of fractions, where:
the denominator of each term is a power of an irreducible (not factorable) polynomial and
the numerator is a polynomial of smaller degree than that irreducible polynomial.

- anonymous

but in the case of "your" expansion: (bx+c)/(x^2-15) term has a reducible denominator and so your expansion is not the FULL partial frac decomposition. Understood?

- anonymous

yes

- anonymous

it is the full fractional decomposition over the integers (or rationals

- watchmath

@Cantorset: can you always integrate (bx + c ) / (x^2-15) without splitting x^2-15

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