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anonymous
 5 years ago
partial fractions fail
anonymous
 5 years ago
partial fractions fail

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok i have 1/ (x (x^24)) = A/ x + (bx+c) / (x^24)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0im breaking the rule of irreducible factoring, but it still works

dumbcow
 5 years ago
Best ResponseYou've already chosen the best response.0you can factor x^24 = (x2)(x+2)

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0i don't think that will always work

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\frac{1}{x(x2)(x+2)}=\frac{A}{x}+\frac{B}{x+2}+\frac{C}{x2}\]\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0myin, can you think of a counterexample

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0here is another example int ( 1/ ( x ( x^2  15))

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0here we have reducible over reals, irreducible over rationals (integers)

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0so this form only right? where you have 1/(a*(x^2b))

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0i mean that a is an x

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0i mean you want me to find a counterexample of that form?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0( 1/ ( x ( x^2  15)) = A / x + (bx + c) / (x^2  15)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0in general 1 / [(xa)^n (x^2+bx+c)^n] = A1/(xa) + A2/ ( xa)^2 + ... An/(xa)^n + (B1x+C1)/ (x^2+bx+c) + (B2x + C2) / (x^2+bx+c)^2 + ... + (Bn x + Cn ) / (x^2+bx+c)^n

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0i guess you may be right

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0see my book keeps saying "irreducible"

dumbcow
 5 years ago
Best ResponseYou've already chosen the best response.0yes it works A = 1/b B = 1/b

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0one must use quadratics that are irreducible, but im finding that is not the case

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now i remember what a pain this is. i got A = 1/4

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0right we have 1/ (x (x^24)) = (1/4) / x + (1/4 * x ) / (x^2  4)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.01/ (x (x^24)) = (1/4) / x + (1/4 * x + 0 ) / (x^2  4)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0B=C and A+B+C=0 1/4+2B=0 making B = 1/8

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but it is late and i could have made a mistake

dumbcow
 5 years ago
Best ResponseYou've already chosen the best response.0no you do not always have to reduce quadratics, as long as there is a product in denominator

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0satellite, you did it out the normal way, we have same solution though

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0whew. glad to know i can still solve a simple system

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i was just making a point, you can relax the rule about quadratics being irreducible

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sure if you use complex you can even do it with \[x^2+1\] and even use it to integrate

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0right but i mean , sometimes you have, say this case int 1 / ( x ( x^2  15)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\int \frac{1}{x^2+1} dx\] can use partial fractions.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0technically, yes over the complex numbers

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i think the issue boils down, what field you are restricting yourself to

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0restricting to rationals , to reals, or to complex

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0myinanay, are you in agrement?

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0yes i just made a proof lol i will scan it

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0for when we were talking about 1/[x(x^2a)] form sorry i doubted you cantorset :)

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0its 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^2a would still give you the same combination of frations if you did not reduce

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so you got the same answer doing it both ways?

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0right! so since we proved we can make it a theorem lol

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0it is no longer a conjunction to us

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how do you know they are equal though?

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0do you in the square i combined the fractions?

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0it came out to be the same as the non reduced way

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you can combine the last two fractions ,

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0i can write it more neatly if you like

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thats fine, just that last step i did on paper

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so why does my book insist on reducing only the irreducible

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0irreducible actually depends on the field i think, since 1 / (x^2 +1) is reducible in C

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0well i only know this true so far for the form 1/[x*(x^2a)] which is reduceable

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you dont think 1 / [(xa)^n (x^2+bx+c)^n] = A1/(xa) + A2/ ( xa)^2 + ... An/(xa)^n + (B1x+C1)/ (x^2+bx+c) + (B2x + C2) / (x^2+bx+c)^2 + ... + (Bn x + Cn ) / (x^2+bx+c)^n

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0maybe it is easier just to memorize just reduce just in case instead of remembering for certain forms it is akay not to reduce

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0thats ugly to me cantorset lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no it has a nice pattern

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The property of irreducibility depends on the field F; a polynomial may be irreducible over some fields but reducible over others

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0right, the field F, but we can just change the field to suit our problem

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes...we are dealing with real analysis here

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so F is the field of R

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0like integral 1 / ( x^2 + 1) = int 1 / (x+ i ) + 1/ (xi) i believe

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.01/ x^2 + 1 = A / x+i + B / x i

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well thats complex analysis cantorset!

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0what is e^(ix) in terms of sine and cosine again i cant remember something like sin(x)+icos(x)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes, ok i see your point

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so you want to stay in field R, ok

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0by the way x+i and xi are not polynomials in real analysis

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0also if we assume int(1/(x^2+1))=int(1/(x+i))+int(1/(xi))=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
 5 years ago
Best ResponseYou've already chosen the best response.0i actually didnt do the A , and B thingy

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.01/ x^2 + 1 = A / x+i + B / x i

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0so A=1/2i and B=1/2i lets see what happens

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0i/2 *ln{(xi)/(x+i)} i don't know if we can do anything with this

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0youre right about the arctangent though

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so my point was, can we relax our condition about being reducible , or irreducible?

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0so far we can relax it for some reducibles like 1/[x*(x^2a)] but not for forms that are irreducible

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0is e^(arctanx)==((xi)/(x+i))^(1/2i) true?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0irreducible over the reals, you mean

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0i don't know saubihik i fail at complex analysis

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i can check my calculator

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but it might be off by a constant

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0how do you plug imaginarys into your calculator?

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0can we just pretend since they are imaginary lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i think its true and may be by some constant and so yah u can do partial fracs with rational functions having COMPLEX VARIABLES

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0only the answer will come in complex terms but dont woory thats same!!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So 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 1stdegree polynomial or a power of such a polynomial, then the numerator is a constant. If the denominator is a 2nddegree polynomial or a power of such a polynomial, then the numerator is a 1stdegree polynomial.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So we have a rule of partial fractions now with the prerequisite: ANY RATIONAL FUNCTION WITH REAL OR COMPLEX VARIABLES. that is very good generalization!!

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0saubhik i didn't realize i made a mistake until u posted the 1/(2i) dont know i put the i on top

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I think there is no field restrictions for partial fracs. its C

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0where did you get that quote from?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0partial 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 1stdegree polynomial or a power of such a polynomial, then the numerator is a constant. If the denominator is a 2nddegree polynomial or a power of such a polynomial, then the numerator is a 1stdegree polynomial.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0But integration is limited to R. Recall the fundamental theorems of calculus!!!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0every function to be integrated/differentiated (if such operation is allowed by the function) has to be realvalued. So we have a doubt here: Applying partial fracs if we get complex fracs. then how we are integrating??

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0for eg. A/(xi) is not real valued so integral of this is not defined.!!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but it should come out the same

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0which type of integral are you using? U cannot use the general NewtonLeibniz integral!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I think if you want to avoid the complex analysis stuff (residue theorem,contour integration) avoid splitting to complex rational functions!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok, but i dont have to reduce things like x^2  15

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0reducing x^215=(xrt(15))(x+rt(15)) is recommended and the PROPER procedure to partial fraction decomposition. i.e. 1/(x^215)=A/(xrt(15))+B/(x+rt(15))

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it makes it much more difficult

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that is the PROPER way. See that x^215 is reducible. So why not use this advantage. But if u use cx+d thing then don't call this partial frac decomposition.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that may be "partial" partial frac decomposition :D

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well try to do partial fract, it will be harder

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0its much faster 1/ (x(x^215) = A/x + (bx + c ) / (x^215)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yah, but that's not partial frac decompositon

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the degree of the rational function (i.e. the degree of both the numerator and denominator polynomials) must be reduced to the utmost extent

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the 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
 5 years ago
Best ResponseYou've already chosen the best response.0but in the case of "your" expansion: (bx+c)/(x^215) term has a reducible denominator and so your expansion is not the FULL partial frac decomposition. Understood?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it is the full fractional decomposition over the integers (or rationals

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0@Cantorset: can you always integrate (bx + c ) / (x^215) without splitting x^215
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