anonymous
  • anonymous
partial fractions fail
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
ok i have 1/ (x (x^2-4)) = A/ x + (bx+c) / (x^2-4)
anonymous
  • anonymous
im breaking the rule of irreducible factoring, but it still works
dumbcow
  • dumbcow
you can factor x^2-4 = (x-2)(x+2)

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
oh i know
myininaya
  • myininaya
i don't think that will always work
anonymous
  • anonymous
\[\frac{1}{x(x-2)(x+2)}=\frac{A}{x}+\frac{B}{x+2}+\frac{C}{x-2}\]\]
anonymous
  • anonymous
myin, can you think of a counterexample
myininaya
  • myininaya
i will think one sec
anonymous
  • anonymous
here is another example int ( 1/ ( x ( x^2 - 15))
anonymous
  • anonymous
here we have reducible over reals, irreducible over rationals (integers)
myininaya
  • myininaya
so this form only right? where you have 1/(a*(x^2-b))
myininaya
  • myininaya
i mean that a is an x
myininaya
  • myininaya
i mean you want me to find a counterexample of that form?
anonymous
  • anonymous
right,
anonymous
  • anonymous
( 1/ ( x ( x^2 - 15)) = A / x + (bx + c) / (x^2 - 15)
anonymous
  • 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
  • myininaya
i guess you may be right
myininaya
  • myininaya
still thinking
anonymous
  • anonymous
see my book keeps saying "irreducible"
dumbcow
  • dumbcow
yes it works A = -1/b B = 1/b
anonymous
  • anonymous
one must use quadratics that are irreducible, but im finding that is not the case
anonymous
  • anonymous
now i remember what a pain this is. i got A = -1/4
anonymous
  • anonymous
right we have 1/ (x (x^2-4)) = (-1/4) / x + (1/4 * x ) / (x^2 - 4)
anonymous
  • anonymous
1/ (x (x^2-4)) = (-1/4) / x + (1/4 * x + 0 ) / (x^2 - 4)
anonymous
  • anonymous
B=C and A+B+C=0 -1/4+2B=0 making B = 1/8
anonymous
  • anonymous
but it is late and i could have made a mistake
dumbcow
  • dumbcow
no you do not always have to reduce quadratics, as long as there is a product in denominator
anonymous
  • anonymous
satellite, you did it out the normal way, we have same solution though
anonymous
  • anonymous
whew. glad to know i can still solve a simple system
anonymous
  • anonymous
i was just making a point, you can relax the rule about quadratics being irreducible
anonymous
  • anonymous
sure if you use complex you can even do it with \[x^2+1\] and even use it to integrate
anonymous
  • anonymous
right but i mean , sometimes you have, say this case int 1 / ( x ( x^2 - 15)
anonymous
  • anonymous
\[\int \frac{1}{x^2+1} dx\] can use partial fractions.
anonymous
  • anonymous
technically, yes over the complex numbers
anonymous
  • anonymous
i think the issue boils down, what field you are restricting yourself to
anonymous
  • anonymous
restricting to rationals , to reals, or to complex
anonymous
  • anonymous
myinanay, are you in agrement?
myininaya
  • myininaya
yes i just made a proof lol i will scan it
myininaya
  • myininaya
for when we were talking about 1/[x(x^2-a)] form sorry i doubted you cantorset :)
1 Attachment
myininaya
  • 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
  • anonymous
ok im back
myininaya
  • myininaya
can you read it?
anonymous
  • anonymous
so you got the same answer doing it both ways?
myininaya
  • myininaya
right! so since we proved we can make it a theorem lol
myininaya
  • myininaya
it is no longer a conjunction to us
anonymous
  • anonymous
how do you know they are equal though?
myininaya
  • myininaya
do you in the square i combined the fractions?
myininaya
  • myininaya
from the reduced way
myininaya
  • myininaya
it came out to be the same as the non reduced way
anonymous
  • anonymous
oh i see
anonymous
  • anonymous
you can combine the last two fractions ,
myininaya
  • myininaya
right
myininaya
  • myininaya
i can write it more neatly if you like
anonymous
  • anonymous
thats fine, just that last step i did on paper
anonymous
  • anonymous
so why does my book insist on reducing only the irreducible
anonymous
  • anonymous
irreducible actually depends on the field i think, since 1 / (x^2 +1) is reducible in C
myininaya
  • myininaya
well i only know this true so far for the form 1/[x*(x^2-a)] which is reduceable
anonymous
  • anonymous
hmmmm
anonymous
  • 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
  • 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
  • myininaya
thats ugly to me cantorset lol
anonymous
  • anonymous
no it has a nice pattern
anonymous
  • anonymous
The property of irreducibility depends on the field F; a polynomial may be irreducible over some fields but reducible over others
anonymous
  • anonymous
right, the field F, but we can just change the field to suit our problem
anonymous
  • anonymous
yes...we are dealing with real analysis here
anonymous
  • anonymous
so F is the field of R
anonymous
  • anonymous
like integral 1 / ( x^2 + 1) = int 1 / (x+ i ) + 1/ (x-i) i believe
myininaya
  • myininaya
tan inverse of x
anonymous
  • anonymous
1/ x^2 + 1 = A / x+i + B / x- i
myininaya
  • myininaya
+C
anonymous
  • anonymous
Well thats complex analysis cantorset!
myininaya
  • myininaya
what is e^(ix) in terms of sine and cosine again i cant remember something like sin(x)+icos(x)
anonymous
  • anonymous
yes, ok i see your point
anonymous
  • anonymous
so you want to stay in field R, ok
anonymous
  • anonymous
by the way x+i and x-i are not polynomials in real analysis
myininaya
  • 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
  • anonymous
i actually didnt do the A , and B thingy
anonymous
  • anonymous
that was a guess
anonymous
  • anonymous
1/ x^2 + 1 = A / x+i + B / x- i
myininaya
  • myininaya
so A=-1/2i and B=1/2i lets see what happens
anonymous
  • anonymous
brb
myininaya
  • myininaya
i/2 *ln{(x-i)/(x+i)} i don't know if we can do anything with this
anonymous
  • anonymous
right
anonymous
  • anonymous
youre right about the arctangent though
anonymous
  • anonymous
so my point was, can we relax our condition about being reducible , or irreducible?
myininaya
  • myininaya
so far we can relax it for some reducibles like 1/[x*(x^2-a)] but not for forms that are irreducible
anonymous
  • anonymous
is e^(arctanx)==((x-i)/(x+i))^(1/2i) true?
anonymous
  • anonymous
irreducible over the reals, you mean
myininaya
  • myininaya
i don't know saubihik i fail at complex analysis
myininaya
  • myininaya
or whatever that is
anonymous
  • anonymous
i can check my calculator
anonymous
  • anonymous
but it might be off by a constant
myininaya
  • myininaya
how do you plug imaginarys into your calculator?
myininaya
  • myininaya
can we just pretend since they are imaginary lol
anonymous
  • anonymous
i have TI 84
anonymous
  • 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
  • anonymous
only the answer will come in complex terms but dont woory thats same!!
anonymous
  • 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
  • myininaya
really? thats neat
myininaya
  • myininaya
man math is deep
anonymous
  • 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
  • anonymous
but whats the field?
myininaya
  • 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
  • anonymous
I think there is no field restrictions for partial fracs. its C
anonymous
  • anonymous
oh,
anonymous
  • anonymous
where did you get that quote from?
anonymous
  • anonymous
which one?
anonymous
  • anonymous
the big one
anonymous
  • 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
  • anonymous
wikipedia rocks!!:)
anonymous
  • anonymous
But integration is limited to R. Recall the fundamental theorems of calculus!!!
anonymous
  • anonymous
hmmmm
anonymous
  • 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
  • anonymous
for eg. A/(x-i) is not real valued so integral of this is not defined.!!
anonymous
  • anonymous
but it should come out the same
anonymous
  • anonymous
which type of integral are you using? U cannot use the general Newton-Leibniz integral!
anonymous
  • anonymous
I think if you want to avoid the complex analysis stuff (residue theorem,contour integration) avoid splitting to complex rational functions!
anonymous
  • anonymous
ok, but i dont have to reduce things like x^2 - 15
anonymous
  • anonymous
you may/maynot
anonymous
  • 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
  • anonymous
actually its not
anonymous
  • anonymous
it makes it much more difficult
anonymous
  • 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
  • anonymous
that may be "partial" partial frac decomposition :D
anonymous
  • anonymous
well try to do partial fract, it will be harder
anonymous
  • anonymous
its much faster 1/ (x(x^2-15) = A/x + (bx + c ) / (x^2-15)
anonymous
  • anonymous
yah, but that's not partial frac decompositon
anonymous
  • 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
  • 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
  • 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
  • anonymous
yes
anonymous
  • anonymous
it is the full fractional decomposition over the integers (or rationals
watchmath
  • watchmath
@Cantorset: can you always integrate (bx + c ) / (x^2-15) without splitting x^2-15

Looking for something else?

Not the answer you are looking for? Search for more explanations.