partial fraction

Onyx · 2010-04-13 04:34:24

Hi, how can I reduce something like the following into partial fractions?

Thanks

ZHero · 2010-04-13 06:44:27

Onyx · 2010-04-13 07:43:38

Where does the A and Cx come from? I thought it was simply:

rzaidan · 2010-04-13 07:54:43

Dear ZHero
I think that it is not easy (or say impossible) to write this fraction into partial fractions as you did,since the fraction which contains in its numerator Cx is not allowed because the denominator contains a linear factor repeated twice , and the appropriate fraction must contain a constant in its numerator, and the purpose of this stratigy is to find the integral of this fraction, and it is easy to find the integral of this fraction by means of substitution u = (aX+b) and continue
Best Wishes
Riad Zaidan

rzaidan · 2010-04-13 07:56:26

Dear Onyx
As you wrote you will have as A =0 and B=1
salam
Riad Zaidan

ZHero · 2010-04-13 17:49:18

rzaidan wrote:

. . . the fraction which contains in its numerator Cx is not allowed because the denominator contains a linear factor repeated twice . . .

well.. i think its perfectly allowed as long as the Degree of Numerator is less than that of the Denominator!
however.. i just forgot to include another D in the expression above!

i'll correct it...

note: A=0

Last edited by ZHero (2010-04-13 17:52:19)

bobbym · 2010-04-13 18:32:21

Hi

Solving we get an answer with 1 free parameter B.

Substituting into ZHero's RHS.

This is clearly true for all B, so this is an identity. It is also not of much use as a partial fraction.

ZHero · 2010-04-13 19:20:43

bobbym wrote:

It is also not of much use as a partial fraction.

this is probably coz Numerator is 1!

Onyx · 2010-04-15 07:01:35

Thanks, it seems this partial fraction doesn't make anything simpler at all, but I found another way to solve my problem. I do however have another partial fraction:

I thought something like this was written as:

I haven't seen partial fractions written with a constant as a term in its own right (As the term 'A' is in your example) before, and also thought the form Bx+C in a numerator was only used for squared variables within the parenthesis such as:

Is the form I have provided not correct?

Last edited by Onyx (2010-04-15 07:02:09)

bobbym · 2010-04-15 07:06:29

Hi Onyx;

That form will work.

Math Is Fun Forum

#1 2010-04-13 04:34:24

partial fraction

#2 2010-04-13 06:44:27

Re: partial fraction

#3 2010-04-13 07:43:38

Re: partial fraction

#4 2010-04-13 07:54:43

Re: partial fraction

#5 2010-04-13 07:56:26

Re: partial fraction

#6 2010-04-13 17:49:18

Re: partial fraction

#7 2010-04-13 18:32:21

Re: partial fraction

#8 2010-04-13 19:20:43

Re: partial fraction

#9 2010-04-15 07:01:35

Re: partial fraction

#10 2010-04-15 07:06:29

Re: partial fraction

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