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Math Help - help with partial fractions

  1. #31
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    I dont have a clue.
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  2. #32
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    Ok, here's the goal. You have to multiply every fraction to be added (just like with the numerical case) by something equal to 1, such that the new denominator is equal to the "least common multiple". You would agree, I hope, that anything of the form x/x is equal to 1, provided x is not zero, right? So, if I'm adding up 1/2+1/3, I determine that the least common multiple is 6, as you've pointed out. Then I multiply each fraction by what's missing (and in order not to change what the fraction is, I multiply by something that looks like x/x) from the LCM. So try to follow this:

    \dfrac{1}{2}+\dfrac{1}{3}=\dfrac{1}{2}\left(\dfrac  {3}{3}\right)+\dfrac{1}{3}\left(\dfrac{2}{2}\right  )=\dfrac{3}{6}+\dfrac{2}{6}.

    The expressions in parentheses are all equal to 1, and since I'm multiplying each fraction by an expression equal to 1, you would agree I haven't changed the fraction numerically, right? This is exactly the same sort of thing I need to do with the addition problem in post # 28. Check this out:

    \displaystyle\dfrac{1}{2x^{2}+x+5}+\dfrac{2}{x+1}=  \dfrac{1}{2x^{2}+x+5}\left(\frac{x+1}{x+1}\right)+  \dfrac{2}{x+1}\left(\frac{2x^{2}+x+5}{2x^{2}+x+5}\  right)

    Again, the expressions in parentheses are all equal to 1 (provided the denominators are not equal to zero). We continue:

    =\dfrac{x+1}{(2x^{2}+x+5)(x+1)}+\dfrac{2(2x^{2}+x+  5)}{(2x^{2}+x+5)(x+1)}.

    Did you follow that? And can you perform the next step in the addition problem?
    Last edited by Ackbeet; November 6th 2010 at 12:35 PM. Reason: Added necessary comma.
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  3. #33
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    the only thing I can think of is multiply out the bracket and thats just a guess. Not sure how this applies to partial fractions, even though it probly does
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  4. #34
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    Addition of fractions is at the very heart of partial fractions. The idea of the method of partial fractions is this: break up a complicated fraction with multiple factors in the denominator into a sum of simpler fractions with unknown coefficients. Use the addition of fractions algorithm (with the least common multiple and so on, like we've been talking about for a while) to get the RHS and LHS denominators looking the same. Then, equate the numerators. This should tell you what the unknown coefficients are.

    Here. Let me show you a partial fractions problem, step by step. Separate out the following fraction

    \dfrac{1}{x(x-1)(x-2)(x+1)}.

    Solution:

    We guess that

    \dfrac{1}{x(x-1)(x-2)(x+1)}=\dfrac{A}{x}+\dfrac{B}{x-1}+\dfrac{C}{x-2}+\dfrac{D}{x+1}.

    The common denominator is, as always, the denominator of the LHS. We multiply each fraction by what it's missing in order to obtain the common denominator:

    \dfrac{1}{x(x-1)(x-2)(x+1)}=\dfrac{A}{x}\,\dfrac{(x-1)(x-2)(x+1)}{(x-1)(x-2)(x+1)}+\dfrac{B}{x-1}\,\dfrac{x(x-2)(x+1)}{x(x-2)(x+1)}

    +\dfrac{C}{x-2}\,\dfrac{x(x-1)(x+1)}{x(x-1)(x+1)}+\dfrac{D}{x+1}\,\dfrac{x(x-1)(x-2)}{x(x-1)(x-2)}.

    Thus, we can equate the following:

    \dfrac{1}{x(x-1)(x-2)(x+1)}=\dfrac{A(x-1)(x-2)(x+1)+Bx(x-2)(x+1)+Cx(x-1)(x+1)+Dx(x-1)(x-2)}{x(x-1)(x-2)(x+1)}.

    Now, we equate the numerators, because the denominators are equal. In addition, I will start multiplying out the RHS's numerator. We get

    1=A(x-1)(x^{2}-x-2)+Bx(x^{2}-x-2)+Cx(x^{2}-1)+Dx(x^{2}-3x+2).

    One more multiplication:

    1=A(x^{3}-x^{2}-x^{2}+x-2x+2)+B(x^{3}-x^{2}-2x)+C(x^{3}-x)+D(x^{3}-3x^{2}+2x), \quad\to

    1=A(x^{3}-2x^{2}-x+2)+B(x^{3}-x^{2}-2x)+C(x^{3}-x)+D(x^{3}-3x^{2}+2x).

    Now, here's another key concept. At this point, we need to set up a system of equations to solve for the coefficients A, B, C, D. The only way this equation can hold is if all the coefficients of like powers of x are the same on both sides of the equation. So, reading off the coefficients, we get the following 4 equations:

    0=A+B+C+D\quad\to\quad x^{3}\;\text{equation}

    0=-2A-B-3D\quad\to\quad x^{2}\;\text{equation}

    0=-A-2B-C+2D\quad\to\quad x\;\text{equation}

    1=2A\quad\to\quad\text{constant equation}.

    Did you see how I got those four equations?

    Ok, next step is to solve these four equations. One slight shortcut I'll take is to notice right away that A=1/2, and plug that into the remaining equations to reduce to a system of 3 equations in 3 unknowns. We get:

    -1/2=B+C+D

    1=-B-3D

    1/2=-2B-C+2D.

    The augmented matrix is then

    \left[\begin{matrix}1&1&1\\-1&0&-3\\-2&-1&2\end{matrix}\left|\begin{matrix}-1/2\\1\\1/2\end{matrix}\right].

    Using the usual Gaussian elimination with back substitution, we go through the following steps:

    \left[\begin{matrix}1&1&1\\-1&0&-3\\-2&-1&2\end{matrix}\left|\begin{matrix}-1/2\\1\\1/2\end{matrix}\right]\quad R_{1}+R_{2}\mapsto R_{2},\;2R_{1}+R_{3}\mapsto R_{3}

    \left[\begin{matrix}1&1&1\\0&1&-2\\0&1&4\end{matrix}\left|\begin{matrix}-1/2\\1/2\\-1/2\end{matrix}\right]\quad -R_{2}+R_{3}\mapsto R_{3}

    \left[\begin{matrix}1&1&1\\0&1&-2\\0&0&6\end{matrix}\left|\begin{matrix}-1/2\\1/2\\-1\end{matrix}\right].

    Using back substitution tells us that

    6D=-1\quad\to\quad D=-1/6.

    C=2D+1/2=2(-1/6)+1/2=-1/3+1/2=1/6.

    B=-C-D-1/2=-1/6+1/6-1/2=(-2+1-4)/8=-1/2.

    Thus, we have

    \dfrac{1}{x(x-1)(x-2)(x+1)}=\dfrac{1/2}{x}+\dfrac{-1/2}{x-1}+\dfrac{1/6}{x-2}+\dfrac{-1/6}{x+1}.

    How can you check this? Well, by getting a common denominator and multiplying it out. That wouldn't be bad practice for you.

    This method of solution I have provided is the most general method for doing partial fractions, as far as I know. There are some shortcuts you can take, particularly in this example. However, I have not used them, because they are not general. Or at least, their generalization gets into more difficult complex analysis.

    Are there any portions of this example which you don't understand?
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  5. #35
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    I know how to deal with linear factors. Im not sure how you could do that with repeated factors. I applied that to 1 + 2s / (s-1)(s-2)(s)^2 and it didnt make sense.
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  6. #36
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    With repeated factors, you have to change your initial guess. You do this:

    \dfrac{1}{x^{2}}=\dfrac{A}{x}+\dfrac{B}{x^{2}}.

    Then go through precisely the same procedure I outlined before.

    Another snag you'll sometimes run across is a factor that does not itself factor over the real numbers. Ones like this:

    \dfrac{1}{2x^{2}+x+5}. In those cases, your initial guess must be

    \dfrac{1}{2x^{2}+x+5}=\dfrac{Ax+B}{2x^{2}+x+5}.

    But you would still, in all these cases, perform the operations I outlined above.
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  7. #37
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    Could you tell me how you put your equations the way you have? I have used word to put an example in your way but I cant insert the image.
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  8. #38
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    Go here.
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  9. #39
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    right I've done this problem here:

    1+2s / (s-1)(s-2)(s)^2

    can you confirm that C = -7/4

    because in my exercise book the answers say C = 7/4

    and for 2 / (s-1)(s-6)(s-2)^2 C = -3/16 because the answers say C = 3/8
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  10. #40
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    I can't tell you whether you're correct or not, because you haven't told me what your guess is. How are you breaking everything out?
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  11. #41
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    I did it your way and got

    1 + 2s = A(s-2)(s)^2 + B(s-1)(s)^2 + C(s-1)(s-2)(s)^2 + D(s-1)(s-2)
    worked from here finding the values when s=2, s=1 s=0 and choose s =-1 to find C
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  12. #42
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    You need to shore up your communication style. You have an interesting habit of not answering the question I asked! I asked what your guess was. And you gave me post # 41, which doesn't answer my question. What I was looking for is something like this (I'm inferring here, so please confirm or correct this):

    \dfrac{1+2s}{s^{2}(s-1)(s-2)}=\dfrac{A}{s-1}+\dfrac{B}{s-2}+\dfrac{C}{s}+\dfrac{D}{s^{2}}.

    Is this correct? Is this your guess?
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  13. #43
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    \dfrac{1+2s}{(s-1)(s-2)s^{2}}=\dfrac{A}{s-1}+\dfrac{B}{s-2}+\dfrac{C}{s}+\dfrac{D}{s^{2}}.
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  14. #44
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    The code didnt work. I put s^2 after (s-1)(s-2) not before
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  15. #45
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    You have to enclose the LaTeX code with math brackets. That is, enclose with (MATH) *code here* (/MATH), only with square braces instead of parentheses. If you click on the Advanced button, you will see the TeX button, which will automatically enclose your selection in math delimiters, or simply insert new delimiters.

    The order of terms in the LHS denominator is unimportant, since regular multiplication is commutative. The important thing is to know which constants A, B, C, or D is lined up with which denominator on the RHS. So, C is the coefficient of the 1/s term.

    It looks like, from post # 41 that you're trying to use the Heaviside cover-up method. That works great if all the terms are linear and first-order. It gets more complicated if you have squared terms, like you have. I would hesitate to use it. The method I have proposed will work in all circumstances, as far as I know.

    Show me your work. How did you arrive at C = -7/4?
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