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Math Help - how do you decompose this into partial fraction?

  1. #1
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    how do you decompose this into partial fraction?

    . I am revising for my end of year exams. Hence, I have been throwing lots of questions lately.

    Express \dfrac{2}{x(x^2+5)} in partial fractions.

    What are the structures for the corresponding partial fractions in the first place?

    Method 1: conventional method of substitution
    Method 2: Cover-up method

    Which method should I use? How?

    PS: my latex skills improved!
    Last edited by stupidguy; September 11th 2010 at 01:11 AM.
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  2. #2
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    Quote Originally Posted by stupidguy View Post
    . I am revising for my end of year exams. Hence, I have been throwing lots of questions lately.

    Express \frac{2}{x(x^2+5)} in partial fractions.

    What are the denominators of the partial fractions in the first place?

    PS: my latex skills improved!

    check those...





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  3. #3
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    Quote Originally Posted by stupidguy View Post
    Express \frac{2}{x(x^2+5)} in partial fractions.
    PS: my latex skills improved!
    [tex]\dfrac{2}{x(x^2+5)}[/tex] gives \dfrac{2}{x(x^2+5)}.
    After the update we need to make some changes.
    Notice the \dfrac tag for \displaystyle
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  4. #4
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    Quote Originally Posted by Plato View Post
    [tex]\dfrac{2}{x(x^2+5)}[/tex] gives \dfrac{2}{x(x^2+5)}.
    After the update we need to make some changes.
    Notice the \dfrac tag for \displaystyle
    Changed!
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  5. #5
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    Let \displaystyle \frac{2}{x(x^2+5)} = \frac{a}{x}+\frac{bx+c}{x^2+5}.
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    Quote Originally Posted by TheCoffeeMachine View Post
    Let \displaystyle \frac{2}{x(x^2+5)} = \frac{a}{x}+\frac{bx+c}{x^2+5}.
    Is it when I see x the corresponding partial fraction is \dfrac{a}{x}?

    you treat x as a linear factor?
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  7. #7
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    Quote Originally Posted by TheCoffeeMachine View Post
    Let \displaystyle \frac{2}{x(x^2+5)} = \frac{a}{x}+\frac{bx+c}{x^2+5}.
    can I still use cover-up method?
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  8. #8
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    Quote Originally Posted by stupidguy View Post
    Is it when I see x the corresponding partial fraction is \frac{a}{x}?
    Yes! The corresponding term for any (unrepeated) linear factor in the denominator is \frac{a}{x}.
    Is it a rule cos my textbook dun show it?
    If f(x) is of less degree than the denominator, then:
    The corresponding partial fraction form of \displaystyle \frac{f(x)}{(x+n_{1})(x+n_{2})(x+n_{3})\cdots(x+n_  {r})}[/Math]<br />
 is [Math]\displaystyle \frac{a_{1}}{x+n_{1}}+\frac{a_{2}}{x+n_{2}}+\frac{  a_{3}}{x+n_{3}}+\cdots +\frac{a_{r}}{x+n_{r}}
    Quote Originally Posted by stupidguy View Post
    can I still use cover-up method?
    Of course.
    Last edited by TheCoffeeMachine; September 11th 2010 at 02:49 AM.
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  9. #9
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    a linear factor has the structure of ax+b. so you are treating a as 1 and b as 0, right?

    If the denominator is 2x, the corresponding partial fraction is \dfrac{a}{2x}?
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  10. #10
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    did you guys get \dfrac{2}{5x}-\dfrac{x+1}{5x^2+25}?
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  11. #11
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    Quote Originally Posted by stupidguy View Post
    a linear factor has the structure of ax+b. so you are treating a as 1 and b as 0, right?
    If you have ax+b in the denominator and it's not repeated then you have its corresponding
    term is of the form Ax+B, with B = 0, x = 1, yes. So it's just a constant, call it just A.
    If the denominator is 2x, the corresponding partial fraction is \dfrac{a}{2x}?
    That's right.
    Quote Originally Posted by stupidguy View Post
    did you guys get \dfrac{2}{5x}-\dfrac{x+1}{5x^2+25}?
    You got a right. I should have cautioned that the cover-up method would only work for finding a.

    Spoiler:
    First, set-up:

    \displaystyle \frac{2}{x(x^2+5)} = \frac{a}{x}+\frac{bx+c}{x^2+5}

    You can see that if you cover-up (x^2+5), then a = \frac{2}{5}. But that is not going to give us b and c, so let's do it in the saver way.

    Put the RHS over a common denomintor:

    \displaystyle \frac{2}{x(x^2+5)} = \frac{a(x^2+5)+(bx+c)(x)}{x(x^2+5)}<br />

    We have 2 = a(x^2+5)+(bx+c)(x)

    [Stop for a moment. Let x = 0: Then we have 2 = 5a. So a = \frac{2}{5}. The cover-up method was really this in disguise. You can see why it doesn't work for finding b and c -- there is no value of x which can make a(x^2+5) equal to zero. So if in general you have a partial fraction form of \frac{Ax+B}{ax^2+bx+c}, then you cannot find A and B by covering-up -- in other words, the denominator have to be linear for the cover-up to work.]

    We have:

    2 = ax^2+5a+bx^2+cx (expanding the right-hand side).

    \Rightarrow 2 = ax^2+bx^2+cx+5a (writing it in terms of descending powers of x).

    <br />
\Rightarrow 2 = (a+b)x^2+(c)x+5a (factoring out the x's out).

    Now, compare the coefficients:

    We have c = 0.
    5a = 2[/Math], so [Math]a = \frac{2}{5}.
    a+b = 0, and since a = \frac{2}{5}, we get \frac{2}{5}+b = 0, hence  b= -\frac{2}{5}.

    Therefore \displaystyle \frac{2}{x(x^2+5)} = \frac{2}{5x}-\frac{2x}{5(x^2+5)}.

    It should of course take a much smaller space to write-up, but that's Mr Beefcake's attempt at being clear.
    Last edited by TheCoffeeMachine; September 12th 2010 at 03:15 AM.
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  12. #12
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    thanks.

    BTW, I did not do cover-up for the quadratic factor. I made a careless mistake during substitution, resulting in the wrong answer.

    Anyway thanks. Your reply will still be useful for others to learn.
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