Fraction decomposition continued

**Greymalkin** · August 14th 2012, 12:25 PM

Decompose:
${x\over {x^4-a^4}}$

answer is:

${-{{{1\over 2a^2}x}\over {x^2+a^2}}}+{{{1\over 4a^2}\over {x-a}}+{{1\over 4a^2}\over {x+a}}$

I solve for b= ${a\over 4a^3}={1\over {4a^2}}$ , however i can't seem to solve for a or c.(if x=-a then -a^2 and -a makes all factors =0)
I try using the quadratic method but i end up with equations that are cubic.

**Plato** · August 14th 2012, 01:16 PM

Originally Posted by Greymalkin

Decompose:
${x\over {x^4-a^4}}$
answer is: ${-{{{1\over 2a^2}x}\over {x^2+a^2}}}+{{{1\over 4a^2}\over {x-a}}+{{1\over 4a^2}\over {x+a}}$
I solve for b= ${a\over 4a^3}={1\over {4a^2}}$ , however i can't seem to solve for a or c.(if x=-a then -a^2 and -a makes all factors =0) I try using the quadratic method but i end up with equations that are cubic.

What exactly is the question? I don't understand what you are asking.

**Prove It** · August 14th 2012, 01:21 PM

Originally Posted by Greymalkin

Decompose:
${x\over {x^4-a^4}}$

answer is:

${-{{{1\over 2a^2}x}\over {x^2+a^2}}}+{{{1\over 4a^2}\over {x-a}}+{{1\over 4a^2}\over {x+a}}$

I solve for b= ${a\over 4a^3}={1\over {4a^2}}$ , however i can't seem to solve for a or c.(if x=-a then -a^2 and -a makes all factors =0)
I try using the quadratic method but i end up with equations that are cubic.

$\displaystyle \begin{align*} \frac{Ax + B}{x^2 + a^2} + \frac{C}{x - a} + \frac{D}{x + a} &\equiv \frac{x}{x^4 - a^4} \\ \frac{(Ax + B)(x - a)(x + a) + C\left(x^2 + a^2\right)(x + a) + D\left(x^2 + a^2\right)(x - a)}{\left(x^2 + a^2\right)(x - a)(x + a)} &\equiv \frac{x}{x^4 - a^4} \\ (Ax + B)(x - a)(x + a) + C\left(x^2 + a^2\right)(x + a) + D\left(x^2 + a^2\right)(x - a) &\equiv x \end{align*}$

Let $\displaystyle \begin{align*} x = a \end{align*}$ to find $\displaystyle \begin{align*} 4a^3 C = a \implies C = \frac{1}{4a^2} \end{align*}$

Let $\displaystyle \begin{align*} x = -a \end{align*}$ to find $\displaystyle \begin{align*} -4a^3 D = -a \implies D = \frac{1}{4a^2} \end{align*}$

Substituting these values into the equation, we have

$\displaystyle \begin{align*} (Ax + B)(x - a)(x + a) + \frac{1}{4a^2}\left(x^2 + a^2\right)(x + a) + \frac{1}{4a^2}\left(x^2 + a^2\right)(x - a) &\equiv x \end{align*}$

So letting x equal any two numbers you like (say 0 and 1) will give you two equations to solve simultaneously for A and B.

**Greymalkin** · August 14th 2012, 02:07 PM

Originally Posted by Plato

What exactly is the question? I don't understand what you are asking.

Apologies, had I not broken the cardinal rule of math help for not showing my work you would have understood. Originally my work did not have 4 variables.

It was along the lines of:

$\frac {A}{x^2+a^2} + \frac {B}{x-a} + \frac {C}{x+a}$

**Greymalkin** · August 14th 2012, 03:01 PM

Originally Posted by Prove It

Let $\displaystyle \begin{align*} x = -a \end{align*}$ to find $\displaystyle \begin{align*} -4a^3 D = -a \implies D = \frac{1}{4a^2} \end{align*}$

Doesnt -a make $D(x^2+a^2)(x-a)$ = $D=D(-a^2+a^2)(-a-a)$ ???

**Prove It** · August 14th 2012, 03:11 PM

Originally Posted by Greymalkin

Doesnt -a make $D(x^2+a^2)(x-a)$ = $D=D(-a^2+a^2)(-a-a)$ ???

No, if $\displaystyle \begin{align*} x = -a \end{align*}$ then $\displaystyle \begin{align*} x^2 = (-a)^2 = a^2 \end{align*}$ .

**Plato** · August 14th 2012, 03:15 PM

Originally Posted by Greymalkin

It was along the lines of:
$\frac {A}{x^2+a^2} + \frac {B}{x-a} + \frac {C}{x+a}$

I don't know why we still teach this topic. I consider is an obsolete topic.
Look at this.

**Prove It** · August 14th 2012, 03:17 PM

Originally Posted by Plato

I don't know why we still teach this topic. I consider is an obsolete topic.
Look at this.

I consider it highly relevant.

**Greymalkin** · August 14th 2012, 03:25 PM

Originally Posted by Plato

I don't know why we still teach this topic. I consider is an obsolete topic.
Look at this.

Yeah I usually use wolfram for finding zeros of high degree polynomials, but only after I forced myself to learn the rational root theorem. I am self teaching myself math so I'm not sure what I'll need what for, so I choose to learn it anyways(Havent started calculus yet, but my goal is to do econometrics). A method that, although time consuming, has proven fruitful in insight. My goal is to have a deep understanding of math, not just to "get by", as I suppose most students tend to do(I used to be one).

**Greymalkin** · August 14th 2012, 03:34 PM

Originally Posted by Prove It

No, if $\displaystyle \begin{align*} x = -a \end{align*}$ then $\displaystyle \begin{align*} x^2 = (-a)^2 = a^2 \end{align*}$ .

How can it be positive if it has not yet been squared though? Isn't that like saying x^2=-x^2, the graphs are reflections of one another. Is it because "a" is a constant and not a variable in cartesian coordinates, so therefore it cannot have a domain/range relationship? think I might have answered my own question but please correct me if I'm wrong.

**Plato** · August 14th 2012, 03:41 PM

Originally Posted by Prove It

I consider it highly relevant.

It is relevant only if one is concerned with teaching basic algebra.
How is it relevant to a calculus course? In fact, I predict that within five to ten years one will hard pressed to find a calculus textbook that includes a chapter on techniques of integration. I have taught calculus on and off since 1964. I never thought that technology would replace basics techniques. But it has and will continue to do so. Partial fractions are gone as another victim of technology.

**Prove It** · August 14th 2012, 05:50 PM

Originally Posted by Plato

It is relevant only if one is concerned with teaching basic algebra.
How is it relevant to a calculus course? In fact, I predict that within five to ten years one will hard pressed to find a calculus textbook that includes a chapter on techniques of integration. I have taught calculus on and off since 1964. I never thought that technology would replace basics techniques. But it has and will continue to do so. Partial fractions are gone as another victim of technology.

I find it hard to believe that you would believe that it SHOULD be replaced though...

**Prove It** · August 14th 2012, 05:51 PM

Originally Posted by Greymalkin

How can it be positive if it has not yet been squared though? Isn't that like saying x^2=-x^2, the graphs are reflections of one another. Is it because "a" is a constant and not a variable in cartesian coordinates, so therefore it cannot have a domain/range relationship? think I might have answered my own question but please correct me if I'm wrong.

What happens when you square a negative value (i.e. multiply a negative value by itself - another negative value)? It becomes positive...

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