Author Topic: 4U Maths Question Thread (Read 665491 times) Tweet Share

katherine123 · « **Reply #150 on:** April 22, 2016, 02:20:28 pm »

- make x/[(x-1)^3(x-2)] into partial fractions

why is it A/(x-1) + (BX+C)/(x-1)^2 + D/(x-2) instead of AX+B/(x-1)^2 + C/(x-2)

- Is the cover up method commonly used and when does it not work?

- Will 4x-3/[x^3(x+1)] decompose into A/x+ B/x^2 + C/x^3 + D/(x+1) and why isnt B linear and C a quadratic

RuiAce · « **Reply #151 on:** April 22, 2016, 07:04:48 pm »

Quote from: katherine123 on April 22, 2016, 02:20:28 pm

- make x/[(x-1)^3(x-2)] into partial fractions

why is it A/(x-1) + (BX+C)/(x-1)^2 + D/(x-2) instead of AX+B/(x-1)^2 + C/(x-2)

- Is the cover up method commonly used and when does it not work?

- Will 4x-3/[x^3(x+1)] decompose into A/x+ B/x^2 + C/x^3 + D/(x+1) and why isnt B linear and C a quadratic

Look at your first question again. It is confusing. All I gathered was this:
$\frac{x}{{\left(x-1\right)}^{3}\left(x-2\right)}$
Which decomposes into
$\frac{A}{x-1}+\frac{B}{{\left(x-1\right)}^{2}}+\frac{C}{{\left(x-1\right)}^{3}} +\frac{D}{x-2}$

As for Heaviside Cover-up, that only works for linear combinations. The instant you throw in a quadratic (or something higher) into the denominator, it won't work as a neater process.

Question two is rightfully so as the idea is the implementation of powers allows ways to abide by the unnecessarily complicated method of introducing a numerator with linear terms in it. To make it seem a bit more obvious however:

(I will ignore the term with x³ as the process is similar)
$\frac { A }{ x } +\frac { Bx+C }{ { x }^{ 2 } } \\ =\frac { A }{ x } +\frac { Bx }{ { x }^{ 2 } } +\frac { C }{ { x }^{ 2 } } \\ =\frac { A }{ x } +\frac { B }{ x } +\frac { C }{ { x }^{ 2 } } \\ =\frac { A+B }{ x } +\frac { C }{ { x }^{ 2 } } \\\text{Hence we have a redundancy as we can just let }E=A+B$

katherine123 · « **Reply #152 on:** April 23, 2016, 09:33:40 pm »

1)how to make 1/(x^2+2x-1) into impartial fractions
is it not decomposible since the denominator cant be factorised

2) how to make 1/[x(x^2-1)^2] into impartial fraction

so will it be
A/x + B/(x-1) + C/(x-1)^2 + D/(x+1) + E/(x+1)^2

RuiAce · « **Reply #153 on:** April 23, 2016, 10:03:05 pm »

Quote from: katherine123 on April 23, 2016, 09:33:40 pm

1)how to make 1/(x^2+2x-1) into impartial fractions
is it not decomposible since the denominator cant be factorised

2) how to make 1/[x(x^2-1)^2] into impartial fraction

so will it be
A/x + B/(x-1) + C/(x-1)^2 + D/(x+1) + E/(x+1)^2

$\text{They're just called "partial" fractions.}$
$\text{The expression generated for Q2 is a correct one.}$

$\text{Q1 isn't exactly the neatest thing to decompose into partial fractions, however it can still be done.}\\\text{Note that the expression can still be factorised over }\mathbb R:\\{x}^{2}+2x-1=\left(x+1+\sqrt{2}\right)\left(x+1-\sqrt{2}\right)\\\text{It just so happens that the surd doesn't exactly give you something pleasant:}\\ \frac{A}{\left(x+1+\sqrt{2}\right)}+\frac{B}{\left(x+1-\sqrt{2}\right)}$

katherine123 · « **Reply #154 on:** April 26, 2016, 12:23:12 am »

how to integrate this x^3/ (x^2-x-3) thanks

RuiAce · « **Reply #155 on:** April 26, 2016, 05:39:11 pm »

Quote from: katherine123 on April 26, 2016, 12:23:12 am

how to integrate this x^3/ (x^2-x-3) thanks

$\text{Normal approach: Start by polynomial long division without hesitation}$
$\text{Cheeky approach: Do the long division by being smart with algebra:}\\\begin{align*}\frac{{x}^{3}}{{x}^{2}-x-3}&=\frac{{x}^{3}-{x}^{2}-3x}{{x}^{2}-x-3}+\frac{{x}^{2}+3x}{{x}^{2}-x-3}\\&=x+\frac{{x}^{2}-x-3}{{x}^{2}-x-3}+\frac{4x+3}{{x}^{2}-x-3}\\&=x+1+\frac{4x+3}{{x}^{2}-x-3}\end{align*}$
$\text{This is much easier to integrate.}$
$\begin{align*}\int{\frac{{x}^{3}}{{x}^{2}-x-3}dx}&=\int{\left(x+1+\frac{4x+3}{{x}^{2}-x-3}\right)dx}\\&=\frac{x^2}{2}+x+\int{\left(\frac{4x-2}{{x}^{2}-x-3}+\frac{5}{{x}^{2}-x-3}\right)dx}\end{align*}$

First one is done by a u-substitution (or just by inspection), whereas the second is done using completing the square.

Happy Physics Land · « **Reply #156 on:** April 26, 2016, 07:55:14 pm »

Quote from: RuiAce on April 26, 2016, 05:39:11 pm

$\text{Normal approach: Start by polynomial long division without hesitation}$
$\text{Cheeky approach: Do the long division by being smart with algebra:}\\\begin{align*}\frac{{x}^{3}}{{x}^{2}-x-3}&=\frac{{x}^{3}-{x}^{2}-3x}{{x}^{2}-x-3}+\frac{{x}^{2}+3x}{{x}^{2}-x-3}\\&=x+\frac{{x}^{2}-x-3}{{x}^{2}-x-3}+\frac{4x+3}{{x}^{2}-x-3}\\&=x+1+\frac{4x+3}{{x}^{2}-x-3}\end{align*}$
$\text{This is much easier to integrate.}$
$\begin{align*}\int{\frac{{x}^{3}}{{x}^{2}-x-3}dx}&=\int{\left(x+1+\frac{4x+3}{{x}^{2}-x-3}\right)dx}\\&=\frac{x^2}{2}+x+\int{\left(\frac{4x-2}{{x}^{2}-x-3}+\frac{5}{{x}^{2}-x-3}\right)dx}\end{align*}$

First one is done by a u-substitution (or just by inspection), whereas the second is done using completing the square.

That was awesome! When I attempted it I had no idea what to do after long division with that 4x - 3 / x² - x - 3 thing

RuiAce · « **Reply #157 on:** April 27, 2016, 09:25:15 pm »

Quote from: Happy Physics Land on April 26, 2016, 07:55:14 pm

That was awesome! When I attempted it I had no idea what to do after long division with that 4x - 3 / x² - x - 3 thing

Oh to just complete the rule of thumb

$\text{Generally, if you have something to integrate in the form }\frac{Ax+B}{ax^2+bx+c}\\ \text{You always want to rearrange it into some form }\frac{2ax+b}{ax^2+bx+c} + \frac{d}{ax^2+bx+c}$

$\text{The former is always done by integration by substitution (or reverse chain rule.} \\ \text{Obviously the latter you'd have to figure out if you should use partial fractions or integrate into tangent inverse.}\\\text{Completing the square may be necessary for tangent inverse.}$

katherine123 · « **Reply #158 on:** April 28, 2016, 12:03:51 am »

How to make x^3/ [(x-1)(x^2+2)^2] into impartial fractions:

ans: (8/9-1/9x)/(x^2+2) + (2/3x-4/3)/(x^2+2)^2 +1/9/(x-1)

doesnt the numerator becomes a constant when
you make something like 1/(x+2)^2 to A/(x+2) + B/(x+2)^2

RuiAce · « **Reply #159 on:** April 28, 2016, 06:31:11 am »

Quote from: katherine123 on April 28, 2016, 12:03:51 am

How to make x^3/ [(x-1)(x^2+2)^2] into impartial fractions:

ans: (8/9-1/9x)/(x^2+2) + (2/3x-4/3)/(x^2+2)^2 +1/9/(x-1)

doesnt the numerator becomes a constant when
you make something like 1/(x+2)^2 to A/(x+2) + B/(x+2)^2

$\text{Like I said, they're just called "PARTIAL" fractions. Not sure who said impartial.}\\\text{This question is an example of an unnecessarily tedious problem so I'll sketch it out:}\\ \frac{{x}^{3}}{\left(x-1\right){\left({x}^{2}+2\right)}^{2}}=\frac{A}{x-1}+\frac{Bx+C}{{x}^{2}+2}+\frac{Dx+E}{{\left({x}^{2}+2\right)}^{2}}\\{x}^{3}=A{\left({x}^{2}+2\right)}^{2}+\left(Bx+C\right)\left(x-1\right)\left({x}^{2}+2\right)+\left(Dx+E\right)\left(x-1\right)$
$\text{Note that your mistake (possibly transcription error) was somehow writing the quadratic denominator as a linear term.}$
$\text{Put }x=1\text{ first as that easily spits out the }A:\\1=A{\left(3\right)}^{2}+0+0\\A=\frac{1}{9}$
$\text{And now things get complicated.}\\\text{It may help a lot to just expand out the entire thing to visualise it more clearly, but I'll attempt to equate coefficients by inspection here:}\\ \text{Take coeff. }{x}^{4}:\\ 0=A+B\\B=-\frac{1}{9}$
Subject to inaccuracy now:
$\text{Coeff. }x^3: \quad 1=-B+C\\\text{Constant term: }\quad 0=4A-2C-E\\\text{etc.}$

jamonwindeyer · « **Reply #160 on:** April 28, 2016, 07:32:17 am »

Quote from: katherine123 on April 28, 2016, 12:03:51 am

How to make x^3/ [(x-1)(x^2+2)^2] into impartial fractions:

ans: (8/9-1/9x)/(x^2+2) + (2/3x-4/3)/(x^2+2)^2 +1/9/(x-1)

doesnt the numerator becomes a constant when
you make something like 1/(x+2)^2 to A/(x+2) + B/(x+2)^2

To clarify on Rui's response:

When you need a partial fraction decomposition of something with a fully factorised quadratic term on the bottom, for example:

$\frac{1}{x(x^2+7)}$

Then it gets broken down like so:

$\frac{1}{x(x^2+7)}=\frac{A}{x}+\frac{Bx+C}{x^2+7}$

For an irreducible quadratic factor, the numerator takes the form (Cx + D)

Also remember this technique only works for proper fractions, where the degree of the denominator is larger than the degree of the numerator.

Hope this helps you a tad!

jamonwindeyer · « **Reply #161 on:** April 28, 2016, 10:54:54 am »

I'll also add, on the topic of Partial Fractions, that over half of my lecture on signal transformations this morning was work on Partial fraction decomposition and the Heaviside method. 2nd year university courses teaching very standard 4U content. Another example of why 4 Unit math is an amazing head start for many Math, Science, and Engineering degrees!!

RuiAce · « **Reply #162 on:** April 28, 2016, 08:34:32 pm »

Quote from: jamonwindeyer on April 28, 2016, 10:54:54 am

I'll also add, on the topic of Partial Fractions, that over half of my lecture on signal transformations this morning was work on Partial fraction decomposition and the Heaviside method. 2nd year university courses teaching very standard 4U content. Another example of why 4 Unit math is an amazing head start for many Math, Science, and Engineering degrees!!

You gotta admit though... without Wolfram or something some really messy partial fractions like that above one are tedious

katherine123 · « **Reply #163 on:** April 28, 2016, 11:30:37 pm »

Quote from: jamonwindeyer on April 28, 2016, 07:32:17 am

To clarify on Rui's response:

When you need a partial fraction decomposition of something with a fully factorised quadratic term on the bottom, for example:

$\frac{1}{x(x^2+7)}$

Then it gets broken down like so:

$\frac{1}{x(x^2+7)}=\frac{A}{x}+\frac{Bx+C}{x^2+7}$

For an irreducible quadratic factor, the numerator takes the form (Cx + D)

Also remember this technique only works for proper fractions, where the degree of the denominator is larger than the degree of the numerator.

Hope this helps you a tad!

so for eg. 1/(x^3+1)^2
the partial fractions will be (Ax^2+Bx+C)/(x^3+1) + (Dx^2+Ex+F)/(x^3+1)^2
since the denominator has irreducible cubic factor , the numerator will be quadratic

RuiAce · « **Reply #164 on:** April 29, 2016, 07:27:09 am »

Quote from: katherine123 on April 28, 2016, 11:30:37 pm

so for eg. 1/(x^3+1)^2
the partial fractions will be (Ax^2+Bx+C)/(x^3+1) + (Dx^2+Ex+F)/(x^3+1)^2
since the denominator has irreducible cubic factor , the numerator will be quadratic

Bad idea.

$\text{Any polynomial of higher than degree 2 can ALWAYS be broken down (by factorisation)}\\\text{into products of linear and/or quadratic components}$
$x^3+1=(x+1)\left(x^2-x+1\right)$

$\frac{1}{{\left(x^3+1\right)}^2}=\frac{A}{x}+\frac{B}{x^2}+\frac{Cx+D}{x^2-x+1}+\frac{Ex+F}{{\left(x^2-x+1\right)}^2}$
$\text{The necessity of 6 unknowns still holds, however this considerably simplifies the algebra.}$
$\textbf{The whole purpose of partial fractions in integration is to decompose into linear and quadratic combinations}\\\textbf{because it is ALWAYS possible to integrate the reciprocal of something linear/quadratic.}$

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