Old HSC Question Help (1 Viewer)

Jonomyster · Oct 2, 2018

So I was doing this question from the 1967 4U HSC which is as follows:

$\text{Express } \dfrac{1-abx^2}{(1-ax)(1-bx)} \text{ in the form: } p + \dfrac{q}{1-ax} + \dfrac{r}{1-bx}$

$\text{Given that } R_n(x) \text{ is a polynomial, and that:}$

$1-abx^2 = (1-ax)(1-bx)(1+u_1x + u_2x^2 + ... + u_nx^n) + x^{n+1}R_n(x)$

$\text{Find } u_r \text{ and } R_n(x)$

Anyway, here is my progress on the problem:

$\text{It can be shown by partial fractions that: }$

$\dfrac{1-abx^2}{(1-ax)(1-bx)} = -1 + \dfrac{1}{1-ax} + \dfrac{1}{1-bx} \text{ (I don't even use this fact...)}$

$\text{Now, } R_n(x) \text{ only affects terms with index n+1 and n+2, therefore: } R_n(x) = Ax + B \text{ and comparing terms with index n+1 and n+2 it can be shown that: }$

$R_n(x) = -abu_nx + (a+b)u_n - abu_{n-1}$

$\text{Now, comparing coefficients of } x^1 \text{ and } x^2 \text{ it can be shown that: }$

$u_1 = a+b \text{ and } u_2 = a^2 + b^2$

$\text{And by comparing the coefficient of } x^r \text{ it can be shown that: }$

$u_r = (a+b)u_{r-1} -abu_{r-2}$

$\text{It follows inductively that: } u_r = a^r + b^r$

So since I have solved the problem, you might be wondering why I am even making this post. The reason why is that the 1967 HSC has no solutions that I could find, and I am pretty sure that such a solution would make use of the partial fractions in the very first line of this post, which I did not. I am also quite interested in what a solution using partial fractions would look like as it could have some interesting techniques, tricks, etc. If anyone has any insights into solving the problem with partial fractions that would be greatly appreciated. Thanks!!

aa180 · Oct 2, 2018

Jonomyster said:
So I was doing this question from the 1967 4U HSC which is as follows:

$\text{Express } \dfrac{1-abx^2}{(1-ax)(1-bx)} \text{ in the form: } p + \dfrac{q}{1-ax} + \dfrac{r}{1-bx}$

$\text{Given that } R_n(x) \text{ is a polynomial, and that:}$

$1-abx^2 = (1-ax)(1-bx)(1+u_1x + u_2x^2 + ... + u_nx^n) + x^{n+1}R_n(x)$

$\text{Find } u_r \text{ and } R_n(x)$

Anyway, here is my progress on the problem:

$\text{It can be shown by partial fractions that: }$

$\dfrac{1-abx^2}{(1-ax)(1-bx)} = -1 + \dfrac{1}{1-ax} + \dfrac{1}{1-bx} \text{ (I don't even use this fact...)}$

$\text{Now, } R_n(x) \text{ only affects terms with index n+1 and n+2, therefore: } R_n(x) = Ax + B \text{ and comparing terms with index n+1 and n+2 it can be shown that: }$

$R_n(x) = -abu_nx + (a+b)u_n - abu_{n-1}$

$\text{Now, comparing coefficients of } x^1 \text{ and } x^2 \text{ it can be shown that: }$

$u_1 = a+b \text{ and } u_2 = a^2 + b^2$

$\text{And by comparing the coefficient of } x^r \text{ it can be shown that: }$

$u_r = (a+b)u_{r-1} -abu_{r-2}$

$\text{It follows inductively that: } u_r = a^r + b^r$

So since I have solved the problem, you might be wondering why I am even making this post. The reason why is that the 1967 HSC has no solutions that I could find, and I am pretty sure that such a solution would make use of the partial fractions in the very first line of this post, which I did not. I am also quite interested in what a solution using partial fractions would look like as it could have some interesting techniques, tricks, etc. If anyone has any insights into solving the problem with partial fractions that would be greatly appreciated. Thanks!!

Basically, you have:

$1 + \sum\limits_{k=1}^{n}u_{k}x^{k} + \frac{x^{n+1}R_{n}(x)}{(1-ax)(1-bx)} \equiv -1 + \frac{1}{1-ax} + \frac{1}{1-bx}$

Use the trick

$\frac{1}{1-ax} = 1 + \frac{ax}{1-ax} = 1 + ax(1+ \frac{ax}{1-ax}) = 1 + ax + a^{2}x^{2}(1+\frac{ax}{1-ax}) = ... = 1 + \sum\limits_{k=1}^{n}a^{k}x^{k} + \frac{a^{n+1}x^{n+1}}{1-ax}.$

Do the same for $\frac{1}{1-bx}$ and combine like terms with the appropriate terms from the $\frac{1}{1-ax}$ expansion to facilitate direct comparison of coefficients with the LHS, as well as bring the trailing terms on both expansions together over a common denominator to get your expression for $R_{n}(x)$ .

Old HSC Question Help (1 Viewer)

Jonomyster

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aa180

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