HSC 2013 MX2 Marathon (archive) (12 Viewers)

Sy123 · Jul 5, 2013

Re: HSC 2013 4U Marathon

$Z_n(\alpha) = \sum_{k=1}^{n}\frac{1}{k^{\alpha}}$

$$Prove that for$ \ \ \alpha > 1$

$\lim_{n \to \infty} Z_n (\alpha) = L \ \ $for some value$ \ L$

$$Also prove that if$ \ \ \alpha \leq 1$

$Z_n(\alpha) \to \infty \ \ \ $as$ \ n \ $increases without bound$$

Sy123 · Jul 5, 2013

Re: HSC 2013 4U Marathon

$A sequence is defined with the recurrence formula$

$F_{n+2} = F_n + F_{n+1} \ \ \ n\geq 0$

$F_0 = 0$

$F_1 = 1$

$$i) Prove with mathematical induction$ \ \ F_n F_{n+3} - F_{n+1}F_{n+2} = (-1)^{n+1}$

$ii) Hence prove that$

$\tan^{-1} \left(\frac{1}{F_{2r+1}} \right ) + \tan^{-1} \left(\frac{1}{F_{2r+2}} \right) = \tan^{-1} \left(\frac{1}{F_{2r}} \right )$

$iii) Hence evaluate the series which you may assume converges to a finite value$

$\sum_{k=1}^{\infty} \tan^{-1 } \left(\frac{1}{F_{2k+1}} \right )$

RealiseNothing · Jul 5, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$Z_n(\alpha) = \sum_{k=1}^{n}\frac{1}{k^{\alpha}}$

$$Prove that for$ \ \ \alpha > 1$

$\lim_{n \to \infty} Z_n (\alpha) = L \ \ $for some value$ \ L$

$$Also prove that if$ \ \ \alpha \leq 1$

$Z_n(\alpha) \to \infty \ \ \ $as$ \ n \ $increases without bound$$

The second part is relatively simple to prove. When $\alpha = 1$ we get the Harmonic Series which $\to \infty$ . You can construct the following bound to show this:

$\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + ... > \frac{1}{1} + \frac{1}{2} + \frac{1}{4} + \frac{1}{4} + \frac{1}{8} + ...$

The RHS $\to \infty$ and so the result falls out.

Now when $\alpha < 1$ we get $\frac{1}{k^{\alpha}} > \frac{1}{k}$ and so for $\alpha < 1$ we get $Z_n(\alpha) \to \infty$

RealiseNothing · Jul 5, 2013

Re: HSC 2013 4U Marathon

braintic said:
Let's light a fire here. No-one knows what school I teach at, right? But if you THINK I might be a teacher at your school, then there is a good chance some questions from this forum might be in your trial. Only 51 pages to sift through.

I feel sorry for your students.

RealiseNothing · Jul 5, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$Z_n(\alpha) = \sum_{k=1}^{n}\frac{1}{k^{\alpha}}$

$$Prove that for$ \ \ \alpha > 1$

$\lim_{n \to \infty} Z_n (\alpha) = L \ \ $for some value$ \ L$

$$Also prove that if$ \ \ \alpha \leq 1$

$Z_n(\alpha) \to \infty \ \ \ $as$ \ n \ $increases without bound$$

$Z_n(2) = \frac{\pi^2}{6}$

so for $\alpha \geq 2$ we get $\lim_{n \to \infty} Z_n(\alpha) = L$ for some value L.

Now all I need is the case where $2 > \alpha > 1$ , but I want to try do a more holistic proof rather than one by cases.

Sy123 · Jul 5, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
$Z_n(2) = \frac{\pi^2}{6}$

so for $\alpha \geq 2$ we get $\lim_{n \to \infty} Z_n(\alpha) = L$ for some value L.

Now all I need is the case where $2 > \alpha > 1$ , but I want to try do a more holistic proof rather than one by cases.

You're first part is correct, for the second though, I'd advise against the case by case basis, it might work though.

There is an easier method however.

Sy123 · Jul 6, 2013

Re: HSC 2013 4U Marathon

$$From a set of$ \ \ r \ \ $even integers$ \ \ x_1, x_2, \dots , x_r \ \ $and$ \ \ s \ \ $odd integers$ \ \ y_1, y_2, \dots , y_s$

$$Consider the set$ \ \left \{ (-1)^{x_1}, (-1)^{x_2} , (-1)^{x_3}, \dots , (-1)^{x_r} , (-1)^{y_1} , (-1)^{y_2} , \dots , (-1)^{y_s} \right \} \ \ \ \ \ (*)$

$$Write down a formula for the sum of the products taken$ \ \ k \ \ $numbers at a time from$ \ \ (*)$

$$Where$ \ \ p, q > k$

Carrotsticks · Jul 6, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
$Z_n(2) = \frac{\pi^2}{6}$

so for $\alpha \geq 2$ we get $\lim_{n \to \infty} Z_n(\alpha) = L$ for some value L.

Now all I need is the case where $2 > \alpha > 1$ , but I want to try do a more holistic proof rather than one by cases.

Consider the curve y=1/x^p in the interval [1,infinity) and partition it into rectangles of unit sub-interval, and construct upper bound rectangles.

RealiseNothing · Jul 6, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$Z_n(\alpha) = \sum_{k=1}^{n}\frac{1}{k^{\alpha}}$

$$Prove that for$ \ \ \alpha > 1$

$\lim_{n \to \infty} Z_n (\alpha) = L \ \ $for some value$ \ L$

$$Also prove that if$ \ \ \alpha \leq 1$

$Z_n(\alpha) \to \infty \ \ \ $as$ \ n \ $increases without bound$$

Constructing rectangle bounds for the curve $\frac{1}{x^{\alpha}}$ gives the inequality:

$\sum_{k=1}^{n} \frac{1}{k^{\alpha}} < \frac{n^{1-\alpha} - \alpha}{1-\alpha}$

Consider the RHS, let $p = \alpha -1$ where $\alpha > 1$

$\frac{n^{-p} - p - 1}{-p}$

$\frac{\frac{-1}{n^p} + p + 1}{p}$

As $n \to \infty$

$\frac{p+1}{p}$

$1 + \frac{1}{p}$

$1 + \frac{1}{\alpha - 1}$

$= L$

There may be an algebraic mistake somewhere as I haven't double checked my working, but it's the same idea nonetheless.

Sy123 · Jul 6, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
Constructing rectangle bounds for the curve $\frac{1}{x^{\alpha}}$ gives the inequality:

$\sum_{k=1}^{n} \frac{1}{k^{\alpha}} < \frac{n^{1-\alpha} - \alpha}{1-\alpha}$

Consider the RHS, let $p = \alpha -1$ where $\alpha > 1$

$\frac{n^{-p} - p - 1}{-p}$

$\frac{\frac{-1}{n^p} + p + 1}{p}$

As $n \to \infty$

$\frac{p+1}{p}$

$1 + \frac{1}{p}$

$1 + \frac{1}{\alpha - 1}$

$= L$

There may be an algebraic mistake somewhere as I haven't double checked my working, but it's the same idea nonetheless.

Yep that is pretty much it.

Sy123 · Jul 7, 2013

Re: HSC 2013 4U Marathon

$i) Prove$

$\int_1^2 f(x) \ dx = \lim_{n \to \infty} \left(\frac{1}{n} \sum_{k=1}^{n}f \left(1 + \frac{k}{n} \right ) \right )$

$ii) Hence evaluate$

$\lim_{n \to \infty} \left(\frac{n}{2n^2 + 2n + 1} + \frac{n}{2n^2 + 4n + 4} + \frac{n}{2n^2 + 6n + 9} + \frac{n}{2n^2 + 8n+ 16} + \dots + \frac{n^2}{2n^3 + 3n^2} \right )$

asianese · Jul 7, 2013

Re: HSC 2013 4U Marathon

Can students be expected to use the fundamental theorem of calculus freely?

Sy123 · Jul 7, 2013

Re: HSC 2013 4U Marathon

asianese said:
Can students be expected to use the fundamental theorem of calculus freely?

I don't see a need to but feel free to do so if it will help.

seanieg89 · Jul 7, 2013

Re: HSC 2013 4U Marathon

For something like part i) of your most recent question, you would need to provide the students with a rigorous definition of integration from which to work ...as this is not assumed knowledge in 4U. You also need to state what kind of functions f you want this to be proven for.

(Unless by "prove" you mean "provide a heuristic argument for why this is true using the informal mx2 notion of an integral").

RealiseNothing · Jul 7, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$i) Prove$

$\int_1^2 f(x) \ dx = \lim_{n \to \infty} \left(\frac{1}{n} \sum_{k=1}^{n}f \left(1 + \frac{k}{n} \right ) \right )$

$ii) Hence evaluate$

$\lim_{n \to \infty} \left(\frac{n}{2n^2 + 2n + 1} + \frac{n}{2n^2 + 4n + 4} + \frac{n}{2n^2 + 6n + 9} + \frac{n}{2n^2 + 8n+ 16} + \dots + \frac{n^2}{2n^3 + 3n^2} \right )$

The first part is taking infinite rectangles between the limits of 1 and 2. Their height will be $f(1+\frac{k}{n})$ with width $\frac{1}{n}$

Just sum this from 1 to infinity to get the area under the curve.

Sy123 · Jul 7, 2013

Re: HSC 2013 4U Marathon

seanieg89 said:
For something like part i) of your most recent question, you would need to provide the students with a rigorous definition of integration from which to work ...as this is not assumed knowledge in 4U. You also need to state what kind of functions f you want this to be proven for.

(Unless by "prove" you mean "provide a heuristic argument for why this is true using the informal mx2 notion of an integral").

Is there a problem with the argument; that considering finite n, the sum of the lower rectangles of width 1/n is approximately the area (which is the integral), and then saying as n increases without bound the rectangles all sum up to become the area?
Similar logic is there when deriving integrals for Volumes question, is this not rigorous/not in syllabus ?

I took the first part only from a STEP paper, it said 'Justify by means of a sketch', is using the word prove changing the meaning the original question had?

seanieg89 · Jul 7, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Is there a problem with the argument; that considering finite n, the sum of the lower rectangles of width 1/n is approximately the area (which is the integral), and then saying as n increases without bound the rectangles all sum up to become the area?
Similar logic is there when deriving integrals for Volumes question, is this not rigorous/not in syllabus ?

That's about as good as you can do with mx2 definitions and notions, but it is certainly not rigorous as the object you are proving things about (the Riemann integral) is not rigorously defined. Other things to think about:

1. What is the definition of area itself? (If you choose to define the integral via area).

2. How do you know that as n->inf the rectangles "area" approaches the "area" under the curve? Beyond the fact that the pictures look more and more alike.

3. You should state an assumption like continuity on f to guarantee those expressions make sense. If you don't assume something like this and your function is highly irregular, then the heuristic of measuring the "area" of things by rectangular approximations breaks down (and indeed the Riemann integral is not defined for such functions).

The main reason I am stressing these things is that typically, the integral is DEFINED via sums of the sort on the RHS, and "area under curves" is defined via this rigorous notion of integration. This puts things on a solid logical basis.

Sy123 · Jul 7, 2013

Re: HSC 2013 4U Marathon

seanieg89 said:
That's about as good as you can do with mx2 definitions and notions, but it is certainly not rigorous as the object you are proving things about (the Riemann integral) is not rigorously defined. Other things to think about:

1. What is the definition of area itself? (If you choose to define the integral via area).

2. How do you know that as n->inf the rectangles "area" approaches the "area" under the curve? Beyond the fact that the pictures look more and more alike.

3. You should state an assumption like continuity on f to guarantee those expressions make sense.

The main reason I am stressing these things is that typically, the integral is DEFINED via sums of the sort on the RHS, and "area under curves" is defined via this rigorous notion of integration. This puts things on a solid logical basis.

Ah alright then :s

Second part is still open though.

Sy123 · Jul 7, 2013

Re: HSC 2013 4U Marathon

$For some function$ \ \ f(t) \ \ $the operation$ \ \ L \ \ $is applied, the resultant function is$

$F(s) = \lim_{z \to \infty} \int_0^z e^{-st} f(t) \ dt \ \ \ \ \ (s>0)$

$Operation$ \ \ L \ \ $can only be applied to a function if its resultant function converges to some finite limit$

$$i) Show that if the operation$ \ \ L \ \ $is applied to $ \ \ f'(t) \ \ $then the resultant function is$ \ \ sF(s) - f(0)$

$ii) Apply the operation$ \ \ L \ \ $to$ \ \ f(t) = \sin t$

asianese · Jul 7, 2013

Re: HSC 2013 4U Marathon

Yay Laplace transforms!

$The resultant function will be$

$\begin{align*} \lim_{z\to \infty}\int_0^z e^{-st}f'(t)\mathop{dt}&= \lim_{z\to\infty} \left[ \left[f(t)e^{-st}\right]_0^z+\int_0^z s f(t)e^{-st} \mathop{dt}\right] \\ &= \lim_{z\to\infty}\left[ f(z)e^{-sz}-f(0)\right] + s\lim_{z\to\infty}\int_0^z f(t)e^{-st}\mathop{dt} \\ \text{Using integration by parts with } &u=e^{-st} \Rightarrow \mathop{du}=-se^{-st}\mathop{dt} \\ \text{and} &\mathop{dv}=f'(t)\mathop{dt} \Rightarrow v=f(t) \\&=0-f(0)+sF(s) \\ &=sF(s)-f(0). \end{align*}$

$$Part two is easily resolved using two applications of integration by parts, rearranging and considering the limit as $ z \to \infty.$

HSC 2013 MX2 Marathon (archive) (12 Viewers)

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