HSC 2013 MX2 Marathon (archive) (1 Viewer)

RealiseNothing · Apr 4, 2013

Re: HSC 2013 4U Marathon

seanieg89 said:
$In this question you may assume:\\ Let $(a_j)$ be a sequence of complex numbers. If there exists an $M>0$ such that:\\ \\ $\sum_{j=0}^n |a_j|\leq M$\\ \\ for all $n$, then the series\\ $\sum_{j=0}^\infty a_j$\\ converges. \\ \\ \\ Let $(a_n)$ be a positive and decreasing sequence with $a_n\rightarrow 0$. Prove that $\sum_{n=0}^\infty a_n$ does not necessarily have to converge, but $\sum_{n=0}^\infty a_n\cos(n\theta)$ converges for every $\theta$ not divisible by $2\pi$. Are all of our conditions on the sequence $a$ necessary for this result to hold?$

Difficulty rating: 4/5.

I'm not sure if this explanation is correct for the first part, but I'll have a go:

Let the sequence $(a_n)$ be written as:

$(a_n)=a_0+a_1+a_2+...+a_{n-1}+a_n$

Now there exists some real number $\epsilon$ such that:

$0 < \epsilon < a_k$ for some $0<k<n$

Hence we can deduce that:

$(a_n) > (n-1)\epsilon$

Now as $n \to \infty$ we get $(a_n) > \lim_{n \to \infty} (n-1)\epsilon$

Hence there exists a sequence $(a_n)$ such that the limit as $n \to \infty$ dominates $\epsilon \to 0$ , which means $(a_n) \to \infty$ and it does not converge.

This will occur in a situation where $\frac{a_k}{a_{k-1}} \to 1$ as $k \to n$

Sy123 · Apr 4, 2013

Re: HSC 2013 4U Marathon

seanieg89 said:
$In this question you may assume:\\ Let $(a_j)$ be a sequence of complex numbers. If there exists an $M>0$ such that:\\ \\ $\sum_{j=0}^n |a_j|\leq M$\\ \\ for all $n$, then the series\\ $\sum_{j=0}^\infty a_j$\\ converges. \\ \\ \\ Let $(a_n)$ be a positive and decreasing sequence with $a_n\rightarrow 0$. Prove that $\sum_{n=0}^\infty a_n$ does not necessarily have to converge, but $\sum_{n=0}^\infty a_n\cos(n\theta)$ converges for every $\theta$ not divisible by $2\pi$. Are all of our conditions on the sequence $a$ necessary for this result to hold?$

Difficulty rating: 4/5.

My attempt:

Part 1

First provide the counter example
Consider the series

$\sum_{n=1}^{\infty} \frac{1}{n} = S$

By drawing the graph y=1/x, and constructing the upper rectangles of width 1 from 1 to some integer m, it follows that:

$\sum_{n=1}^{m} \frac{1}{n} > \int_{1}^m \frac{1}{x} \ dx = \ln m$

Take m to infinity,

$S > \lim_{m \to \infty} \ln (m)$

$\therefore S \to \infty$

Despite the fact that:

$a_n = \frac{1}{n} \to 0 \ \ $as$ \ \ n \to \infty$

Part II

Since the sequence is decreasing, we can form the inequality:

$\sum_{n=0}^{\infty} a_n \cos (n\theta) < \lim_{m \to \infty}a_0 \sum_{n=0}^{m} \cos (n\theta)$

So let the LHS be = S
And sum the RHS using complex numbers and equating the real parts, it follows that:

$S < a_0 \lim_{m \to \infty} \frac{1}{2} + \frac{\cos(n-1) \theta - \cos(n\theta)}{2-2\cos \theta}$

Now, we know that:

$-1 \leq \cos n\theta \leq 1$

Therefore the last term on the RHS:

$u \leq \lim_{n \to \infty}\frac{\cos (n-1)\theta - \cos(n\theta)}{2-2\cos \theta} \leq v$

For some finite values v and u, therefore we arrive at:

$S < a_0 \times M$ for some finite limit M

Therefore due to the initial assumption that you have given us, S must converge. However it cannot converge for when $\theta = 2\pi k$ (integral values of k), since if this is the case, then

$\cos(2\pi kn) \in (-1, 1)$ (This is probably wrong notation but I'm saying that cos(2pikn) can only be one of those 2 values)

And this results in:

$S < a_0 \sum_{k=1}^{\infty} (-1)^{f(k)}$ for some function of k f(k), which is indeterminable, so the test is not applicable here.

Part III

I'm not sure here, but I think all the conditions involved in a are necessary.

======

Probably wrong

seanieg89 · Apr 4, 2013

Re: HSC 2013 4U Marathon

Essentially correct although your notation hurts my head... (a_n) typically refers to the sequence (a_0,a_1,...) rather than the corresponding series.

For things like that it suffices to provide an example, in this case the harmonic series would be adequate.

seanieg89 · Apr 4, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
My attempt:

Part 1

First provide the counter example
Consider the series

$\sum_{n=1}^{\infty} \frac{1}{n} = S$

By drawing the graph y=1/x, and constructing the upper rectangles of width 1 from 1 to some integer m, it follows that:

$\sum_{n=1}^{m} \frac{1}{n} > \int_{1}^m \frac{1}{x} \ dx = \ln m$

Take m to infinity,

$S > \lim_{m \to \infty} \ln (m)$

$\therefore S \to \infty$

Despite the fact that:

$a_n = \frac{1}{n} \to 0 \ \ $as$ \ \ n \to \infty$

Part II

Since the sequence is decreasing, we can form the inequality:

$\sum_{n=0}^{\infty} a_n \cos (n\theta) < \lim_{m \to \infty}a_0 \sum_{n=0}^{m} \cos (n\theta)$

So let the LHS be = S
And sum the RHS using complex numbers and equating the real parts, it follows that:

$S < a_0 \lim_{m \to \infty} \frac{1}{2} + \frac{\cos(n-1) \theta - \cos(n\theta)}{2-2\cos \theta}$

Now, we know that:

$-1 \leq \cos n\theta \leq 1$

Therefore the last term on the RHS:

$u \leq \lim_{n \to \infty}\frac{\cos (n-1)\theta - \cos(n\theta)}{2-2\cos \theta} \leq v$

For some finite values v and u, therefore we arrive at:

$S < a_0 \times M$ for some finite limit M

Therefore due to the initial assumption that you have given us, S must converge. However it cannot converge for when $\theta = 2\pi k$ (integral values of k), since if this is the case, then

$\cos(2\pi kn) \in (-1, 1)$ (This is probably wrong notation but I'm saying that cos(2pikn) can only be one of those 2 values)

And this results in:

$S < a_0 \sum_{k=1}^{\infty} (-1)^{f(k)}$ for some function of k f(k), which is indeterminable, so the test is not applicable here.

Part III

I'm not sure here, but I think all the conditions involved in a are necessary.

======

Probably wrong

Your first step in II doesn't work unfortunately, remember that cos(blah) isn't always positive.

Sy123 · Apr 4, 2013

Re: HSC 2013 4U Marathon

seanieg89 said:
Your first step in II doesn't work unfortunately, remember that cos(blah) isn't always positive.

Oops :/

twinklegal19 · Apr 5, 2013

RealiseNothing said:
This makes me proud :')

Don't get too used to it

Can barely answer the other 99% of sy's questions

Carrotsticks said:
$\\ $Alternatively:$ \\\\ u^2=x \Rightarrow 2udu=dx \\\\ I = 2 \int \frac{u}{1+u} du = 2 \int 1 - \frac{1}{1+u}du = 2 (u - \ln(1+u)) = 2 (\sqrt{x} - \ln|1+\sqrt{x}|)$

Carrotsticks said:
Plus C.

Ah that's a much easier substitution, thanks

seanieg89 · Apr 5, 2013

Re: HSC 2013 4U Marathon

A hint for my earlier question:

The corresponding result holds true for differentiable functions and is probably easier to understand for an mx2 student. The ideas behind the proofs are the same though.

$Let $f$ be a positive, continuously differentiable, decreasing function with $f(x)\rightarrow 0$ as $x\rightarrow \infty$.\\ Then $\lim_{L\rightarrow\infty}\int_0^L f(x)\cos(x)\,dx$ exists.\\ \\You may assume that if $\int_0^L |g(x)|\, dx\leq M$ for some $M>0$ for all $L$ then $\lim_{L\rightarrow\infty}\int_0^L g(x)\, dx$ exists.$

shongaponga · Apr 6, 2013

Re: HSC 2013 4U Marathon

$\\ $ For any given function $f $ let, $ \\ \\I = \int [f'(x)]^2 [f(x)]^n dx; $ \\ \\ where n is a positive integer. Show that, if $ f(x)$ satisfies $ \\ f''(x)= kf(x)f'(x)$, for some constant $ k$, then $ I$ can be integrated to obtain an expression in terms of $ f(x), f'(x), k $ and $ n. \\ \\ $ Hence find, $ \\ \\ \int sec^2x(secx + tanx)^6 dx.$

Sy123 · Apr 7, 2013

Re: HSC 2013 4U Marathon

shongaponga said:
$\\ $ For any given function $f $ let, $ \\ \\I = \int [f'(x)]^2 [f(x)]^n dx; $ \\ \\ where n is a positive integer. Show that, if and only if $ f(x)$ satisfies $ \\ f''(x)= kf(x)f'(x)$, for some constant $ k$, then $ I$ can be integrated to obtain an expression in terms of $ f(x), f'(x), k $ and $ n. \\ \\ $ Hence find, $ \\ \\ \int sec^2x(secx + tanx)^6 dx.$

$f''(x) = k f(x) f'(x) \ \ \ \ (1)$

$f'(x) = k \int f(x) f'(x) \ dx \ \ \ \ (2)$

$f'(x) = \frac{k}{2} [f(x)]^2 \ \ \ \ (3)$

$I= \int f'(x)^2 f(x)^n \ dx$

Using IBP

$u=f'(x) f(x)^{n-1}$

$du = k f'(x) f(x)^n + (n-1) \frac{k}{2} f(x)^n$ due to identity (3)

$dv = f(x) f'(x)$

$v=\frac{1}{k} f'(x)$ due to identity (2)

We then arrive at:

$I = \frac{1}{k} f'(x)^2 f(x)^{n-1} - \frac{1}{k} \int f'(x)^2 f(x)^{n} \ dx - \frac{n-1}{2} \int f'(x) f(x)^{n} \ dx$

We arrive at:

$I = \frac{1}{k} f'(x)^2 f(x)^{n-1} - I - \frac{n-1}{2(n+1)} f(x)^{n+1}$

$I = \frac{1}{2} \left (\frac{1}{k}f'(x) - \frac{n-1}{2(n+1)} f(x)^2 \right )$

And this can only arise due to our use of identities (1), (2) and (3) hence satisfying the iff condition.

I hope I haven't made a mistake with the algebra.

============

$f(x) = \sec x + \tan x$

$f'(x) = \sec x \tan x + \sec^2 x$

$f'(x) = \sec x (\sec x + \tan x)$

$f'(x) = \sec x f(x)$

$f''(x) = \sec x \tan x f(x) + f'(x) \sec x$

$f''(x) = \sec x (\sec x \tan x + \tan^2 x + \sec^2 x + \sec x \tan x)$

$f''(x) = \sec x (\sec x + \tan x)^2$

$f''(x) = 1 \cdot f'(x) \cdot f(x)$

$\therefore k = 1$

$n = 4$

$\therefore I = \frac{1}{2} (\sec x + \tan x)^{3} \left (\sec x (\sec x + \tan x) - \frac{3}{10} (\sec x + \tan x)^2 \right )$

Might of made a mistake but I'm sure the logic is right.

shongaponga · Apr 7, 2013

Re: HSC 2013 4U Marathon

Your logic and method is correct Sy, but you've made a mistake with your algebra.

$\\ $ To begin with we must suitably split up the integrand so that it is easily integratable via IBP. $$

$$ Thus the simplest way is to re-write $ I $ as \int f'(x)*[f'(x)f(x)^{n}]dx$

$$ Let $ u = f'(x) $ and $ dv = f'(x)f(x)^n$

$\Rightarrow du = f''(x) $ and $ v = \frac {f(x)^{n+1}}{(n+1)}$

$.'. I = \frac {f'(x)f(x)^{n+1}}{(n+1)} - \int \frac {f''(x)f(x)^{n+1}}{(n+1)} dx.$

$\\ $ Now, $ f''(x) = kf(x)f'(x). $ Subbing this value into the integrand in yields:$$

$I = \frac {f'(x)f(x)^{n+1}}{(n+1)} - \int \frac {kf'(x)f(x)^{n+2}}{(n+1)} dx.$

$\\ $ This is now directly integratable to yield the desired result in terms of $ f(x), f'(x), k $ and $ n.$

$I = \frac {f'(x)f(x)^{n+1}}{(n+1)} - \frac {kf(x)^{n+3}}{(n+1)(n+3)} + C$

For the second part:

$\\ $ If $ f(x) = \sec x + \tan x$

$\\ $ Then $ f'(x) = \sec x \tan x + \sec^2 x$

$\Rightarrow f'(x) = \sec x (\sec x + \tan x)$

$\\ $ And $ f''(x) = \sec^2 x(\sec x + \tan x) + \sec x \tan x(\sec x + \tan x)$

$\Rightarrow f''(x) = \sec x(\sec x + \tan x)^2 = kf(x)f'(x) $ with $ k = 1$

$\\ $ Then $ \int \sec^2 x(\sec x + \tan x)^6 = \int {\sec x(\sec x + \tan x)}^2 * (\sec x + \tan x)^4 dx$

$= \frac {\sec x(\sec x + \tan x)^6}{5} - \frac { (\sec x + \tan x)^7}{35} + C$

seanieg89 · Apr 7, 2013

Re: HSC 2013 4U Marathon

shongaponga said:
$\\ $ For any given function $f $ let, $ \\ \\I = \int [f'(x)]^2 [f(x)]^n dx; $ \\ \\ where n is a positive integer. Show that, if and only if $ f(x)$ satisfies $ \\ f''(x)= kf(x)f'(x)$, for some constant $ k$, then $ I$ can be integrated to obtain an expression in terms of $ f(x), f'(x), k $ and $ n. \\ \\ $ Hence find, $ \\ \\ \int sec^2x(secx + tanx)^6 dx.$

If and only if? I don't see how it follows from your working that an f which does not satisfy f''=kff' cannot give rise to an I which has a primitive in terms of f,f',k and n...

shongaponga · Apr 7, 2013

Re: HSC 2013 4U Marathon

seanieg89 said:
If and only if? I don't see how it follows from your working that an f which does not satisfy f''=kff' cannot give rise to an I which has a primitive in terms of f,f',k and n...

My apologies. Should simply read "if f(x)" not if and only if. I was in the middle of doing another question when i posted that one up haha; fixed it now.

seanieg89 · Apr 7, 2013

Re: HSC 2013 4U Marathon

Cool, good question by the way.

Sy123 · Apr 8, 2013

Re: HSC 2013 4U Marathon

$Consider the integral$

$I_n(t) = \int_{0}^{t} x^n e^{-x} \ dx$

$i) Prove that$

$I_n(t) = n! \left (1 - e^{-t} \sum_{k=0}^{n} \frac{t^{k}}{k!} \right )$

$ii) Prove that$

$e = \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \frac{1}{5!} + \frac{1}{6!} + \dots$

Trebla · Apr 8, 2013

Re: HSC 2013 4U Marathon

HSC 2005

Sy123 · Apr 8, 2013

Re: HSC 2013 4U Marathon

Trebla said:
HSC 2005

Well I guess, but I'm not spoon feeding

, and its very much reasonable to do this without the spoon feeding I think.

Makematics · Apr 8, 2013

Re: HSC 2013 4U Marathon

Hey guys, here's a five marker from my school paper from last year.

The complex number z is given by z=(t-1)/(t+1) where t=a+ib and |t|= 4. Find and describe the locus of the point P which represents z.

Not the hardest of questions but still worth 5 marks so i thought id include it. you uni guys leave this one alone and leave to us 2013ers

cutemouse · Apr 9, 2013

Re: HSC 2013 4U Marathon

Makematics said:
Hey guys, here's a five marker from my school paper (fort street) from last year.

The complex number z is given by z=(t-1)/(t+1) where t=a+ib and |t|= 4. Find and describe the locus of the point P which represents z.

Not the hardest of questions but still worth 5 marks so i thought id include it. you uni guys leave this one alone and leave to us 2013ers

It's worth noting that these types are not in the 4U syllabus. Still interesting though.

RealiseNothing · Apr 9, 2013

Re: HSC 2013 4U Marathon

Makematics said:
Hey guys, here's a five marker from my school paper (fort street) from last year.

The complex number z is given by z=(t-1)/(t+1) where t=a+ib and |t|= 4. Find and describe the locus of the point P which represents z.

Not the hardest of questions but still worth 5 marks so i thought id include it. you uni guys leave this one alone and leave to us 2013ers

$z = \frac{t-1}{t+1}$

$z(t+1) = (t-1)$

$zt + z = t-1$

$zt - t + z + 1 = 0$

$t(z-1) = -(z+1)$

$t = \frac{-(z+1)}{(z-1)}$

$|t| = |\frac{-(z+1)}{(z-1)}|$

$4 = \frac{|-(z+1)|}{|(z-1)|}$

$4|z-1| = |z+1|$

Sy123 · Apr 9, 2013

Re: HSC 2013 4U Marathon

$Find$

$\int \frac{dx}{x\sqrt{2x-1}}$

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