HSC 2013 MX2 Marathon (archive) (3 Viewers)

seanieg89 · Dec 31, 2012

Re: HSC 2013 4U Marathon

Couldn't find a non-bashy solution to ii), will leave it for Sy to finish

.

Sy123 · Dec 31, 2012

Re: HSC 2013 4U Marathon

GoldyOrNugget said:
Sounds good

Hm ok... but what about the arrangement (3 grapes, 3 apples)? This is an arrangement for 6 fruits, and it's not proportional to any arrangement of 3 fruits , but yet it isn't legal...

You are almost there go deeper!

This problem only arises with choosing an even number of fruit, we will disregard the the situation with 6 fruits as you have mentioned as it will be reflected in 12 fruits and hence cancels out the arrangements anyway (but I get there is a problem).

However the problem only arises for even numbers, that when we half it we get an odd number, so in this case, it will be 2, 6, 10.
We already disregard 2 so that doesnt matter, and we disregard 6 since it cancels out with arrangements with 12 anyway.

Look at 10 fruits being chosen, we can split it into 5 G, 5 A and no arrangement of 5 fruits can reflect this proportion. Hence we disregard that arrangement,

Hence 5726 - 5C2

But there is more to be excluded, we now look at numbers of fruits, such that we can separate them into groups of 3. i.e. multiples of 3:
3, 6, 9, 12

3, 6, 9 disregarded (odd numbers)

Look at 12: We can seperate it into 4 groups of 3, and no arrangement of 6 fruits can reflect this proportionality.

Hence 5726 - 5C2 - 5C4

Now lets check multiples of 4: 4, 8, 12

For 8: we can split into 2 groups of 4, however we can replicate this with 4, and same thing for 12, we can replicate this with 6. The reason being is that for 8, we seperate into 2 groups and 2 is a factor of 4 (whereas earlier 2 is not a factor of 5), and similarly for 12, 3 is a factor of 6.

Check multiples of 5: 5, 10

We already covered this. So the final answer which I hope to be right is:

5711

GoldyOrNugget · Dec 31, 2012

Re: HSC 2013 4U Marathon

Oops, sorry. Wasn't thinking carefully enough

Sy123 said:
Let us establish the fact that if we pick an odd number of fruits, we cannot retain proportionality. Hence we keep all the odd numbered fruits arrangements (odd numbers above 6, I will get to this later)

Not always true. Picking 9 apples is proportional to picking 3 apples, for example.

Sy123 · Dec 31, 2012

Re: HSC 2013 4U Marathon

GoldyOrNugget said:
Oops, sorry. Wasn't thinking carefully enough

Not always true. Picking 9 apples is proportional to picking 3 apples, for example.

, I will have a go at it later:

As for another question:

$$Find the limiting sum$ \\ \\ \sum_{n=1}^{\infty} n \left(\frac{1}{4}\right) ^n$

RealiseNothing · Jan 1, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
, I will have a go at it later:

As for another question:

$$Find the limiting sum$ \\ \\ \sum_{n=1}^{\infty} n \left(\frac{1}{4}\right) ^n$

Summing each one individually etc ends up with:

$\frac{4}{9}$

I hope I've done this right considering it's 4am, I should be in bed argh.

Sy123 · Jan 1, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
Summing each one individually etc ends up with:

$\frac{4}{9}$

I hope I've done this right considering it's 4am, I should be in bed argh.

This is correct, I am wondering what you mean by summing each individually, I was aiming at a geometric series differentiation approach

Carrotsticks · Jan 1, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
This is correct, I am wondering what you mean by summing each individually, I was aiming at a geometric series differentiation approach

$\\ \frac{1}{4} + \left ( \frac{1}{4} \right )^2 + \left ( \frac{1}{4} \right )^3 + \left ( \frac{1}{4} \right )^4 + ... \\\\ + \left ( \frac{1}{4} \right )^2 +\left ( \frac{1}{4} \right )^3 + \left ( \frac{1}{4} \right )^4 + ... \\\\ +\left ( \frac{1}{4} \right )^3 + \left ( \frac{1}{4} \right )^4 + ... \\\\ ...$

Sy123 · Jan 1, 2013

Re: HSC 2013 4U Marathon

Carrotsticks said:
$\\ \frac{1}{4} + \left ( \frac{1}{4} \right )^2 + \left ( \frac{1}{4} \right )^3 + \left ( \frac{1}{4} \right )^4 + ... \\\\ + \left ( \frac{1}{4} \right )^2 +\left ( \frac{1}{4} \right )^3 + \left ( \frac{1}{4} \right )^4 + ... \\\\ +\left ( \frac{1}{4} \right )^3 + \left ( \frac{1}{4} \right )^4 + ... \\\\ ...$

Well my working out was

$1+x^2+...+x^{n-1}= \frac{1-x^n}{1-x}$

Differentiate:

$1+2x+3x^2+...+(n-1)x^{n-2}=\frac{-nx^{n-1}+(n-1)x^n+1}{(1-x)^2}$

Sub in x=1/4, then make n approach infinity, here we do assume that $\lim_{n \to \infty} n \cdot k^{-n} = 0 \ \ \ \ \ k>1$
And I probably should of stated that in the question.

$1+2(1/4)+3(1/4)^2+... = \frac{1}{6/19}$

Multiply everything by 1/4

$\sum_{n=1}^{\infty} n (1/4)^n = 4/9$

Sy123 · Jan 1, 2013

Re: HSC 2013 4U Marathon

$$Show that$ \\ \\ \sum_{k=1}^{\infty} \frac{1}{k^2} = \lim_{n \to \infty} \int_{0}^1 \left (\frac{1}{x} \int \frac{1-x^n}{1-x} \ dx \right) \ dx$

GoldyOrNugget · Jan 1, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$$Show that$ \\ \\ \sum_{k=1}^{\infty} \frac{1}{k^2} = \lim_{n \to \infty} \int_{0}^1 \left (\frac{1}{x} \int \frac{1-x^n}{1-x} \ dx \right) \ dx$

$\lim_{n \to \infty} \int_{0}^1 \left (\frac{1}{x} \int \frac{1-x^n}{1-x} \ dx \right) \ dx = \lim_{n \to \infty} \int_{0}^1 \left (\frac{1}{x} \int (1 + x + x^2 + ... + x^{n-1}) \ dx \right) \ dx = \lim_{n \to \infty} \int_{0}^1 \left (\frac{1}{x} \int (x + \frac{x^2}{2} + \frac{x^3}{3} + ... + \frac{x^n}{n}) \ dx \right) \ dx = \lim_{n \to \infty} \int_{0}^1 \left (\int (1 + \frac{x}{2} + \frac{x^2}{3} + ... + \frac{x^{n-1}}{n}) \ dx \right) \ dx = \lim_{n \to \infty} \int_{0}^1 \left (x + \frac{x^2}{4} + \frac{x^3}{9} + ... + \frac{x^{n}}{n^2} \right) \ dx = \lim_{n \to \infty} \left[ x + \frac{x^2}{4} + \frac{x^3}{9} + ... + \frac{x^{n}}{n^2} \ \right]_{0}^{1} = \lim_{n \to \infty} \left[ 1 + \frac{1}{4} + \frac{1}{9} + ... + \frac{1}{n^2} \ \right] = \sum_{k=1}^{\infty} \frac{1}{k^2} \quad \blacksquare$

GoldyOrNugget · Jan 1, 2013

Re: HSC 2013 4U Marathon

Hope people don't mind that I'm not a 2013-er -- just trying to keep sharp over the holidays.

Solve for n:

$\ln (\prod_{k=1}^n 13^k) = 11325 \ln 13$

HeroicPandas · Jan 1, 2013

Re: HSC 2013 4U Marathon

GoldyOrNugget said:
Hope people don't mind that I'm not a 2013-er -- just trying to keep sharp over the holidays.

Solve for n:

$\ln (\prod_{k=1}^n 13^k) = 11325 \ln 13$

n = 150?

GoldyOrNugget · Jan 1, 2013

Re: HSC 2013 4U Marathon

HeroicPandas said:
n = 150?

Yep. Out of interest, what method did you use?

HeroicPandas · Jan 1, 2013

Re: HSC 2013 4U Marathon

GoldyOrNugget said:
Yep. Out of interest, what method did you use?

$\prod_{k=1}^{n}13^k = 13^{11325}$ by logs \\ \\ $ \therefore 13.13^2.13^3...13^n = 13^{11325} \\ \\ $ \therefore 13^{1+2+3+...+n} = 13^{11325} \\ \\$ $Equate powers $\\ \\ $ \therefore 1 + 2+3+...+n = 11325 \\ \\ $ $Solve for n by using the S_n = \frac{n}{2}(a + l)$

Sy123 · Jan 1, 2013

Re: HSC 2013 4U Marathon

GoldyOrNugget said:
Hope people don't mind that I'm not a 2013-er -- just trying to keep sharp over the holidays.

Solve for n:

$\ln (\prod_{k=1}^n 13^k) = 11325 \ln 13$

No problem man.

$E(x)=\sum_{r=0}^{n} \frac{x^{r}}{r!}$

$i) Show that E(x) does not have a double root$ \\ \\ $ii) Show that E(x) does not have any real roots if n is even$

RealiseNothing · Jan 1, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
No problem man.

$E(x)=\sum_{r=0}^{n} \frac{x^{r}}{r!}$

$i) Show that E(x) does not have a double root$ \\ \\ $ii) Show that E(x) does not have any real roots if n is even$

I've tried to do part (i) using a different approach than what you normally would, just to see if it works. Curious to see how justified this method is because I'm not sure lol:

First assume that there is a double root, that is:

$E(x) = (x-\alpha)^2Q(x)$ for some root $\alpha$

Differentiating both sides gives:

$E'(x) = 2(x-\alpha)Q(x) + (x-\alpha)^2Q'(x)$

Now it is obvious that:

$E'(x) = E(x) - \frac{x^n}{n!}$

So substituting this in gives:

$E(x) - \frac{x^n}{n!} = 2(x-\alpha)Q(x) + (x-\alpha)^2Q'(x)$

Solving the following equations simulatenously:

$E(x) = (x-\alpha)^2Q(x)$

$E(x) - \frac{x^n}{n!} = 2(x-\alpha)Q(x) + (x-\alpha)^2Q'(x)$

By subtraction of the second from the first gives:

$\frac{x^n}{n!} = (x-\alpha)^2Q(x) - 2(x-\alpha)Q(x) - (x-\alpha)^2Q'(x)$

$\frac{x^n}{n!} = (x-\alpha)[(x-\alpha)Q(x) - 2Q(x) - (x-\alpha)Q'(x)]$

Now letting $x=\alpha$ gives:

$\frac{\alpha^n}{n!} = 0$

Hence:

$\alpha = 0$ is a root of $E(x)$

But substituting this into $E(x)$ gives the result of $1$

Hence $\alpha \neq 0$

Now we have come to a contradiction, and hence our assumption must be wrong. That is, $E(x)$ does not have a double root.

GoldyOrNugget · Jan 1, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
No problem man.

$E(x)=\sum_{r=0}^{n} \frac{x^{r}}{r!}$

$i) Show that E(x) does not have a double root$ \\ \\ $ii) Show that E(x) does not have any real roots if n is even$

$i) Double roots occur when $E(x) = E'(x) = 0$ for some real $x$. Now $ E(x) = 1 + \frac{x}{1!} + \frac{x^2}{2!} + ... + \frac{x^n}{n!} \\ E'(x) = 1 + \frac{x}{1!} + \frac{x^2}{2!} + ... + \frac{x^{n-1}}{(n-1)!} = E(x) - \frac{x^n}{n!}$, so $E(x) = E'(x)$ iff $x=0$, but E(0) = 1, so there is no real $x$ where $E(x) = E'(x) = 0$, so $E(x)$ does not have a double root.$

Sy123 · Jan 1, 2013

Re: HSC 2013 4U Marathon

Nice work both of you:

As for a hint to the second part if anyone wants it, I have it in white here:

Consider E(alpha) where alpha is the stationary point of E(x) (this is how I proved it anyway)

GoldyOrNugget · Jan 1, 2013

Re: HSC 2013 4U Marathon

The hint helped -- that question was (and still is) making me feel pretty stupid

hope I didn't make a mistake here.

$If $\alpha$ is a stationary point of $E(x)$, then $E'(\alpha) = 0$. But we established that $E'(x) = E(x) - \frac{x^n}{n!}$ so $E(\alpha) = \frac{\alpha^n}{n!} > 0$ as $n$ is even. So all stationary points of $E(x)$ are above the $x$-axis. Furthermore the double derivative $E''(\alpha) = E'(\alpha) - \frac{\alpha^{n-1}}{(n-1)!} = -\frac{\alpha^{n-1}}{(n-1)!}$ is negative for any negative $\alpha$ so this positive stationary point is a minimum if $\alpha$ is negative, and $E(x)$ is clearly monotonic increasing for positive values of $x$ and thus has no stationary points for these values. So $E(x)$ is always positive and thus has no real roots for even $n$.$

Sy123 · Jan 1, 2013

Re: HSC 2013 4U Marathon

GoldyOrNugget said:
The hint helped -- that question was (and still is) making me feel pretty stupid hope I didn't make a mistake here.

$If $\alpha$ is a stationary point of $E(x)$, then $E'(\alpha) = 0$. But we established that $E'(x) = E(x) - \frac{x^n}{n!}$ so $E(\alpha) = \frac{\alpha^n}{n!} > 0$ as $n$ is even. So all stationary points of $E(x)$ are above the $x$-axis. Furthermore the double derivative $E''(\alpha) = E'(\alpha) - \frac{\alpha^{n-1}}{(n-1)!} = -\frac{\alpha^{n-1}}{(n-1)!}$ is negative for any negative $\alpha$ so this positive stationary point is a minimum if $\alpha$ is negative, and $E(x)$ is clearly monotonic increasing for positive values of $x$ and thus has no stationary points for positive $\alpha$. So $E(x)$ is always positive and thus has no real roots for even $n$.$

Nice work

$$i) Show that $ \ \ y=e^x \ \ $is the only function which satisfies$ \ \ \ y'=y \ \ \ \ \ \fbox{1}$

$$ii) Hence show that $ \\ \\ $If$ \ \ \ \ E(x) = \lim_{n \to \infty} \sum_{k=0}^{n} \frac{x^n}{n!} \\ \\ E(x) = e^x \ \ \ \ \ \fbox{2}$

$$iii) You are given that $ \ \ \ \ y= \sin x \ \ \ \ y=\cos x \\ \\ $are the only functions which satisfy$ \ \ \ \ y'' + y =0 \\ \\ $Develop similar infinite power series to part ii for the sine and cosine functions$ \ \ \ \ \ \fbox{3}$

$$iv) Hence show that$ \\ \\ e^{i\theta} = \cos \theta + i\sin \theta \ \ \ \ \ \ \fbox{4}$

(probably a billion rigour issues with this)

HSC 2013 MX2 Marathon (archive) (3 Viewers)

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