HSC 2013 MX2 Marathon (archive) (10 Viewers)

seanieg89 · Jan 1, 2013

Re: HSC 2013 4U Marathon

Carrotsticks said:
If you do manage to properly prove the Taylor Series formula (for all R and C), then Euler's formula should follow immediately with no issues because the radius of convergence for sin(x) and cos(x) are infinity.

However, the issue is HOW you are going to properly prove the formula using methods within Extension 2?

Same issue though, you would be trying to prove that the infinite series converge to e,sin,cos respectively. But to talk about this happening on C we need to define e,sin and cos on C in the first place!

Carrotsticks · Jan 1, 2013

Re: HSC 2013 4U Marathon

The only treatment of Taylor Series I have ever seen in the HSC has only been in R because it has been very well-defined in the course. So the infinite series for e = 1+1/1!+1/2! +... is fine, since it's in R.

Sy123, might want to stick within R when having problems to do with functions, though polynomial functions have been defined in C.

seanieg89 · Jan 1, 2013

Re: HSC 2013 4U Marathon

Carrotsticks said:
The only treatment of Taylor Series I have ever seen in the HSC has only been in R because it has been very well-defined in the course. So the infinite series for e = 1+1/1!+1/2! +... is fine, since it's in R.

Sy123, might want to stick within R when having problems to do with functions, though polynomial functions have been defined in C.

Agreed.

You would need something like the mean value theorem to actually prove convergence of the taylor series though.

Shadowdude · Jan 1, 2013

Re: HSC 2013 4U Marathon

Plus no doubt you'll learn about them when you continue your mathematical study at tertiary level...

hint hint

Sy123 · Jan 1, 2013

Re: HSC 2013 4U Marathon

Heh don't worry I won't do anything that extravagant anymore, here is a neat question I just solved:

$$The positive integers in this sequence are as follows$ \ \ \ \ (1) , (2,3) , (4, 5, 6) , (7, 8, 9, 10) , ... \\ \\ $Prove that the sum of the integers in the nth bracket is$ \ \ \ \ \frac{n}{2}(n^2+1)$

It isn't that difficult I guess but I reckon it was neat.

bleakarcher · Jan 1, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Heh don't worry I won't do anything that extravagant anymore, here is a neat question I just solved:

$$The positive integers in this sequence are as follows$ \ \ \ \ (1) , (2,3) , (4, 5, 6) , (7, 8, 9, 10) , ... \\ \\ $Prove that the sum of the integers in the nth bracket is$ \ \ \ \ \frac{n}{2}(n^2+1)$

It isn't that difficult I guess but I reckon it was neat.

Very nice question Sy.

RealiseNothing · Jan 2, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Heh don't worry I won't do anything that extravagant anymore, here is a neat question I just solved:

$$The positive integers in this sequence are as follows$ \ \ \ \ (1) , (2,3) , (4, 5, 6) , (7, 8, 9, 10) , ... \\ \\ $Prove that the sum of the integers in the nth bracket is$ \ \ \ \ \frac{n}{2}(n^2+1)$

It isn't that difficult I guess but I reckon it was neat.

Let $t_j = \frac{j}{2}(j+1})$ which denotes the j'th triangular number. Then the sum in the n'th bracket is just:

$\sum_{k=1}^{t_n}k - \sum_{k=1}^{t_{n-1}}k$

Now when we write this out we get:

$[\frac{n}{2}(n-1)+1] + [\frac{n}{2}(n-1)+2] + ... + [\frac{n}{2}(n+1)-1] + [\frac{n}{2}(n+1)]$

Usin the sum of a series formula we get:

$\frac{n}{2}[\frac{n}{2}(n-1)+1+\frac{n}{2}(n+1)]$

$\frac{n}{2}[\frac{n}{2}(n-1+n+1)+1]$

$\frac{n}{2}[\frac{n}{2}(2n)+1]$

$\frac{n}{2}[n^2+1]$

As required.

Sy123 · Jan 2, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
Let $t_j = \frac{j}{2}(j+1})$ which denotes the j'th triangular number. Then the sum in the n'th bracket is just:

$\sum_{k=1}^{t_n}k - \sum_{k=1}^{t_{n-1}}k$

Now when we write this out we get:

$[\frac{n}{2}(n-1)+1] + [\frac{n}{2}(n-1)+2] + ... + [\frac{n}{2}(n+1)-1] + [\frac{n}{2}(n+1)]$

Usin the sum of a series formula we get:

$\frac{n}{2}[\frac{n}{2}(n-1)+1+\frac{n}{2}(n+1)]$

$\frac{n}{2}[\frac{n}{2}(n-1+n+1)+1]$

$\frac{n}{2}[\frac{n}{2}(2n)+1]$

$\frac{n}{2}[n^2+1]$

As required.

Nice work guys, I will post another question soon. (unless someone else could contribute heh, instead of me just giving questions out)

GoldyOrNugget · Jan 2, 2013

Re: HSC 2013 4U Marathon

$Which is greater: $9999^{10000}$ or $10000^{9999}$? Justify your answer$

Sy123 · Jan 2, 2013

Re: HSC 2013 4U Marathon

GoldyOrNugget said:
$Which is greater: $9999^{10000}$ or $10000^{9999}$? Justify your answer$

$a=9999^ {10 000} = 9999 \cdot 9999^{9999}$

$b= 10 000^ {9999} = (9999 \cdot \frac{10 000}{9999})^{9999} = 9999^{9999} \cdot (10000/9999)^{9999}$

$a > b$

I can justify this, since I know the limit: $\lim_{n \to \infty} (1+\frac{1}{n})^{n} = e$

Hence the product (10000/9999)^ 9999 is always less than e < 9999

RealiseNothing · Jan 2, 2013

Re: HSC 2013 4U Marathon

GoldyOrNugget said:
$Which is greater: $9999^{10000}$ or $10000^{9999}$? Justify your answer$

Without working anything out or anything I'm going with $9999^{10000}$ off intuition.

But will try find a proof or justification etc.

GoldyOrNugget · Jan 2, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$a=9999^ {10 000} = 9999 \cdot 9999^{9999}$

$b= 10 000^ {9999} = (9999 \cdot \frac{10 000}{9999})^{9999} = 9999^{9999} \cdot (10000/9999)^{9999}$

$a > b$

I can justify this, since I know the limit: $\lim_{n \to \infty} (1+\frac{1}{n})^{n} = e$

Hence the product (10000/9999)^ 9999 is always less than e < 9999

Sneaky... I haven't seen that approach before. What if I changed it to $9999^{10001} $ vs $ 10001^{9999}$ ? The method I have in mind can compare any pair of large exponents regardless of the values of the bases and exponents.

Sy123 · Jan 2, 2013

Re: HSC 2013 4U Marathon

GoldyOrNugget said:
Sneaky... I haven't seen that approach before. What if I changed it to $9999^{10001} $ vs $ 10001^{9999}$ ? The method I have in mind can compare any pair of large exponents regardless of the values of the bases and exponents.

Well we can define the actual definition of e^x to do something similar:

$e^x = \lim_{n \to \infty}(1+\frac{x}{n})^n$

Now looking back at the new thing you gave us:

$9999^{10001} = 9999^{9999} \cdot 9999^2 = a$

$10001^{9999}= (10001/9999)^ {9999} \cdot 9999^{9999} = b$

$(1+ \frac{2}{9999})^{9999} \doteq e^2 < 9999^2$

$\therefore a > b$

So I too can do it for any large exponents

GoldyOrNugget · Jan 2, 2013

Re: HSC 2013 4U Marathon

Hmph. Fine, I'll accept it

the solution I have in mind uses logs.

Sy123 · Jan 2, 2013

Re: HSC 2013 4U Marathon

$$Prove that$ \ \ \ \frac{1}{n!} < \frac{1}{2^ {n-1}} \ \ \ $for $ \ \ \ n \geq 3$

$WITHOUT using a proof by induction$

Sy123 · Jan 2, 2013

Re: HSC 2013 4U Marathon

GoldyOrNugget said:
Hmph. Fine, I'll accept it the solution I have in mind uses logs.

I originally thought of using a^x > log_a x somehow but this came to me first.

Do you mind posting your method? I'm curious

cutemouse · Jan 2, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$$Prove that$ \ \ \ \frac{1}{n!} < \frac{1}{2^ {n-1}} \ \ \ $for $ \ \ \ n \geq 3$

$WITHOUT using a proof by induction$

1/n! = 1/[n(n-1)(n-2)...(3)(2)1] < 1/[2.2.2...2.2.1] (as n, n-1, ... , 3 > 2 for n >= 3) = 1/2^(n-1)

New question: Expand and simplify (x-a)(x-b)(x-c)...(x-z).

Sy123 · Jan 2, 2013

Re: HSC 2013 4U Marathon

cutemouse said:
1/n! = 1/[n(n-1)(n-2)...(3)(2)1] < 1/[2.2.2...2.2.1] (as n, n-1, ... , 3 > 2 for n >= 3) = 1/2^(n-1)

Precisely.

$$Find the values of $ \ \ a, b, x, y \ \ $that satisfy the equations$ \\ \\ a+b=1 \\ ax+by=\frac{1}{3} \\ ax^2+by^2=\frac{1}{5} \\ ax^3+by^3=\frac{1}{7}$

$$You may wish to start by multiplying the second equation by$ \ \ x+y$

Sy123 · Jan 2, 2013

Re: HSC 2013 4U Marathon

cutemouse said:
1/n! = 1/[n(n-1)(n-2)...(3)(2)1] < 1/[2.2.2...2.2.1] (as n, n-1, ... , 3 > 2 for n >= 3) = 1/2^(n-1)

New question: Expand and simplify (x-a)(x-b)(x-c)...(x-z).

$x$ \neq x $Anyway, if I were to expand it, I will first assume that there are n brackets$

$(x-a)(x-b)(x-c)...(x-z)=\sum_{k=0}^{n} x^{k} (-1)^{k} \sum \prod_{i=0}^{n-k}a_i$

Carrotsticks · Jan 2, 2013

Re: HSC 2013 4U Marathon

cutemouse said:
1/n! = 1/[n(n-1)(n-2)...(3)(2)1] < 1/[2.2.2...2.2.1] (as n, n-1, ... , 3 > 2 for n >= 3) = 1/2^(n-1)

New question: Expand and simplify (x-a)(x-b)(x-c)...(x-z).

0

Sy123 said:
$x$ \neq x $Anyway, if I were to expand it, I will first assume that there are n brackets$

$(x-a)(x-b)(x-c)...(x-z)=\sum_{k=0}^{n} x^{k} (-1)^{k} \sum \prod_{i=0}^{n-k}a_i$

You are thinking too hard.

EDIT: Just saw your x =/= x, but I think that was the intended result.

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