HSC 2013 MX2 Marathon (archive) (19 Viewers)

bleakarcher · Dec 18, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
Nope, CSSA Q16 was a real finding the Basel Problem rather than just the expression in this case.

Though for the result at the end of Heroic's question, I have a question for the maths pros

If I take the limit to infinity of the RHS of Heroic's last result. I get the sum of squares of the reciprocals of the natural numbers (basel problem) (since the fraction on the very right converges to 1). How would I prove that at m infinite the inequality becomes equality?

If you're trying to conclude that the sum of 1/k^2 from k=0 to infinity is (pi^2)/6, I don't think you can. You would need the upper bound of the inequality to then apply Squeeze law.

Sy123 · Dec 18, 2012

Re: HSC 2013 4U Marathon

bleakarcher said:
If you're trying to conclude that the sum of 1/k^2 from k=0 to infinity is (pi^2)/6, I don't think you can. You would need the upper bound of the inequality to then apply Squeeze law.

Ah yeah that makes sense. Technically we could find an inequality like cot x < 1 / x to but instead
cot x > f( x ) to create an upperbound right?

bleakarcher · Dec 18, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
Ah yeah that makes sense. Technically we could find an inequality like cot x < 1 / x to but instead
cot x > f( x ) to create an upperbound right?

yeah I think so.

seanieg89 · Dec 18, 2012

Re: HSC 2013 4U Marathon

Yeah the other bound would be needed. I think it is about the same difficulty.

Sy123 · Dec 19, 2012

Re: HSC 2013 4U Marathon

$$i) By showing that $ \ \ \ \ \frac{1}{x}=\frac{1}{1-(1-x)} \\ \\ $Show that$ \ \ \frac{1}{x}= \sum_{n=0}^{\infty} (1-x)^n \ \ \ \ $for$ \ \ \ 0 \leq x \leq 2 \ \ \ \ \ \ \ \fbox{1}$

$$ii) Using integration, show that$ \\ \\ \ln 2 = \frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+... \equiv \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \ \ \ \ \ \fbox{3}$

Carrotsticks · Dec 19, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
$$i) By showing that $ \ \ \ \ \frac{1}{x}=\frac{1}{1-(1-x)} \\ \\ $Show that$ \ \ \frac{1}{x}= \sum_{n=0}^{\infty} (1-x)^n \ \ \ \ $for$ \ \ \ 0 \leq x \leq 2 \ \ \ \ \ \ \ \fbox{1}$

$$ii) Using integration, show that$ \\ \\ \ln 2 = \frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+... \equiv \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \ \ \ \ \ \fbox{3}$

The general idea is correct, but you are assuming that an infinite number of terms can be integrated.

What would be better, is to use the sum of GP formula for an n'th partial sum, integrate both sides (since it is a finite sum), then take the limit as n -> infinity. You will also need to show that the error term converges to 0 (usually by squeeze law).

Sy123 · Dec 19, 2012

Re: HSC 2013 4U Marathon

Carrotsticks said:
The general idea is correct, but you are assuming that an infinite number of terms can be integrated.

What would be better, is to use the sum of GP formula for an n'th partial sum, integrate both sides (since it is a finite sum), then take the limit as n -> infinity. You will also need to show that the error term converges to 0 (usually by squeeze law).

I used ln(1-x) as it was easier. I ended up with:

$\ln(1-x)= -\sum_{k=1}^n \frac{x^k}{k} - \int \frac{x^n}{1-x} \ dx + C$

The integral cannot be evaluted in terms of elementary function (that's the term right?)
How would I show that as n increases without bound the integral approaches zero?
(Finding C is trivial through subbing x=0)

EDIT: Hang on hang on hang on, I think I have a good justification. We initially defined the interval for x as -1 < x < 1 right?
So we can only have the function x^n/(1-x) in that inteval.

As n increases without bound, x^n will reduce to zero since it is in that interval -1 < x < 1
Hence when we integrate as n is infintie, we get constant from which we can evaluate.

Yes?

Carrotsticks · Dec 19, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
I used ln(1-x) as it was easier. I ended up with:

$\ln(1-x)= -\sum_{k=1}^n \frac{x^k}{k} - \int \frac{x^n}{1-x} \ dx + C$

The integral cannot be evaluted in terms of elementary function (that's the term right?)
How would I show that as n increases without bound the integral approaches zero?
(Finding C is trivial through subbing x=0)

$\\ 1 - x + x^2 - x^3 + ... + (-1)^{n-1} x^{n-1} = \frac{1-(-1)^n x^{n}}{1+x} = \frac{1}{1+x} + (-1)^{n+1} \frac{x^n}{1+x} \\\\ $Integrate both sides w.r.t $ x $ in the interval $ 0 \leq x \leq 1 \\\\ $Afterwards, show that as $ n \rightarrow \infty, (-1)^{n+1} \int_{0}^{1} \frac{x^n}{1+x}dx \rightarrow 0 $ by considering the fact that $ 0 < \int_{0}^{1} \frac{x^n}{1+x}dx < \int_{0}^{1} x^n dx$

seanieg89 · Dec 19, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
I used ln(1-x) as it was easier. I ended up with:

$\ln(1-x)= -\sum_{k=1}^n \frac{x^k}{k} - \int \frac{x^n}{1-x} \ dx + C$

The integral cannot be evaluted in terms of elementary function (that's the term right?)
How would I show that as n increases without bound the integral approaches zero?
(Finding C is trivial through subbing x=0)

EDIT: Hang on hang on hang on, I think I have a good justification. We initially defined the interval for x as -1 < x < 1 right?
So we can only have the function x^n/(1-x) in that inteval.

As n increases without bound, x^n will reduce to zero since it is in that interval -1 < x < 1
Hence when we integrate as n is infintie, we get constant from which we can evaluate.

Yes?

You can't really use indefinite integrals here, the value of an indefinite integral is a class of functions differing by a constant, it is NOT a number so we cannot make claims about its limit. (at least not the sort of claims you are trying for here). Try to derive a similar expression using definite integrals and then you will be able to apply something like your bolded edit.

Carrotsticks · Dec 19, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
I used ln(1-x) as it was easier. I ended up with:

$\ln(1-x)= -\sum_{k=1}^n \frac{x^k}{k} - \int \frac{x^n}{1-x} \ dx + C$

The integral cannot be evaluted in terms of elementary function (that's the term right?)
How would I show that as n increases without bound the integral approaches zero?
(Finding C is trivial through subbing x=0)

EDIT: Hang on hang on hang on, I think I have a good justification. We initially defined the interval for x as -1 < x < 1 right?
So we can only have the function x^n/(1-x) in that inteval.

As n increases without bound, x^n will reduce to zero since it is in that interval -1 < x < 1
Hence when we integrate as n is infintie, we get constant from which we can evaluate.

Yes?

Not sure how you have ln(1-x) = etc etc, because it cannot be expressed as a finite polynomial, only approximated.

seanieg89 · Dec 19, 2012

Re: HSC 2013 4U Marathon

Exactly as carrot has done lol. A similar thing proves Gregory's series for pi.

seanieg89 · Dec 19, 2012

Re: HSC 2013 4U Marathon

Carrotsticks said:
Not sure how you have ln(1-x) = etc etc, because it cannot be expressed as a finite polynomial, only approximated.

The error term is the integral, slight modifications to his working would make it legit and analogous to yours just with a sign change.

Sy123 · Dec 19, 2012

Re: HSC 2013 4U Marathon

Ok how about this:

$1+x+x^2+...+x^{n-1}=\frac{1-x^n}{1-x} \ \ \ |x|<1$
Implementing the definite integral
$\int_{-1}^{0} 1+x+x^2+...+x^{n-1} \ dx = \int_{-1}^0 \frac{1}{1-x} \ dx - \int_{-1}^0 \frac{x^n}{1-x} \ dx$

$-\sum_{k=1}^{n} \frac{(-1)^n}{n} + \int_{-1}^0 \frac{x^n}{1-x} = [-\ln(1-x)]_{-1}^{0}$

$\lim_{n \to \infty} \sum_{k=1}^n \frac{(-1)^{n+1}}{n} + \lim_{n\to \infty} \int_{-1}^0 \frac{x^n}{1-x} = \ln 2$

Now we observe that we define |x|< 1

$\therefore \lim_{n \to \infty} x^n = 0$

$\therefore \lim_{n \to \infty} \int_{-1}^0 \frac{x^n}{1-x} = 0$

It is zero due to it being a definite integral unlike before

$\therefore 1-1/2+1/3-1/4+...=\ln 2$

Justification good now?

seanieg89 · Dec 19, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
Ok how about this:

$1+x+x^2+...+x^{n-1}=\frac{1-x^n}{1-x} \ \ \ |x|<1$
Implementing the definite integral
$\int_{-1}^{0} 1+x+x^2+...+x^{n-1} \ dx = \int_{-1}^0 \frac{1}{1-x} \ dx - \int_{-1}^0 \frac{x^n}{1-x} \ dx$

$-\sum_{k=1}^{n} \frac{(-1)^n}{n} + \int_{-1}^0 \frac{x^n}{1-x} = [-\ln(1-x)]_{-1}^{0}$

$\lim_{n \to \infty} \sum_{k=1}^n \frac{(-1)^{n+1}}{n} + \lim_{n\to \infty} \int_{-1}^0 \frac{x^n}{1-x} = \ln 2$

Now we observe that we define |x|< 1

$\therefore \lim_{n \to \infty} x^n = 0$

$\therefore \lim_{n \to \infty} \int_{-1}^0 \frac{x^n}{1-x} = 0$

It is zero due to it being a definite integral unlike before

$\therefore 1-1/2+1/3-1/4+...=\ln 2$

Justification good now?

Not quite, the fact that $x^n\rightarrow 0$ for |x|<1 is not quite enough. For example, if the thing we were multiplying $x^n$ by in our integrand was something like $1/x^n$ , then obviously our integral would not tend to zero. The rest of the integrand is important too, see Carrot's formulation as a question for a concrete outline to proving what you want.

Sy123 · Dec 19, 2012

Re: HSC 2013 4U Marathon

seanieg89 said:
Not quite, the fact that $x^n\rightarrow 0$ for |x|<1 is not quite enough. For example, if the thing we were multiplying $x^n$ by in our integrand was something like $1/x^n$ , then obviously our integral would not tend to zero. The rest of the integrand is important too, see Carrot's formulation as a question for a concrete outline to proving what you want.

I see. Time to think of a new question considering that other failed haha (I learnt something from it I guess). I think I will stay away from infinite series for a while, too many rigour issues at a High School level I think

seanieg89 · Dec 19, 2012

Re: HSC 2013 4U Marathon

Yep, once you learn things like the rigorous definitions of limits it will be much easier for you to determine if your manipulations are legit I promise.

Sy123 · Dec 19, 2012

Re: HSC 2013 4U Marathon

$A cube has 6 faces, on each of these faces I paint each one with either Blue, Green, Yellow, Red, White or Black. The cube must have all 6 of these colours. How many different cubes can I paint?$

yasminee96 · Dec 19, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
$A cube has 6 faces, on each of these faces I paint each one with either Blue, Green, Yellow, Red, White or Black. The cube must have all 6 of these colours. How many different cubes can I paint?$

6C4*4! * 2C2*2!
?

confidence=zero. don't judge me if im wrong haha

Sy123 · Dec 19, 2012

Re: HSC 2013 4U Marathon

yasminee96 said:
6C4*4! * 2C2*2!
?

confidence=zero. don't judge me if im wrong haha

Not quite, think of the cube as a net

i.e

seanieg89 · Dec 19, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
$A cube has 6 faces, on each of these faces I paint each one with either Blue, Green, Yellow, Red, White or Black. The cube must have all 6 of these colours. How many different cubes can I paint?$

30.

Fix the cube, count bijections from {1,2,3,4,5,6} to it's faces, then divide by the size of the symmetry group of the cube (because the first two steps count possible cubes with multiplicity).

HSC 2013 MX2 Marathon (archive) (19 Viewers)

Active Member

This too shall pass

Active Member

Well-Known Member

This too shall pass

Retired

This too shall pass

Retired

Well-Known Member

Retired

Well-Known Member

Well-Known Member

This too shall pass

Well-Known Member

This too shall pass

Well-Known Member

This too shall pass

Active Member

This too shall pass

Well-Known Member

Users Who Are Viewing This Thread (Users: 0, Guests: 19)