HSC 2013 MX2 Marathon (archive) (2 Viewers)

Sy123 · Dec 21, 2012

Re: HSC 2013 4U Marathon

$$Given that$\\ \\ \frac{\pi^2}{8}=1+\frac{1}{3^2}+\frac{1}{5^2}+ \frac{1}{7^2}+\frac{1}{9^2}+...$

$$Find (displaying full working) the value of$ \\ \\ 1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...$

RealiseNothing · Dec 21, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
$$Given that$\\ \\ \frac{\pi^2}{8}=1+\frac{1}{3^2}+\frac{1}{5^2}+ \frac{1}{7^2}+\frac{1}{9^2}+...$

$$Find (displaying full working) the value of$ \\ \\ 1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...$

CSSA 2012 iirc?

Sy123 · Dec 21, 2012

Re: HSC 2013 4U Marathon

RealiseNothing said:
CSSA 2012 iirc?

Last question - last part yeah, I thought the intuition you needed would be great for a marathon question (short and sweet).

seanieg89 · Dec 21, 2012

Re: HSC 2013 4U Marathon

$Prove by induction or otherwise that a function $f:\mathbb{R}\rightarrow\mathbb{R}$ is $n$-times differentiable at $a\in\mathbb{R}$ if and only if there exists a polynomial $P$ of degree $n$ with: \\ \\ $\lim_{x\rightarrow a}\frac{f(x)-P(x-a)}{|x-a|^n}=0 \textrm{ and } f(a)=P(0).$ \\ (So $P$ is a good polynomial approximation to $f$ near $a$.) \\ \\Find an expression for this polynomial in terms of the derivatives of $f(x)$ at $a$.$

Sy123 · Dec 21, 2012

Re: HSC 2013 4U Marathon

This is my attempt:

$$Step 1: n=1$ \\ \\ $By definition$ f'(a)=\lim_{x\to a} \frac{f(x)-f(a)}{x-a} \\ $There must exist a polynomial P, of degree 1, in the form$ \ \ P(x-a) \ \ $such that$ \\ P(x-a)=f(a) \ \ \ $IFF$ \ \ \ x \to a$

$$i.e. a polynomial must exist such that$ \ \ P(0)=f(a) \ \ $This polynomial is a degree 1, in the form$ \ \ \ P(x)=kx+f(a)$

$Hence we can substitute $ \ \ f(a) \ \ $for$ \ \ P(x-a) \ \ $Where P is a degree 1 polynomial$

$$\therefore \lim_{x \to a} \frac{f(x)-P(x-a)}{|x-a|} = f'(a) \\ \\ \text{Now, if f(x) is only differentiable once at x=a, then f'(a)=0 as shown above. Hence statement is true for n=1}$

$$Step 2: n=k$ \\ \\ $Assume true that a function $ \ \ f(x) \ \ $ is ONLY k times differentiable at x=a, IFF$ \\ \\ \lim_{x \to a} \frac{f(x)-P(x-a)}{|x-a|^k} = 0$
$Where P is a polynomial degree k$

$$Step 3: n=k+1$ \ \\ $Define the polynomial$ \ \ P(x)= \sum_{m=0}^{k}c_m x^m \ \ \ \ $AND$ \ \ \ \ R(x)=\sum_{m=0}^{k+1} c_m x^m = P(x) + c_{k+1}x^{k+1}$

$\lim_{x\to a} \frac{f(x)-P(x-a)-c_{k+1}(x-a)^{k+1}+c_{k+1}(x-a)^{k+1}}{|x-a|^k} = 0 \\$From Step 2$$

$\lim_{x \to a} \frac{f(x)-R(x-a)}{|x-a|^k} + \frac{c_{m+1}(x-a)^{k+1}}{|x-a|^k} = 0$

$\lim_{x \to a} \frac{f(x)-R(x-a)}{|x-a|^k} + 0 = 0$

$\therefore \lim_{x \to a} \frac{f(x)-R(x-a)}{|x-a|^{k+1}} = 0$

$Hence for a polynomial of degree k+1, f is only k+1 times differentiable since the limit expression above, is an expression for$ f^{(k+1)}(a) $ which happens to be 0$

$$Proof complete by Mathematical Induction$ \square$

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

I don't think my proof is absolutely correct though.

Rezen · Dec 21, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
This is my attempt:

$$Step 1: n=1$ \\ \\ $By definition$ f'(a)=\lim_{x\to a} \frac{f(x)-f(a)}{x-a} \\ $There must exist a polynomial P, of degree 1, in the form$ \ \ P(x-a) \ \ $such that$ \\ P(x-a)=f(a) \ \ \ $IFF$ \ \ \ x \to a$

$$i.e. a polynomial must exist such that$ \ \ P(0)=f(a) \ \ $This polynomial is a degree 1, in the form$ \ \ \ P(x)=kx+f(a)$

$Hence we can substitute $ \ \ f(a) \ \ $for$ \ \ P(x-a) \ \ $Where P is a degree 1 polynomial$

$$\therefore \lim_{x \to a} \frac{f(x)-P(x-a)}{|x-a|} = f'(a) \\ \\ \text{Now, if f(x) is only differentiable once at x=a, then f'(a)=0 as shown above. Hence statement is true for n=1}$

I didn't read the rest but i can already tell your confused. Unsurprising since this question is definitely out of the scope of the hsc. (note that you said f'(a)=0, which is note true in general).

Here's a further hint:

$\text{by definition, f is differentiable at a means } f'(a) = \lim_{x \to a} \frac{f(x)-f(a)}{x-a} \\ \\ \text{ We can rethink this as } \lim_{x\to a}\frac{f(x)-f(a) - f'(a)(x-a)}{x-a}=0 \\ \\\text{ Note that f(a) and f'(a) are constant so f(a) + f'(a)(x-a) is a polynomial in x-a}$

EDIT: So I read the second step and the idea of adding and subtracting the same term was good and is a pretty standard technique in analysis. However, it still has some subtle errors. Firstly, while this is mostly a superficial error, in your assumption your said f is only k times differentiable, this condition is too strong and infact contradicts what your trying to prove. The assumption should be f is k times differentiable. Secondly, there is a more subtle error and I leave it as an exercise to find it. but note that as it stands, if your n=k+1 step was correct then this would imply that any function that is differentiable once would be differentiable to all orders. (While this is obviously wrong to me, I'm unsure if a HSC student would have run across a counter example since this tends to be true for most functions encountered at a hsc level.)

Sy123 · Dec 21, 2012

Re: HSC 2013 4U Marathon

Rezen said:
I didn't read the rest but i can already tell your confused. Unsurprising since this question is definitely out of the scope of the hsc. (note that you said f'(a)=0, which is note true in general).

Here's a further hint:

$\text{by definition, f is differentiable at a means } f'(a) = \lim_{x \to a} \frac{f(x)-f(a)}{x-a} \\ \\ \text{ We can rethink this as } \lim_{x\to a}\frac{f(x)-f(a) - f'(a)(x-a)}{x-a}=0 \\ \\\text{ Note that f(a) and f'(a) are constant so f(a) + f'(a)(x-a) is a polynomial in x-a}$

EDIT: So I read the second step and the idea of adding and subtracting the same term was good and is a pretty standard technique in analysis. However, it still has some subtle errors. Firstly, while this is mostly a superficial error, in your assumption your said f is only k times differentiable, this condition is too strong and infact contradicts what your trying to prove. The assumption should be f is k times differentiable. Secondly, there is a more subtle error and I leave it as an exercise to find it. but note that as it stands, if your n=k+1 step was correct then this would imply that any function that is differentiable once would be differentiable to all orders. (While this is obviously wrong to me, I'm unsure if a HSC student would have run across a counter example since this tends to be true for most functions encountered at a hsc level.)

Ah I see there, well you just posted the solution for Step 1, if I get rid of the strength of the assumption and loosen it up is it correct?

EDIT: And I am guessing that:

$P(x)=\sum_{k=0}^{n} f^{(k)}(a) x^k$
??

Sy123 · Dec 21, 2012

Re: HSC 2013 4U Marathon

$\text{Given that} \ \ x^m y^n = (x+y)^{m+n} \\ \\ \text{Show that} \ \ \ \frac{dy}{dx}=\frac{y}{x}$

==========

Is latex broken or something?

seanieg89 · Dec 22, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
Ah I see there, well you just posted the solution for Step 1, if I get rid of the strength of the assumption and loosen it up is it correct?

EDIT: And I am guessing that:

$P(x)=\sum_{k=0}^{n} f^{(k)}(a) x^k$
??

Missing a factor of k!. Perhaps I should have given the form of the polynomial in my question but anyway, I think you should be able to prove the result if you know what the poly is.

Rezen · Dec 22, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
Ah I see there, well you just posted the solution for Step 1, if I get rid of the strength of the assumption and loosen it up is it correct?

EDIT: And I am guessing that:

$P(x)=\sum_{k=0}^{n} f^{(k)}(a) x^k$
??

Google taylor polynomials.

Drongoski · Dec 22, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
$\text{Given that} \ \ x^m y^n = (x+y)^{m+n} \\ \\ \text{Show that} \ \ \ \frac{dy}{dx}=\frac{y}{x}$

==========

Is latex broken or something?

Done it!

But solution too horrendous to type out in LaTeX.

Synchronised · Dec 22, 2012

Re: HSC 2013 4U Marathon

Can you guys please post some difficult integration and conics questions?

Sy123 · Dec 22, 2012

Re: HSC 2013 4U Marathon

Drongoski said:
Done it!

But solution too horrendous to type out in LaTeX.

Just the main points will do, and like a one sentence explanation between each 'major' line of working (i.e. skip the non important algebra)

deswa1 · Dec 22, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
Just the main points will do, and like a one sentence explanation between each 'major' line of working (i.e. skip the non important algebra)

I haven't done it but iirc, this was a question in a Grammar paper a while ago and the easiest way to do it was to take logs of both sides and then differentiate implicitly.

Drongoski · Dec 22, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
Just the main points will do, and like a one sentence explanation between each 'major' line of working (i.e. skip the non important algebra)

Basically using implicit differentiation.

Steps:

1. - differentiate both sides wrt x

2. - expand as required & simplify

3. - express dy/dx as approp fraction

4. - multiply numerator and denominator of said fraction by: xy(x+y)

5. - replace (x+y)ⁿ with x^myⁿ

6. - simplify, then factorise and simplify further

7. - voila

Using logs lots easier. See below.

Drongoski · Dec 22, 2012

Re: HSC 2013 4U Marathon

Using logs, as suggested by deswa1, is heaps easier:

ln(LHS) = ln(RHS)

mlnx + nlny = (m+n)ln(x+y)

.: m/x + n/y . dy/dx = (m+n)/(x+y) .(1+dy/dx)

.: ((m+n)/(x+y) - n/y) . dy/dx = m/x - (m+n)/(x+y)

.: dy/dx = []/[] = [m(x+y)-(m+n)x]/[(m+n)y - n(x+y)] . y/x = y/x

Sy123 · Dec 23, 2012

Re: HSC 2013 4U Marathon

$The plane curve defined by the relation$ \\ \\ x^{\frac{2}{3}}+y^{\frac{2}{3}}=a^{\frac{2}{3}} \\ \\ $is called an astroid curve$

Note, replace the 1 with a. The image in the graph sets a=1

$Prove that the length of the tangent to the astroid curve at the point$ \ \ \ (l,m) \ \ \ $cut off by the co-ordinate axes is of constant value$

Sy123 · Dec 23, 2012

Re: HSC 2013 4U Marathon

Synchronised said:
Can you guys please post some difficult integration and conics questions?

SpiralFlex said:
$(a) The equation of the ellipse E is$\; \frac{(x-h)^2}{a^2}+\frac{y^2}{b^2}=1\; $where a,b,h are real and a,b is greater than zero.$

$(i) By integration find the area enclosed by E.$

$$(ii) If$\; y=mx \;$is a tangent to$\; E\; $show that$\; m^2=\frac{b^2}{h^2-a^2}$

$(b) If n belongs to the set of natural numbers, let$ \;E_{n} \;$be the ellipse given by$

$\frac{(x-h_{n})^2}{(a_{n})^2}+\frac{y^2}{p^2(a_{n})^2}=1\; $where$\; p>0, h_{n}>h_{n+1}, h_{n}>a_{n}>0\;$

$$Supposing for all natural numbers that$\; E_{n}\;$and$\; E_{n+1}\; $touch each other externally and$\; y=mx\;$is the common tangent to all$\; E_{n}.$

$$(i) Express$\; h_{n}-h_{n+1} \;$in terms of$\; a_{n}\;$and$\; a_{n+1}.$

$Hence show,$

$$(ii)$\;\frac{a_{n+1}}{a_{n}}=\frac{h_{1}-a_{1}}{h_{1}+a_{1}}$

$$(iii)$ $If$\; S_{n}\; $is the area enclosed by the ellipse$\; E_{n}. \;$Evaluate$\; \sum_{n=1}^{\infty } S_{n}\; $in terms of$\; a_{1}, h_{1}, p.$

I haven't learnt conics yet but this was on the second page (and no one answered all of it yet)

Carrotsticks · Dec 23, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
$The plane curve defined by the relation$ \\ \\ x^{\frac{2}{3}}+y^{\frac{2}{3}}=a^{\frac{2}{3}} \\ \\ $is called an astroid curve$

Note, replace the 1 with a. The image in the graph sets a=1

$Prove that the tangent to the astroid curve at the point$ \ \ \ (l,m) \ \ \ $cut off by the co-ordinate axes is of constant value$

Did you mean the length of the tangent cut off by the coordinate axes?

Sy123 · Dec 23, 2012

Re: HSC 2013 4U Marathon

Carrotsticks said:
Did you mean the length of the tangent cut off by the coordinate axes?

Yes, edited now.

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