HSC 2014 MX2 Marathon ADVANCED (archive) (1 Viewer)

Chlee1998 · Nov 30, 2014

Re: HSC 2014 4U Marathon - Advanced Level

what is the value of the infinite series, 1/1+ 1/2 +....+1/8+ 1/10 +...+1/18+ 1/20+.......
Btw this is the 1/1+1/2+1/3+....+1/n except without any terms with a 9 in the denominator

Sy123 · Nov 30, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Chlee1998 said:
what is the value of the infinite series, 1/1+ 1/2 +....+1/8+ 1/10 +...+1/18+ 1/20+.......
Btw this is the 1/1+1/2+1/3+....+1/n except without any terms with a 9 in the denominator

So do you want a finite sum or infinite sum?

RealiseNothing · Nov 30, 2014

Re: HSC 2014 4U Marathon - Advanced Level

SilentWaters said:
do I detect sarcasm

bro if you really think it's negative then good luck

Sy123 · Nov 30, 2014

Re: HSC 2014 4U Marathon - Advanced Level

$\\ $Prove for integers$ \ m,n > 1 \ $that$ \ (1+m)^{-\frac{1}{n}} + (1+n)^{- \frac{1}{m}} > 1$

Chlee1998 · Nov 30, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Sy123 said:
So do you want a finite sum or infinite sum?

It converges.

Chlee1998 · Nov 30, 2014

Re: HSC 2014 4U Marathon - Advanced Level

just to clarify, if the term has at least 1 digit which is 9, it is excluded from the series. So 1/998, 1/89 etc are all excluded

Chlee1998 · Nov 30, 2014

Re: HSC 2014 4U Marathon - Advanced Level

actually u can't find the exact value it converges to.. Change the question to 'explain why the series converges and roughly where it converges'

SilentWaters · Nov 30, 2014

Re: HSC 2014 4U Marathon - Advanced Level

RealiseNothing said:
bro if you really think it's negative then good luck

My bad. Question's asked incorrectly then.

Sy123 said:
(HSC Methods only)

$$Hence show that$ {\bf\ \frac{1}{1} - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \dots} = \ln 2$

ftfy

FrankXie · Nov 30, 2014

Re: HSC 2014 4U Marathon - Advanced Level

SilentWaters said:
Also I think the third line should read $H_n - \ln n.$

oh yep my bad

FrankXie · Nov 30, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Chlee1998 said:
actually u can't find the exact value it converges to.. Change the question to 'explain why the series converges and roughly where it converges'

$$ yep the series converges, i can show that it is bounded above by $ 10(1+\frac12+\frac13+\cdots+\frac18)$

glittergal96 · Nov 30, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Let $S_k$ be the set of reciprocals of k-digit positive integers that don't contain 9's.

Then the sum is

$\sum_{k=1}^\infty \sum_{j\in S_k}j\leq \sum_{k=1}^\infty \max_{j\in S_k}(j)\cdot |S_k|=\sum_{k=1}^\infty 10^{-k+1}\cdot 8\cdot 9^{k-1}=8\sum_{k=1}^\infty (9/10)^{k-1}<\infty.$

The monotone convergence theorem completes the proof of convergence.

To obtain an accurate estimate, exactly sum the contributions of the first few S_k and use the same GP to bound the error. (This takes a while though, because 9/10 is pretty close to 1.)

Sy123 · Nov 30, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Chlee1998 said:
actually u can't find the exact value it converges to.. Change the question to 'explain why the series converges and roughly where it converges'

Chlee1998 · Nov 30, 2014

Re: HSC 2014 4U Marathon - Advanced Level

In trapezium (quadrilateral with one side parallel) with sides XY and ZU parallel, Angle UXY=6 degrees, Angle XYZ= 42 degrees. Point B is one side XY such that angle XBU = 78 degrees and angle ZBY = 66 degrees. If XY and Zu are 1 cm apart (vertically) prove that XU- YZ +UB-ZB = 8cm

SilentWaters · Nov 30, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Chlee1998 said:
In trapezium (quadrilateral with one side parallel) with sides XY and ZU parallel, Angle UXY=6 degrees, Angle XYZ= 42 degrees. Point B is one side XY such that angle XBU = 78 degrees and angle ZBY = 66 degrees. If XY and Zu are 1 cm apart (vertically) prove that XU- YZ +UB-ZB = 8cm

Drop perpendiculars from $U$ and $Z$ to $XY$ . From a consideration of the right angled triangles formed thereby, it should be easy trigonometry that $XU = (sin\,6^{\circ})^{-1}, YZ = (sin\,42^{\circ})^{-1}, UB = (sin\,78^{\circ})^{-1}$ and $ZB = (sin\,66^{\circ})^{-1}.$ Plug these into the left-hand side and you have your eight.

glittergal96 · Nov 30, 2014

Re: HSC 2014 4U Marathon - Advanced Level

SilentWaters said:
It should suffice to prove that each component of the sum exceeds $\frac{1}{2}.$ By the conditions, $1+m>2$ so any $n^{th}$ root of this where $n>1$ should be less than two. It follows that the reciprocal must exceed a half.

Why does the n-th root of a number bigger than two have to be less than two? This doesn't look right to me.

Sy123 · Nov 30, 2014

Re: HSC 2014 4U Marathon - Advanced Level

SilentWaters said:
It should suffice to prove that each component of the sum exceeds $\frac{1}{2}.$ By the conditions, $1+m>2$ so any $n^{th}$ root of this where $n>1$ should be less than two. It follows that the reciprocal must exceed a half.

At this part of your reasoning, are you saying that since:

$1 + m > 2 \Rightarrow \ 2 > \sqrt[n]{1+m}$ because that does not seem to be a valid inference

glittergal96 · Nov 30, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Sy123 said:
$\\ $Prove for integers$ \ m,n > 1 \ $that$ \ (1+m)^{-\frac{1}{n}} + (1+n)^{- \frac{1}{m}} > 1$

$(1+m)^{1/n}=((1+m)\cdot 1^{n-1})^{1/n}<\frac{m+n}{n}$

by AM-GM. (Noting that not all terms are equal, so we cannot have equality.)

Similarly, we have

$(1+n)^{1/m}<\frac{m+n}{m}.$

Invert these inequalities and sum them to complete the proof.

SilentWaters · Nov 30, 2014

Re: HSC 2014 4U Marathon - Advanced Level

woah what the hell am I doing, wishful thinking. sorry guys, ignore that

Axio · Nov 30, 2014

Re: HSC 2014 4U Marathon - Advanced Level

glittergal96 said:
$(1+m)^{1/n}=((1+m)\cdot 1^{n-1})^{1/n}<\frac{m+n}{n}$

by AM-GM. (Noting that not all terms are equal, so we cannot have equality.)

Similarly, we have

$(1+n)^{1/m}<\frac{m+n}{m}.$

Invert these inequalities and sum them to complete the proof.

Noice.

Could you explain to me how you get that first line to fit the AM-GM inequality?

Chlee1998 · Nov 30, 2014

Re: HSC 2014 4U Marathon - Advanced Level

SilentWaters said:
Drop perpendiculars from $U$ and $Z$ to $XY$ . From a consideration of the right angled triangles formed thereby, it should be easy trigonometry that $XU = (sin\,6^{\circ})^{-1}, YZ = (sin\,42^{\circ})^{-1}, UB = (sin\,78^{\circ})^{-1}$ and $ZB = (sin\,66^{\circ})^{-1}.$ Plug these into the left-hand side and you have your eight.

how do you know that plugging those values into the LHS gives you 8 without using a calculator? You cannot use a calculator for this question

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